Drorbn - User contributions [en]

0708-1300/Class notes for Tuesday, November 20

2007-11-26T18:52:11Z

Aptikuis: /* Second Hour */

{{0708-1300/Navigation}}

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===First Hour===

Recall we are ultimately attempting to understand and prove Stokes theorem. Currently we are investigating the meaning of <math>d\omega</math>.

Recall we had that <math>d(\sum f_I dx^I) = \sum (df_I)\wedge dx^I = \sum_{I,j}\frac{\partial f_I}{\partial x^j} dx^j\wedge dx^I = \sum dx^j\frac{\partial\omega}{\partial x_j}</math>

Now we want to compute <math>d\omega</math> on the parallelepiped formed from k+1 tangent vectors. For instance let us suppose k= 2 then are interested in the parallelepiped formed from the three basis vectors <math>v_1, v_2, v_3</math>.

Feeding in the parallelepiped we get <math>\sum_I df_I(v_1)dx^I(v_2,v_3) - \sum_I df_I(v_2)dx^I(v_1,v_3) + \sum_I df_I(v_3)dx^I(v_1,v_2) </math><math>= \sum_I (v_1 f_I)dx^I(v_2,v_3) - \sum_I (v_2 f_I)dx^I(v_1,v_3) + \sum_I (v_3 f_I)dx^I(v_1,v_2)</math>

<math>= (v_1 w)(v_2,v_3) - (v_2 w)(v_1,v_3) + (v_3 w)(v_1,v_2)</math>

Now, <math>v_1 f = lim\frac{(p+\epsilon v_1) - f(p)}{\epsilon}</math>, or, loosely <math>f(p+v_1) - f(p)</math>

So this corresponds to the difference between <math>\omega</math> calculated on each of the two faces parallel to the <math>v_3,v_2</math> plane.

Hence, our <math>d\omega</math> on the parallelepiped is just the sum of parallelograms making the boundary of the parallelepiped counted with some signs.

We can see the loose idea of how the proof of stokes theorem is going to work: dividing the manifold up into little parallelepiped like this, <math>d\omega</math> will just be the faces of the parallelepipeds and when summing over the whole manifold all of faces will cancel except those on the boundary thus just leaving the integral of <math>\omega</math> along the boundary.

We note that this is similar to the proof of the fundamental theorem of calculus, where we take an integral and compute the value of f' at many little subintervals. But the value of f' is just the difference of f' at the boundary of each sub interval so when we add everything up everything cancels except the values of the function at the endpoint of the big interval.

'''Claim'''

d exists if <math>M = \mathbb{R}^n</math> and is unique.

Define <math>d(\omega) = d(\sum f_I dx^I) :=\sum_{j,I}\frac{\partial f_I}{\partial x_j}dx^j\wedge dx^I</math>

''Proof''

We need to check that this satisfies the properties:

1) <math>df(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} dx_j(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} \delta_{ji} = \partial_i f</math> and so satisfies

''Note''

We now adopt ''Einstein Summation Convention'' which means that if in a term there is an index that is repeated, once as a subscript and once as a superscript, it is meant as implicit that we are summing over this index. This just cleans up the notation so we don't have to have sums everywhere.

2) <math>d(df_I dx^I) = d(\frac{\partial f_I}{\partial dx^j} dx^j\wedge dx^I) = \frac{\partial^2 f_I}{\partial x^j \partial x^{j'}}dx^{j'}\wedge dx^j\wedge dx^I = 0</math> because the mixed partial is symmetric under exchange of indices but the wedge product is antisymmetric under exchange of indices. That is, each term cancels with the one where j and j' are exchanged.

3) let <math>\omega=f_I dx^I</math> and <math>\lambda = g_J dx^J</math> then <math>\omega\wedge\lambda = f_I g_J dx^I\wedge dx^J</math>

so <math>d(f_I g_J dx^I\wedge dx^J) = \frac{\partial f_I g_I}{\partial x^j} dx^j\wedge dx^I\wedge dx^J = (\frac{\partial f_I}{\partial x^j}g_J + f_I\frac{\partial g_I}{\partial x^j})dx^j\wedge dx^I\wedge dx^J = d\omega\wedge\lambda \pm\omega\wedge d\lambda</math>

Via assignment 3 this is unique.

''Q.E.D.''

Now, we can extend this definition on manifolds by using coordinate charts.

'''Claim'''

Properties 1-3 imply that on any M, d is ''local''. That is, if <math>\omega|_U = \lambda|_U</math> then <math>d\omega|_U = d\lambda|_U</math>

''Proof:'' Exercise.

'''Definition'''

For <math>\omega\in\Omega^k(M)</math> the <math>supp\ \omega = \overline{\{p\in M\ :\ w|_p\neq0\}}</math>

Then ''<math>\omega</math> has compact support'' if the supp <math>\omega</math> is compact.

Define <math>\Omega^*_c(M)</math> := the compactly support <math>w\in\Omega^*(M)</math>

'''Definition'''

We define <math>\int_{\mathbb{R}^n}: \Omega^n_c(\mathbb{R}^n)\rightarrow\mathbb{R}</math> by <math>\omega\in\Omega^n_c(\mathbb{R}^n)</math> and thus <math>\omega = fdx^1\wedge\cdots\wedge dx^n</math> then

<math>\int_{\mathbb{R}^n}\omega := \int_{\mathbb{R}^n}f</math>

I.e., <math>\int_{\mathbb{R}^n}fdx^1\wedge\cdots\wedge dx^n = \int_{\mathbb{R}^n}fdx^1\ldots dx^n</math>

=== Second Hour===

In general if we have a diffeomorphism <math>\phi:\mathbb{R}^n \rightarrow\mathbb{R}^n</math> then the normal integral

<math>\int \phi^*f = \int f\circ\phi</math> is not equal to <math>\int f</math>

However we claim that this IS true for differential forms. I.e.,

'''Claim'''

<math>\int \phi^*\omega = \pm\int \omega</math> as forms

This is very important because it essentially means we can integrate in whatever charts we like and get the same thing.

''Proof''

<math>\phi^*\omega = \phi^*(fdx^1\wedge\ldots\wedge dx^n) = \phi^* f\phi^* dx^1\wedge\ldots\wedge \phi^* dx^n</math>

Now <math>\phi^* (dg) = d\phi^* g</math> by chain rule and so we extend to <math>\phi^*(d\omega) = d(\phi^*\omega)</math>

Hence,

<math>\phi^*\omega = (f\circ\phi) d(x^1\circ\phi)\wedge\ldots\wedge d(x^n\circ\phi) = (f\circ\phi) d\phi^1\wedge\ldots\wedge d\phi^n</math>

<math>= (f\circ\phi)\left(\frac{\partial\phi^1}{\partial y^{i_1}}\right)\wedge\ldots\wedge\left(\frac{\partial\phi^n}{\partial y^{i_n}}\right) = (f\circ\phi)\sum_{i_1,\ldots,i_n =1}^n \frac{\partial\phi^1}{\partial y^{i_1}}\ldots\frac{\partial\phi^n}{\partial y^{i_n}}dy^{1_1}\wedge\ldots\wedge dy^{i_n}</math>

but the wedge product is zero unless <math>(i_1,\ldots,i_n)=\sigma\in S^n</math> and in that case yields <math>(-1)^{\sigma} dy^1\wedge\ldots\wedge dy^n</math>

Hence we get,

<math>=(f\circ\phi)\sum_{\sigma\in S^n}(-1)^{\sigma}\Pi_{\alpha} \frac{\partial\phi^{\alpha}}{\partial y^{\sigma(\alpha)}}dy^1\wedge\ldots\wedge dy^n = (f\circ\phi) det (d\phi)dy^1\wedge\ldots\wedge dy^n = (f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n</math>

where <math>J_{\phi}</math> is the determinant of the Jacobian matrix.

Hence, <math>\int_{\mathbb{R}^n} \phi^*\omega = \int_{\mathbb{R}^n}(f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n = \int_{Old\ sense} (f\circ\phi)J_{\phi} =\pm \int (f\circ\phi)|J_{\phi}| = \pm\int f = \pm\int \omega</math>

''Q.E.D''

If we restrict our attention to just the ''orientation preserving'' <math>\phi</math>'s so that <math>J_{\phi}>0</math> then we will always get the + in the end.

'''Definition'''

An ''orientation'' of M is an assignment of the charts to <math>\pm 1</math>, <math>\phi\mapsto S_{\phi}</math>

Then if the domains on <math>\phi</math> and <math>\psi</math> overlap then <math>S_{\psi} = (sign J_{\psi^{-1}\phi})S_{\phi}</math>

By definition, M is orientable if we can find an orientation.

'''Examples'''

1) <math>\mathbb{R}^n</math>. We declare the identity positive and then all other designations follow from this.

2) The finite cylinder <math>S^1\times I</math>

We can put two charts on the cylinder by considering two rectangles which overlap by a little bit that cover the whole cylinder. If we denote one of these as positive, the overlap makes the other positive. We compare any other chart to these two.

3) Consider the mobius strip and the same attempted charts as for the cylinder. If we label one section positive, the other section must be positive due to the overlap on one side, but must be negative due to the overlap on the other side. Thus the mobius strip is not orientable.

'''Definition'''

An ''orientation of a vector space V'' is an equivalence class of order bases <math>v_{\alpha}</math> of V under <math>v_{\alpha}\sim w_{\beta}</math> if the determinant of the transition matrix between these two bases is positive.

'''Definition'''

An ''orientation of M'' is a continuous choice of orientations for <math>T_p M</math> for any p. We haven't technically defined what it means to be continuous in this sense but the meaning is clear.

'''Definition'''

let <math>M^n</math> be an oriented manifold and let <math>\omega\in\Omega^n_c(M)</math>. Let <math>\phi_{\alpha}:U_{\alpha}\mapsto\mathbb{R}^n</math> be a collection of positive charts that cover M. Let <math>\lambda^{\alpha}</math> be a partition of unity subordinate to this cover then

<math>\int_M\omega = \int_M 1\omega = \int_M \sum_{\alpha}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{U_{\alpha}}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{\mathbb{R}^n} (\phi^{-1}_{\alpha})^*(\lambda_{\alpha}\omega)</math>

Note all the intermediate steps were merely properties we would LIKE the integral to have, the actual definition is the equality of the left most and right most expressions.

'''Theorem'''

<math>\int_M\omega</math> is independent of the choices.

0708-1300/Class notes for Tuesday, November 20

2007-11-26T18:51:31Z

Aptikuis: /* Second Hour */

{{0708-1300/Navigation}}

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===First Hour===

Recall we are ultimately attempting to understand and prove Stokes theorem. Currently we are investigating the meaning of <math>d\omega</math>.

Recall we had that <math>d(\sum f_I dx^I) = \sum (df_I)\wedge dx^I = \sum_{I,j}\frac{\partial f_I}{\partial x^j} dx^j\wedge dx^I = \sum dx^j\frac{\partial\omega}{\partial x_j}</math>

Now we want to compute <math>d\omega</math> on the parallelepiped formed from k+1 tangent vectors. For instance let us suppose k= 2 then are interested in the parallelepiped formed from the three basis vectors <math>v_1, v_2, v_3</math>.

Feeding in the parallelepiped we get <math>\sum_I df_I(v_1)dx^I(v_2,v_3) - \sum_I df_I(v_2)dx^I(v_1,v_3) + \sum_I df_I(v_3)dx^I(v_1,v_2) </math><math>= \sum_I (v_1 f_I)dx^I(v_2,v_3) - \sum_I (v_2 f_I)dx^I(v_1,v_3) + \sum_I (v_3 f_I)dx^I(v_1,v_2)</math>

<math>= (v_1 w)(v_2,v_3) - (v_2 w)(v_1,v_3) + (v_3 w)(v_1,v_2)</math>

Now, <math>v_1 f = lim\frac{(p+\epsilon v_1) - f(p)}{\epsilon}</math>, or, loosely <math>f(p+v_1) - f(p)</math>

So this corresponds to the difference between <math>\omega</math> calculated on each of the two faces parallel to the <math>v_3,v_2</math> plane.

Hence, our <math>d\omega</math> on the parallelepiped is just the sum of parallelograms making the boundary of the parallelepiped counted with some signs.

We can see the loose idea of how the proof of stokes theorem is going to work: dividing the manifold up into little parallelepiped like this, <math>d\omega</math> will just be the faces of the parallelepipeds and when summing over the whole manifold all of faces will cancel except those on the boundary thus just leaving the integral of <math>\omega</math> along the boundary.

We note that this is similar to the proof of the fundamental theorem of calculus, where we take an integral and compute the value of f' at many little subintervals. But the value of f' is just the difference of f' at the boundary of each sub interval so when we add everything up everything cancels except the values of the function at the endpoint of the big interval.

'''Claim'''

d exists if <math>M = \mathbb{R}^n</math> and is unique.

Define <math>d(\omega) = d(\sum f_I dx^I) :=\sum_{j,I}\frac{\partial f_I}{\partial x_j}dx^j\wedge dx^I</math>

''Proof''

We need to check that this satisfies the properties:

1) <math>df(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} dx_j(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} \delta_{ji} = \partial_i f</math> and so satisfies

''Note''

We now adopt ''Einstein Summation Convention'' which means that if in a term there is an index that is repeated, once as a subscript and once as a superscript, it is meant as implicit that we are summing over this index. This just cleans up the notation so we don't have to have sums everywhere.

2) <math>d(df_I dx^I) = d(\frac{\partial f_I}{\partial dx^j} dx^j\wedge dx^I) = \frac{\partial^2 f_I}{\partial x^j \partial x^{j'}}dx^{j'}\wedge dx^j\wedge dx^I = 0</math> because the mixed partial is symmetric under exchange of indices but the wedge product is antisymmetric under exchange of indices. That is, each term cancels with the one where j and j' are exchanged.

3) let <math>\omega=f_I dx^I</math> and <math>\lambda = g_J dx^J</math> then <math>\omega\wedge\lambda = f_I g_J dx^I\wedge dx^J</math>

so <math>d(f_I g_J dx^I\wedge dx^J) = \frac{\partial f_I g_I}{\partial x^j} dx^j\wedge dx^I\wedge dx^J = (\frac{\partial f_I}{\partial x^j}g_J + f_I\frac{\partial g_I}{\partial x^j})dx^j\wedge dx^I\wedge dx^J = d\omega\wedge\lambda \pm\omega\wedge d\lambda</math>

Via assignment 3 this is unique.

''Q.E.D.''

Now, we can extend this definition on manifolds by using coordinate charts.

'''Claim'''

Properties 1-3 imply that on any M, d is ''local''. That is, if <math>\omega|_U = \lambda|_U</math> then <math>d\omega|_U = d\lambda|_U</math>

''Proof:'' Exercise.

'''Definition'''

For <math>\omega\in\Omega^k(M)</math> the <math>supp\ \omega = \overline{\{p\in M\ :\ w|_p\neq0\}}</math>

Then ''<math>\omega</math> has compact support'' if the supp <math>\omega</math> is compact.

Define <math>\Omega^*_c(M)</math> := the compactly support <math>w\in\Omega^*(M)</math>

'''Definition'''

We define <math>\int_{\mathbb{R}^n}: \Omega^n_c(\mathbb{R}^n)\rightarrow\mathbb{R}</math> by <math>\omega\in\Omega^n_c(\mathbb{R}^n)</math> and thus <math>\omega = fdx^1\wedge\cdots\wedge dx^n</math> then

<math>\int_{\mathbb{R}^n}\omega := \int_{\mathbb{R}^n}f</math>

I.e., <math>\int_{\mathbb{R}^n}fdx^1\wedge\cdots\wedge dx^n = \int_{\mathbb{R}^n}fdx^1\ldots dx^n</math>

=== Second Hour===

In general if we have a diffeomorphism <math>\phi:\mathbb{R}^n \rightarrow\mathbb{R}^n</math> then the normal integral

<math>\int \phi^*f = \int f\circ\phi</math> is not equal to <math>\int f</math>

However we claim that this IS true for differential forms. I.e.,

'''Claim'''

<math>\int \phi^*\omega = \pm\int \omega</math> as forms

This is very important because it essentially means we can integrate in whatever charts we like and get the same thing.

''Proof''

<math>\phi^*\omega = \phi^*(fdx^1\wedge\ldots\wedge dx^n) = \phi^* f\phi^* dx^1\wedge\ldots\wedge \phi^* dx^n</math>

Now <math>\phi^* (dg) = d\phi^* g</math> by chain rule and so we extend to <math>\phi^*(d\omega) = d(\phi^*\omega)</math>

Hence,

<math>\phi^*\omega = (f\circ\phi) d(x^1\circ\phi)\wedge\ldots\wedge d(x^n\circ\phi) = (f\circ\phi) d\phi^1\wedge\ldots\wedge d\phi^n</math>

<math>= (f\circ\phi)\left(\frac{\partial\phi^1}{\partial y^{i_1}}\right)\wedge\ldots\wedge\left(\frac{\partial\phi^n}{\partial y^{i_n}}\right) = (f\circ\phi)\sum_{i_1,\ldots,i_n =1}^n \frac{\partial\phi^1}{\partial y^{i_1}}\ldots\frac{\partial\phi^n}{\partial y^{i_n}}dy^{1_1}\wedge\ldots\wedge dy^{i_n}</math>

but the wedge product is zero unless <math>(i_1,\ldots,i_n)=\sigma\in S^n</math> and in that case yields <math>(-1)^{\sigma} dy^1\wedge\ldots\wedge dy^n</math>

Hence we get,

<math>=(f\circ\phi)\sum_{\sigma\in S^n}(-1)^{\sigma}\Pi_{\alpha} \frac{\partial\phi^{\alpha}}{\partial y^{\sigma(\alpha)}}dy^1\wedge\ldots\wedge dy^n = (f\circ\phi) det (d\phi)dy^1\wedge\ldots\wedge dy^n = (f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n</math>

where <math>J_{\phi}</math> is the determinant of the Jacobian matrix.

Hence, <math>\int_{\mathbb{R}^n} \phi^*\omega = \int_{\mathbb{R}^n}(f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n = \int_{Old\ sense} (f\circ\phi)J_{\phi} =\pm \int (f\circ\phi)|J_{\phi}| = \pm\int f = \pm\int \omega</math>

''Q.E.D''

If we restrict our attention to just the ''orientation preserving'' <math>\phi</math>'s so that <math>J_{\phi}>0</math> then we will always get the + in the end.

'''Definition'''

An ''orientation'' of M is an assignment of the charts to <math>\pm 1</math>, <math>\phi\mapsto S_{\phi}</math>

Then if the domains on <math>\phi</math> and <math>\psi</math> overlap then <math>S_{\psi} = (sign J_{\psi^{-1}\phi})S_{\phi}</math>

By definition, M is orientable if we can find an orientation.

'''Examples'''

1) <math>\mathbb{R}^n</math>. We declare the identity positive and then all other designations follow from this.

2) The finite cylinder <math>S^1\times I</math>

We can put two charts on the cylinder by considering two rectangles which overlap by a little bit that cover the whole cylinder. If we denote one of these as positive, the overlap makes the other positive. We compare any other chart to these two.

3) Consider the mobius strip and the same attempted charts as for the cylinder. If we label one section positive, the other section must be positive due to the overlap on one side, but must be negative due to the overlap on the other side. Thus the mobius strip is not orientable.

'''Definition'''

An ''orientation of a vector space V'' is an equivalence class of order bases <math>v_{\alpha}</math> of V under <math>v_{\alpha}\sim w_{\beta}</math> if the determinant of the transition matrix between these two bases is positive.

'''Definition'''

An ''orientation of M'' is a continuous choice of orientations for <math>T_p M</math> for any p. We haven't technically defined what it means to be continuous in this sense but the meaning is clear.

'''Definition'''

let <math>M^n</math> be an oriented manifold and let <math>\omega\in\Omega^n_c(M)</math>. Let <math>\phi_{\alpha}:U_{\alpha}\mapsto\mathbb{R}^n</math> be a collection of positive charts that cover M. Let <math>\lambda^{\alpha}</math> be a partition of unity subordinate to this cover then

<math>\int_M\omega = \int_M 1\omega = \int_M \sum_{\alpha}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{U_{\alpha}}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{\mathbb{R}^n} \phi^{-1}_{\alpha}(\lambda_{\alpha}\omega)</math>

Note all the intermediate steps were merely properties we would LIKE the integral to have, the actual definition is the equality of the left most and right most expressions.

'''Theorem'''

<math>\int_M\omega</math> is independent of the choices.

0708-1300/Class notes for Tuesday, November 20

2007-11-26T18:27:11Z

Aptikuis: /* First Hour */

{{0708-1300/Navigation}}

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===First Hour===

Recall we are ultimately attempting to understand and prove Stokes theorem. Currently we are investigating the meaning of <math>d\omega</math>.

Recall we had that <math>d(\sum f_I dx^I) = \sum (df_I)\wedge dx^I = \sum_{I,j}\frac{\partial f_I}{\partial x^j} dx^j\wedge dx^I = \sum dx^j\frac{\partial\omega}{\partial x_j}</math>

Now we want to compute <math>d\omega</math> on the parallelepiped formed from k+1 tangent vectors. For instance let us suppose k= 2 then are interested in the parallelepiped formed from the three basis vectors <math>v_1, v_2, v_3</math>.

Feeding in the parallelepiped we get <math>\sum_I df_I(v_1)dx^I(v_2,v_3) - \sum_I df_I(v_2)dx^I(v_1,v_3) + \sum_I df_I(v_3)dx^I(v_1,v_2) </math><math>= \sum_I (v_1 f_I)dx^I(v_2,v_3) - \sum_I (v_2 f_I)dx^I(v_1,v_3) + \sum_I (v_3 f_I)dx^I(v_1,v_2)</math>

<math>= (v_1 w)(v_2,v_3) - (v_2 w)(v_1,v_3) + (v_3 w)(v_1,v_2)</math>

Now, <math>v_1 f = lim\frac{(p+\epsilon v_1) - f(p)}{\epsilon}</math>, or, loosely <math>f(p+v_1) - f(p)</math>

So this corresponds to the difference between <math>\omega</math> calculated on each of the two faces parallel to the <math>v_3,v_2</math> plane.

Hence, our <math>d\omega</math> on the parallelepiped is just the sum of parallelograms making the boundary of the parallelepiped counted with some signs.

We can see the loose idea of how the proof of stokes theorem is going to work: dividing the manifold up into little parallelepiped like this, <math>d\omega</math> will just be the faces of the parallelepipeds and when summing over the whole manifold all of faces will cancel except those on the boundary thus just leaving the integral of <math>\omega</math> along the boundary.

We note that this is similar to the proof of the fundamental theorem of calculus, where we take an integral and compute the value of f' at many little subintervals. But the value of f' is just the difference of f' at the boundary of each sub interval so when we add everything up everything cancels except the values of the function at the endpoint of the big interval.

'''Claim'''

d exists if <math>M = \mathbb{R}^n</math> and is unique.

Define <math>d(\omega) = d(\sum f_I dx^I) :=\sum_{j,I}\frac{\partial f_I}{\partial x_j}dx^j\wedge dx^I</math>

''Proof''

We need to check that this satisfies the properties:

1) <math>df(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} dx_j(\partial_i) = \sum_j \frac{\partial f}{\partial x_j} \delta_{ji} = \partial_i f</math> and so satisfies

''Note''

We now adopt ''Einstein Summation Convention'' which means that if in a term there is an index that is repeated, once as a subscript and once as a superscript, it is meant as implicit that we are summing over this index. This just cleans up the notation so we don't have to have sums everywhere.

2) <math>d(df_I dx^I) = d(\frac{\partial f_I}{\partial dx^j} dx^j\wedge dx^I) = \frac{\partial^2 f_I}{\partial x^j \partial x^{j'}}dx^{j'}\wedge dx^j\wedge dx^I = 0</math> because the mixed partial is symmetric under exchange of indices but the wedge product is antisymmetric under exchange of indices. That is, each term cancels with the one where j and j' are exchanged.

3) let <math>\omega=f_I dx^I</math> and <math>\lambda = g_J dx^J</math> then <math>\omega\wedge\lambda = f_I g_J dx^I\wedge dx^J</math>

so <math>d(f_I g_J dx^I\wedge dx^J) = \frac{\partial f_I g_I}{\partial x^j} dx^j\wedge dx^I\wedge dx^J = (\frac{\partial f_I}{\partial x^j}g_J + f_I\frac{\partial g_I}{\partial x^j})dx^j\wedge dx^I\wedge dx^J = d\omega\wedge\lambda \pm\omega\wedge d\lambda</math>

Via assignment 3 this is unique.

''Q.E.D.''

Now, we can extend this definition on manifolds by using coordinate charts.

'''Claim'''

Properties 1-3 imply that on any M, d is ''local''. That is, if <math>\omega|_U = \lambda|_U</math> then <math>d\omega|_U = d\lambda|_U</math>

''Proof:'' Exercise.

'''Definition'''

For <math>\omega\in\Omega^k(M)</math> the <math>supp\ \omega = \overline{\{p\in M\ :\ w|_p\neq0\}}</math>

Then ''<math>\omega</math> has compact support'' if the supp <math>\omega</math> is compact.

Define <math>\Omega^*_c(M)</math> := the compactly support <math>w\in\Omega^*(M)</math>

'''Definition'''

We define <math>\int_{\mathbb{R}^n}: \Omega^n_c(\mathbb{R}^n)\rightarrow\mathbb{R}</math> by <math>\omega\in\Omega^n_c(\mathbb{R}^n)</math> and thus <math>\omega = fdx^1\wedge\cdots\wedge dx^n</math> then

<math>\int_{\mathbb{R}^n}\omega := \int_{\mathbb{R}^n}f</math>

I.e., <math>\int_{\mathbb{R}^n}fdx^1\wedge\cdots\wedge dx^n = \int_{\mathbb{R}^n}fdx^1\ldots dx^n</math>

=== Second Hour===

In general if we have a diffeomorphism <math>\phi:\mathbb{R}^n \rightarrow\mathbb{R}^n</math> then the normal integral

<math>\int \phi^*f = \int f\circ\phi</math> is not equal to <math>\int f</math>

However we claim that this IS true for differential forms. I.e.,

'''Claim'''

<math>\int \phi^*\omega = \pm\int \omega</math> as forms

This is very important because it essentially means we can integrate in whatever charts we like and get the same thing.

''Proof''

<math>\phi^*\omega = \phi^*(fdx^1\wedge\ldots\wedge dx^n) = \phi^* f\phi^* dx^1\wedge\ldots\wedge \phi^* dx^n</math>

Now <math>\phi^* (dg) = d\phi^* g</math> by chain rule and so we extend to <math>\phi^*(d\omega) = d(\phi^*\omega)</math>

Hence,

<math>\phi^*\omega = (f\circ\phi) d(x^1\circ\phi)\wedge\ldots\wedge d(x^n\circ\phi) = (f\circ\phi) d\phi^1\wedge\ldots\wedge d\phi^n</math>

<math>= (f\circ\phi)\left(\frac{\partial\phi^1}{\partial y^{i_1}}\right)\wedge\ldots\wedge\left(\frac{\partial\phi^n}{\partial y^{i_n}}\right) = (f\circ\phi)\sum_{i_1,\ldots,i_n =1}^n \frac{\partial\phi^1}{\partial y^{i_1}}\ldots\frac{\partial\phi^n}{\partial y^{i_n}}dy^{1_1}\wedge\ldots\wedge dy^{i_n}</math>

but the wedge product is zero unless <math>(i_1,\ldots,i_n)=\sigma\in S^n</math> and in that case yields <math>(-1)^{\sigma} dy^1\wedge\ldots\wedge dy^n</math>

Hence we get,

<math>=(f\circ\phi)\sum_{\sigma\in S^n}(-1)^{\sigma}\Pi_{\alpha} \frac{\partial\phi^{\alpha}}{\partial y^{\sigma(\alpha)}}dy^1\wedge\ldots\wedge dy^n = (f\circ\phi) det (d\phi)dy^1\wedge\ldots\wedge dy^n = (f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n</math>

where <math>J_{\phi}</math> is the determinant of the Jacobian matrix.

Hence, <math>\int_{\mathbb{R}^n} \phi^*\omega = \int_{\mathbb{R}^n}(f\circ\phi)J_{\phi}dy^1\wedge\ldots\wedge dy^n = \int_{Old\ sense} (f\circ\phi)J_{\phi} =\pm \int (f\circ\phi)|J_{\phi}| = \pm\int f = \pm\int \omega</math>

''Q.E.D''

If we restrict our attention to just the ''orientation preserving'' <math>\phi</math>'s so that <math>J_{\phi}>0</math> then we will always get the + in the end.

'''Definition'''

An ''orientation'' of M is an assignment of the charts to <math>\pm 1</math>, <math>\phi\mapsto S_{\phi}</math>

Then if the domains on <math>\phi</math> and <math>\psi</math> overlap then <math>S_{\psi} = (sign J_{\psi^{-1}\phi})S_{\phi}</math>

By definition, M is orientable if we can find an orientation.

'''Examples'''

1) <math>\mathbb{R}^n</math>. We declare the identity positive and then all other designations follow from this.

2) The finite cylinder <math>S^1\times I</math>

We can put two charts on the cylinder by considering two rectangles which overlap by a little bit that cover the whole cylinder. If we denote one of these as positive, the overlap makes the other positive. We compare any other chart to these two.

3) Consider the mobius strip and the same attempted charts as for the cylinder. If we label one section positive, the other section must be positive due to the overlap on one side, but must be negative due to the overlap on the other side. Thus the mobius strip is not orientable.

'''Definition'''

An ''orientation of a vector space V'' is an equivalence class of order bases <math>v_{\alpha}</math> of V under <math>v_{\alpha}\sim w_{\beta}</math> if the determinant of the transition matrix between these two bases is positive.

'''Definition'''

An ''orientation of M'' is a continuous choice of orientations for <math>T_p M</math> for any p. We haven't technically defined what it means to be continuous in this sense but the meaning is clear.

'''Definition'''

let <math>M^n</math> be an oriented manifold and let <math>\omega\in\Omega^n_c(M)</math>. Let <math>\phi_{\alpha}:U_{\alpha}\mathbb{R}^n</math> be a collection of positive charts that cover M. Let <math>\lambda^{\alpha}</math> be a partition of unity subordinate to this cover then

<math>\int_M\omega = \int_M 1\omega = \int_M \sum_{\alpha}\lambda_{\alpha}\omega = \sum_{\alpha}\int_{U_{\alpha}}\lambda_{\alpha}\omega = \sum_{\alpha}\int_\mathbb{R}^n \phi^{-1}_{\alpha}(\lambda_{\alpha}\omega)</math>

Note all the intermediate steps were merely properties we would LIKE the integral to have, the actual definition is the equality of the left most and right most expressions.

'''Theorem'''

<math>\int_M\omega</math> is independent of the choices.

0708-1300/Class notes for Tuesday, November 13

2007-11-25T22:59:01Z

Aptikuis: /* First Hour */

{{0708-1300/Navigation}}

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===First Hour===

We begin with a review of last class. Since no one has typed up the notes for last class yet, I will do the review here.

Recall had an association <math>M\rightarrow\Omega^k(M)</math> which was the "k forms on M" which equaled <math>\{w:\ p\in M\rightarrow A^k(T_p M)\}</math>

where

<math>A^k(V):=\{w:\underbrace{V\times\ldots\times V}_{k\ times} \rightarrow\mathbb{R}\}</math>
which is

1) Multilinear

2) Alternating

We had proved that :

1) <math>A^k(V)</math> is a vector space

2) there was a wedge product <math>\wedge:A^k(V)\times A^l(V)\rightarrow A^{k+l}(V)</math> via <math>\omega\wedge\lambda(v_1,\ldots,v_{k+l}) =

</math><math>\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}(-1)^{\sigma} \omega(v_{\sigma(1)},\ldots,v_{\sigma(k)})\lambda(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})</math>
that is

a) bilinear

b) associative

c) supercommutative, i.e., <math>\omega\wedge\lambda = (-1)^{deg(w)deg(\lambda)}\lambda\wedge\omega</math>

From these definitions we can define for <math>\omega\in\Omega^k(M)</math> and <math>\lambda\in\Omega^l(M)</math> that <math>\omega\wedge\lambda\in\Omega^{k+l}(M)</math> with the same properties as above.

''Claim''

If <math>\omega_1,\ldots,\omega_n</math> is a basis of <math>A^1(V) = V^* </math> then <math> \{\omega_I = w_{i_1}\wedge\ldots\wedge\omega_{i_k}\ :\ I=(i_1,\ldots,i_k)\ with\ 1\leq i_1<\ldots<i_k\leq n\}</math> is a basis of <math>A^k(V)</math> and <math>dim A^k(V) = nCk</math>

If <math>\omega_1,\ldots,\omega_n\in\Omega^1(M)</math> and <math>w_1 |_p,\ldots,\omega_n |_p</math> a basis of <math>(T_p M)^*\ \forall p\in M</math> then any <math>\lambda</math> can be written as <math>\lambda = \sum_I a_I(p)\omega_I</math> where <math>a_I:M\rightarrow\mathbb{R}</math> are smooth.

The equivalence of these is left as an exercise.

''Example''

Let us investigate <math>\Omega^*(\mathbb{R}^3)</math> (the * just means "anything").

Now, <math>(T_p(\mathbb{R}^3))^* = <dx_1,dx_2,dx_3></math> where <math>x_i:\mathbb{R}^3\rightarrow\mathbb{R}</math> and so <math>dx_i|_p:T_p\mathbb{R}^3\rightarrow T_{x_i(p)}\mathbb{R} =\mathbb{R}</math>

Hence, <math>dx_i|_p\in (T_p\mathbb{R}^3)^*</math>

Now, <math>dx_2(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}x_2 =\delta_{i2}</math> and hence we get a basis.

So, <math>\Omega^1(\mathbb{R}^3) = \{g_1 dx_1 + g_2 dx_2 + g_3 dx_3\}\approx\{g_1,g_2,g_3\}\approx</math> {vector fields on <math>\mathbb{R}^3</math>}

where the <math>g_i:\mathbb{R}^3\rightarrow\mathbb{R}</math> are smooth.

<math>\Omega^0(\mathbb{R}^3) = \{f:\mathbb{R}^3\rightarrow\mathbb{R}\}</math>

This is because to each point p we associate something that takes zero copies of the tangent space into the real numbers. Thus to each p we associate a number.

<math>\Omega^3(\mathbb{R}^3) = \{kdx_1\wedge dx_2\wedge dx_3\} \approx</math> {functions} where again the k is just a smooth function from <math>\mathbb{R}^3</math> to <math>\mathbb{R}</math>.

<math>\Omega^2(\mathbb{R}^3) = \{h_1 dx_2\wedge dx_3 + h_2 dx_3\wedge dx_1 + h_3 dx_1\wedge dx_2\}\approx \{h_1,h_2,h_3\}\approx</math> {vector fields}

''Aside''

Recall our earlier discussion of how points and things like points (curves, equivalence classes of curves) pushfoward while things dual to points (functions) pullback and that things dual to functions (such as derivations) push forward. See earlier for the precise definitions.

Now differential forms pull back, i.e., for <math>\phi:M\rightarrow N</math> then <math>\phi_*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)</math>
via
<math>\phi_*(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_* v_1,\ldots \phi_* v_k)</math>

The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via <math>\phi_*(\omega\wedge\lambda)=\phi_*(\omega)\wedge\phi_*(\lambda)</math>

'''Theorem-Definition'''

Given M, <math>\exists</math> ! linear map <math>d:\Omega^*(M)\rightarrow\Omega^{k*+1}(M)</math> satisfies

1) If <math>f\in\Omega^0(M)</math> then <math>df(X) = X(f)</math> for <math>X\in TM</math>

2) <math>d^2 = 0</math>. I.e. if <math>d_k:\Omega^k\rightarrow\Omega^{k+1}</math> and <math>d_{k+1}:\Omega^{k+1}\rightarrow\Omega^{k+2}</math> then <math>d_{k+1}\circ d_k = 0</math>.

3) <math>d(\omega\wedge\lambda) = d\omega\wedge\lambda + (-1)^{deg\omega}\omega\wedge d\lambda</math>

===Second Hour===

Some notes about the above definition:

1) When we restrict our ''d'' to functions we just get the old meaning for d.

2) Philosophically, there is a duality between differential forms and manifolds and that duality is given by integration. In this duality, d is the adjoint of the boundary operator on manifolds. For manifolds, the boundary of the boundary is empty and hence it is reasonable that <math>d^2=0</math> on differential forms.

3) To remember the formula in 3 given above and others like it, it helps to keep in mind what objects are "odd" and what are "even" and thus when commuting such operators we will get the signs as you would expect from multiplying objects that are either odd or even.

''Example''

Let us aim for a formula for d on <math>\Omega^*(\mathbb{R}^n)</math>.

Lets compute <math>d(\sum_I f_I dx_I)</math> where <math>dx_I = dx_{i_1}\wedge\ldots\wedge dx_{i_k}</math> and <math>I = (i_1,\ldots,i_k)</math>

Then, <math>d(\sum_I f_I dx_I) = \sum_I d(f_I \wedge dx_{i_1}\wedge\ldots\wedge dx_{i_k}) = \sum_I df_I\wedge(dx_I) + f_I\wedge d(dx_I)</math>

The last term vanishes because of (2) in the theorem (proving uniqueness!)

Now, as an aside, we claim that for <math>f\in\Omega^0(M),\ df = \sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j</math>

Indeed, we know <math>(df)(\frac{\partial}{\partial x_i}) = \frac{\partial}{\partial x_i}f</math>

However, <math>(\sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j)(\frac{\partial}{\partial x_i}) = \sum_j \frac{\partial f}{\partial x_j}\delta_{ij} = \frac{\partial f}{\partial x_i}</math> which is the same.

Returning, we thus get <math>d(\sum_I f_I dx_I) = \sum_{i,J} \frac{\partial f_I}{\partial x_j} dx_k\wedge dx_I</math>

Thus our d takes functions to vector fields by <math>f\mapsto (\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3})</math>

This is just the grad operator from calculus and we can see that the d operator appropriately takes things from <math>\Omega^0(M)</math> to <math>\Omega^1(M)</math>.

Now let us compute <math>d(h_1 dx_2\wedge dx_3+h_2 dx_3\wedge dx_1+h_3 dx_1\wedge dx_2) =\frac{\partial h_1}{\partial x_1} dx_1\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_2} dx_3\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_3} dx_3\wedge dx_2\wedge dx_3</math> + 6 more terms representing the 3 partials of each of the last 2 terms.

As each <math>dx_i\wedge dx_i</math> term vanishes we are left with just,

<math>= \frac{\partial h_1}{\partial x_1} + \frac{\partial h_2}{\partial x_2} + \frac{\partial h_3}{\partial x_3})dx_1\wedge dx_2\wedge dx_3</math>

I.e., d takes <math>(h_1,h_2,h_3)\mapsto \sum_i\frac{\partial h_i}{\partial dx_i}</math>

this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from <math>\Omega^2(M)</math> to <math>\Omega^3(M)</math>.

We are left with computing d from <math>\Omega^2(M)</math> to <math>\Omega^3(M)</math>

Computing, <math>d(g_1 dx_1 + g_2 dx_2 + g_3 dx_3) = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1})dx_3\wedge dx_1 + (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2})dx_1\wedge dx_2</math>

I.e., we just have the curl operator.

Note that the well known calculus laws that curl grad = 0 and div curl = 0 are just the expression that <math>d^2 =0</math>.

To provide some physical insight to the meanings of these operators:

1) The gradient represents the direction of maximum descent. I.e. if you had a function on the plane the graph would look like the surface of a mountain range and the direction that water would run would be the gradient.

2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point.

3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the x_2, x_1 plane the tenancy for it to rotate clockwise would be given by <math>\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2}</math>

0708-1300/Class notes for Tuesday, November 13

2007-11-25T22:58:15Z

Aptikuis: /* First Hour */

{{0708-1300/Navigation}}

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

===First Hour===

We begin with a review of last class. Since no one has typed up the notes for last class yet, I will do the review here.

Recall had an association <math>M\rightarrow\Omega^k(M)</math> which was the "k forms on M" which equaled <math>\{w:\ p\in M\rightarrow A^k(T_p M)\}</math>

where

<math>A^k(V):=\{w:\underbrace{V\times\ldots\times V}_{k\ times} \rightarrow\mathbb{R}\}</math>
which is

1) Multilinear

2) Alternating

We had proved that :

1) <math>A^k(V)</math> is a vector space

2) there was a wedge product <math>\wedge:A^k(V)\times A^l(V)\rightarrow A^{k+l}(V)</math> via <math>\omega\wedge\lambda(v_1,\ldots,v_{k+l}) =

</math><math>\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}}(-1)^{\sigma} \omega(v_{\sigma(1)},\ldots,v_{\sigma(k)})\lambda(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})</math>
that is

a) bilinear

b) associative

c) supercommutative, i.e., <math>\omega\wedge\lambda = (-1)^{deg(w)deg(\lambda)}\lambda\wedge\omega</math>

From these definitions we can define for <math>\omega\in\Omega^k(M)</math> and <math>\lambda\in\Omega^l(M)</math> that <math>\omega\wedge\lambda\in\Omega^{k+l}(M)</math> with the same properties as above.

''Claim''

If <math>\omega_1,\ldots,\omega_n</math> is a basis of <math>A^1(V) = V^* </math> then <math> \{\omega_I = w_{i_1}\wedge\ldots\wedge\omega_{i_k}\ :\ I=(i_1,\ldots,i_k)\ with\ 1\leq i_1<\ldots<i_k\leq n\}</math> is a basis of <math>A^k(V)</math> and <math>dim A^k(V) = nCk</math>

If <math>\omega_1,\ldots,\omega_n\in\Omega^1(M)</math> and <math>w_1 |_p,\ldots,\omega_n |_p</math> a basis of <math>(T_p M)^*\ \forall p\in M</math> then any <math>\lambda</math> can be written as <math>\lambda = \sum_I a_I(p)\omega_I</math> where <math>a_I:M\rightarrow\mathbb{R}</math> are smooth.

The equivalence of these is left as an exercise.

''Example''

Let us investigate <math>\Omega^*(\mathbb{R}^3)</math> (the * just means "anything").

Now, <math>(T_p(\mathbb{R}^3))^* = <dx_1,dx_2,dx_3></math> where <math>x_i:\mathbb{R}^3\rightarrow\mathbb{R}</math> and so <math>dx_i|_p:T_p\mathbb{R}^3\rightarrow T_{x_i(p)}\mathbb{R} =\mathbb{R}</math>

Hence, <math>dx_i|_p\in (T_p\mathbb{R}^3)^*</math>

Now, <math>dx_2(\frac{\partial}{\partial x}) = \frac{\partial}{\partial x_i}x_2 =\delta_{i2}</math> and hence we get a basis.

So, <math>\Omega^1(\mathbb{R}^3) = \{g_1 dx_1 + g_2 dx_2 + g_3 dx_3\}\approx\{g_1,g_2,g_3\}\approx</math> {vector fields on <math>\mathbb{R}^3</math>}

where the <math>g_i:\mathbb{R}^3\rightarrow\mathbb{R}</math> are smooth.

<math>\Omega^0(\mathbb{R}^3) = \{f:\mathbb{R}^3\rightarrow\mathbb{R}\}</math>

This is because to each point p we associate something that takes zero copies of the tangent space into the real numbers. Thus to each p we associate a number.

<math>\Omega^3(\mathbb{R}^3) = \{kdx_1\wedge dx_2\wedge dx_3\} \approx</math> {functions} where again the k is just a smooth function from <math>\mathbb{R}^3</math> to <math>\mathbb{R}</math>.

<math>\Omega^2(\mathbb{R}^3) = \{h_1 dx_2\wedge dx_3 + h_2 dx_3\wedge dx_1 + h_3 dx_1\wedge dx_2\}\approx \{h_1,h_2,h_3\}\approx</math> {vector fields}

''Aside''

Recall our earlier discussion of how points and things like points (curves, equivalence classes of curves) pushfoward while things dual to points (functions) pullback and that things dual to functions (such as derivations) push forward. See earlier for the precise definitions.

Now differential forms pull back, i.e., for <math>\phi:M\rightarrow N</math> then <math>\phi_*(\lambda)\in\Omega^k(M)\leftarrow\lambda\in\Omega^k(N)</math>
via
<math>\phi_*(\lambda)(v_1,\ldots,v_k)=\lambda(\phi_* v_1,\ldots \phi_* v_k)</math>

The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via <math>\phi^*(\omega\wedge\lambda)=\phi^*(\omega)\wedge\phi^*(\lambda)</math>

'''Theorem-Definition'''

Given M, <math>\exists</math> ! linear map <math>d:\Omega^*(M)\rightarrow\Omega^{k*+1}(M)</math> satisfies

1) If <math>f\in\Omega^0(M)</math> then <math>df(X) = X(f)</math> for <math>X\in TM</math>

2) <math>d^2 = 0</math>. I.e. if <math>d_k:\Omega^k\rightarrow\Omega^{k+1}</math> and <math>d_{k+1}:\Omega^{k+1}\rightarrow\Omega^{k+2}</math> then <math>d_{k+1}\circ d_k = 0</math>.

3) <math>d(\omega\wedge\lambda) = d\omega\wedge\lambda + (-1)^{deg\omega}\omega\wedge d\lambda</math>

===Second Hour===

Some notes about the above definition:

1) When we restrict our ''d'' to functions we just get the old meaning for d.

2) Philosophically, there is a duality between differential forms and manifolds and that duality is given by integration. In this duality, d is the adjoint of the boundary operator on manifolds. For manifolds, the boundary of the boundary is empty and hence it is reasonable that <math>d^2=0</math> on differential forms.

3) To remember the formula in 3 given above and others like it, it helps to keep in mind what objects are "odd" and what are "even" and thus when commuting such operators we will get the signs as you would expect from multiplying objects that are either odd or even.

''Example''

Let us aim for a formula for d on <math>\Omega^*(\mathbb{R}^n)</math>.

Lets compute <math>d(\sum_I f_I dx_I)</math> where <math>dx_I = dx_{i_1}\wedge\ldots\wedge dx_{i_k}</math> and <math>I = (i_1,\ldots,i_k)</math>

Then, <math>d(\sum_I f_I dx_I) = \sum_I d(f_I \wedge dx_{i_1}\wedge\ldots\wedge dx_{i_k}) = \sum_I df_I\wedge(dx_I) + f_I\wedge d(dx_I)</math>

The last term vanishes because of (2) in the theorem (proving uniqueness!)

Now, as an aside, we claim that for <math>f\in\Omega^0(M),\ df = \sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j</math>

Indeed, we know <math>(df)(\frac{\partial}{\partial x_i}) = \frac{\partial}{\partial x_i}f</math>

However, <math>(\sum_{j=1}^{n}\frac{\partial f}{\partial x_j} dx_j)(\frac{\partial}{\partial x_i}) = \sum_j \frac{\partial f}{\partial x_j}\delta_{ij} = \frac{\partial f}{\partial x_i}</math> which is the same.

Returning, we thus get <math>d(\sum_I f_I dx_I) = \sum_{i,J} \frac{\partial f_I}{\partial x_j} dx_k\wedge dx_I</math>

Thus our d takes functions to vector fields by <math>f\mapsto (\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3})</math>

This is just the grad operator from calculus and we can see that the d operator appropriately takes things from <math>\Omega^0(M)</math> to <math>\Omega^1(M)</math>.

Now let us compute <math>d(h_1 dx_2\wedge dx_3+h_2 dx_3\wedge dx_1+h_3 dx_1\wedge dx_2) =\frac{\partial h_1}{\partial x_1} dx_1\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_2} dx_3\wedge dx_2\wedge dx_3 + \frac{\partial h_1}{\partial x_3} dx_3\wedge dx_2\wedge dx_3</math> + 6 more terms representing the 3 partials of each of the last 2 terms.

As each <math>dx_i\wedge dx_i</math> term vanishes we are left with just,

<math>= \frac{\partial h_1}{\partial x_1} + \frac{\partial h_2}{\partial x_2} + \frac{\partial h_3}{\partial x_3})dx_1\wedge dx_2\wedge dx_3</math>

I.e., d takes <math>(h_1,h_2,h_3)\mapsto \sum_i\frac{\partial h_i}{\partial dx_i}</math>

this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from <math>\Omega^2(M)</math> to <math>\Omega^3(M)</math>.

We are left with computing d from <math>\Omega^2(M)</math> to <math>\Omega^3(M)</math>

Computing, <math>d(g_1 dx_1 + g_2 dx_2 + g_3 dx_3) = (\frac{\partial g_3}{\partial x_2} - \frac{\partial g_2}{\partial x_3})dx_2\wedge dx_3 + (\frac{\partial g_1}{\partial x_3} - \frac{\partial g_3}{\partial x_1})dx_3\wedge dx_1 + (\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2})dx_1\wedge dx_2</math>

I.e., we just have the curl operator.

Note that the well known calculus laws that curl grad = 0 and div curl = 0 are just the expression that <math>d^2 =0</math>.

To provide some physical insight to the meanings of these operators:

1) The gradient represents the direction of maximum descent. I.e. if you had a function on the plane the graph would look like the surface of a mountain range and the direction that water would run would be the gradient.

2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point.

3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the x_2, x_1 plane the tenancy for it to rotate clockwise would be given by <math>\frac{\partial g_2}{\partial x_1} - \frac{\partial g_1}{\partial x_2}</math>

0708-1300/Class notes for Thursday, October 25

2007-11-02T00:59:40Z

Aptikuis: /* Typed Notes */

{{0708-1300/Navigation}}

==Typed Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

'''General Comments'''

1) Exam is on Nov 8th

2) Specs for the exam will be given next thursday.

'''Aside 2'''

''Classification of Surfaces''

1) Every 2 dimensional manifold that is compact connected and without boundary is smoothly equivalent to one of the following:

a) There is a family of such surfaces consisting of the sphere <math>\mathbb{S}^2</math>, the torus <math>\mathbb{R}^2</math> and various connected sums of many tori.

b) A connected sum of <math>\mathbb{RP}^2</math>'s, the real projective plane. This is formed by taking the sphere <math>\mathbb{S}^2</math> and identifying antipodal points.

''Definition''

A manifold without boundary has every point locally homeomorphic to a (relatively) open subset of <math>H^n=\{(x_1,\ldots,x_n)| x_n\geq 0\}</math>

2) If we now consider such manifolds WITH boundary they will be equivalent to something in case one only with open disks removed.

'''Aside 1'''

''Classification of 1-manifolds''

1) A compact 1-manifold without boundary is smoothly equivalent to a circle

2) A compact 1-manifold with boundary is smoothly equivalent to a closed interval

3) A non compact 1-manifold with boundary is smoothly equivalent to a half open interval

4) A non compact 1-manifold without boundary is smoothly equivalent to an open interval

'''Continuing with the Proof of the Whitney Embedding Theorem'''

We recall we are in part 3a of the proof.

The main steps we did in this last class were repeated again and so I won't do that here; however, there was an analogy to the proof presented that I will comment on.

Recall we have the idea of a ''remoteness function'' (that last class we called s) that we call r. If we were to consider the slice of the manifold with a certain value of r we will have problems with the fact that an embedding on one slice won't smoothly transition to an embedding on an adjacent slice. This problem is not solved from moving from a slice to disjoint intervals of values of r. However we get a nice analogy for this case:

We consider a roll of film on which each section (being the subset (not submanifold) of the manifold with values of r in some interval) occurring on each image in the roll of film. Thus as we move along the film we get to see the section with larger and larger r values. Indeed we could cut each image on the film up separately and lay them on top of each other in a stack. We would like this stack to correspond to out atlas. The problem of course is in the smoothness of the transition from one image to another.

The way we resolve this problem is to think of it as follows. We take an image and then shrink it down to zero and when we blow it back up again it is a new image. This is akin to turning the bulb off, then changing the film and then turning the bulb back on again.

In computer graphics, there is a problem where one tries to draw an image but that the processing time to draw the image is greater then the rate at which images are displayed. And hence part of the image is redrawn while part remains the same resulting in a distorted image. The solution is two have two buffers in memory. The image is displayed from one buffer while the next image is calculated and stored in the second buffer. Once the new image is completely drawn in it displayed on the screen and the original buffer is used to draw/compute the next image.

We now return to last class where we have defined the function

<math>\Phi(p) = (\Phi_{even},\ \Phi_{odd},\ r(p))</math>
that went from the manifold into <math>\mathbb{R}^{4m+3}</math>. (as each <math>\Phi</math> has dimension 2m+1 and an extra dimension for the r)

Here the <math>\Phi_{even}</math> and <math>\Phi_{odd}</math> correspond to the two buffers in the analogy.

Now this will turn out to be an embedding and we can reproduce our point on the manifold through use of the r. Indeed, the function is 1:1 as if we are near the same value of n then we are in the same <math>\phi</math> and so 1:1. If we are not near the same n, then we will have different r's necessarily.

'''Claim'''

A proper 1:1 continuous function is a homeomorphism onto its image.

i.e., we need to show a proper function <math>\Rightarrow</math> closed.

''Note''

This claim shows that our map <math>\Phi</math> is an embedding (provided such a proper r exists)

''Summary of what is left to do:''

1) Prove this claim

2) Prove the existence of such an r

3) Show how we can reduce from <math>\mathbb{R}^{4m+3}</math> to <math>\mathbb{R}^{2m+1}</math>

''Proof of 3''

We can repeat the arguments used in part 2 of this proof again (the lack of compactness is not problem, as Sard's theorem and the dimension reducing argument doesn't depend on it)

However, we do have to make the following modification. In addition to not choosing a projection direction in im<math>\beta_1</math> <math>\cup</math> im<math>\beta_2</math> we have to also make sure not to project in the vertical direction down into the r axis. But this is still quite possible by Sard's Theorem and hence we can reduce down to an embedding into <math>\mathbb{R}^{2m+1}</math>

''Proof of 2''

Consider <math>\{\lambda_{\alpha}\}</math> a partition of unity subordinant to a countable cover by sets with compact closure. We can make this partition be countable as each chart with compact closure only needs finitely many <math>\lambda_{\alpha}</math>'s.

Hence we get a partition of unity <math>\{\lambda_k\}_{k\in\mathbb{N}}</math> with supp<math>\lambda_k</math> being compact.

Set <math>r:=\sum k\lambda_k</math>

''Claim:''
r is proper

''Proof'

Assume <math>p\notin\bigcup_{k<n} supp \lambda_k</math>

then <math>1=\sum_{k=1}^{\infty}\lambda_k (p) = \sum_{k=n}^{\infty}\lambda_k (p)</math>

<math>\Rightarrow</math>

<math>\sum_{k=1}^{\infty}k\lambda_k(p) =\sum_{k=n}^{\infty}k\lambda_k(p)\geq \sum_{k=n}^{\infty}n\lambda_k(p) = n\sum_{k=n}^{\infty}\lambda_k(p) =n</math>

so, <math>r^{-1}([0,n])\subset\bigcup_{k=1}^{n-1}supp\lambda_k</math>.

So compact

''Q.E.D''

0708-1300/Class notes for Tuesday, October 16

2007-10-31T12:52:58Z

Aptikuis: /* Second Hour */

{{0708-1300/Navigation}}
==Dror's Computer Program for C+<math>\alpha</math> C==

Our handout today is a printout of a Mathematica notebook that computes the measure of the projection of <math>C\times C</math> in a direction <math>t\in[0,\pi/2]</math> (where <math>C</math> is the standard Cantor set). Here's the {{Home Link|classes/0708/GeomAndTop/CCShadow.nb|notebook}}, and here's a {{Home Link|classes/0708/GeomAndTop/CCShadow.pdf|PDF}} version. Also, here's the main picture on that notebook:

[[Image:0708-1300-CCShadow.png|center|540px]]

==Typed Class Notes==

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

=== First Hour===

'''Today's Agenda:'''

Proof of Sard's Theorem. That is, for <math>f:M^m\rightarrow N^n</math> being smooth, the measure <math>\mu</math>(critical values of f) = 0.

'''Comments regarding last class'''

1) In our counterexample to Sard's Theorem for the case of <math>C^1</math> functions it was emphasized that there are functions f from ''a'' Cantor set C' to ''the'' Cantor set C. We then let <math>g(x,y) = f(x) + f(y)</math> and hence the critical values are <math>C+C = [0,2]</math> as was shown last time. The sketch of such an f was the same as last class.

Furthermore, in general we can find a <math>C^n</math> such function where we make the "bumps" in f smoother as needed and so <math>f(C') = C''</math> where <math>C''</math> is a "very thin" Cantor set. But now let <math>g(x,y,z,\ldots) = f(x) + f(y) + f(z) +\ldots</math> which will have an image of <math>C'' + C'' + C'' + \ldots =</math> an interval.

2) The code, and what the program does, for Dror's program (above) was described. It is impractical to describe it here in detail and so I will only comment that it computes the measure of <math>C+\alpha C</math> for various alpha and that the methodology relied on the 2nd method of proof regarding C+C done last class.

'''Proof of Sard's Theorem'''

Firstly, it is enough to argue locally.

Further, the technical assumption about manifolds that until now has been largely ignored is that our M must be second countable. Recall that this means that there is a countable basis for the topology on M.

As a counterexample to Sard's Theorem when M is NOT second countable consider the real line with the discrete topology, a zero dimensional manifold, mapping via the identity onto the real line with the normal topology. Every point in the real line is thus a critical point and the real line has non zero measure.

We can restrict our neighborhoods so that we can assume <math>M=\mathbb{R}^m</math> and <math>N=\mathbb{R}^n</math>

The general idea here is that if we consider a function g=f' that is nonzero at p but that f is zero at p, the inverse image is (in some chart) a straight line (a manifold). As such, we will inductively reduce the dimension from m down to zero. For m=0 there is nothing to prove. Hence we assume true for m-1.

Now, set <math>D_k = \{p |</math> all partial derivities of f of order <math>\leq k\}</math> for <math>k\geq 1</math>

Set <math>D_0 = \{p\ |\ df_p</math> is not onto }. This is just the critical points.

Clearly <math>D_0\supset D_1\supset\ldots\supset D_i\supset\ldots\supset D_m</math>

We will show by backwards induction that <math>\mu(F(D_k)) = 0</math>

''Comment:''

We have not actually defined the measure <math>\mu</math>. We use it merely to denote that <math>F(D_k))</math> has measure zero, a concept that we DID define.

===Second Hour===

'''Claim 1'''

<math>F(D_m)</math> has measure 0.

''Proof''

W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size <math>\epsilon</math> in 1D becomes a smaller set of size <math>\epsilon^2</math> in 2D etc. More precisely, for <math>f = (f_1,\ldots,f_n)</math>, <math>f(D_n)\subset f_1(D_m)\times{R}^{n-1}</math>. Applying the proposition that if A is of measure zero in <math>\mathbb{R}</math> then <math>A\times\mathbb{R}^{n-1}</math> is measure zero in <math>\mathbb{R}^n</math> we now see that assuming n=1 is justified.

''Reminder''

Taylor's Theorem: for smooth enough <math>g:\mathbb{R}\rightarrow\mathbb{R}</math> and some <math>x_0</math> then <math>g(x) = \sum_{j=0}^{m} \frac{g^{(j)}(x_0)}{j!}(x-x_0)^j + \frac{g^{(m+1)}(t)}{(m+1)!}(x-x_0)^{m+1}</math>
for some t between x and <math>x_0</math>.

For <math>x_0\in D_m</math> all but the last term vanishes and so we can conclude that f(x) is bounded by a constant times <math>(x-x_0)^{m+1}</math>.

Now let us consider a box B in <math>\mathbb{R}^m</math> containing a section of <math>D_m</math>. We divide B into <math>C_1\lambda^m</math> boxes of side <math>1/\lambda</math>.

By Taylor's Theorem, <math>f(B_i)\subset</math> of an interval of length <math>C_2\frac{1}{\lambda^{m+1}}</math> where the constant <math>C_2</math> is determined by Taylor's Theorem. Call this interval <math>I_i</math>

Hence,

<math>f(D_m)\subset\bigcup_{i:B_i\cap D_m \neq 0} f(B_i)\subset \bigcup_{i:B_i\cap D_m \neq 0}I_i</math>

But <math>\sum_{i} length(I_i)\leq C_1\lambda^m C_2\frac{1}{\lambda^{m+1}}</math> which tends to zero as <math>\lambda</math> tends to infinity.

''Q.E.D for Claim 1''

'''Claim 2'''

<math>f(D_k)</math> has measure zero for <math>k\geq 1</math>. We just proved this for k=m.

Now, W.L.O.G. <math>D_{k+1}</math> is the empty set. If not, just consider <math>M^m - D_{k+1}</math> which is still a manifold as <math>D_{k+1}</math> is closed (as it is determined by the "closed" condition that a determinant equals zero)

So, there is some kth derivative g of f such that <math>dg\neq 0</math>.

So <math>D_k\subset g^{-1}(0)</math> but <math>g^{-1}(0)</math> is at least locally a manifold of dimension 1 less. So, <math>f(D_k)\subset f(D_k\cap g^{-1}(0))</math> which has measure zero due to our induction hypothesis.

''Q.E.D for Claim 2''

'''Claim 3'''

<math>f(D_0)</math> is of measure zero.

Recall <math>D_0</math> is defined differently from the <math>D_k</math> and so requires a different technique to prove.

W.L.O.G. lets assume that <math>D_1</math> is the empty set. So, some derivative of f is not zero. W.L.O.G. <math>\frac{\partial f_1}{\partial x_1}</math> is non zero near some point p. We can simply move to a coordinate system where this is true.

The idea here is to prove that the intersection with any "slice" has measure zero where we will then invoke a theorem that will claim everything has measure zero.

So, let U be an open neighborhood of a point <math>p\in M</math>. Consider <math>f_1:U\rightarrow\mathbb{R}</math> and let <math>df_1</math> be onto. Using our previous theorem for the local structure of such a submersion W.L.O.G. let us assume <math>f_1:\mathbb{R}^m\rightarrow\mathbb{R}</math> via <math>(x_1,\ldots,x_m)\mapsto x_1</math>. That is, <math>f_1 = x_1</math>.

Our differential df then is just the matrix whose first row consists of <math>(1,0,\ldots,0)</math>. Then df is onto if the submatrix consisting of all but the first row and first column is invertible.

For notational convenience let us say <math>f =(f_1,f_{rest})</math>.

Now define <math>U_t = \{t\} \times\mathbb{R}^{m-1}</math>
Also lets denote "critical points of f'' by CP(f) and "critical values of f" by CV(f)

'''Claim 4'''

The <math>CP(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CP(f_{rest}|_{U_t})</math>

<math>\Rightarrow</math>

<math>CV(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CV(f_{rest}|_{U_t})</math>.

But <math>CV(f_{rest}|_{U_t})</math> has measure zero by our induction.

'''Lemma 1'''

If <math>A\subset I^2</math> is closed and has <math>\mu(A\cap(\{t\}\times I)) = 0\ \forall t</math>
then <math>\mu(A)=0</math>.

''Proof''

Note: We prove this significantly differently then in Bredon

''Sublemma''

If <math>\{t\}\times U</math> for an open U cover <math>\{t\}\times I\cap A</math> for a closed A then <math>\exists\epsilon>0</math> such that <math>(t-\epsilon,t+\epsilon)\times U\supset (t-\epsilon,t+\epsilon)\times I\cap A</math>

Indeed, let <math>d:A-(I\times U)\rightarrow\mathbb{R}</math> be <math>d(x) = |x_1-t|</math> then d is a continuous function of a compact set and so obtains a minimum and since d>0 then <math>min(d)>0 \rightarrow d>\epsilon>0</math>. But this <math>\epsilon</math> works for the claim. ''Q.E.D''

''The rest of the proof of Lemma 1, and of Sard's Theorem will be left until next class''

0708-1300/Class notes for Tuesday, October 9

2007-10-31T00:48:39Z

Aptikuis: /* Typed Notes - Second Hour */

{{0708-1300/Navigation}}

==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

=== Scanned Notes ===
Scanned notes for today's class can be found [[0708-1300/Scanned Class notes for Tuesday, October 9| here]]

(see discussion)

===Typed Notes - First Hour===

'''Reminder:'''

1) An immersion locally looks like <math>\mathbb{R}^n\rightarrow\mathbb{R}^m</math> given by <math>x\mapsto(x,0)</math>

2) A submersion locally looks like <math>\mathbb{R}^m\rightarrow\mathbb{R}^n</math> given by <math>(x,y)\mapsto x</math>

'''Today's Goals'''

1) More about "locally things look like their differential"

2) The tricky Sard's Theorem: "Evil points are rare, good points everywhere"

'''Definition 1'''

Let <math>f:M^m\rightarrow N^n</math> be smooth. A point <math>p\in M^m</math> is ''critical'' if <math>df_p</math> is not onto <math>\Leftrightarrow</math> rank <math>df_p <n</math>. Otherwise, p is ''regular.

'''Definition 2'''
A point <math>y\in N^n</math> is a ''critical value'' of f if <math>\exists p\in M^m</math> such that p is critical and <math>f(p) = y</math>. Otherwise, y is a ''regular value''

'''Example 1'''

Consider the map <math>f:S^2\rightarrow\mathbb{R}^2</math> given by <math>(x,y,z)\mapsto (x,y)</math>. I.e., the projection map. The regular points are all the points on <math>S^2</math> except the equator. The regular values, however, are all <math>(x,y)</math> such that <math>x^2+y^2 \neq 1</math>

'''Example 2'''

Consider <math>f:\mathbb{R}^3\rightarrow\mathbb{R}</math> given by <math>p\mapsto ||p||^2</math>.
That is, <math>(x,y,z)\mapsto x^2+y^2+z^2</math>. Clearly <math>df|_{p=(x,y,z)} = (2x,2y,2z)</math> and so p is regular <math>\Leftrightarrow df_p\neq 0 \Leftrightarrow p\neq 0</math>

So, the critical values are the image of zero, thus only zero. All other <math>x\in\mathbb{R}</math> are regular values.

''Note:'' In both the last two examples there were points in the target space that were NOT hit by the function and thus are vacuously regular. In the previous example these are the point x<0.

'''Example 3'''

Consider a function <math>\gamma</math> from a segment in <math>\mathbb{R}</math> onto a curve in <math>\mathbb{R}^2</math> such that <math>d\gamma</math> is never zero. Thus, rank(<math>d\gamma = 1</math>) and so <math>d\gamma</math> is never onto. Hence, ALL points are critical in the segment. The points on the curve are critical values, as they are images of critical points, and all points in <math>\mathbb{R}^2</math> NOT on the curve are vacuously regular.

'''Theorem 1'''

''Sard's Theorem''

Almost every <math>y\in N^n</math> is regular <math>\Leftrightarrow</math> the set of critical values of f is of measure zero.

''Note:'' The measure is not specified (indeed, for a topological space there is no canonical measure defined). However the statement will be true for any measure.

'''Theorem 2'''

If <math>f:M^m\rightarrow N^n</math> is smooth and y is a regular value then <math>f^{-1}(y)</math> is an embedded submanifold of <math>M^m</math> of dimension m-n.

'''Re: Example 2'''

<math>f^{-1}(y)</math> is a sphere and hence (again!) the sphere is a manifold

'''Re: Example 3'''

<math>f^{-1}(y)</math> for regular y is empty and hence we get the trivial result that the empty set is a manifold

'''Proof of Theorem 2'''

Let <math>f:M^m\rightarrow N^n</math> is smooth and y is a regular value. Pick a <math>p\in f^{-1}(y)</math>. p is a regular point and thus <math>df_p</math> is onto. Hence, by the submersion property (Reminder 2) we can find a "good charts" thats maps a neighborhood U of p by projection to a neighborhood V about y. Indeed, on U f looks like <math>\mathbb{R}^n\times\mathbb{R}^{m-n}\rightarrow \mathbb{R}^n</math> by <math>(x,z)\mapsto x</math>.

So<math> f^{-1}(0) = \{(0,z)\} = \mathbb{R}^{m-n}</math>. ''Q.E.D''

'''Diversion'''

Arbitrary objects can be described in two ways:

1) With a constructive definition

2) with an implicit definition

For example, a constructive definition of lines in <math>\mathbb{R}^3</math> is given by <math>\{v_1 + tv_2\}</math>
but implicitly they are the solutions to the equations <math>ax+by+cz = d</math> and <math>ez+fy+gz+h</math>.

Hence in general, a constructive definition can be given in terms of an image and an implicit definition can be given in terms of a kernal.

Homological algebra is concerned with the difference between these philosophical approaches.

'''Remark'''

For submanifolds of smooth manifolds, there is no difference between the methods of definition.

'''Definition 3 '''

Loosely we have the idea that a concave and convex curve which just touch at a tangent point is a "bad" intersection as it is unstable under small perturbation where as the intersection point in an X (thought of as being in <math>\mathbb{R}^2</math>) is a "good" intersection as it IS stable under small perturbations.

Precisely,

Let <math>N_1^{n_1}, N_2^{n_2} \subset M</math> be smooth submanifolds. Let <math>p\in N_1^{n_1} \bigcap N_2^{n_2}</math>

We say <math>N_1</math> is ''transverse'' to <math>N_2</math> in M at p if <math>T_p N_1\subset T_p M</math> and<math> T_p N_2 \subset T_p M</math> satisfy <math>T_p N_1 + T_p N_2 = T_p M</math>

'''Example 4'''

Our concave intersecting with convex curve example intersecting tangentially has both of their tangent spaces at the intersection point being the same line and thus does not intersect transversally as the sum of the tangent spaces is not all of <math>\mathbb{R}^2</math>.

Our X example does however work.

===Typed Notes - Second Hour===

''Assistance needed'': There is a symbol for "intersects transversally" but I am not sure of the latex command. See the scanned notes for what this looks like, I will just write it.

'''Definition 4'''

<math>N_1</math> intersects <math>N_2</math> transversally if <math>N_1</math> is transversal to <math>N_2</math> at every point

'''Theorem 3'''

If <math>N_1^{n_1}\cap N_2^{n_2} \subset M</math> transversally then

1) <math>N_1\cap N_2</math> is a manifold of dimension <math>n_1 + n_2 - m</math>

2) Locally can find charts so that <math>N_1 = \mathbb{R}^{n_1} \times 0^{m-n_1}</math> and <math>N_2 = 0^{m-n_2} \times \mathbb{R}^{n_2}</math> with <math>N_1\cap N_2 = 0^{m-n_2} \times \mathbb{R}^{n_1+n_2 - 2m} \times 0^{m-n_1}</math>

''Recall a Thm from Linear Algebra''

If <math>W_1</math> and <math>W_2</math> subspaces of V then <math>dim(W_1 + W_2) + dim(W_1\cap W_2) = dim(W_1) + dim)W_2)</math>

In particular if <math>W_1 + W_2 = V</math> then <math>dim(W_1\cap W_2) = -dim(V) + dim(W_1) + dim(W_2)</math>

'''Proof Scheme for Theorem 3'''

We can write <math>N_i = \varphi_i^{-1}(0)</math> for some such <math>\varphi_i:U\subset M \rightarrow\mathbb{R}^{m-n_i = cod(N_i)}</math>

We then write <math>\varphi = \varphi_1\times\varphi_2: M\rightarrow \mathbb{R}^{m-n_1}\times\mathbb{R}^{m-n_2} = \mathbb{R}^{2m - n_1 - n_2}</math>

Hence, <math>N_1\cap N_2 = \varphi^{-1}(0)</math>

We want <math>rank(d\varphi_p) = 2m-n_1-n_2</math>. To prove this, we consider the aforementioned theorem from linear algebra with respect to the vector spaces obeying <math>ker(d\varphi) = ker(d\varphi_1)\cap ker(d\varphi_2)</math>

hence, and by rank nullity, <math>rank(d\varphi_p) = m - dim(ker(d\varphi)) = m - \left( dim(TN_1) + dim(TN_2) - dim(TN_1 + TN_2)\right) = </math>

<math>m - (n_1 + n_2 - m) = 2m- n_1 - n_2</math>
as we wanted.

This shows that 0 is a regular value and hence by our previous theorem <math>N_1\cap N_2</math> is a submanifold.

Now, we know we can construct the following diagram,

<math>\begin{matrix}
N_1\cap N_2 & \rightarrow^{\varphi} & \mathbb{R}^{2m-n_1-n_2}\\
\downarrow &&\downarrow^{\iota}\\
M&\rightarrow^{\lambda} & \mathbb{R}^m\\
\end{matrix}</math>

where <math>\iota(x) = (x,0)</math>

We then set <math>\psi =</math> (the function that takes the first <math>n_1 +n_2-m</math> coordinates only)<math>\circ\lambda</math>

hence, <math>\psi: M\rightarrow\mathbb{R}^{n_1 + n_2 - m}</math> and <math>\psi |_{N_1\cap N_2}</math> is a chart for <math>N_1\cap N_2</math>

We now consider <math>\zeta: M\rightarrow\mathbb{R}^m</math> given by <math>(\varphi_2,\psi, \varphi_1)</math>

i.e. operating via the following table,

<math>\begin{matrix}
&N_1&N_2\\
\varphi_2&\zeta&0\\
\psi&\zeta&\zeta\\
\varphi_1&0&\zeta\\
\end{matrix}</math>

Then,

<math>d\zeta = \begin{bmatrix}
I&0\\
I&I\\
0&I\\
\end{bmatrix}</math>

for blocks I of the appropriate sizes.

Thus (loosely) ''Q.E.D''

Now on to some examples and comments about why Sard's Theorem is expected, but not obvious:

'''Example 5'''

Consider a standard "first year" smooth function <math>f:\mathbb{R}\rightarrow\mathbb{R}</math>. The "critical points" are in the first year calculus sense where the derivative is zero and the critical values are the images of these points. hence, the set of critical values we expect to be "small"

'''Example 6'''

Consider the function that folds the plane in half. The critical points are along the fold, as are the critical values and this line has 1 dimension and so of trivial measure in the plane (not that we have not given it a measure yet!)

'''Claims:'''

1) <math>\exists f:\mathbb{R}\rightarrow\mathbb{R}</math> whose critical values are homeomorphic to a cantor set.

2) <math>\exists</math> cantor sets with measure arbitrarily close to 1

3) <math>\exists g\in C^1:\mathbb{R}^2\rightarrow\mathbb{R}</math> whose critical points are a cantor set cross a cantor set and whose critical values are everything. Hence we will need our functions to be <math>C^{\infty}</math> in the theorem.

==Evil functions==

===Example 1===
'''Nota benne:''' Here we are using the name Cantor set for any perfect set with empty interior.

There exists a function <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> smooth such that its set of critical values is homeomorphic to a Cantor set.

Remember that
<math>g_{a,b}:[a,b]\rightarrow \mathbb{R}</math>

<math>g_{a,b}(x)=e^{-1/(x-a)^2}e^{-1/(x-b)^2}</math>

is a smooth function such that <math>g_{a,b}(a)=g_{a,b}(b)=0</math>

We can define the function <math>h</math> in the complement of a Cantor set using the appropriate <math>g_{a,b}</math> in the intervals of the complement.

Notices that <math>f(x)=\int_{0}^{x}h(t)dt</math> holds the conditions of the example.

We can divide the new <math>g_{a,b}</math> in each step of the construction by <math>1/2^n</math> just to make the integral converge. And of course define h(t)=0 on the Cantor set.

Observe that, since <math>h</math> is non negative, <math>f</math> is increasing (observe it is strictly increasing)(it is continuous too!).
Since increasing continuous functions have continuous inverses it is a homeomorphism.

0708-1300/Class notes for Tuesday, October 9

2007-10-31T00:33:10Z

Aptikuis: /* Typed Notes - First Hour */

{{0708-1300/Navigation}}

==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

=== Scanned Notes ===
Scanned notes for today's class can be found [[0708-1300/Scanned Class notes for Tuesday, October 9| here]]

(see discussion)

===Typed Notes - First Hour===

'''Reminder:'''

1) An immersion locally looks like <math>\mathbb{R}^n\rightarrow\mathbb{R}^m</math> given by <math>x\mapsto(x,0)</math>

2) A submersion locally looks like <math>\mathbb{R}^m\rightarrow\mathbb{R}^n</math> given by <math>(x,y)\mapsto x</math>

'''Today's Goals'''

1) More about "locally things look like their differential"

2) The tricky Sard's Theorem: "Evil points are rare, good points everywhere"

'''Definition 1'''

Let <math>f:M^m\rightarrow N^n</math> be smooth. A point <math>p\in M^m</math> is ''critical'' if <math>df_p</math> is not onto <math>\Leftrightarrow</math> rank <math>df_p <n</math>. Otherwise, p is ''regular.

'''Definition 2'''
A point <math>y\in N^n</math> is a ''critical value'' of f if <math>\exists p\in M^m</math> such that p is critical and <math>f(p) = y</math>. Otherwise, y is a ''regular value''

'''Example 1'''

Consider the map <math>f:S^2\rightarrow\mathbb{R}^2</math> given by <math>(x,y,z)\mapsto (x,y)</math>. I.e., the projection map. The regular points are all the points on <math>S^2</math> except the equator. The regular values, however, are all <math>(x,y)</math> such that <math>x^2+y^2 \neq 1</math>

'''Example 2'''

Consider <math>f:\mathbb{R}^3\rightarrow\mathbb{R}</math> given by <math>p\mapsto ||p||^2</math>.
That is, <math>(x,y,z)\mapsto x^2+y^2+z^2</math>. Clearly <math>df|_{p=(x,y,z)} = (2x,2y,2z)</math> and so p is regular <math>\Leftrightarrow df_p\neq 0 \Leftrightarrow p\neq 0</math>

So, the critical values are the image of zero, thus only zero. All other <math>x\in\mathbb{R}</math> are regular values.

''Note:'' In both the last two examples there were points in the target space that were NOT hit by the function and thus are vacuously regular. In the previous example these are the point x<0.

'''Example 3'''

Consider a function <math>\gamma</math> from a segment in <math>\mathbb{R}</math> onto a curve in <math>\mathbb{R}^2</math> such that <math>d\gamma</math> is never zero. Thus, rank(<math>d\gamma = 1</math>) and so <math>d\gamma</math> is never onto. Hence, ALL points are critical in the segment. The points on the curve are critical values, as they are images of critical points, and all points in <math>\mathbb{R}^2</math> NOT on the curve are vacuously regular.

'''Theorem 1'''

''Sard's Theorem''

Almost every <math>y\in N^n</math> is regular <math>\Leftrightarrow</math> the set of critical values of f is of measure zero.

''Note:'' The measure is not specified (indeed, for a topological space there is no canonical measure defined). However the statement will be true for any measure.

'''Theorem 2'''

If <math>f:M^m\rightarrow N^n</math> is smooth and y is a regular value then <math>f^{-1}(y)</math> is an embedded submanifold of <math>M^m</math> of dimension m-n.

'''Re: Example 2'''

<math>f^{-1}(y)</math> is a sphere and hence (again!) the sphere is a manifold

'''Re: Example 3'''

<math>f^{-1}(y)</math> for regular y is empty and hence we get the trivial result that the empty set is a manifold

'''Proof of Theorem 2'''

Let <math>f:M^m\rightarrow N^n</math> is smooth and y is a regular value. Pick a <math>p\in f^{-1}(y)</math>. p is a regular point and thus <math>df_p</math> is onto. Hence, by the submersion property (Reminder 2) we can find a "good charts" thats maps a neighborhood U of p by projection to a neighborhood V about y. Indeed, on U f looks like <math>\mathbb{R}^n\times\mathbb{R}^{m-n}\rightarrow \mathbb{R}^n</math> by <math>(x,z)\mapsto x</math>.

So<math> f^{-1}(0) = \{(0,z)\} = \mathbb{R}^{m-n}</math>. ''Q.E.D''

'''Diversion'''

Arbitrary objects can be described in two ways:

1) With a constructive definition

2) with an implicit definition

For example, a constructive definition of lines in <math>\mathbb{R}^3</math> is given by <math>\{v_1 + tv_2\}</math>
but implicitly they are the solutions to the equations <math>ax+by+cz = d</math> and <math>ez+fy+gz+h</math>.

Hence in general, a constructive definition can be given in terms of an image and an implicit definition can be given in terms of a kernal.

Homological algebra is concerned with the difference between these philosophical approaches.

'''Remark'''

For submanifolds of smooth manifolds, there is no difference between the methods of definition.

'''Definition 3 '''

Loosely we have the idea that a concave and convex curve which just touch at a tangent point is a "bad" intersection as it is unstable under small perturbation where as the intersection point in an X (thought of as being in <math>\mathbb{R}^2</math>) is a "good" intersection as it IS stable under small perturbations.

Precisely,

Let <math>N_1^{n_1}, N_2^{n_2} \subset M</math> be smooth submanifolds. Let <math>p\in N_1^{n_1} \bigcap N_2^{n_2}</math>

We say <math>N_1</math> is ''transverse'' to <math>N_2</math> in M at p if <math>T_p N_1\subset T_p M</math> and<math> T_p N_2 \subset T_p M</math> satisfy <math>T_p N_1 + T_p N_2 = T_p M</math>

'''Example 4'''

Our concave intersecting with convex curve example intersecting tangentially has both of their tangent spaces at the intersection point being the same line and thus does not intersect transversally as the sum of the tangent spaces is not all of <math>\mathbb{R}^2</math>.

Our X example does however work.

===Typed Notes - Second Hour===

''Assistance needed'': There is a symbol for "intersects transversally" but I am not sure of the latex command. See the scanned notes for what this looks like, I will just write it.

'''Definition 4'''

<math>N_1</math> intersects <math>N_2</math> transversally if <math>N_1</math> is transversal to <math>N_2</math> at every point

'''Theorem 3'''

If <math>N_1^{n_1}\cap N_2^{n_2} \subset M</math> transversally then

1) <math>N_1\cap N_2</math> is a manifold of dimension <math>n_1 + n_2 - m</math>

2) Locally can find charts so that <math>N_1 = \mathbb{R}^{n_1} \times 0^{m-n_1}</math> and <math>N_2 = 0^{m-n_2} \times \mathbb{R}^{n_2}</math> with <math>N_1\cap N_2 = 0^{m-n_2} \times \mathbb{R}^{n_1+n_2 - 2m} \times 0^{m-n_1}</math>

''Recall a Thm from Linear Algebra''

If <math>W_1</math> and <math>W_2</math> subspaces of V then <math>dim(W_1 + W_2) + dim(W_1\cap W_2) = dim(W_1) + dim)W_2)</math>

In particular if <math>W_1 + W_2 = V</math> then <math>dim(W_1\cap W_2) = -dim(V) + dim(W_1) + dim(W_2)</math>

'''Proof Scheme for Theorem 3'''

We can write <math>N_i = \varphi_i^{-1}(0)</math> for some such <math>\varphi_i:U\subset M \rightarrow\mathbb{R}^{m-n_1 = cod(N_i)}</math>

We then write <math>\varphi = \varphi_1\times\varphi_2: M\rightarrow \mathbb{R}^{m-n_1}\times\mathbb{R}^{m-n_2} = \mathbb{R}^{2m - n_1 - n_2}</math>

Hence, <math>N_1\cap N_2 = \varphi^{-1}(0)</math>

We want <math>rank(d\varphi_p) = 2m-n_1-n_2</math>. To prove this, we consider the aforementioned theorem from linear algebra with respect to the vector spaces obeying <math>ker(d\varphi) = ker(d\varphi_1)\cap ker(d\varphi_2)</math>

hence, and by rank nullity, <math>rank(d\varphi_p) = m - dim(ker(d\varphi)) = m - \left( dim(TN_1) + dim(TN_2) - dim(TN_1 + TN_2)\right) = </math>

<math>m - (n_1 + n_2 - m) = 2m- n_1 - n_2</math>
as we wanted.

This shows that 0 is a regular point and hence by our previous theorem <math>N_1\cap N_2</math> is a submanifold.

Now, we know we can construct the following diagram,

<math>\begin{matrix}
N_1\cap N_2 & \rightarrow^{\varphi} & \mathbb{R}\\
\downarrow &&\downarrow^{\iota}\\
M&\rightarrow^{\lambda} & \mathbb{R}^m\\
\end{matrix}</math>

where <math>\iota(x) = (x,0)</math>

We then set <math>\psi =</math> (the function that takes the first <math>n_1 +n_2-m</math> coordinates only)<math>\circ\lambda</math>

hence, <math>\psi: M\rightarrow\mathbb{R}^{n_1 + n_2 - m}</math> and <math>\psi |_{N_1\cap N_2}</math> is a chart for <math>N_1\cap N_2</math>

We now consider <math>\zeta: M\rightarrow\mathbb{R}^m</math> given by <math>(\varphi_2,\psi, \varphi_1)</math>

i.e. operating via the following table,

<math>\begin{matrix}
&N_1&N_2\\
\varphi_2&\zeta&0\\
\psi&\zeta&\zeta\\
\varphi_1&0&\zeta\\
\end{matrix}</math>

Then,

<math>d\zeta = \begin{bmatrix}
I&0\\
I&I\\
0&I\\
\end{bmatrix}</math>

for blocks I of the appropriate sizes.

Thus (loosely) ''Q.E.D''

Now on to some examples and comments about why Sard's Theorem is expected, but not obvious:

'''Example 5'''

Consider a standard "first year" smooth function <math>f:\mathbb{R}\rightarrow\mathbb{R}</math>. The "critical points" are in the first year calculus sense where the derivative is zero and the critical values are the images of these points. hence, the set of critical values we expect to be "small"

'''Example 6'''

Consider the function that folds the plane in half. The critical points are along the fold, as are the critical values and this line has 1 dimension and so of trivial measure in the plane (not that we have not given it a measure yet!)

'''Claims:'''

1) <math>\exists f:\mathbb{R}\rightarrow\mathbb{R}</math> whose critical values are homeomorphic to a cantor set.

2) <math>\exists</math> cantor sets with measure arbitrarily close to 1

3) <math>\exists g\in C^1:\mathbb{R}^2\rightarrow\mathbb{R}</math> whose critical points are a cantor set cross a cantor set and whose critical values are everything. Hence we will need our functions to be <math>C^{\infty}</math> in the theorem.

==Evil functions==

===Example 1===
'''Nota benne:''' Here we are using the name Cantor set for any perfect set with empty interior.

There exists a function <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> smooth such that its set of critical values is homeomorphic to a Cantor set.

Remember that
<math>g_{a,b}:[a,b]\rightarrow \mathbb{R}</math>

<math>g_{a,b}(x)=e^{-1/(x-a)^2}e^{-1/(x-b)^2}</math>

is a smooth function such that <math>g_{a,b}(a)=g_{a,b}(b)=0</math>

We can define the function <math>h</math> in the complement of a Cantor set using the appropriate <math>g_{a,b}</math> in the intervals of the complement.

Notices that <math>f(x)=\int_{0}^{x}h(t)dt</math> holds the conditions of the example.

We can divide the new <math>g_{a,b}</math> in each step of the construction by <math>1/2^n</math> just to make the integral converge. And of course define h(t)=0 on the Cantor set.

Observe that, since <math>h</math> is non negative, <math>f</math> is increasing (observe it is strictly increasing)(it is continuous too!).
Since increasing continuous functions have continuous inverses it is a homeomorphism.

0708-1300/Class notes for Tuesday, October 9

2007-10-11T01:21:14Z

Aptikuis: /* Typed Notes - Second Hour */

{{0708-1300/Navigation}}

==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

=== Scanned Notes ===
Scanned notes for today's class can be found [[0708-1300/Scanned Class notes for Tuesday, October 9| here]]

(see discussion)

===Typed Notes - First Hour===

'''Reminder:'''

1) An immersion locally looks like <math>\mathbb{R}^n\rightarrow\mathbb{R}^m</math> given by <math>x\mapsto(x,0)</math>

2) A submersion locally looks like <math>\mathbb{R}^m\rightarrow\mathbb{R}^n</math> given by <math>(x,y)\mapsto x</math>

'''Today's Goals'''

1) More about "locally things look like their differential"

2) The tricky Sard's Theorem: "Evil points are rare, good points everywhere"

'''Definition 1'''

Let <math>f:M^m\rightarrow N^n</math> be smooth. A point <math>p\in M^m</math> is ''critical'' if <math>df_p</math> is not onto <math>\Leftrightarrow</math> rank <math>df_p <n</math>. Otherwise, p is ''regular.

'''Definition 2'''
A point <math>y\in N^n</math> is a ''critical value'' of f if <math>\exists p\in M^m</math> such that p is critical and <math>f(p) = y</math>. Otherwise, y is a ''regular value''

'''Example 1'''

Consider the map <math>f:S^2\rightarrow\mathbb{R}^2</math> given by <math>(x,y,z)\mapsto (x,y)</math>. I.e., the projection map. The regular points are all the points on <math>S^2</math> except the equator. The regular values, however, are all <math>(x,y)</math> such that <math>x^2+y^2 \neq 1</math>

'''Example 2'''

Consider <math>f:\mathbb{R}^3\rightarrow\mathbb{R}</math> given by <math>p\mapsto ||p||^2</math>.
That is, <math>(x,y,z)\mapsto x^2+y^2+z^2</math>. Clearly <math>df|_{p=(x,y,z)} = (2x,2y,2z)</math> and so p is regular <math>\Leftrightarrow df_p\neq 0 \Leftrightarrow p\neq 0</math>

So, the critical values are the image of zero, thus only zero. All other <math>x\in\mathbb{R}</math> are regular values.

''Note:'' In both the last two examples there were points in the target space that were NOT hit by the function and thus are vacuously regular. In the previous example these are the point x<0.

'''Example 3'''

Consider a function <math>\gamma</math> from a segment in <math>\mathbb{R}</math> onto a curve in <math>\mathbb{R}^2</math> such that <math>d\gamma</math> is never zero. Thus, rank(<math>d\gamma = 1</math>) and so <math>d\gamma</math> is never onto. Hence, ALL points are critical in the segment. The points on the curve are critical values, as they are images of critical points, and all points in <math>\mathbb{R}^2</math> NOT on the curve are vacuously regular.

'''Theorem 1'''

''Sard's Theorem''

Almost every <math>y\in N^n</math> is regular <math>\Leftrightarrow</math> the set of critical values of f is of measure zero.

''Note:'' The measure is not specified (indeed, for a topological space there is no canonical measure defined). However the statement will be true for any measure.

'''Theorem 2'''

If <math>f:M^m\rightarrow N^n</math> is smooth and y is a regular value then <math>f^{-1}(y)</math> is an embedded submanifold of <math>M^m</math> of dimension m-n.

'''Re: Example 2'''

<math>f^{-1}(y)</math> is a sphere and hence (again!) the sphere is a manifold

'''Re: Example 3'''

<math>f^{-1}(y)</math> for regular y is empty and hence we get the trivial result that the empty set is a manifold

'''Proof of Theorem 2'''

Let <math>f:M^m\rightarrow N^n</math> is smooth and y is a regular value. Pick a <math>p\in f^{-1}(y)</math>. p is a regular point and thus <math>df_p</math> is onto. Hence, by the submersion property (Reminder 2) we can find a "good charts" thats maps a neighborhood U of p by projection to a neighborhood V about y. Indeed, on U f looks like <math>\mathbb{R}^n\times\mathbb{R}^{m-1}\rightarrow \mathbb{R}^n</math> by <math>(x,z)\mapsto x</math>.

So<math> f^{-1}(0) = \{(0,z)\} = \mathbb{R}^{m-n}</math>. ''Q.E.D''

'''Diversion'''

Arbitrary objects can be described in two ways:

1) With a constructive definition

2) with an implicit definition

For example, a constructive definition of lines in <math>\mathbb{R}^3</math> is given by <math>\{v_1 + tv_2\}</math>
but implicitly they are the solutions to the equations <math>ax+by+cz = d</math> and <math>ez+fy+gz+h</math>.

Hence in general, a constructive definition can be given in terms of an image and an implicit definition can be given in terms of a kernal.

Homological algebra is concerned with the difference between these philosophical approaches.

'''Remark'''

For submanifolds of smooth manifolds, there is no difference between the methods of definition.

'''Definition 3 '''

Loosely we have the idea that a concave and convex curve which just touch at a tangent point is a "bad" intersection as it is unstable under small perturbation where as the intersection point in an X (thought of as being in <math>\mathbb{R}^2</math>) is a "good" intersection as it IS stable under small perturbations.

Precisely,

Let <math>N_1^{n_1}, N_2^{n_2} \subset M</math> be smooth submanifolds. Let <math>p\in N_1^{n_1} \bigcap N_2^{n_2}</math>

We say <math>N_1</math> is ''transverse'' to <math>N_2</math> in M at p if for <math>T_p N_1\subset T_p M</math> and<math> T_p N_2 \subset T_p M</math> satisfies <math>T_p N_1 + T_p N_2 = T_p M</math>

'''Example 4'''

Our concave intersecting with convex curve example intersecting tangentially has both of their tangent spaces at the intersection point being the same line and thus does not intersect transversally as the sum of the tangent spaces is not all of <math>\mathbb{R}^2</math>.

Our X example does however work.

===Typed Notes - Second Hour===

''Assistance needed'': There is a symbol for "intersects transversally" but I am not sure of the latex command. See the scanned notes for what this looks like, I will just write it.

'''Definition 4'''

<math>N_1</math> intersects <math>N_2</math> transversally if <math>N_1</math> is transversal to <math>N_2</math> at every point

'''Theorem 3'''

If <math>N_1^{n_1}\cap N_2^{n_2} \subset M</math> transversally then

1) <math>N_1\cap N_2</math> is a manifold of dimension <math>n_1 + n_2 - m</math>

2) Locally can find charts so that <math>N_1 = \mathbb{R}^{n_1} \times 0^{m-n_1}</math> and <math>N_2 = 0^{m-n_2} \times \mathbb{R}^{n_2}</math> with <math>N_1\cap N_2 = 0^{m-n_2} \times \mathbb{R}^{n_1+n_2 - 2m} \times 0^{m-n_1}</math>

''Recall a Thm from Linear Algebra''

If <math>W_1</math> and <math>W_2</math> subspaces of V then <math>dim(W_1 + W_2) + dim(W_1\cap W_2) = dim(W_1) + dim)W_2)</math>

In particular if <math>W_1 + W_2 = V</math> then <math>dim(W_1\cap W_2) = -dim(V) + dim(W_1) + dim(W_2)</math>

'''Proof Scheme for Theorem 3'''

We can write <math>N_i = \varphi_i^{-1}(0)</math> for some such <math>\varphi_i:U\subset M \rightarrow\mathbb{R}^{m-n_1 = cod(N_i)}</math>

We then write <math>\varphi = \varphi_1\times\varphi_2: M\rightarrow \mathbb{R}^{m-n_1}\times\mathbb{R}^{m-n_2} = \mathbb{R}^{2m - n_1 - n_2}</math>

Hence, <math>N_1\cap N_2 = \varphi^{-1}(0)</math>

We want <math>rank(d\varphi_p) = 2m-n_1-n_2</math>. To prove this, we consider the aforementioned theorem from linear algebra with respect to the vector spaces obeying <math>ker(d\varphi) = ker(d\varphi_1)\cap ker(d\varphi_2)</math>

hence, and by rank nullity, <math>rank(d\varphi_p) = m - dim(ker(d\varphi)) = m - \left( dim(TN_1) + dim(TN_2) - dim(TN_1 + TN_2)\right) = </math>

<math>m - (n_1 + n_2 - m) = 2m- n_1 - n_2</math>
as we wanted.

This shows that 0 is a regular point and hence by our previous theorem <math>N_1\cap N_2</math> is a submanifold.

Now, we know we can construct the following diagram,

<math>\begin{matrix}
N_1\cap N_2 & \rightarrow^{\varphi} & \mathbb{R}\\
\downarrow &&\downarrow^{\iota}\\
M&\rightarrow^{\lambda} & \mathbb{R}^m\\
\end{matrix}</math>

where <math>\iota(x) = (x,0)</math>

We then set <math>\psi =</math> (the function that takes the first <math>n_1 +n_2-m</math> coordinates only)<math>\circ\lambda</math>

hence, <math>\psi: M\rightarrow\mathbb{R}^{n_1 + n_2 - m}</math> and <math>\psi |_{N_1\cap N_2}</math> is a chart for <math>N_1\cap N_2</math>

We now consider <math>\zeta: M\rightarrow\mathbb{R}^m</math> given by <math>(\varphi_2,\psi, \varphi_1)</math>

i.e. operating via the following table,

<math>\begin{matrix}
&N_1&N_2\\
\varphi_2&\zeta&0\\
\psi&\zeta&\zeta\\
\varphi_1&0&\zeta\\
\end{matrix}</math>

Then,

<math>d\zeta = \begin{bmatrix}
I&0\\
I&I\\
0&I\\
\end{bmatrix}</math>

for blocks I of the appropriate sizes.

Thus (loosely) ''Q.E.D''

Now on to some examples and comments about why Sard's Theorem is expected, but not obvious:

'''Example 5'''

Consider a standard "first year" smooth function <math>f:\mathbb{R}\rightarrow\mathbb{R}</math>. The "critical points" are in the first year calculus sense where the derivative is zero and the critical values are the images of these points. hence, the set of critical values we expect to be "small"

'''Example 6'''

Consider the function that folds the plane in half. The critical points are along the fold, as are the critical values and this line has 1 dimension and so of trivial measure in the plane (not that we have not given it a measure yet!)

'''Claims:'''

1) <math>\exists f:\mathbb{R}\rightarrow\mathbb{R}</math> whose critical values are homeomorphic to a cantor set.

2) <math>\exists</math> cantor sets with measure arbitrarily close to 1

3) <math>\exists g\in C^1:\mathbb{R}^2\rightarrow\mathbb{R}</math> whose critical points are a cantor set cross a cantor set and whose critical values are everything. Hence we will need our functions to be <math>C^{\infty}</math> in the theorem.

==Evil functions==

===Example 1===
'''Nota benne:''' Here we are using the name Cantor set for any perfect set with empty interior.

There exists a function <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> smooth such that its set of critical values is homeomorphic to a Cantor set.

Remember that
<math>g_{a,b}:[a,b]\rightarrow \mathbb{R}</math>

<math>g_{a,b}(x)=e^{-1/(x-a)^2}e^{-1/(x-b)^2}</math>

is a smooth function such that <math>g_{a,b}(a)=g_{a,b}(b)=0</math>

We can define the function <math>h</math> in the complement of a Cantor set using the appropriate <math>g_{a,b}</math> in the intervals of the complement.

Notices that <math>f(x)=\int_{0}^{x}h(t)dt</math> holds the conditions of the example.

We can divide the new <math>g_{a,b}</math> in each step of the construction by <math>1/2^n</math> just to make the integral converge. And of course define h(t)=0 on the Cantor set.

Observe that, since <math>h</math> is non negative, <math>f</math> is increasing (observe it is strictly increasing)(it is continuous too!).
Since increasing continuous functions have continuous inverses it is a homeomorphism.

0708-1300/Class notes for Tuesday, October 2

2007-10-06T02:41:48Z

Aptikuis: /* Second Hour */

{{0708-1300/Navigation}}
==English Spelling==
Many interesting rules about [[0708-1300/English Spelling]]
==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

'''General class comments'''

1) The class photo is up, please add yourself

2) A questionnaire was passed out in class

3) Homework one is due on thursday

===First Hour===

Today's Theme: Locally a function looks like its differential

'''Pushforward/Pullback'''

Let <math>\theta:X\rightarrow Y</math> be a smooth map.

We consider various objects, defined with respect to X or Y, and see in which direction it makes sense to consider corresponding objects on the other space. In general <math>\theta_*</math> will denote the push forward, and <math>\theta^*</math> will denote the pullback.

1) points ''pushforward'' <math>x\mapsto\theta_*(x) := \theta(x)</math>

2) Paths <math>\gamma:\mathbb{R}\rightarrow X</math>, ie a bunch of points, ''pushforward'', <math>\gamma\rightarrow \theta_*(\gamma):=\theta\circ\gamma</math>

3) Sets <math>B\subset Y</math> ''pullback'' via <math>B\rightarrow \theta^*(B):=\theta^{-1}(B)</math>
Note that if one tried to pushforward sets A in X, the set operations compliment and intersection would not commute appropriately with the map <math>\theta</math>

4) A measures <math>\mu</math> ''pushforward'' via <math>\mu\rightarrow (\theta_*\mu)(B) :=\mu(\theta^*B)</math>

5)In some sense, we consider functions, "dual" to points and thus should go in the opposite direction of points, namely <math>\theta^*f = f\circ\theta</math>

6) Tangent vectors, defined in the sense of equivalence classes of paths, [<math>\gamma</math>] ''pushforward'' as we would expect since each path pushes forward. <math>[\gamma]\rightarrow \theta_*[\gamma]:=[\theta_*\gamma] = [\theta\circ\gamma]</math>

CHECK: This definition is well defined, that is, independent of the representative choice of <math>\gamma</math>

7) We can consider operators on functions to be in a sense dual to the functions and hence should go in the opposite direction. Hence, tangent vectors, defined in the sense of derivations, ''pushforward'' via <math>D\rightarrow (\theta_*D)(f):= D(\theta^*f)</math>

CHECK: This definition satisfies linearity and Liebnitz property.

'''Theorem 1'''

The two definitions for the pushforward of a tangent vector coincide.

''Proof:''

Given a <math>\gamma</math> we can construct <math>\theta_{*}\gamma</math> as above. However from both <math>\gamma</math> and <math>\theta_*\gamma</math> we can also construct <math>D_{\gamma}f</math> and <math>D_{\theta_*\gamma}f</math> because we have previously shown our two definitions for the tangent vector are equivalent. We can then ''pushforward'' <math>D_{\gamma}f</math> to get <math>\theta_*D_{\gamma}f</math>. The theorem is reduced to the claim that:

<math>\theta_*D_{\gamma}f = D_{\theta_*\gamma}f</math>

for functions <math>f:Y\rightarrow \mathbb{R}</math>

Now, <math>D_{\theta_*\gamma}f = \frac{d}{dt}f\circ(\theta_*\gamma)|_{t=0} = \frac{d}{dt}f\circ(\theta\circ\gamma)|_{t=0} = \frac{d}{dt}(f\circ\theta)\circ\gamma |_{t=0} = D_{\gamma}(f\circ\theta) =\theta_*D_{\gamma}f</math>

''Q.E.D''

'''Functorality'''

let <math>\theta:X\rightarrow Y, \lambda:Y\rightarrow Z</math>

Consider some "object" s defined with respect to X and some "object u" defined with respect to Z. Something has the property of functorality if

<math>\lambda_*(\theta_*s) = (\lambda\circ\theta)_*s</math>

and

<math>\theta^*(\lambda^*u) = (\lambda\circ\theta)^*u</math>

Claim: All the classes we considered previously have the functorality property; in particular, the pushforward of tangent vectors does.

Let us consider <math>\theta_*</math> on <math>T_pM</math> given a <math>\theta:M\rightarrow N</math>

We can arrange for charts <math>\varphi</math> on a subset of M into <math>\mathbb{R}^m</math> (with coordinates denoted <math>(x_1,\dots,x_m)</math>)and <math>\psi</math> on a subset of N into <math>\mathbb{R}^n</math> (with coordinates denoted <math>(y_1,\dots,y_n)</math>)such that <math>\varphi(p) = 0</math> and <math>\psi(\theta(p))=0</math>

Define <math>\theta^o = \psi\circ\theta\circ\varphi^{-1} = (\theta_1(x_1,\dots,x_m),\dots,\theta_n(x_1,\dots,x_m))</math>

Now, for a <math>D\in T_pM</math> we can write <math>D=\sum_{i=1}^m a_i\frac{\partial}{\partial x_i}</math>

So, <math>(\theta_*D)(f) = \sum_{i=1}^m a_i\frac{\partial}{\partial x_i}(f\circ\varphi) = \sum_{i=1}^m a_i \sum_{j=1}^n\frac{\partial f}{\partial y_j}\frac{\partial\theta_j}{\partial x_i}=</math>
<math>=\begin{bmatrix}
\frac{\partial f}{\partial y_1} & \cdots & \frac{\partial f}{\partial y_n}\\
\end{bmatrix}
\begin{bmatrix}

\frac{\partial \theta_1}{\partial x_1} & \cdots & \frac{\partial \theta_1}{\partial x_m}\\
\vdots& & \vdots\\
\frac{\partial \theta_n}{\partial x_1} & \cdots & \frac{\partial \theta_n}{\partial x_m}\\
\end{bmatrix}
\begin{bmatrix}
a_1\\
\vdots\\
a_m\\
\end{bmatrix}</math>

Now, we want to write <math>\theta_*D = \sum b_j\frac{\partial}{\partial y_j}</math>

and so, <math>b_k = (\theta_*D)y_k =\begin{bmatrix}
0&\cdots & 1 & \cdots &0\\
\end{bmatrix}
\begin{bmatrix}

\frac{\partial \theta_1}{\partial x_1} & \cdots & \frac{\partial \theta_1}{\partial x_m}\\
\vdots& & \vdots\\
\frac{\partial \theta_n}{\partial x_1} & \cdots & \frac{\partial \theta_n}{\partial x_m}\\
\end{bmatrix}
\begin{bmatrix}
a_1\\
\vdots\\
a_m\\
\end{bmatrix}</math>

where the 1 is at the kth location.

So, <math>\theta_* = d\theta_p</math>, i.e., <math>\theta_*</math> is the differential of <math>\theta</math> at p

We can check the functorality, <math>(\lambda\circ\theta)_* = \lambda_*\circ\theta_*</math>, then <math>d(\lambda\circ\theta) = d\lambda\circ d\theta</math>
This is just the chain rule.

===Second Hour===

'''Defintion 1'''

An ''immersion'' is a (smooth) map <math>\theta:M^m\rightarrow N^n</math> such that <math>\theta_*</math> of tangent vectors is 1:1. More precisely, <math>d\theta_p: T_pM\rightarrow T_{\theta(p)}N</math> is 1:1 <math>\forall p\in M</math>

'''Example 1'''

Consider the canonical immersion, for m<n given by <math>\iota:(x_1,...,x_m)\mapsto (x_1,...,x_m,0,...,0)</math> with n-m zeros.

'''Example 2'''

This is the map from <math>\mathbb{R}</math> to <math>\mathbb{R}^2</math> that looks like a loop-de-loop on a roller coaster (but squashed into the plane of course!) The map <math>\theta</math> itself is NOT 1:1 (consider the crossover point) however <math>\theta_*</math> IS 1:1, hence an immersion.

'''Example 3'''

Consider the map from <math>\mathbb{R}</math> to <math>\mathbb{R}^2</math> that looks like a check mark. While this map itself is 1:1, <math>\theta_*</math> is NOT 1:1 (at the cusp in the check mark) and hence is not an immersion.

'''Example 4'''

Can there be objects, such as the graph of |x| that are NOT an immersion, but are constructed from a smooth function?

Consider the function <math>\lambda(x) = e^{-1/x^2}</math> for x>0 and 0 otherwise.

Then the map <math>x\mapsto \begin{bmatrix}
(\lambda(x),\lambda(x))& x>0\\
(0,0)& x=0\\
(-\lambda(-x),\lambda(-x)) & x<0\\
\end{bmatrix}</math>

is a smooth mapping with the graph of |x| as its image, but is NOT an immersion.

'''Example 5'''

The torus, as a subset of <math>\mathbb{R}^3</math> is an immersion

Now, consider the 1:1 linear map <math>T:V\rightarrow W</math> where V,W are vector spaces that takes <math>(v_1,...,v_m)\mapsto (Tv_1,...,Tv_m) = (w_1,..,w_m,w_{m+1},...,w_n)</math>

From linear algebra we know that we can choose a basis such that T is represented by a matrix with 1's along the first m diagonal locations and zeros elsewhere.

'''Theorem 2'''

Locally, every immersion looks like the inclusion <math>\iota:\mathbb{R}^m\rightarrow \mathbb{R}^n</math>.

More precisely, if <math>\theta:M^m\rightarrow N^n</math> and <math>d\theta_p</math> is 1:1 then there exists charts <math>\varphi</math> acting on <math>U\subset M</math> and <math>\phi</math> acting on <math>V\subset N</math> such that for <math>p\in U, \phi(p) = \psi(\theta(p)) = 0</math> such that the following diagram commutes:

<math>\begin{matrix}
U&\rightarrow^{\phi}&U'\subset \mathbb{R}^m\\
\downarrow_{\theta} &&\downarrow_{\iota} \\
V& \rightarrow^{\psi}& V'\subset \mathbb{R}^n\\
\end{matrix}</math>

that is, <math>\iota\circ\varphi = \psi\circ\theta</math>

'''Definition 2'''

M is a ''submanifold'' of N if there exists a mapping <math>\theta:M\rightarrow N</math> such that <math>\theta</math> is a 1:1 immersion.

'''Example 6'''

Our previous example of the graph of a "loop-de-loop", while an immersion, the function is not 1:1 and hence the graph is not a sub manifold.

'''Example 6'''

The torus is a submanifold as the natural immersion into <math>\mathbb{R}^3</math> is 1:1

'''Definition 3'''

The map <math>\theta:M\rightarrow N</math> is an embedding if the subset topology on <math>\theta(M)</math> coincides with the topology induced from the original topology of M.

'''Example 7'''

Consider the map from <math>\mathbb{R}\rightarrow \mathbb{R}^2</math> whose graph looks like the open interval whose two ends have been wrapped around until they just touch (would intersect at one point if they were closed) the points 1/3 and 2/3rds of the way along the interval respectively.
The map is both 1:1 and an immersion. However, any neighborhood about the endpoints of the interval will ALSO include points near the 1/3rd and 2/3rd spots on the line, i.e., the topology is different and hence this is not an embedding.

'''Corollary 1 to Theorem 2'''

The functional structure on an embedded manifold induced by the functional structure on the containing manifold is equal to its original functional structure.

Indeed, for all smooth <math>f:M\rightarrow \mathbb{R}</math> and <math>\forall p\in M</math> there exists a neighborhood V of <math>\theta(p)</math> and a smooth <math>g:N\rightarrow \mathbb{R}</math> such that <math>g|_{\theta(M)\bigcap U} = f|_U</math>

''Proof of Corollary 1''

Loosely (and a sketch is most useful to see this!) we consider the embedded submanifold M in N and consider its image, under the appropriate charts, to a subset of <math>\mathbb{R}^m\subset \mathbb{R}^n</math>. We then consider some function defined on M, and hence on the subset in <math>\mathbb{R}^n</math> which we can extend canonically as a constant function in the "vertical" directions. Now simply pullback into N to get the extended member of the functional structure on N.

''Proof of Theorem 2''

We start with the normal situation of <math>\theta:M\rightarrow N</math> with M,N manifolds with atlases containing <math>(\varphi_0,U_)0)</math> and <math>(\psi_0, V_0)</math> respectively. We also expect that for <math>p\in U_0, \varphi_0(p) = \psi_0(\theta(p)) = 0</math>. I will first draw the diagram and will subsequently justify the relevant parts. The proof reduces to showing a certain part of the diagram commutes appropriately.

<math>
\begin{matrix}
M\supset U_0 & \rightarrow^{\varphi_0} & U_1\subset \mathbb{R}^m & \rightarrow^{Id} & U_2 = U_1 \\
\downarrow_{\theta} & &\downarrow_{\theta_1} & &\downarrow_{\iota}\\
N\supset V_0 & \rightarrow^{\psi_0} & V_1\subset \mathbb{R}^n & \leftarrow^{\xi} & V_2\\
\end{matrix}
</math>

It is very important to note that the <math>\varphi_0</math> and <math>\psi_0</math> are NOT the charts we are looking for , they are merely one of the ones that happen to act about the point p.

In the diagram above, <math>\theta_1 = \psi_0\circ\theta\circ\varphi^{-1}</math>. So, <math>\theta_1(0) = 0</math> and <math>(d\theta_1)_0 = i</math>. Note the <math>\theta_1</math>, being merely the normal composition with the appropriate charts, does not fundamentally add anything. What makes this theorem work is the function <math>\xi</math>

We consider the map <math>\xi:V_2\rightarrow V_1</math> given <math>(x,y)\mapsto \theta_1(x) + (0,y)</math>. We note that <math>\xi</math> corresponds with the idea of "lifting" a flattened image back to its original height.

Claims:

1) <math>\xi</math> is invertible near zero. Indeed, computing <math>d\xi_0 = I</math> which is invertible as a matrix and hence <math>\xi</math> is invertible as a function near zero.

2) Take an <math>x\in U_2</math>. There are two routes to get to <math>V_1</math> and upon computing both ways yields the same result. Hence, the diagram commutes.

Hence, our immersion looks (locally) like the standard immersion between real spaces given by <math>\iota</math> and the charts are the compositions going between <math>U_0</math> to <math>U_2</math> and <math>V_0</math> to <math>V_2</math>

''Q.E.D''

0708-1300/Class notes for Tuesday, October 2

2007-10-06T01:23:49Z

Aptikuis: /* First Hour */

{{0708-1300/Navigation}}
==English Spelling==
Many interesting rules about [[0708-1300/English Spelling]]
==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

'''General class comments'''

1) The class photo is up, please add yourself

2) A questionnaire was passed out in class

3) Homework one is due on thursday

===First Hour===

Today's Theme: Locally a function looks like its differential

'''Pushforward/Pullback'''

Let <math>\theta:X\rightarrow Y</math> be a smooth map.

We consider various objects, defined with respect to X or Y, and see in which direction it makes sense to consider corresponding objects on the other space. In general <math>\theta_*</math> will denote the push forward, and <math>\theta^*</math> will denote the pullback.

1) points ''pushforward'' <math>x\mapsto\theta_*(x) := \theta(x)</math>

2) Paths <math>\gamma:\mathbb{R}\rightarrow X</math>, ie a bunch of points, ''pushforward'', <math>\gamma\rightarrow \theta_*(\gamma):=\theta\circ\gamma</math>

3) Sets <math>B\subset Y</math> ''pullback'' via <math>B\rightarrow \theta^*(B):=\theta^{-1}(B)</math>
Note that if one tried to pushforward sets A in X, the set operations compliment and intersection would not commute appropriately with the map <math>\theta</math>

4) A measures <math>\mu</math> ''pushforward'' via <math>\mu\rightarrow (\theta_*\mu)(B) :=\mu(\theta^*B)</math>

5)In some sense, we consider functions, "dual" to points and thus should go in the opposite direction of points, namely <math>\theta^*f = f\circ\theta</math>

6) Tangent vectors, defined in the sense of equivalence classes of paths, [<math>\gamma</math>] ''pushforward'' as we would expect since each path pushes forward. <math>[\gamma]\rightarrow \theta_*[\gamma]:=[\theta_*\gamma] = [\theta\circ\gamma]</math>

CHECK: This definition is well defined, that is, independent of the representative choice of <math>\gamma</math>

7) We can consider operators on functions to be in a sense dual to the functions and hence should go in the opposite direction. Hence, tangent vectors, defined in the sense of derivations, ''pushforward'' via <math>D\rightarrow (\theta_*D)(f):= D(\theta^*f)</math>

CHECK: This definition satisfies linearity and Liebnitz property.

'''Theorem 1'''

The two definitions for the pushforward of a tangent vector coincide.

''Proof:''

Given a <math>\gamma</math> we can construct <math>\theta_{*}\gamma</math> as above. However from both <math>\gamma</math> and <math>\theta_*\gamma</math> we can also construct <math>D_{\gamma}f</math> and <math>D_{\theta_*\gamma}f</math> because we have previously shown our two definitions for the tangent vector are equivalent. We can then ''pushforward'' <math>D_{\gamma}f</math> to get <math>\theta_*D_{\gamma}f</math>. The theorem is reduced to the claim that:

<math>\theta_*D_{\gamma}f = D_{\theta_*\gamma}f</math>

for functions <math>f:Y\rightarrow \mathbb{R}</math>

Now, <math>D_{\theta_*\gamma}f = \frac{d}{dt}f\circ(\theta_*\gamma)|_{t=0} = \frac{d}{dt}f\circ(\theta\circ\gamma)|_{t=0} = \frac{d}{dt}(f\circ\theta)\circ\gamma |_{t=0} = D_{\gamma}(f\circ\theta) =\theta_*D_{\gamma}f</math>

''Q.E.D''

'''Functorality'''

let <math>\theta:X\rightarrow Y, \lambda:Y\rightarrow Z</math>

Consider some "object" s defined with respect to X and some "object u" defined with respect to Z. Something has the property of functorality if

<math>\lambda_*(\theta_*s) = (\lambda\circ\theta)_*s</math>

and

<math>\theta^*(\lambda^*u) = (\lambda\circ\theta)^*u</math>

Claim: All the classes we considered previously have the functorality property; in particular, the pushforward of tangent vectors does.

Let us consider <math>\theta_*</math> on <math>T_pM</math> given a <math>\theta:M\rightarrow N</math>

We can arrange for charts <math>\varphi</math> on a subset of M into <math>\mathbb{R}^m</math> (with coordinates denoted <math>(x_1,\dots,x_m)</math>)and <math>\psi</math> on a subset of N into <math>\mathbb{R}^n</math> (with coordinates denoted <math>(y_1,\dots,y_n)</math>)such that <math>\varphi(p) = 0</math> and <math>\psi(\theta(p))=0</math>

Define <math>\theta^o = \psi\circ\theta\circ\varphi^{-1} = (\theta_1(x_1,\dots,x_m),\dots,\theta_n(x_1,\dots,x_m))</math>

Now, for a <math>D\in T_pM</math> we can write <math>D=\sum_{i=1}^m a_i\frac{\partial}{\partial x_i}</math>

So, <math>(\theta_*D)(f) = \sum_{i=1}^m a_i\frac{\partial}{\partial x_i}(f\circ\varphi) = \sum_{i=1}^m a_i \sum_{j=1}^n\frac{\partial f}{\partial y_j}\frac{\partial\theta_j}{\partial x_i}=</math>
<math>=\begin{bmatrix}
\frac{\partial f}{\partial y_1} & \cdots & \frac{\partial f}{\partial y_n}\\
\end{bmatrix}
\begin{bmatrix}

\frac{\partial \theta_1}{\partial x_1} & \cdots & \frac{\partial \theta_1}{\partial x_m}\\
\vdots& & \vdots\\
\frac{\partial \theta_n}{\partial x_1} & \cdots & \frac{\partial \theta_n}{\partial x_m}\\
\end{bmatrix}
\begin{bmatrix}
a_1\\
\vdots\\
a_m\\
\end{bmatrix}</math>

Now, we want to write <math>\theta_*D = \sum b_j\frac{\partial}{\partial y_j}</math>

and so, <math>b_k = (\theta_*D)y_k =\begin{bmatrix}
0&\cdots & 1 & \cdots &0\\
\end{bmatrix}
\begin{bmatrix}

\frac{\partial \theta_1}{\partial x_1} & \cdots & \frac{\partial \theta_1}{\partial x_m}\\
\vdots& & \vdots\\
\frac{\partial \theta_n}{\partial x_1} & \cdots & \frac{\partial \theta_n}{\partial x_m}\\
\end{bmatrix}
\begin{bmatrix}
a_1\\
\vdots\\
a_m\\
\end{bmatrix}</math>

where the 1 is at the kth location.

So, <math>\theta_* = d\theta_p</math>, i.e., <math>\theta_*</math> is the differential of <math>\theta</math> at p

We can check the functorality, <math>(\lambda\circ\theta)_* = \lambda_*\circ\theta_*</math>, then <math>d(\lambda\circ\theta) = d\lambda\circ d\theta</math>
This is just the chain rule.

===Second Hour===

'''Defintion 1'''

An ''immersion'' is a (smooth) map <math>\theta:M^m\rightarrow N^n</math> such that <math>\theta_*</math> of tangent vectors is 1:1. More precisely, <math>d\theta_p: T_pM\rightarrow T_{\theta(p)}N</math> is 1:1 <math>\forall p\in M</math>

'''Example 1'''

Consider the canonical immersion, for m<n given by <math>\iota:(x_1,...,x_m)\mapsto (x_1,...,x_m,0,...,0)</math> with n-m zeros.

'''Example 2'''

This is the map from R to <math>R^2</math> that looks like a loop-de-loop on a roller coaster (but squashed into the plane of course!) The map <math>\theta</math> itself is NOT 1:1 (consider the crossover point) however <math>\theta_*</math> IS 1:1, hence an immersion.

'''Example 3'''

Consider the map from R to <math>R^2</math> that looks like a check mark. While this map itself is 1:1, <math>\theta_*</math> is NOT 1:1 (at the cusp in the check mark) and hence is not an immersion.

'''Example 4'''

Can there be objects, such as the graph of |x| that are NOT an immersion, but are constructed from a smooth function?

Consider the function <math>\lambda(x) = e^{-1/x^2}</math> for x>0 and 0 otherwise.

Then the map <math>x\mapsto \begin{bmatrix}
(\lambda(x),\lambda(x))& x>0\\
(0,0)& x=0\\
(-\lambda(-x),\lambda(-x)) & x<0\\
\end{bmatrix}</math>

is a smooth mapping with the graph of |x| as its image, but is NOT an immersion.

'''Example 5'''

The torus, as a subset of <math>R^3</math> is an immersion

Now, consider the 1:1 linear map <math>T:V\rightarrow W</math> where V,W are vector spaces that takes <math>(v_1,...,v_m)\mapsto (Tv_1,...,Tv_m) = (w_1,..,w_m,w_{m+1},...,w_n)</math>

From linear algebra we know that we can choose a basis such that T is represented by a matrix with 1's along the first m diagonal locations and zeros elsewhere.

'''Theorem 2'''

Locally, every immersion looks like the inclusion <math>\iota:R^m\rightarrow R^n</math>.

More precisely, if <math>\theta:M^m\rightarrow N^n</math> and <math>d\theta_p</math> is 1:1 then there exists charts <math>\varphi</math> acting on <math>U\subset M</math> and <math>\phi</math> acting on <math>V\subset N</math> such that for <math>p\in U, \phi(p) = \psi(\theta(p)) = 0</math> such that the following diagram commutes:

<math>\begin{matrix}
U&\rightarrow^{\phi}&U'\subset R^n\\
\downarrow_{\theta} &&\downarrow_{\iota} \\
V& \rightarrow^{\psi}& V'\subset R^n\\
\end{matrix}</math>

that is, <math>\iota\circ\varphi = \psi\circ\theta</math>

'''Definition 2'''

M is a ''submanifold'' of N if there exists a mapping <math>\theta:M\rightarrow N</math> such that <math>\theta</math> is a 1:1 immersion.

'''Example 6'''

Our previous example of the graph of a "loop-de-loop", while an immersion, the function is not 1:1 and hence the graph is not a sub manifold.

'''Example 6'''

The torus is a submanifold as the natural immersion into <math>R^3</math> is 1:1

'''Definition 3'''

The map <math>\theta:M\rightarrow N</math> is an embedding if the subset topology on <math>\theta(M)</math> coincides with the topology induced from the original topology of M.

'''Example 7'''

Consider the map from <math>R\rightarrow R^2</math> whose graph looks like the open interval whose two ends have been wrapped around until they just touch (would intersect at one point if they were closed) the points 1/3 and 2/3rds of the way along the interval respectively.
The map is both 1:1 and an immersion. However, any neighborhood about the endpoints of the interval will ALSO include points near the 1/3rd and 2/3rd spots on the line, i.e., the topology is different and hence this is not an embedding.

'''Corollary 1 to Theorem 2'''

The functional structure on an embedded manifold induced by the functional structure on the containing manifold is equal to its original functional structure.

Indeed, for all smooth <math>f:M\rightarrow R</math> and <math>\forall p\in M</math> there exists a neighborhood V of <math>\theta(p)</math> and a smooth <math>g:N\rightarrow R</math> such that <math>g|_{\theta(M)\bigcap U} = f|_U</math>

''Proof of Corollary 1''

Loosely (and a sketch is most useful to see this!) we consider the embedded submanifold M in N and consider its image, under the appropriate charts, to a subset of <math>R^m\subset R^n</math>. We then consider some function defined on M, and hence on the subset in R^n which we can extend canonically as a constant function in the "vertical" directions. Now simply pullback into N to get the extended member of the functional structure on N.

''Proof of Theorem 2''

We start with the normal situation of <math>\theta:M\rightarrow N</math> with M,N manifolds with atlases containing <math>(\varphi_0,U_)0)</math> and <math>(\psi_0, V_0)</math> respectively. We also expect that for <math>p\in U_0, \varphi_0(p) = \psi_0(\theta(p)) = 0</math>. I will first draw the diagram and will subsequently justify the relevant parts. The proof reduces to showing a certain part of the diagram commutes appropriately.

<math>
\begin{matrix}
M\supset U_0 & \rightarrow^{\varphi_0} & U_1\subset R^m & \rightarrow^{Id} & U_2 = U_1 \\
\downarrow_{\theta} & &\downarrow_{\theta_1} & &\downarrow_{\iota}\\
N\supset V_0 & \rightarrow^{\psi_0} & V_1\subset R^n & \leftarrow^{\xi} & V_2\\
\end{matrix}
</math>

It is very important to note that the <math>\varphi_0</math> and <math>\psi_0</math> are NOT the charts we are looking for , they are merely one of the ones that happen to act about the point p.

In the diagram above, <math>\theta_1 = \psi_0\circ\theta\circ\varphi^{-1}</math>. So, <math>\theta_1(0) = 0</math> and <math>(d\theta_1)_0 = i</math>. Note the <math>\theta_1</math>, being merely the normal composition with the appropriate charts, does not fundamentally add anything. What makes this theorem work is the function <math>\xi</math>

We consider the map <math>\xi:V_2\rightarrow V_1</math> given <math>(x,y)\mapsto \theta_1(x) + (0,y)</math>. We note that <math>\xi</math> corresponds with the idea of "lifting" a flattened image back to its original height.

Claims:

1) <math>\xi</math> is invertible near zero. Indeed, computing <math>d\xi_0 = I</math> which is invertible as a matrix and hence <math>\xi</math> is invertible as a function near zero.

2) Take an <math>x\in U_2</math>. There are two routes to get to <math>V_1</math> and upon computing both ways yields the same result. Hence, the diagram commutes.

Hence, our immersion looks (locally) like the standard immersion between real spaces given by <math>\iota</math> and the charts are the compositions going between <math>U_0</math> to <math>U_2</math> and <math>V_0</math> to <math>V_2</math>

''Q.E.D''

0708-1300/Class notes for Thursday, September 27

2007-10-06T00:58:35Z

Aptikuis: /* Class Notes */

{{0708-1300/Navigation}}

==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

'''General comments regarding the wiki page'''

1) Use the history/recent changes to track your own work

2) Never post/upload without linking

'''Comments on Problem 4, page 71, Assignment 1'''

Dror gave three hints towards a solution to this this problem:

1) Consider the analogy with a (smooth) car which must stop when approaching a sharp bend. When it does stop, everything around the car, such as a tree, stops moving relative to the car as well

2) There is a map h going from the restriction of <math>\mathbb{R}^{2}</math> to our set into <math>\mathbb{R}</math> as well as a map (f,g) going in reverse that satisfies <math>h\circ(f,g)=I_{d}</math>. We can then apply the chain rule (think about why!) to get <math>h_{x}f' + h_{y}g' = 1</math>. However, <math>f=\pm g</math> and both cases occur at adjacent points, resulting in <math>f' =\pm g'</math> at adjacent point and thus establishing the contradiction.

3) This hint uses methods from beyond page 71. It is possible to find two linearly independent directional derivatives on functions on our set A near zero. However this is a contradiction as a one dimensional space cannot have a two dimensional tangent space.

At this point, the discussion returned to the previous days class regarding the theorem of the equivalence of our two definitions of a tangent vector. It was reiterated that a major point in proving the bijection between the two types of vectors was indeed onto is that it was possible, as a result of Hadamard's Lemma, to determine D by the n constants <math>Dx_{i}</math>

It is easily checked that the tangent space <math>T_{0}\mathbb{R}^{n}</math> forms an n dimensional ''vector space''. This is because the D's are linear and because the D is determined by the n constants <math>Dx_{i}</math>.

We wish to generalize this concept to show that <math>T_{p}M^{n}</math> is a vector space. This is easily done as there is a canonical isomorphism between <math>T_{p}M^{n}</math> and <math>T_{0}\mathbb{R}^{n}</math> via the chart <math>\varphi</math>

'''Proof of Hadamard's Lemma'''

<math>f(p)-f(0)=\int_0^1 \frac{d}{dt}f(tp)\, dt </math>
<math>=\int_0^1 \sum_{i=1}^{n} \frac{\partial f}{\partial x_{i}}(tp)x_{i}dt</math>
<math>=\sum_{i=1}^{n}x_{i}\int_0^1\frac{\partial f}{\partial x_{i}}(tp)dt</math>
<math>=\sum_{i=1}^{n}x_{i}g_i (p)</math>

where <math>g_i (p)=\int_0^1\frac{\partial f}{\partial x_{i}}(tp)dt</math>

f is smooth with respect to p and so <math>g_i</math> is, as derivatives with respect to p can pass through the integral which is with respect to t.

''QED''

'''Corollary''':

<math>g_i(0)=\frac{\partial f}{\partial x_{i}}(0)</math>

'''Local Coordinates'''

<math>
\mathbb{R}^n</math> possesses canonical functions <math>(x_1,...,x_n)</math> that are merely the levels <math>x_i = const</math>.

The pullback of these into the manifold under <math>\varphi^{-1}</math> yields a similar 'grid' of lines on the manifolds only these lines are curves. Formally, we equip the manifold with functions <math>x^{o}_1 = x_1\circ\varphi~</math>, <math>x^{o}_2 = x_2\circ\varphi~</math>, etc...

Now, <math>\forall f:M\rightarrow \mathbb{R}\ \exists g:\mathbb{R}^n\rightarrow \mathbb{R}</math> such that <math>f=g\circ\varphi</math> and <math>f(p) = g(\varphi(p)) = g((x_1(\varphi(p)),...,x_n(\varphi(p))) = g(x_1^{o},...,x_m^{o})</math>

Conventionally the distinction between x and <math>x^{0}</math> is not made.

''Question:''
How do you express <math>D\in T_p(M)</math> using the local coordinates?

'''Claim'''

1) <math>\frac{\partial}{\partial x_i}</math> is a tangent vector; <math>\frac{\partial}{\partial x_i}(f):=\frac{\partial}{\partial x_i}(g)</math> where <math>g=f\circ\varphi^{-1}</math>

2) <math>D = \sum (Dx_i)\frac{\partial}{\partial x_i}</math>

'''Proof'''

1) We need to check linearity and liebnitz's rule (easy)

2) We only need to check this on an arbitrary <math>x_j</math> as they span all such functions.
So, <math>Dx_j = \sum Dx_i \frac{\partial x_j}{\partial x_i}
= \sum Dx_i \delta_{i,j} = Dx_j</math>

0708-1300/Class notes for Tuesday, September 25

2007-10-06T00:40:27Z

Aptikuis: /* Second Hour */

{{0708-1300/Navigation}}

==Dror's Notes==
* Class photo is on Thursday, show up and be at your best! More seriously -
** The class photo is of course not mandatory, and if you are afraid of google learning about you, you should not be in it.
** If you want to be in the photo but can't make it on Thursday, I'll take a picture of you some other time and add it as an inset to the main picture.

* I just got the following email message, which some of you may find interesting:

NSERC - CMS Math in Moscow Scholarships

The Natural Sciences and Engineering Research Council (NSERC) and the
Canadian Mathematical Society (CMS) support scholarships at $9,000
each. Canadian students registered in a mathematics or computer
science program are eligible.

The scholarships are to attend a semester at the small elite Moscow
Independent University.

Math in Moscow program
www.mccme.ru/mathinmoscow/
Application details
www.cms.math.ca/bulletins/Moscow_web/

For additional information please see your department or call the CMS
at 613-562-5702.

Deadline September 30, 2007 to attend the Winter 2008 semester.

{{0708-1300/Class Notes}}
===First hour===
Recall from last class we were proving the equivalence of the two definitions for a smooth manifold. The only nontrivial point that remained to be proved was that if we started with the definition of a manifold in the sense of functional structures and produced charts <math>\varphi,\psi</math> that these charts would satisfy the property of a manifold, defined in the atlas sense, that <math>\psi\circ\varphi^{-1}</math> is smooth where defined.

''Proof''

<math>\psi\circ\varphi^{-1}:\mathbb{R}^n\rightarrow \mathbb{R}^n</math>
is smooth <math>\Leftrightarrow</math> <math>(\psi\circ\varphi^{-1})_i:\mathbb{R}^n\rightarrow \mathbb{R}</math> is smooth <math>\forall</math> i
<math>\Leftrightarrow</math> <math>\pi_i\circ\psi\circ\varphi^{-1}</math> is smooth where <math>\pi_i</math> is the <math>i^{th}</math> coordinate projection map.

Now, since <math>\pi_i</math> is always smooth, <math>\pi_i\in F_{\mathbb{R}^{n}}(U^{'}_{\psi})</math>

But then we have <math>\pi_{i}\circ\psi\in F_{M}(U_{\psi})</math>
and so, by a property of functional structures,
<math>\pi_{i}\circ\psi |_{U_{\varphi}\bigcap U_{\psi}}\in F_{M}(U_{\varphi}\bigcap U_{\psi})</math>
and hence <math>\pi_{i}\circ\psi\circ\varphi^{-1}\in F_{\mathbb{R}^{n}}</math> where it is defined and thus is smooth.
''QED''

'''Definition 1''' (induced structure)
Suppose <math>\pi:X\rightarrow Y</math> and suppose Y is equipped with a functional structure <math>F_Y</math> then the "induced functional structure" on X is

<math>F_{X}(U) = \{f:U\rightarrow \mathbb{R}\ |\ \exists g\in F_{Y}(V)\ such\ that\ V\supset \pi(U)\ and\ f=g\circ\pi\}</math>

Claim: this does in fact define a functional structure on X

'''Definition 2'''
This is the reverse definition of that given directly above. Let <math>\pi:X\rightarrow Y</math> and let X be equipped with a functional structure <math>F_{X}</math>. Then we get a functional structure on Y by
<math>F_{Y}(V) = \{g:V\rightarrow \mathbb{R}\ |\ g\circ\pi\in F_{X}(\pi^{-1}(V)\}
</math>
Claim: this does in fact define a functional structure on X

'''Example 1'''
Let <math>S^{2} = \mathbb{R}^{3}-\{0\}/</math>~
where the equivalence relation ~ is given by x~<math>\alpha</math>x for <math>\alpha</math>>0
We thus get a canonical projection map <math>\pi:\mathbb{R}^{3}-\{0\}\rightarrow S^{2}</math>
and hence, there is an induced functional structure on <math>S^{2}</math>.
Claim:
1) This induced functional structure makes <math>S^{2}</math> into a manifold
2) This resulting manifold is the ''same'' manifold as from the atlas definition given previously

'''Example 2'''
Consider the torus thought of as <math>T^{2} = \mathbb{R}^{2}/\mathbb{Z}^{2}</math>, i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in <math>\mathbb{R}^{2}</math>and (n,m) in <math>\mathbb{Z}^{2}</math>

As in the previous example, the torus inherits a functional structure from the real plane we must again check that
1) We get a manifold
2) This is the ''same'' manifold as we had previously with the atlas definition

'''Example 3'''
Let <math>CP^{n}</math> denote the n dimensional complex projective space, that is,
<math>CP^{n} = \mathbb{C}^{n+1}-\{0\}/</math>~ where <math>[z_{0},...,z_{n}]</math>
~ <math>[\alpha z_{0},...,\alpha z_{n}]</math>
where <math>\alpha\in \mathbb{C}</math>

Again, this space inherits a functional structure from <math>\mathbb{C}^{n+1}</math> and we again need to claim that this yields a manifold.

''Proof of Claim''

We consider the subsets <math>CP^{n}\supset U_{i} = \{[z_{0},...,z_{n}]\ |\ z_{i}\neq 0\}</math> for <math>0\leq i \leq n</math>

Clearly <math>\bigcup U_{i} = CP^{n}</math>

Now, for each <math>p\in U_{i}</math> there is a unique representative for its equivalence class of the form <math>[z_{0},...,1,...,z_{n}]</math> where the 1 is at the ith location.

We thus can get a map from <math>\varphi_{i}:U_{i}\rightarrow \mathbb{C}^{n} = \mathbb{R}^{2n}</math> by
<math>p\mapsto [z_{0}/z_{i},...,z_{i-1}/z_{i},z_{i+1}/z_{i},...,z_{n}/z_{i}]</math>
Hence we have shown (loosely) that our functional structure is locally isormorphic to <math>(\mathbb{R},C^\infty)</math>

'''Definition 3''' Product Manifolds

Suppose <math>M^{m}</math> and <math>N^{n}</math> are manifolds. Then the product manifold, on the set MxN has an atlas given by
<math>\{\varphi \times \psi: U\times V\rightarrow U'\times V'\in \mathbb{R}^{m}\times \mathbb{R}^{n}\ | \varphi: U\rightarrow U'\subset \mathbb{R}^{m}\ and\ \psi:V\rightarrow V'\subset \mathbb{R}^{n}</math> are charts in resp. manifolds}

Claim: This does in fact yield a manifold

'''Example 4'''
It can be checked that <math>T^{2} = S^{1}\times S^{1}</math> gives the torus a manifold structure, by the product manifold, that is indeed the same as the normal structure given previously.

===Second Hour===
Aim: We consider two manifolds, M and N, and a function f between them. We aim for the analogous idea of the tangent in the reals, namely that every smooth <math>f:M \rightarrow N</math> has a good linear approximation. Of course, we will need to define what is meant by such a ''smooth'' function, as well as what this linear approximation is.

'''Definition 1''' Smooth Function, Atlas Sense

Let <math>M^{m}</math> and <math>N^{n}</math> be manifolds. We say that a function <math>f:M^{m}\rightarrow N^{n}</math> is smooth if <math>\psi\circ f\circ\varphi^{-1}</math> is smooth, where it makes sense, <math>\forall</math> charts <math>\psi,\varphi</math>

'''Definition 2''' Smooth Function, Functional Structure Sense

Let <math>M^{m}</math> and <math>N^{n}</math> be manifolds. We say that a function <math>f:M^{m}\rightarrow N^{n}</math> is smooth if <math>\forall h:V\rightarrow \mathbb{R}</math> such that h is smooth, <math>h\circ f</math> is smooth on subsets of M where it is defined.

'''Claim 1'''

The two definitions are equivalent

'''Definition 3''' Category (Loose Definition)

A category is a collection of "objects" (such as sets, topological spaces, manifolds, etc.) such that for any two objects, x and y, there exists a set of "morphisms" denoted mor(<math>x\rightarrow y</math>) (these would be functions for sets, continuous functions for topological spaces, smooth functions for manifolds, etc.) along with

1)Composition maps <math>\circ</math>:mor(<math>x\rightarrow y</math>) <math>\times</math> mor(<math>y\rightarrow z</math>) <math>\rightarrow</math> mor (<math>x\rightarrow z</math>)

2) For any x there exists an element <math>1_{x}\in</math>mor(<math>x \rightarrow x</math>)

such that several axioms are satisfied, including

a) Associativity of the composition map

b) <math>1_{x}</math> behaves like an identity should.

'''Claim 2'''

The collection of smooth manifolds with smooth functions form a category.
This (loosely) amounts to (easily) checking the two claims that

1) if <math>f:M\rightarrow N</math> and <math>g:N\rightarrow L</math> are smooth maps between manifolds then <math>g\circ f:M\rightarrow L</math> is smooth

2) <math>1_{M}</math> is smooth

We now provide two alternate definition of the tangent vector, one from each definition of a manifold.

'''Definition 4''' Tangent Vector, Atlas Sense

Consider the set of curves on the manifold, that is
<math>\{\gamma:\mathbb{R}\rightarrow M\ |\ s.t.\ \gamma\ smooth,\ \gamma(0) = p\}</math>

We define an equivalence relation between paths by <math>\gamma_{1}</math> ~ <math>\gamma_{2}</math> if <math>\varphi\circ\gamma_{1}</math> is tangent to <math>\varphi\circ\gamma_{2}</math> as paths in <math>\mathbb{R}^{n}\ \forall\varphi</math>.

Then, a ''tangent vector'' is an equivalence class of curves.

'''Definition 5''' Tangent Vector, Functional Structure Sense

A tangent vector D at p, often called a directional derivative in this sense, is an operator that takes the smooth real valued functions near p into <math>\mathbb{R}</math>.

such that,

1) D(af +bg) = aDf + bDg for a,b constants in <math>\mathbb{R}</math> and f,g such functions

2) D(fg) = (Df)g(p) + f(p)D(g) Leibniz' Rules

'''Theorem 1'''
These two definitions are equivalent.

'''Definition 6''' Tangent Space

The tangent space at a point, <math>T_{p}M

</math> = the set of all tangent vectors at a point p

'''Claim 3'''

We will show later that in fact <math>T_{p}M

</math> is a vector space

Consider a map <math>f:M\rightarrow N</math>. We are interested in defining an associated map between the tangent spaces, namely, <math>df_{p}:T_{p}M\rightarrow T_{f(p)}N</math>

We get two different definitions from the two different definitions of a tangent vector

'''Definition 7''' (atlas sense)

<math>(df_{p})([\gamma]):=[f\circ\gamma]</math>

'''Definition 8''' (functional structure sense)

<math>((df_{p})(D))(h):=D(h\circ f)</math>

'''Proof of Theorem 1'''

We consider a curve <math>\gamma</math> and denote its equivalence class of curves by <math>[\gamma]</math>

We consider the map <math>\Phi:[\gamma]\rightarrow D_{\gamma}</math> defined by
<math>D_{\gamma}(f) = (f\circ\gamma)'(0)</math>

It is clear that this map is linear is f and satisfies the leibnitz rule.

Claim: <math>\Phi</math> is a bijection.

We will need to use the following lemma, which will not be proved now:

''Hadamard's Lemma''

If <math>f:\mathbb{R}^{m}\rightarrow \mathbb{R}</math> is smooth near 0 then there exists smooth <math>g_{i}:\mathbb{R}^{m}\rightarrow \mathbb{R}</math> such that
<math>
f(x) = f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)</math>

Now, let D be a tangent vector. One can quickly see that D(const) = 0.

So, by Hadamard's lemma,
<math>Df = D(f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)) = D(\Sigma_{i=1}^{n}x_{i}g_{i}(x)) = \Sigma_{i=1}^{n}(Dx_{i})g_{i}(0) + \Sigma_{i=1}^{n}x_{i}(0)Dg_{i}</math> = <math>\Sigma_{i=1}^{n}(Dx_{i})g_{i}(0)</math>
as <math>x_{i}(0) = 0</math>

and hence, Df is a linear comb of fixed quantities.

Now, let <math>\gamma:\mathbb{R}\rightarrow \mathbb{R}^{n}</math> be such that <math>\gamma(0) = 0</math>

<math>D_{\gamma}f = \Sigma_{i=1}^{n}(D_{\gamma}x_{i})g_{i}(0)</math>

but <math>D_{\gamma}(x_{i}) = (x_{i}\circ\gamma)'(0) = \gamma_{i}'(0)</math>

We can now claim that <math>\Phi</math> is onto because given a D take <math>\gamma_{i}(t)=D(x_{i})t</math>

<math>\Phi</math> is trivially 1:1.

==Smoot==
In this class It was riced the question of what a "Smoot" is. Here it is [[0708-1300/Smoot]]

0708-1300/Class notes for Tuesday, September 25

2007-10-05T02:25:48Z

Aptikuis: /* Second Hour */

{{0708-1300/Navigation}}

==Dror's Notes==
* Class photo is on Thursday, show up and be at your best! More seriously -
** The class photo is of course not mandatory, and if you are afraid of google learning about you, you should not be in it.
** If you want to be in the photo but can't make it on Thursday, I'll take a picture of you some other time and add it as an inset to the main picture.

* I just got the following email message, which some of you may find interesting:

NSERC - CMS Math in Moscow Scholarships

The Natural Sciences and Engineering Research Council (NSERC) and the
Canadian Mathematical Society (CMS) support scholarships at $9,000
each. Canadian students registered in a mathematics or computer
science program are eligible.

The scholarships are to attend a semester at the small elite Moscow
Independent University.

Math in Moscow program
www.mccme.ru/mathinmoscow/
Application details
www.cms.math.ca/bulletins/Moscow_web/

For additional information please see your department or call the CMS
at 613-562-5702.

Deadline September 30, 2007 to attend the Winter 2008 semester.

{{0708-1300/Class Notes}}
===First hour===
Recall from last class we were proving the equivalence of the two definitions for a smooth manifold. The only nontrivial point that remained to be proved was that if we started with the definition of a manifold in the sense of functional structures and produced charts <math>\varphi,\psi</math> that these charts would satisfy the property of a manifold, defined in the atlas sense, that <math>\psi\circ\varphi^{-1}</math> is smooth where defined.

''Proof''

<math>\psi\circ\varphi^{-1}:\mathbb{R}^n\rightarrow \mathbb{R}^n</math>
is smooth <math>\Leftrightarrow</math> <math>(\psi\circ\varphi^{-1})_i:\mathbb{R}^n\rightarrow \mathbb{R}</math> is smooth <math>\forall</math> i
<math>\Leftrightarrow</math> <math>\pi_i\circ\psi\circ\varphi^{-1}</math> is smooth where <math>\pi_i</math> is the <math>i^{th}</math> coordinate projection map.

Now, since <math>\pi_i</math> is always smooth, <math>\pi_i\in F_{\mathbb{R}^{n}}(U^{'}_{\psi})</math>

But then we have <math>\pi_{i}\circ\psi\in F_{M}(U_{\psi})</math>
and so, by a property of functional structures,
<math>\pi_{i}\circ\psi |_{U_{\varphi}\bigcap U_{\psi}}\in F_{M}(U_{\varphi}\bigcap U_{\psi})</math>
and hence <math>\pi_{i}\circ\psi\circ\varphi^{-1}\in F_{\mathbb{R}^{n}}</math> where it is defined and thus is smooth.
''QED''

'''Definition 1''' (induced structure)
Suppose <math>\pi:X\rightarrow Y</math> and suppose Y is equipped with a functional structure <math>F_Y</math> then the "induced functional structure" on X is

<math>F_{X}(U) = \{f:U\rightarrow \mathbb{R}\ |\ \exists g\in F_{Y}(V)\ such\ that\ V\supset \pi(U)\ and\ f=g\circ\pi\}</math>

Claim: this does in fact define a functional structure on X

'''Definition 2'''
This is the reverse definition of that given directly above. Let <math>\pi:X\rightarrow Y</math> and let X be equipped with a functional structure <math>F_{X}</math>. Then we get a functional structure on Y by
<math>F_{Y}(V) = \{g:V\rightarrow \mathbb{R}\ |\ g\circ\pi\in F_{X}(\pi^{-1}(V)\}
</math>
Claim: this does in fact define a functional structure on X

'''Example 1'''
Let <math>S^{2} = \mathbb{R}^{3}-\{0\}/</math>~
where the equivalence relation ~ is given by x~<math>\alpha</math>x for <math>\alpha</math>>0
We thus get a canonical projection map <math>\pi:\mathbb{R}^{3}-\{0\}\rightarrow S^{2}</math>
and hence, there is an induced functional structure on <math>S^{2}</math>.
Claim:
1) This induced functional structure makes <math>S^{2}</math> into a manifold
2) This resulting manifold is the ''same'' manifold as from the atlas definition given previously

'''Example 2'''
Consider the torus thought of as <math>T^{2} = \mathbb{R}^{2}/\mathbb{Z}^{2}</math>, i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in <math>\mathbb{R}^{2}</math>and (n,m) in <math>\mathbb{Z}^{2}</math>

As in the previous example, the torus inherits a functional structure from the real plane we must again check that
1) We get a manifold
2) This is the ''same'' manifold as we had previously with the atlas definition

'''Example 3'''
Let <math>CP^{n}</math> denote the n dimensional complex projective space, that is,
<math>CP^{n} = \mathbb{C}^{n+1}-\{0\}/</math>~ where <math>[z_{0},...,z_{n}]</math>
~ <math>[\alpha z_{0},...,\alpha z_{n}]</math>
where <math>\alpha\in \mathbb{C}</math>

Again, this space inherits a functional structure from <math>\mathbb{C}^{n+1}</math> and we again need to claim that this yields a manifold.

''Proof of Claim''

We consider the subsets <math>CP^{n}\supset U_{i} = \{[z_{0},...,z_{n}]\ |\ z_{i}\neq 0\}</math> for <math>0\leq i \leq n</math>

Clearly <math>\bigcup U_{i} = CP^{n}</math>

Now, for each <math>p\in U_{i}</math> there is a unique representative for its equivalence class of the form <math>[z_{0},...,1,...,z_{n}]</math> where the 1 is at the ith location.

We thus can get a map from <math>\varphi_{i}:U_{i}\rightarrow \mathbb{C}^{n} = \mathbb{R}^{2n}</math> by
<math>p\mapsto [z_{0}/z_{i},...,z_{i-1}/z_{i},z_{i+1}/z_{i},...,z_{n}/z_{i}]</math>
Hence we have shown (loosely) that our functional structure is locally isormorphic to <math>(\mathbb{R},C^\infty)</math>

'''Definition 3''' Product Manifolds

Suppose <math>M^{m}</math> and <math>N^{n}</math> are manifolds. Then the product manifold, on the set MxN has an atlas given by
<math>\{\varphi \times \psi: U\times V\rightarrow U'\times V'\in \mathbb{R}^{m}\times \mathbb{R}^{n}\ | \varphi: U\rightarrow U'\subset \mathbb{R}^{m}\ and\ \psi:V\rightarrow V'\subset \mathbb{R}^{n}</math> are charts in resp. manifolds}

Claim: This does in fact yield a manifold

'''Example 4'''
It can be checked that <math>T^{2} = S^{1}\times S^{1}</math> gives the torus a manifold structure, by the product manifold, that is indeed the same as the normal structure given previously.

===Second Hour===
Aim: We consider two manifolds, M and N, and a function f between them. We aim for the analogous idea of the tangent in the reals, namely that every smooth <math>f:M \rightarrow N</math> has a good linear approximation. Of course, we will need to define what is meant by such a ''smooth'' function, as well as what this linear approximation is.

'''Definition 1''' Smooth Function, Atlas Sense

Let <math>M^{m}</math> and <math>N^{n}</math> be manifolds. We say that a function <math>f:M^{m}\rightarrow N^{n}</math> is smooth if <math>\psi\circ f\circ\varphi^{-1}</math> is smooth, where it makes sense, <math>\forall</math> charts <math>\psi,\varphi</math>

'''Definition 2''' Smooth Function, Functional Structure Sense

Let <math>M^{m}</math> and <math>N^{n}</math> be manifolds. We say that a function <math>f:M^{m}\rightarrow N^{n}</math> is smooth if <math>\forall h:V\rightarrow \mathbb{R}</math> such that h is smooth, <math>h\circ f</math> is smooth on subsets of M where it is defined.

'''Claim 1'''

The two definitions are equivalent

'''Definition 3''' Category (Loose Definition)

A category is a collection of "objects" (such as sets, topological spaces, manifolds, etc.) such that for any two objects, x and y, there exists a set of "morphisms" denoted mor(<math>x\rightarrow y</math>) (these would be functions for sets, continuous functions for topological spaces, smooth functions for manifolds, etc.) along with

1)Composition maps <math>\circ</math>:mor(<math>x\rightarrow y</math>) <math>\times</math> mor(<math>y\rightarrow z</math>) <math>\rightarrow</math> mor (<math>x\rightarrow z</math>)

2) For any x there exists an element <math>1_{x}\in</math>mor(<math>x \rightarrow x</math>)

such that several axioms are satisfied, including

a) Associativity of the composition map

b) <math>1_{x}</math> behaves like an identity should.

'''Claim 2'''

The collection of smooth manifolds with smooth functions form a category.
This (loosely) amounts to (easily) checking the two claims that

1) if <math>f:M\rightarrow N</math> and <math>g:N\rightarrow L</math> are smooth maps between manifolds then <math>g\circ f:M\rightarrow L</math> is smooth

2) <math>1_{M}</math> is smooth

We now provide two alternate definition of the tangent vector, one from each definition of a manifold.

'''Definition 4''' Tangent Vector, Atlas Sense

Consider the set of curves on the manifold, that is
<math>\{\gamma:\mathbb{R}\rightarrow M\ |\ s.t.\ \gamma\ smooth,\ \gamma(0) = p\}</math>

We define an equivalence relation between paths by <math>\gamma_{1}</math> ~ <math>\gamma_{2}</math> if <math>\varphi\circ\gamma_{1}</math> is tangent to <math>\varphi\circ\gamma_{2}</math> as paths in <math>\mathbb{R}^{n}\ \forall\varphi</math>.

Then, a ''tangent vector'' is an equivalence class of curves.

'''Definition 5''' Tangent Vector, Functional Structure Sense

A tangent vector D at p, often called a directional derivative in this sense, is an operator that takes the smooth real valued functions near p into <math>\mathbb{R}</math>.

such that,

1) D(af +bg) = aDf + bDg for a,b constants in <math>\mathbb{R}</math> and f,g such functions

2) D(fg) = (Df)g(p) + f(p)D(g) Leibniz' Rules

'''Theorem 1'''
These two definitions are equivalent.

'''Definition 6''' Tangent Space

The tangent space at a point, <math>T_{p}M

</math> = the set of all tangent vectors at a point p

'''Claim 3'''

We will show later that in fact <math>T_{p}M

</math> is a vector space

Consider a map <math>f:M\rightarrow N</math>. We are interested in defining an associated map between the tangent spaces, namely, <math>df_{p}:T_{p}M\rightarrow T_{f(p)}N</math>

We get two different definitions from the two different definitions of a tangent vector

'''Definition 7''' (atlas sense)

<math>(df_{p})([\gamma]):=[f\circ\gamma]</math>

'''Definition 8''' (functional structure sense)

<math>((df_{p})(D))(h):=D(h\circ f)</math>

'''Proof of Theorem 1'''

We consider a curve <math>\gamma</math> and denote its equivalence class of curves by <math>[\gamma]</math>

We consider the map <math>\Phi:[\gamma]\rightarrow D_{\gamma}</math> defined by
<math>D_{\gamma}(f) = (f\circ\gamma)'(0)</math>

It is clear that this map is linear is f and satisfies the leibnitz rule.

Claim: <math>\Phi</math> is a bijection.

We will need to use the following lemma, which will not be proved now:

''Hadamard's Lemma''

If <math>f:\mathbb{R}^{m}\rightarrow \mathbb{R}</math> is smooth near 0 then there exists smooth <math>g_{i}:\mathbb{R}^{m}\rightarrow \mathbb{R}</math> such that
<math>
f(x) = f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)</math>

Now, let D be a tangent vector. One can quickly see that D(const) = 0.

So, by Hadamard's lemma,
<math>Df = D(f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)) = D(\Sigma_{i=1}^{n}x_{i}g_{i}(x)) = \Sigma_{i=1}^{n}(Dx_{i})g_{i}(0) + \Sigma_{i=1}^{n}x_{i}(0)Dg_{i}</math> = <math>\Sigma_{i=1}^{n}(Dx_{i})g_{i}(0)</math>
as <math>x_{i}(0) = 0</math>

and hence, Df is a linear comb of fixed quantities.

Now, let <math>\gamma:\mathbb{R}\rightarrow \mathbb{R}^{n}</math> be such that <math>\gamma(0) = 0</math>

<math>D_{\gamma}f = \Sigma_{i=1}^{n}(D_{\gamma}x_{i}))g_{i}(0)</math>

but <math>D_{\gamma}(x_{i}) = (x_{i}\circ\gamma)'(0) = \gamma_{i}'(0)</math>

We can now claim that <math>\Phi</math> is onto because given a D take <math>\gamma_{i}(t)=D(x_{i})t</math>

<math>\Phi</math> is trivially 1:1.

==Smoot==
In this class It was riced the question of what a "Smoot" is. Here it is [[0708-1300/Smoot]]

0708-1300/Class notes for Tuesday, September 25

2007-10-04T23:56:25Z

Aptikuis: /* First hour */

{{0708-1300/Navigation}}

==Dror's Notes==
* Class photo is on Thursday, show up and be at your best! More seriously -
** The class photo is of course not mandatory, and if you are afraid of google learning about you, you should not be in it.
** If you want to be in the photo but can't make it on Thursday, I'll take a picture of you some other time and add it as an inset to the main picture.

* I just got the following email message, which some of you may find interesting:

NSERC - CMS Math in Moscow Scholarships

The Natural Sciences and Engineering Research Council (NSERC) and the
Canadian Mathematical Society (CMS) support scholarships at $9,000
each. Canadian students registered in a mathematics or computer
science program are eligible.

The scholarships are to attend a semester at the small elite Moscow
Independent University.

Math in Moscow program
www.mccme.ru/mathinmoscow/
Application details
www.cms.math.ca/bulletins/Moscow_web/

For additional information please see your department or call the CMS
at 613-562-5702.

Deadline September 30, 2007 to attend the Winter 2008 semester.

{{0708-1300/Class Notes}}
===First hour===
Recall from last class we were proving the equivalence of the two definitions for a smooth manifold. The only nontrivial point that remained to be proved was that if we started with the definition of a manifold in the sense of functional structures and produced charts <math>\varphi,\psi</math> that these charts would satisfy the property of a manifold, defined in the atlas sense, that <math>\psi\circ\varphi^{-1}</math> is smooth where defined.

''Proof''

<math>\psi\circ\varphi^{-1}:\mathbb{R}^n\rightarrow \mathbb{R}^n</math>
is smooth <math>\Leftrightarrow</math> <math>(\psi\circ\varphi^{-1})_i:\mathbb{R}^n\rightarrow \mathbb{R}</math> is smooth <math>\forall</math> i
<math>\Leftrightarrow</math> <math>\pi_i\circ\psi\circ\varphi^{-1}</math> is smooth where <math>\pi_i</math> is the <math>i^{th}</math> coordinate projection map.

Now, since <math>\pi_i</math> is always smooth, <math>\pi_i\in F_{\mathbb{R}^{n}}(U^{'}_{\psi})</math>

But then we have <math>\pi_{i}\circ\psi\in F_{M}(U_{\psi})</math>
and so, by a property of functional structures,
<math>\pi_{i}\circ\psi |_{U_{\varphi}\bigcap U_{\psi}}\in F_{M}(U_{\varphi}\bigcap U_{\psi})</math>
and hence <math>\pi_{i}\circ\psi\circ\varphi^{-1}\in F_{\mathbb{R}^{n}}</math> where it is defined and thus is smooth.
''QED''

'''Definition 1''' (induced structure)
Suppose <math>\pi:X\rightarrow Y</math> and suppose Y is equipped with a functional structure <math>F_Y</math> then the "induced functional structure" on X is

<math>F_{X}(U) = \{f:U\rightarrow \mathbb{R}\ |\ \exists g\in F_{Y}(V)\ such\ that\ V\supset \pi(U)\ and\ f=g\circ\pi\}</math>

Claim: this does in fact define a functional structure on X

'''Definition 2'''
This is the reverse definition of that given directly above. Let <math>\pi:X\rightarrow Y</math> and let X be equipped with a functional structure <math>F_{X}</math>. Then we get a functional structure on Y by
<math>F_{Y}(V) = \{g:V\rightarrow \mathbb{R}\ |\ g\circ\pi\in F_{X}(\pi^{-1}(V)\}
</math>
Claim: this does in fact define a functional structure on X

'''Example 1'''
Let <math>S^{2} = \mathbb{R}^{3}-\{0\}/</math>~
where the equivalence relation ~ is given by x~<math>\alpha</math>x for <math>\alpha</math>>0
We thus get a canonical projection map <math>\pi:\mathbb{R}^{3}-\{0\}\rightarrow S^{2}</math>
and hence, there is an induced functional structure on <math>S^{2}</math>.
Claim:
1) This induced functional structure makes <math>S^{2}</math> into a manifold
2) This resulting manifold is the ''same'' manifold as from the atlas definition given previously

'''Example 2'''
Consider the torus thought of as <math>T^{2} = \mathbb{R}^{2}/\mathbb{Z}^{2}</math>, i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in <math>\mathbb{R}^{2}</math>and (n,m) in <math>\mathbb{Z}^{2}</math>

As in the previous example, the torus inherits a functional structure from the real plane we must again check that
1) We get a manifold
2) This is the ''same'' manifold as we had previously with the atlas definition

'''Example 3'''
Let <math>CP^{n}</math> denote the n dimensional complex projective space, that is,
<math>CP^{n} = \mathbb{C}^{n+1}-\{0\}/</math>~ where <math>[z_{0},...,z_{n}]</math>
~ <math>[\alpha z_{0},...,\alpha z_{n}]</math>
where <math>\alpha\in \mathbb{C}</math>

Again, this space inherits a functional structure from <math>\mathbb{C}^{n+1}</math> and we again need to claim that this yields a manifold.

''Proof of Claim''

We consider the subsets <math>CP^{n}\supset U_{i} = \{[z_{0},...,z_{n}]\ |\ z_{i}\neq 0\}</math> for <math>0\leq i \leq n</math>

Clearly <math>\bigcup U_{i} = CP^{n}</math>

Now, for each <math>p\in U_{i}</math> there is a unique representative for its equivalence class of the form <math>[z_{0},...,1,...,z_{n}]</math> where the 1 is at the ith location.

We thus can get a map from <math>\varphi_{i}:U_{i}\rightarrow \mathbb{C}^{n} = \mathbb{R}^{2n}</math> by
<math>p\mapsto [z_{0}/z_{i},...,z_{i-1}/z_{i},z_{i+1}/z_{i},...,z_{n}/z_{i}]</math>
Hence we have shown (loosely) that our functional structure is locally isormorphic to <math>(\mathbb{R},C^\infty)</math>

'''Definition 3''' Product Manifolds

Suppose <math>M^{m}</math> and <math>N^{n}</math> are manifolds. Then the product manifold, on the set MxN has an atlas given by
<math>\{\varphi \times \psi: U\times V\rightarrow U'\times V'\in \mathbb{R}^{m}\times \mathbb{R}^{n}\ | \varphi: U\rightarrow U'\subset \mathbb{R}^{m}\ and\ \psi:V\rightarrow V'\subset \mathbb{R}^{n}</math> are charts in resp. manifolds}

Claim: This does in fact yield a manifold

'''Example 4'''
It can be checked that <math>T^{2} = S^{1}\times S^{1}</math> gives the torus a manifold structure, by the product manifold, that is indeed the same as the normal structure given previously.

===Second Hour===
Aim: We consider two manifolds, M and N, and a function f between them. We aim for the analogous idea of the tangent in the reals, namely that every smooth f:M -> N has a good linear approximation. Of course, we will need to define what is meant by such a ''smooth'' function, as well as what this linear approximation is.

'''Definition 1''' Smooth Function, Atlas Sense

Let <math>M^{m}</math> and <math>N^{n}</math> be manifolds. We say that a function <math>f:M^{m}\rightarrow N^{n}</math> is smooth if <math>\psi\circ f\circ\varphi^{-1}</math> is smooth, where it makes sense, <math>\forall</math> charts <math>\psi,\varphi</math>

'''Definition 2''' Smooth Function, Functional Structure Sense

Let <math>M^{m}</math> and <math>N^{n}</math> be manifolds. We say that a function <math>f:M^{m}\rightarrow N^{n}</math> is smooth if <math>\forall h:V\rightarrow R</math> such that h is smooth, <math>h\circ f</math> is smooth on subsets of M where it is defined.

'''Claim 1'''

The two definitions are equivalent

'''Definition 3''' Category (Loose Definition)

A category is a collection of "objects" (such as sets, topological spaces, manifolds, etc.) such that for any two objects, x and y, there exists a set of "morphisms" denoted mor(x-> y) (these would be functions for sets, continuous functions for topological spaces, smooth functions for manifolds, etc.) along with

1)Composition maps <math>\circ</math>:mor(x->y) X mor(y->z) --> mor (x->z)

2) For any x there exists an element <math>1_{x}\in</math>mor(x-> x)

such that several axioms are satisfied, including

a) Associativity of the composition map

b) <math>1_{x}</math> behaves like an identity should.

'''Claim 2'''

The collection of smooth manifolds with smooth functions form a category.
This (loosely) amounts to (easily) checking the two claims that

1) if f:M-> N and g:N->L are smooth maps between manifolds then gof:M->L is smooth

2) <math>1_{M}</math> is smooth

We now provide two alternate definition of the tangent vector, one from each definition of a manifold.

'''Definition 4''' Tangent Vector, Atlas Sense

Consider the set of curves on the manifold, that is
<math>\{\gamma:R\rightarrow M\ |\ s.t.\ \gamma\ smooth,\ \gamma(0) = p\}</math>

We define an equivalence relation between paths by <math>\gamma_{1}</math> ~ <math>\gamma_{2}</math> if <math>\varphi\circ\gamma_{1}</math> is tangent to <math>\varphi\circ\gamma_{2}</math> as paths in <math>R^{n}\ \forall\varphi</math>.

Then, a ''tangent vector'' is an equivalence class of curves.

'''Definition 5''' Tangent Vector, Functional Structure Sense

A tangent vector D at p, often called a directional derivative in this sense, is an operator that takes the smooth real valued functions near p into '''R'''.

such that,

1) D(af +bg) = aDf + bDg for a,b constants in '''R''' and f,g such functions

2) D(fg) = (Df)g(p) + f(p)D(g) Leibniz' Rules

'''Theorem 1'''
These two definitions are equivalent.

'''Definition 6''' Tangent Space

The tangent space at a point, <math>T_{p}M

</math> = the set of all tangent vectors at a point p

'''Claim 3'''

We will show later that in fact <math>T_{p}M

</math> is a vector space

Consider a map <math>f:M\rightarrow N</math>. We are interested in defining an associated map between the tangent spaces, namely, <math>df_{p}:T_{p}M\rightarrow T_{f(p)}N</math>

We get two different definitions from the two different definitions of a tangent vector

'''Definition 7''' (atlas sense)

<math>(df_{p})([\gamma]):=[f\circ\gamma]</math>

'''Definition 8''' (functional structure sense)

<math>((df_{p})(D))(h):=D(h\circ f)</math>

'''Proof of Theorem 1'''

We consider a curve <math>\gamma</math> and denote its equivalence class of curves by <math>[\gamma]</math>

We consider the map <math>\Phi:[\gamma]\rightarrow D_{\gamma}</math> defined by
<math>D_{\gamma}(f) = (f\circ\gamma)'(0)</math>

It is clear that this map is linear is f and satisfies the leibnitz rule.

Claim: <math>\Phi</math> is a bijection.

We will need to use the following lemma, which will not be proved now:

''Hadamard's Lemma''

If <math>f:R^{m}\rightarrow R</math> is smooth near 0 then there exists smooth <math>g_{i}:R^{m}\rightarrow R</math> such that
<math>
f(x) = f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)</math>

Now, let D be a tangent vector. One can quickly see that D(const) = 0.

So, by hadamard's lemma,
<math>Df = D(f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)) = D(\Sigma_{i=1}^{n}x_{i}g_{i}(x)) = \Sigma_{i=1}^{n}(Dx_{i})g_{i}(0) + \Sigma_{i=1}^{n}x_{i}(0)Dg_{i}</math> = <math>\Sigma_{i=1}^{n}(Dx_{i})g_{i}(0)</math>
as <math>x_{i}(0) = 0</math>

and hence, Df is a linear comb of fixed quantities.

Now, let <math>\gamma:R\rightarrow R^{n}</math> be such that <math>\gamma(0) = 0</math>

<math>D_{\gamma}f = \Sigma_{i=1}^{n}(D_{\gamma}x_{i}))g_{i}(0)</math>

but <math>D_{\gamma}(x_{i}) = (x_{i}\circ\gamma)'(0) = \gamma_{i}'(0)</math>

We can now claim that <math>\Phi</math> is onto because given a D take <math>\gamma_{i}(t)=D(x_{i})t</math>

<math>\Phi</math> is trivially 1:1.
==Smoot==
In this class It was riced the question of what a "Smoot" is. Here it is [[0708-1300/Smoot]]

0708-1300/Class notes for Tuesday, October 2

2007-10-03T22:42:30Z

Aptikuis: /* First Hour */

{{0708-1300/Navigation}}
==English Spelling==
Many interesting rules about [[0708-1300/English Spelling]]
==Class Notes==
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

'''General class comments'''

1) The class photo is up, please add yourself

2) A questionnaire was passed out in class

3) Homework one is due on thursday

===First Hour===

Today's Theme: Locally a function looks like its differential

'''Pushforward/Pullback'''

Let <math>\theta:X\rightarrow Y</math> be a smooth map.

We consider various objects, defined with respect to X or Y, and see in which direction it makes sense to consider corresponding objects on the other space. In general <math>\theta_*</math> will denote the push forward, and <math>\theta^*</math> will denote the pullback.

1) points ''pushforward'' <math>x\mapsto\theta_*(x) := \theta(x)</math>

2) Paths <math>\gamma:R\rightarrow X</math>, ie a bunch of points, ''pushforward'', <math>\gamma\rightarrow \theta_*(\gamma):=\theta\circ\gamma</math>

3) Sets <math>B\subset Y</math> ''pullback'' via <math>B\rightarrow \theta^*(B):=\theta^{-1}(B)</math>
Note that if one tried to pushforward sets A in X, the set operations compliment and intersection would not commute appropriately with the map <math>\theta</math>

4) A measures <math>\mu</math> ''pushforward'' via <math>\mu\rightarrow (\theta_*\mu)(B) :=\mu(\theta^*B)</math>

5)In some sense, we consider functions, "dual" to points and thus should go in the opposite direction of points, namely <math>\theta^*f = f\circ\theta</math>

6) Tangent vectors, defined in the sense of equivalence classes of paths, [<math>\gamma</math>] ''pushforward'' as we would expect since each path pushes forward. <math>[\gamma]\rightarrow \theta_*[\gamma]:=[\theta_*\gamma] = [\theta\circ\gamma]</math>

CHECK: This definition is well defined, that is, independent of the representative choice of <math>\gamma</math>

7) We can consider operators on functions to be in a sense dual to the functions and hence should go in the opposite direction. Hence, tangent vectors, defined in the sense of derivations, ''pushforward'' via <math>D\rightarrow (\theta_*D)(f):= D(\theta^*f)</math>

CHECK: This definition satisfies linearity and Liebnitz property.

'''Theorem 1'''

The two definitions for the pushforward of a tangent vector coincide.

''Proof:''

Given a <math>\gamma</math> we can construct <math>\theta_{*}\gamma</math> as above. However from both <math>\gamma</math> and <math>\theta_*\gamma</math> we can also construct <math>D_{\gamma}f</math> and <math>D_{\theta_*\gamma}f</math> because we have previously shown our two definitions for the tangent vector are equivalent. We can then ''pushforward'' <math>D_{\gamma}f</math> to get <math>\theta_*D_{\gamma}f</math>. The theorem is reduced to the claim that:

<math>\theta_*D_{\gamma}f = D_{\theta_*\gamma}f</math>

for functions <math>f:Y\rightarrow R</math>

Now, <math>D_{\theta_*\gamma}f = \frac{d}{dt}f\circ(\theta_*\gamma)|_{t=0} = \frac{d}{dt}f\circ(\theta\circ\gamma)|_{t=0} = \frac{d}{dt}(f\circ\theta\gamma |_{t=0} = D_{\gamma}(f\circ\theta) =\theta_*D_{\gamma}f</math>

''Q.E.D''

'''Functorality'''

let <math>\theta:X\rightarrow Y, \lambda:Y\rightarrow Z</math>

Consider some "object" s defined with respect to X and some "object u" defined with respect to Z. Something has the property of functorality if

<math>\lambda_*(\theta_*s) = (\lambda\circ\theta)_*s</math>

and

<math>\theta^*(\lambda^*u) = (\lambda\circ\theta)^*u</math>

Claim: All the classes we considered previously have the functorality property; in particular, the pushforward of tangent vectors does.

Let us consider <math>\theta_*</math> on <math>T_pM</math> given a <math>\theta:M\rightarrow N</math>

We can arrange for charts <math>\varphi</math> on a subset of M into <math>R^m</math> (with coordinates denoted <math>(x_1,\dots,x_m)</math>)and <math>\psi</math> on a subset of N into <math>R^n</math> (with coordinates denoted <math>(y_1,\dots,y_n)</math>)such that <math>\varphi(p) = 0</math> and <math>\psi(\theta(p)=0</math>

Define <math>\theta^o = \psi\circ\theta\circ\varphi^{-1} = (\theta_1(x_1,\dots,x_m),\dots,\theta_n(x_1,\dots,x_m))</math>

Now, for a <math>D\in T_pM</math> we can write <math>D=\sum_i a_{i=1}^m\frac{\partial}{\partial x_i}</math>

So, <math>(\theta_*D)(f) = \sum_{i=1}^m a_i\frac{\partial}{\partial x_i}(f\circ\varphi) = \sum_{i=1}^m a_i \sum_{j=1}^n\frac{\partial f}{\partial y_j}\frac{\partial\theta_j}{\partial x_i}=</math>
<math>=\begin{bmatrix}
\frac{\partial f}{\partial y_1} & \cdots & \frac{\partial f}{\partial y_n}\\
\end{bmatrix}
\begin{bmatrix}

\frac{\partial \theta_1}{\partial x_1} & \cdots & \frac{\partial \theta_1}{\partial x_m}\\
\vdots& & \vdots\\
\frac{\partial \theta_n}{\partial x_1} & \cdots & \frac{\partial \theta_n}{\partial x_m}\\
\end{bmatrix}
\begin{bmatrix}
a_1\\
\vdots\\
a_m\\
\end{bmatrix}</math>

Now, we want to write <math>\theta_*D = \sum b_j\frac{\partial}{\partial y_j}</math>

and so, <math>b_k = (\theta_*D)y_k =\begin{bmatrix}
0&\cdots & i & \cdots &0\\
\end{bmatrix}
\begin{bmatrix}

\frac{\partial \theta_1}{\partial x_1} & \cdots & \frac{\partial \theta_1}{\partial x_m}\\
\vdots& & \vdots\\
\frac{\partial \theta_n}{\partial x_1} & \cdots & \frac{\partial \theta_n}{\partial x_m}\\
\end{bmatrix}
\begin{bmatrix}
a_1\\
\vdots\\
a_m\\
\end{bmatrix}</math>

where the i is at the kth location.

So, <math>\theta_* = d\theta_p</math>, i.e., <math>\theta_*</math> is the differential of <math>\theta</math> at p

We can check the functorality, <math>(\lambda\circ\theta)_* = \lambda_*\circ\theta_*</math>, then <math>d(\lambda\circ\theta) = d\lambda\circ d\theta</math>
This is just the chain rule.

===Second Hour===

'''Defintion 1'''

An ''immersion'' is a (smooth) map <math>\theta:M^m\rightarrow N^n</math> such that <math>\theta_*</math> of tangent vectors is 1:1. More precisely, <math>d\theta_p: T_pM\rightarrow T_{\theta(p)}N</math> is 1:1 <math>\forall p\in M</math>

'''Example 1'''

Consider the canonical immersion, for m<n given by <math>\iota:(x_1,...,x_m)\mapsto (x_1,...,x_m,0,...,0)</math> with n-m zeros.

'''Example 2'''

This is the map from R to <math>R^2</math> that looks like a loop-de-loop on a roller coaster (but squashed into the plane of course!) The map <math>\theta</math> itself is NOT 1:1 (consider the crossover point) however <math>\theta_*</math> IS 1:1, hence an immersion.

'''Example 3'''

Consider the map from R to <math>R^2</math> that looks like a check mark. While this map itself is 1:1, <math>\theta_*</math> is NOT 1:1 (at the cusp in the check mark) and hence is not an immersion.

'''Example 4'''

Can there be objects, such as the graph of |x| that are NOT an immersion, but are constructed from a smooth function?

Consider the function <math>\lambda(x) = e^{-1/x^2}</math> for x>0 and 0 otherwise.

Then the map <math>x\mapsto \begin{bmatrix}
(\lambda(x),\lambda(x))& x>0\\
(0,0)& x=0\\
(-\lambda(-x),\lambda(-x)) & x<0\\
\end{bmatrix}</math>

is a smooth mapping with the graph of |x| as its image, but is NOT an immersion.

'''Example 5'''

The torus, as a subset of <math>R^3</math> is an immersion

Now, consider the 1:1 linear map <math>T:V\rightarrow W</math> where V,W are vector spaces that takes <math>(v_1,...,v_m)\mapsto (Tv_1,...,Tv_m) = (w_1,..,w_m,w_{m+1},...,w_n)</math>

From linear algebra we know that we can choose a basis such that T is represented by a matrix with 1's along the first m diagonal locations and zeros elsewhere.

'''Theorem 2'''

Locally, every immersion looks like the inclusion <math>\iota:R^m\rightarrow R^n</math>.

More precisely, if <math>\theta:M^m\rightarrow N^n</math> and <math>d\theta_p</math> is 1:1 then there exists charts <math>\varphi</math> acting on <math>U\subset M</math> and <math>\phi</math> acting on <math>V\subset N</math> such that for <math>p\in U, \phi(p) = \psi(\theta(p)) = 0</math> such that the following diagram commutes:

<math>\begin{matrix}
U&\rightarrow^{\phi}&U'\subset R^n\\
\downarrow_{\theta} &&\downarrow_{\iota} \\
V& \rightarrow^{\psi}& V'\subset R^n\\
\end{matrix}</math>

that is, <math>\iota\circ\varphi = \psi\circ\theta</math>

'''Definition 2'''

M is a ''submanifold'' of N if there exists a mapping <math>\theta:M\rightarrow N</math> such that <math>\theta</math> is a 1:1 immersion.

'''Example 6'''

Our previous example of the graph of a "loop-de-loop", while an immersion, the function is not 1:1 and hence the graph is not a sub manifold.

'''Example 6'''

The torus is a submanifold as the natural immersion into <math>R^3</math> is 1:1

'''Definition 3'''

The map <math>\theta:M\rightarrow N</math> is an embedding if the subset topology on <math>\theta(M)</math> coincides with the topology induced from the original topology of M.

'''Example 7'''

Consider the map from <math>R\rightarrow R^2</math> whose graph looks like the open interval whose two ends have been wrapped around until they just touch (would intersect at one point if they were closed) the points 1/3 and 2/3rds of the way along the interval respectively.
The map is both 1:1 and an immersion. However, any neighborhood about the endpoints of the interval will ALSO include points near the 1/3rd and 2/3rd spots on the line, i.e., the topology is different and hence this is not an embedding.

'''Corollary 1 to Theorem 2'''

The functional structure on an embedded manifold induced by the functional structure on the containing manifold is equal to its original functional structure.

Indeed, for all smooth <math>f:M\rightarrow R</math> and <math>\forall p\in M</math> there exists a neighborhood V of <math>\theta(p)</math> and a smooth <math>g:N\rightarrow R</math> such that <math>g|_{\theta(M)\bigcap U} = f|_U</math>

''Proof of Corollary 1''

Loosely (and a sketch is most useful to see this!) we consider the embedded submanifold M in N and consider its image, under the appropriate charts, to a subset of <math>R^m\subset R^n</math>. We then consider some function defined on M, and hence on the subset in R^n which we can extend canonically as a constant function in the "vertical" directions. Now simply pullback into N to get the extended member of the functional structure on N.

''Proof of Theorem 2''

We start with the normal situation of <math>\theta:M\rightarrow N</math> with M,N manifolds with atlases containing <math>(\varphi_0,U_)0)</math> and <math>(\psi_0, V_0)</math> respectively. We also expect that for <math>p\in U_0, \varphi_0(p) = \psi_0(\theta(p)) = 0</math>. I will first draw the diagram and will subsequently justify the relevant parts. The proof reduces to showing a certain part of the diagram commutes appropriately.

<math>
\begin{matrix}
M\supset U_0 & \rightarrow^{\varphi_0} & U_1\subset R^m & \rightarrow^{Id} & U_2 = U_1 \\
\downarrow_{\theta} & &\downarrow_{\theta_1} & &\downarrow_{\iota}\\
N\supset V_0 & \rightarrow^{\psi_0} & V_1\subset R^n & \leftarrow^{\xi} & V_2\\
\end{matrix}
</math>

It is very important to note that the <math>\varphi_0</math> and <math>\psi_0</math> are NOT the charts we are looking for , they are merely one of the ones that happen to act about the point p.

In the diagram above, <math>\theta_1 = \psi_0\circ\theta\circ\varphi^{-1}</math>. So, <math>\theta_1(0) = 0</math> and <math>(d\theta_1)_0 = i</math>. Note the <math>\theta_1</math>, being merely the normal composition with the appropriate charts, does not fundamentally add anything. What makes this theorem work is the function <math>\xi</math>

We consider the map <math>\xi:V_2\rightarrow V_1</math> given <math>(x,y)\mapsto \theta_1(x) + (0,y)</math>. We note that <math>\xi</math> corresponds with the idea of "lifting" a flattened image back to its original height.

Claims:

1) <math>\xi</math> is invertible near zero. Indeed, computing <math>d\xi_0 = I</math> which is invertible as a matrix and hence <math>\xi</math> is invertible as a function near zero.

2) Take an <math>x\in U_2</math>. There are two routes to get to <math>V_1</math> and upon computing both ways yields the same result. Hence, the diagram commutes.

Hence, our immersion looks (locally) like the standard immersion between real spaces given by <math>\iota</math> and the charts are the compositions going between <math>U_0</math> to <math>U_2</math> and <math>V_0</math> to <math>V_2</math>

''Q.E.D''

0708-1300/Class notes for Thursday, September 20

2007-09-23T13:21:02Z

Aptikuis: /* Exercise */

{{0708-1300/Navigation}}
==Dror's Note==
Come to {{Home Link|Talks/Fields-0709/|my talk}} today at 4:30PM at the Fields Institute!

==Class Notes==
PDF file of the class notes typed up in latex can be located [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.pdf here]

Tex version of the file is also avaliable [http://individual.utoronto.ca/tbazett/mat1300/0708-1300-Class_Notes_latex_20-09.tex here] so that people can easily make changes and repost here if they wish.

==Exercise==
Configurations of a Generalized Cockroach (not entirely rigourous)

Let <math>C_n</math> be the manifold of configurations of a "cockroach" with <math>n</math> legs:
[[Image:0708-1300-cockroach-labelling.jpg||center|200px|]]

Q: What is <math>C_n</math>?
In particular, what is <math>C_6</math>?

<math>n=2</math>: Consider <math>C_2</math>.
[[Image:0708-1300-cockroach-C2.jpg||left||200px]]
As in the picture, label the angles of the joints <math>\theta_1</math> and <math>\theta_2</math> and the distances from the body to the feet <math>d_1</math> and <math>d_2</math> respectively.

The value of <math>\theta_i</math> determines the value of <math>d_i</math>.
So, given values <math>(\theta_1, \theta_2)</math>, possible configurations are given by positions of the body, which must have distance <math>d_i</math> from the <math>i</math>th foot.

That is, the body must lie on the intersection of the two circles of radius <math>d_i</math> centred at the <math>i</math>th foot:
[[Image:0708-1300-cockroach-twocircles.jpg||center||]]

There are <math>0, 1, </math> or <math>2</math> solutions for the body position.
If we look only at the top body position, the pair <math>(\theta_1, \theta_2)</math> determines a unique configuration.
So, we can plot the subset on <math>\R^2_{(\theta_1,\theta_2)}</math>:
[[Image:0708-1300-cockroach-C2region.jpg||center||]]
The boundary points are where the "top" solution is in fact the unique solution.

By symmetry, taking the bottom solution gives us a similar region, and since the boundaries are where the top and bottom solutions coincide (there is only one solution along the boundary), the entire manifold is given by gluing the boundaries together.
This gives a sphere.

<math>n=3</math>: Configurations with <math>3</math> legs consist of a configuration with <math>2</math> legs plus the configuration of the third leg.
The configuration of the first <math>2</math> legs fixes the position of the body - and thus, the distance <math>d_3</math> from the third foot to the body.

For certain configurations of the first two legs, the body is too far from the third foot, so these are not found as part of configurations with three feet.
When the distance from the body to the third foot equals the length of the third leg completely extended, this gives a unique configuration of the <math>3</math>-legged cockroach.
Any closer and there are two possible configurations, corresponding to the two ways that the third joint can bend.

Let <math>R</math> be the region in <math>C_2 = S_2</math> where the distance to the third leg is close enough to give solutions.
The boundary of <math>R</math> is a curve, consisting of all the points at which there is a unique solution:
[[Image:0708-1300-cockroach-regionR.jpg||center||]]
So the manifold <math>C_3</math> is given by taking two copies of <math>R</math> and gluing their boundaries together.
This gives a sphere.

<math>n \geq 3</math>: Likewise, given that <math>C_n = S_2</math>, it follows that <math>C_{n+1} = S_2</math> also.
In particular, <math>C_6 = S_2</math>.

File:0708-1300-cockroach-C2region.jpg

2007-09-23T13:14:59Z

Aptikuis:

File:0708-1300-cockroach-regionR.jpg

2007-09-23T13:07:52Z

Aptikuis:

File:0708-1300-cockroach-twocircles.jpg

2007-09-23T13:07:46Z

Aptikuis:

File:0708-1300-cockroach-labelling.jpg

2007-09-23T13:07:41Z

Aptikuis:

File:0708-1300-cockroach-C2.jpg

2007-09-23T13:07:33Z

Aptikuis:

0708-1300/Class notes for Thursday, September 20

2007-09-22T17:31:31Z

Aptikuis:

0708-1300/Class notes for Tuesday, September 11

2007-09-20T01:51:55Z

Aptikuis: /* Implicit Function Theorem */

{{0708-1300/Navigation}}
==In Small Scales, Everything's Linear==
{| align=center
|- align=center
|[[Image:06-240-QuiltBeforeMap.png|200px]]
|<math>\longrightarrow</math>
|[[Image:06-240-QuiltAfterMap.png|200px]]
|- align=center
|<math>z</math>
|<math>\mapsto</math>
|<math>z^2</math>
|}

Code in [http://www.wolfram.com Mathematica]:

<pre>
QuiltPlot[{f_,g_}, {x_, xmin_, xmax_, nx_}, {y_, ymin_, ymax_, ny_}] :=
Module[
{dx, dy, grid, ix, iy},
SeedRandom[1];
dx=(xmax-xmin)/nx;
dy=(ymax-ymin)/ny;
grid = Table[
{x -> xmin+ix*dx, y -> ymin+iy*dy},
{ix, 0, nx}, {iy, 0, ny}
];
grid = Map[({f, g} /. #)&, grid, {2}];
Show[
Graphics[Table[
{
RGBColor[Random[], Random[], Random[]],
Polygon[{
grid[[ix, iy]],
grid[[ix+1, iy]],
grid[[ix+1, iy+1]],
grid[[ix, iy+1]]
}]
},
{ix, nx}, {iy, ny}
]],
Frame -> True
]
]

QuiltPlot[{x, y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
QuiltPlot[{x^2-y^2, 2*x*y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
</pre>

See also [[06-240/Linear Algebra - Why We Care]].

{{0708-1300/Class Notes}}
===Differentiability===
Let <math>U</math>, <math>V</math> and <math>W</math> be two normed finite dimensional vector spaces and let <math>f:V\rightarrow W</math> be a function defined on a neighborhood of the point <math>x</math>.

'''Definition:'''

We say that <math>f</math> is differentiable (''diffable'') at <math>x</math> if there is a linear map <math>L</math> so that

<math>\lim_{h\rightarrow0}\frac{|f(x+h)-f(x)-L(h)|}{|h|}=0.</math>

In this case we will say that <math>L</math> is a differential of <math>f</math> at <math>x</math> and will denote it by <math>df_{x}</math>.

'''Theorem'''

If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following asertions holds:

# <math>df_{x}</math> is unique.
# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>
# If <math>f</math> is linear then <math>df_{x}=f</math>
# <math>d(f\circ g)_{x}=df_{g(x)}\circ dg_{x}</math>
# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>

===Implicit Function Theorem===

'''Example'''
Although <math>x^2+y^2=1</math> does not defines <math>y</math> as a function of <math>x</math>, in a neighborhood of <math>(0;-1)</math> we can define <math>g(x)=-\sqrt{1-x^2}</math> so that <math>x^2+g(x)^2=1</math>. Furthermore, <math>g</math> is differentiable with differential <math>dg_{x}=\frac{x}{\sqrt{1-x^2}}</math>. This is a motivation for the following theorem.

'''Notation'''

If <math>f:X\times Y\rightarrow Z</math> then given <math>x\in X</math> we will define <math>f_{[x]}:Y\rightarrow Z</math> by <math>f_{[x]}(y)=f(x;y).</math>

'''Definition'''

<math>C^{p}(V)</math> will be the class of all functions defined on <math>V</math> with continuous partial derivatives up to order <math>p.</math>

'''Theorem'''(''Implicit function theorem'')

Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a <math>C^{1}(\mathbb{R}^n \times \mathbb{R}^m)</math> function defined on a neighborhood <math>U</math> of the point <math>(x_0;y_0)</math> and such that <math>f(x_0;y_0)=0</math> and suppose that <math>d(f_{[x]})_{y}</math> is non-singular then, the following results holds:

There is an open neighborhood of <math>x_0</math>, <math>V\subset U</math>, and a ''diffable'' function <math>g:V\rightarrow\mathbb{R}^m</math> such that <math>g(x_0)=y_0</math> and for every <math>x\in V</math> <math>f(x;g(x))=0.</math>.

0708-1300/Class notes for Tuesday, September 11

2007-09-18T04:59:35Z

Aptikuis:

{{0708-1300/Navigation}}
==In Small Scales, Everything's Linear==
{| align=center
|- align=center
|[[Image:06-240-QuiltBeforeMap.png|200px]]
|<math>\longrightarrow</math>
|[[Image:06-240-QuiltAfterMap.png|200px]]
|- align=center
|<math>z</math>
|<math>\mapsto</math>
|<math>z^2</math>
|}

Code in [http://www.wolfram.com Mathematica]:

<pre>
QuiltPlot[{f_,g_}, {x_, xmin_, xmax_, nx_}, {y_, ymin_, ymax_, ny_}] :=
Module[
{dx, dy, grid, ix, iy},
SeedRandom[1];
dx=(xmax-xmin)/nx;
dy=(ymax-ymin)/ny;
grid = Table[
{x -> xmin+ix*dx, y -> ymin+iy*dy},
{ix, 0, nx}, {iy, 0, ny}
];
grid = Map[({f, g} /. #)&, grid, {2}];
Show[
Graphics[Table[
{
RGBColor[Random[], Random[], Random[]],
Polygon[{
grid[[ix, iy]],
grid[[ix+1, iy]],
grid[[ix+1, iy+1]],
grid[[ix, iy+1]]
}]
},
{ix, nx}, {iy, ny}
]],
Frame -> True
]
]

QuiltPlot[{x, y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
QuiltPlot[{x^2-y^2, 2*x*y}, {x, -10, 10, 8}, {y, 5, 10, 8}]
</pre>

See also [[06-240/Linear Algebra - Why We Care]].

{{0708-1300/Class Notes}}
===Differentiability===
Let <math>U</math>, <math>V</math> and <math>W</math> be two normed finite dimensional vector spaces and let <math>f:V\rightarrow W</math> be a function defined on a neighborhood of the point <math>x</math>.

'''Definition:'''

We say that <math>f</math> is differentiable (''diffable'') at <math>x</math> if there is a linear map <math>L</math> so that

<math>\lim_{h\rightarrow0}\frac{|f(x+h)-f(x)-L(h)|}{|h|}=0.</math>

In this case we will say that <math>L</math> is a differential of <math>f</math> at <math>x</math> and will denote it by <math>df_{x}</math>.

'''Theorem'''

If <math>f:V\rightarrow W</math> and <math>g:U\rightarrow V</math> are ''diffable'' maps then the following asertions holds:

# <math>df_{x}</math> is unique.
# <math>d(f+g)_{x}=df_{x}+dg_{x}</math>
# If <math>f</math> is linear then <math>df_{x}=f</math>
# <math>d(f\circ g)_{x}=df_{g(x)}\circ dg_{x}</math>
# For every scalar number <math>\alpha</math> it holds <math>d(\alpha f)_{x}=\alpha df_{x}</math>

===Implicit Function Theorem===

'''Example'''
Although <math>x^2+y^2=1</math> does not defines <math>y</math> as a function of <math>x</math>, in a neighborhood of <math>(0;-1)</math> we can define <math>g(x)=-\sqrt{1-x^2}</math> so that <math>x^2+g(x)^2=1</math>. Furthermore, <math>g</math> is differentiable with differential <math>dg_{x}=\frac{x}{\sqrt{1-x^2}}</math>. This is a motivation for the following theorem.

'''Notation'''

If <math>f:X\times Y\rightarrow Z</math> then given <math>x\in X</math> we will define <math>f_{[x]}:Y\rightarrow Z</math> by <math>f_{[x]}(y)=f(x;y).</math>

'''Definition'''

<math>C^{p}(V)</math> will be the class of all functions defined on <math>V</math> with continuous partial derivatives up to order <math>p.</math>

'''Theorem'''(''Implicit function theorem'')

Let <math>f:\mathbb{R}^n \times \mathbb{R}^m\rightarrow \mathbb{R}^m</math> be a <math>C^{1}(\mathbb{R}^n \times \mathbb{R}^m)</math> function defined on a neighborhood <math>U</math> of the point <math>(x_0;y_0)</math> and such that <math>f(x_0;y_0)=0</math> and suppose that <math>d(f_{[x]})_{y}</math> is non-singular then, the following results holds:

There is an open neighborhood of <math>x</math>, <math>V\subset U</math>, and a ''diffable'' function <math>g:V\rightarrow\mathbb{R}^m</math> such that for every <math>x\in V</math> <math>f(x;g(x))=0.</math>.