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0708-1300/Class notes for Tuesday, April 1

2008-04-24T19:30:02Z

Jvoltz:

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==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===

'''Theorem'''

If <math>\phi:I^k\rightarrow S^n</math> is an embedding then <math>\tilde{H}_*(S^n-\phi(I^k)) = 0</math>

''Proof:''

By induction on k.

For k=0 this is easy, <math>I^k</math> is a single point and <math>S^n</math> - a point is just <math>\mathbb{R}^n</math>

Now suppose we know the theorem is true for k-1, let <math>\phi:I^k\rightarrow S^n</math> be an embedding. Write the cube <math>I^k = C</math> as <math>C=C_L\cup C_R</math> where <math>C_L=\{x\in C:x_1\leq 1/2\}</math> and <math>C_R=\{x\in C:x_1\geq 1/2\}</math>

<math>C_M:= C_L\cap C_R\cong I^{k-1}</math>

So, <math>(S^n-\phi(C_m)) = (S^n-\phi(C_l))\cup(S^n-\phi(C_R))</math>

where both sets in the union are open. Recall for sets <math>X = A\cup B</math> where A and B are open we have the Mayer Vietoris Sequence.

Note: It is usefull to "open the black box" and think about the Mayer Vietoris Sequence from the veiwpoint of singular homology, ie using maps of simplexes and the like. This was briefly sketched in class.

We get a sequence <math>\rightarrow\tilde{H}_{p+1}(S^n-\phi(C_m))\rightarrow \tilde{H}_{p}(S^n-\phi(C))\rightarrow \tilde{H}_{p}(S^n-\phi(C_L))\oplus \tilde{H}_{p}(S^n-\phi(C_R))\rightarrow \tilde{H}_{p}(S^n-\phi(C_M))\rightarrow</math>

The first and last terms in this sequence (at least the small part of it we have written) vanish by induction hypothesis.

Note: Technically we need to check at Mayer Vietoris sequence also works for reduced homology.

We thus get the isomorphism: <math>\tilde{H}_{p}(S^n-\phi(C))\cong \tilde{H}_{p}(S^n-\phi(C_L))\oplus \tilde{H}_{p}(S^n-\phi(C_R))</math>

Assume <math>0\neq[\beta]\in \tilde{H}_{p}(S^n-\phi(C))
</math>
As the above isormorphism is induced by the inclusion maps at the chain level, we get that <math>[\beta]\neq 0</math> either in <math>\tilde{H}_{p}(S^n-\phi(C_L))</math> or in <math>\tilde{H}_{p}(S^n-\phi(C_R))</math>

Now repeat, find a sequence of intervals <math>I_j</math> such that

1) <math>\cap I_j = \{x\}</math>

2) <math>[\beta]\neq 0</math> in <math>\tilde{H}_{p}(S^n-\phi(I_j\times I^{k-1}))</math>

Pictorially, we cut C in half, find which half <math>[\beta]</math> is non zero, then cut this half in half and find the one the <math>[\beta]</math> is non zero in and again cut it in half, etc...

But, <math>[\beta] = 0</math> in <math>\tilde{H}_{p}(S^n-\phi(\{x\}\times I^{k-1}))</math> by induction. Hence, <math>\exists\gamma\in C_{p+1}(S^n-\phi(\{x\}\times I^{k-1}))</math> such that <math>\partial\gamma=\beta</math>.

But, <math>supp\gamma</math> = { union of images of simplexes in U} is compact, so, <math>supp\gamma\subset S^n-\phi(I_j\times I^{k-1})</math> for some j. But this contradicts assumption 2)

''Q.E.D''

We now prove an analogous theorem for spheres opposed to cubes

'''Theorem'''

For <math>\phi:S^k\rightarrow S^n</math> an embedding then <math>\tilde{H}_{p}(S^n-\phi(S^k)) = \mathbb{Z}</math> if p = n-k-1 and zero otherwise.

Intuitively,<math> S^n-S^k = S^{n-k-1}</math>

Recall we had seen this explicitly for <math>S^3-S^1</math> when we wrote <math>S^3</math> as the union of two tori (an inflated circle)

''Proof:''

As in the previous proof, we use Mayer Vietoris. Let <math>S^k = D^k_-\cup D^k_+</math>

Then, <math>S^{k-1} = D^k_-\cap D^k_+</math>
<math>
\tilde{H}_{p}(S^n-\phi(S^k))\rightarrow \tilde{H}_{p}(S^n-\phi(D^k_-))\oplus\tilde{H}_{p}(S^n-\phi(D^k_+))\rightarrow \tilde{H}_{p}(S^n-\phi(S^{k-1}))\rightarrow \tilde{H}_{p-1}(S^n-\phi(S^k))</math>

The first and last term written vanish by induction hypothesis and so have an isomorphism <math>\tilde{H}_{p}(S^n-\phi(D^k_-))\oplus\tilde{H}_{p}(S^n-\phi(D^k_+))\cong \tilde{H}_{p}(S^n-\phi(S^{k-1}))</math>

Hence, it is enough to prove the theorem for k=0. Well in this case we have <math>\tilde{H}_{p}(S^n-\{2\ pts\})</math>

Since <math>S^n-\{2\ pts\}\sim S^{n-1}</math>, this is <math>\mathbb{Z}</math> for p = n-1 and 0 otherwise

''Q.E.D''

Take k=n-1
<math>
\tilde{H}_{0}(S^n-\phi(S^{n-1})) =\mathbb{Z}</math> so <math>H_{0}(S^n-\phi(S^{n-1}))=\mathbb{Z}^2</math>

Hence the compliment of an n-1 sphere in <math>S^n</math> has two connected components. This is the Jordan Curve Theorem.

===Second Hour===

'''Excersize'''

Consider the embedding of a line in <math>\mathbb{R}^3</math> which wraps arround somewhat like a helix only each loop makes a link with each previous loop and in addition the helical object gets smaller and converges to two points at the end.

The question is, should we put a little circle L about a part of the helical object can we make this L the boundary of a disk embedded in <math>R^3</math>.

Sketch of Answer: Consider the bubble which has L as the boundary and goes all the way arround the object at one end. This is not an embedding since there are two intersection points with the helical object. Cut two small holes from the bubble at each of these points. This is still not quite it since the boundary also has these two small holes. Hence connect the two holes via a tubular neighbourhood going along the helical object and we have the result.

The real excersize here is to figure out how the Mayer Vietoris sequence actually determines this particular surface.

'''Proposition'''

An open path connected set in <math>\mathbb{R}^n</math> is clopen connected.

''Proof'' This is easy, just using general topology

Hence, for open sets in <math>\mathbb{R}^n</math> path components are the same as clopen components

'''Theorem'''

"Invariance of Domain" or "Openness in <math>\mathbb{R}^n</math> is intrinsic"

If <math>V\subset\mathbb{R}^n</math> is homeomorphic to an open set in <math>\mathbb{R}^n</math> then it is an open set in <math>\mathbb{R}^n</math>

Here are two examples that illustrate why this might not be obvious

1) Replacing open with dense would not be true, for instance, the rationals in [0,1] are not dence in <math>\mathbb{R}</math> but they ARE homeomorphic to the rationals which ARE dense in R.

2) Replacing open with closed is also not true, for instance, <math>\mathbb{R}^n</math> is closed in <math>\mathbb{R}^n</math>

but it is also homeomorphic to the open ball which is not closed.

''Proof''

Let <math>\phi:U\rightarrow B</math>

Consider a ball B in U, with boundary the circle S and we get a disk <math>D=B\cup S</math>

We have that <math>\phi(S)</math> devides <math>\mathbb{R}^n_V</math> into two connected components. Call them <math>A_1</math> and <math>A_2</math>.

We have both <math>\phi(B)</math> and <math>\mathbb{R}^m - \phi(D)</math> but we need to show these ARE the two connected components. We first note that both are in fact connected.

<math>(R^n -\phi(D))\cup\phi(B) = \mathbb{R}^n-\phi(S) = A_1\cup A_2</math>

Hence <math>\phi(B)</math> is one of the components and so it is open. ''Q.E.D''

'''Theorem''' Borsuk-Ulam

If <math>g:S^n\rightarrow R^n</math> is continuous then <math>\exists x\in S^n</math> such that g(x) = g(-x)

'''Corrollary''' The Ham Sandwich Theorem (or Salad Bowl Theorem, according to Dr. Bar Natan)

Loosely this says that if we have n objects in a bounded domain of <math>\mathbb{R}^n</math> such as items in a salad or sandwhich then we can find an n-1 dimensional hyperplane such that the hyperplane precisely cuts each item in half.

Formally, If <math>\mu_1,\cdots,\mu_n</math> are densities in <math>\mathbb{R}^n</math> with bounded support then <math>\exists</math> a hyperplane <math>H\subset\mathbb{R}^n</math> deviding <math>\mathbb{R}^n</math> into <math>H_+\cup H_-</math> such that <math>\forall j</math>, <math>\mu_j(H_+) = \mu_j(H_-)</math>

''Proof:''

Consider the set of sided hyperplanes in <math>\mathbb{R}^n</math> which equals <math>S^{n-1}_v\times R_t</math> where the first component determines the normal vector v and the second component determines how far away this plane is front the origin in the direction of v.

Define <math>\bar{g}:S^{n-1}\times\mathbb{R}\rightarrow\mathbb{R}^n</math> via

<math>(v,t)\mapsto (\mu_1(H_+),\cdots, \mu_n(H_-))</math>

Now, if t>>0, clearly <math>\bar{g}(v,t) = 0</math> and if t<<0 then <math>\bar{g}(v,t) = (\mu_1(\mathbb{R}^n),\cdots,\mu_n(\mathbb{R}^n))</math>

Hence we have a cylinder mapping into <math>\mathbb{R}^n</math> such that it is constant along the top and bottom of the cylindar. Hence, we can pinch the top and bottom of the cylinder. We thus get a map g from <math>S^n</math> to <math>\mathbb{R}^n</math> and, applying Borsuk-Ulam, deduce there is a point such that g(x) = g(-x) and so <math>\bar{g}(v,t) = \bar{g}(-v,-t)</math> for some point (v,t). Hence <math>\mu_i(H_+)=\mu_i(H_-)</math> for all i

''Q.E.D''

0708-1300/Class notes for Tuesday, November 6

2007-11-20T14:18:03Z

Jvoltz: /* Definition */

{{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.

We will now shift our attention to the theory of integration on smooth manifolds. The first thing that we need to construct is a means of measuring volumes on manifolds. To accomplish this goal, we begin by imagining that we want to measure the volume of the "infinitiesimal" parallelepiped [http://en.wikipedia.org/wiki/Parallelepiped] defined by a set of vectors <math>X_1 , \ldots ,X_k \in T_pM\!</math> by feeding these vectors into some function <math>\omega : (T_pM)^k \to \mathbb{R}\!</math>. We would like <math>\omega\!</math> to satisfy a few properties:

<ol>
<li> <math>\omega\!</math> should be linear in each argument: for example, if we double the length of one of the sides, the volume should double.
<li> If two of the vectors fed to <math>\omega\!</math> are parallel, the volume assigned by <math>\omega\!</math> should be zero because the parallelepiped collapses to something with lower dimenion in this case.
</ol>

 Inspired by these requirements, we make the following definition:

===Definition===



Let <math>V\!</math> be a real vector space, let <math>p \in \mathbb{N}\!</math> and let <math>L(V^p; \mathbb{R})</math> denote the collection of maps from <math>V^p\!</math> to <math>\mathbb{R}</math> that are linear in each argument separately. We set 

<div align="center"><math>A^p(V) = \left\{ \omega \in L(V^p; \mathbb{R}) : \omega(\ldots,v,\ldots,v,\ldots) = 0\ \forall v \in V \right\}</math></div>

 and if <math>\omega \in A^p(V)\!</math>, we say that the degree of <math>\omega\!</math> is <math>p\!</math> and write <math>\mathrm{deg}(\omega) = p\!</math>. <math>\Box\!</math>



===Proposition===



Suppose that <math>\omega \in A^p(V)\!</math> and <math>v_1,\ldots,v_p \in V\!</math>. The following statements hold:

<ol>
<li> <math>A^p(V)\!</math> has a natural vector space structure
<li> <math>A^0(V)\!</math> is <math>\mathbb{R}</math>
<li> <math>A^1(V) = V^*\!</math> is the dual space of <math>V\!</math>
<li> <math>\omega(v_1,\ldots,v_j,\ldots,v_k,\ldots,v_p) = - \omega(v_1,\ldots,v_k,\ldots,v_j,\ldots,v_p)\!</math> for every <math>j<k \in \{1,\ldots,p\}\!</math>
<li> If <math>\sigma \in S_p\!</math> is a permutation, then <math>\omega(v_{\sigma(1)},\ldots,v_{\sigma(p)}) = (-1)^\sigma \omega(v_1,\ldots,v_p)</math>
</ol>



====Proof====

The first statement is easy to show and is left as an exercise. The second statement is more of a convenient definition. Note that <math>A^0(V)\!</math> consists of all maps that take no vectors and return a real number since the other properties are vacuous when the domain is empty. We can thus interpret an element in this space simply as a real number. The third statement is clear as the defintions of <math>A^1(V)\!</math> and <math>V^*\!</math> coincide.

As for the fourth, note that <math>0 = \omega(v_1, \ldots, v_j + v_k, \ldots, v_j+v_k, \ldots, v_p)</math> so that using linearity we obtain

<math>0= \omega(v_1, \ldots, v_j , \ldots, v_j, \ldots, v_p) +\omega(v_1, \ldots, v_k , \ldots, v_j, \ldots, v_p) + \omega(v_1, \ldots, v_j , \ldots, v_k, \ldots, v_p) +\omega(v_1, \ldots, v_k , \ldots, v_k, \ldots, v_p) </math>

 and hence <math>\omega(v_1, \ldots, v_k , \ldots, v_j, \ldots, v_p) + \omega(v_1, \ldots, v_j , \ldots, v_k, \ldots, v_p) = 0</math>.

 The fifth statement then follows from repeated application of the fourth. <math>\Box\!</math>

 

Our computation in the previous proof shows that we could equally well have defined <math>A^p(V)\!</math> to consist of all those multilinear maps from <math>V^k\!</math> to <math>\mathbb{R}\!</math> that change sign when two arguments are interchanged.

 One of the nicest things about these spaces is that we can define a sort of multiplication of elements of <math>A^p(V)\!</math> with <math>A^q(V)\!</math>. This multiplication is called the wedge product and is defined as follows. 

===Definition===



For each <math>p,q \in \mathbb{N}\!</math> the wedge product is the map <math>\wedge : A^p(V) \times A^q(V) \to A^{p+q}(V), (\omega,\lambda) \mapsto \omega \wedge \lambda</math> defined by

<div align="center"><math>(\omega \wedge \lambda) (v_1,\ldots,v_{p+q}) = \sum_{\sigma \in S_{p,q}} (-1)^\sigma\omega(v_{\sigma(1)},\ldots,v_{\sigma(p)})\lambda(v_{\sigma(p+1)},\ldots,v_{\sigma(p+q)})</math> </div>

 for every <math>v_1 ,\ldots,v_{p+q} \in V</math>, where <math>S_{p,q} = \{ \sigma \in S_{p+q} | \sigma(1) < \ldots < \sigma(p), \sigma(p+1) < \ldots < \sigma(p+q)\}</math>.<math>\Box</math>




 

The idea behind this definition is to feed vectors to <math>\omega \wedge \lambda\!</math> in as many ways as possible. We could equally well have set 

<div align="center"> <math>(\omega \wedge \lambda) (v_1,\ldots,v_{p+q}) = \frac{1}{p!q!} \sum_{\sigma \in S_{p+q}} (-1)^\sigma\omega(v_{\sigma(1)},\ldots,v_{\sigma(p)})\lambda(v_{\sigma(p+1)},\ldots,v_{\sigma(p+q)})</math>. </div>

The factor of <math>\frac{1}{p!q!}</math> compensates for the overcounting that we do by summing over all permutations, since their are <math>p!\!</math> ways of rearranging the <math>p\!</math> vectors fed to <math>\omega\!</math> if we don't care about order, but only one way if we do care. The same argument accounts for the <math>q!\!</math>.

Of course, as we have defined it, it is not immediately clear that <math>\omega \wedge \lambda\in A^{p+q}(V)\!</math>. However, multilinearity is obvious and it is fairly clear that the <math>(-1)^\sigma\!</math> takes care of the skew-symmetry.

In fact, <math>\wedge\!</math> has a number of nice properties:

===Proposition===



The following statements hold:

<ol>
<li> <math>\wedge\!</math> is a bilinear map.
<li> <math>\wedge\!</math> is associative.
<li> <math>\wedge\!</math> is supercommutative: <math>\omega \wedge \lambda = (-1)^{\mathrm{deg}(\omega)\mathrm{deg}(\lambda)} \lambda \wedge \omega\!</math>.
</ol>



====Proof====

 Bilinearity is clear. Associativity and supercommutativity follow from some combinatorial arguments. <math>\Box\!</math> 

 

 It turns out that we can use the wedge product to find bases for <math>A^p(V)\!</math>:
===Proposition===



 If <math>\{\omega_1,\ldots,\omega_n \}\subset V^* \!</math> is a basis for <math>V^*\!</math> then <math>\{\omega_{i_1}\wedge\cdots\wedge\omega_{i_p} \in A^p(V) | i_1 < \ldots < i_p \}\!</math> is a basis for <math>A^p(V)\!</math>



====Proof====

Let <math>\{v_1,\ldots,v_n \}\subset V \!</math> be the dual basis to <math>\{\omega_1,\ldots,\omega_n \}</math>, so that <math>\omega_i(v_j) = \delta_{ij}\!</math>. Let <math>\rho_p = \{(i_1,\ldots,i_p) \in \mathbb{N}^p | i_1 < \ldots < i_p\}\!</math>. For <math>I,J \in \rho_p\!</math> with <math>I = (i_1,\ldots,i_p)\!</math> and <math>J= (j_1,\ldots,j_p)\!</math>, let <math>\omega_I = \omega_{i_1} \wedge \cdots \wedge \omega_{i_p}</math>, and let <math>v_J = (v_{j_1},\ldots,v_{j_p})</math>. Then <math>\omega_I(v_J) = 1\!</math> if <math>I=J\!</math> and <math>\omega_I(v_J) = 0\!</math> otherwise.

We claim that if <math>\omega \in A^p(V)\!</math>, then <math>\omega = \sum_{I\in\rho_p} \omega(v_I) \omega_I\!</math>. But <math>\sum_{I\in\rho_p} \omega(v_I) \omega_I(v_J) = \sum_{I\in\rho_p} \omega(v_I) \delta_{IJ} = \omega(v_J)\!</math>, so equality holds for ordered sequences of basis vectors. Equality then holds for any sequence of vectors by skew-symmetry and linearity. We claim further that the <math>\omega_I\!</math> are linearly independent. But if <math>0 = \sum_{I \in \rho_p} \alpha_I \omega_I\!</math>, then <math>\alpha_J = 0 \! </math> by applying <math>\sum_{I \in \rho_p} \alpha_I \omega_I\!</math> to <math>v_J\!</math>. Hence the <math>\omega_I\!</math> form a basis.<math>\Box\!</math>

===Corollary===



 <math>\mathrm{dim}(A^p(V)) = \frac{n!}{p!(n-p)!}\!</math>, where <math>n=\mathrm{dim}(V)\!</math>. <math>\Box\!</math>



 

 We may now define differential forms. The idea is to smoothly assign to each point <math>x\!</math> in a manifold <math>M\!</math> an element of <math>A^p(T_xM)\!</math>. 

===Definition===



Let <math> M\!</math> be a smooth manifold of dimension <math>m\!</math>. For <math>0 \le p \le m\!</math>, a differential <math>p\!</math>-form on <math>M\!</math> (or simply a p-form) is an assignment to each <math>x \in M\!</math> an element <math>\omega_x \in A^p(T_x M)\!</math> that is smooth in the sense that if <math>X_1,\ldots,X_p\!</math> are smooth vector fields on <math>M\!</math> then the map <math>M \ni x \mapsto \omega_x(X_1(x),\ldots,X_p(x)) \in \mathbb{R}\!</math> is <math>C^\infty\!</math>.

The collection of <math>p\!</math>-forms on <math>M\!</math> will be denoted by <math>\Omega^p(M)\!</math>.<math>\Box\!</math>



 

If <math>\omega_1,\ldots,\omega_n \in \Omega^1(M)\!</math> are such that <math>(\omega_1)_x,\ldots,(\omega_n)_x\!</math> form a basis for <math>(T_xM)^*\!</math> for each <math>x\in U\!</math> with <math>U \subset M\!</math> open, then <math>\lambda \in \Omega^k(M)\!</math> can be written (for <math>x\in U\!</math>) as

<div align="center"> <math> \lambda_x = \sum_{I \in \rho_k} a_I(x) (\omega_I)_x </math> </div>

where the maps <math>a_I : U \to \mathbb{R}\!</math> are smooth. In fact, we could have taken this property as our definition of smoothness on <math>U\!</math>.

0708-1300/Class notes for Tuesday, November 6

2007-11-20T14:17:06Z

Jvoltz: /* 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.

We will now shift our attention to the theory of integration on smooth manifolds. The first thing that we need to construct is a means of measuring volumes on manifolds. To accomplish this goal, we begin by imagining that we want to measure the volume of the "infinitiesimal" parallelepiped [http://en.wikipedia.org/wiki/Parallelepiped] defined by a set of vectors <math>X_1 , \ldots ,X_k \in T_pM\!</math> by feeding these vectors into some function <math>\omega : (T_pM)^k \to \mathbb{R}\!</math>. We would like <math>\omega\!</math> to satisfy a few properties:

<ol>
<li> <math>\omega\!</math> should be linear in each argument: for example, if we double the length of one of the sides, the volume should double.
<li> If two of the vectors fed to <math>\omega\!</math> are parallel, the volume assigned by <math>\omega\!</math> should be zero because the parallelepiped collapses to something with lower dimenion in this case.
</ol>

 Inspired by these requirements, we make the following definition:

===Definition===



Let <math>V\!</math> be a real vector space, let <math>p \in \mathbb{N}\!</math> and let <math>L(V^p; \mathbb{R})</math> denote the collection maps from <math>V^p\!</math> to <math>\mathbb{R}</math> that are linear in each argument separately. We set 

<div align="center"><math>A^p(V) = \left\{ \omega \in L(V^p; \mathbb{R}) : \omega(\ldots,v,\ldots,v,\ldots) = 0\ \forall v \in V \right\}</math></div>

 and if <math>\omega \in A^p(V)\!</math>, we say that the degree of <math>\omega\!</math> is <math>p\!</math> and write <math>\mathrm{deg}(\omega) = p\!</math>. <math>\Box\!</math>



===Proposition===



Suppose that <math>\omega \in A^p(V)\!</math> and <math>v_1,\ldots,v_p \in V\!</math>. The following statements hold:

<ol>
<li> <math>A^p(V)\!</math> has a natural vector space structure
<li> <math>A^0(V)\!</math> is <math>\mathbb{R}</math>
<li> <math>A^1(V) = V^*\!</math> is the dual space of <math>V\!</math>
<li> <math>\omega(v_1,\ldots,v_j,\ldots,v_k,\ldots,v_p) = - \omega(v_1,\ldots,v_k,\ldots,v_j,\ldots,v_p)\!</math> for every <math>j<k \in \{1,\ldots,p\}\!</math>
<li> If <math>\sigma \in S_p\!</math> is a permutation, then <math>\omega(v_{\sigma(1)},\ldots,v_{\sigma(p)}) = (-1)^\sigma \omega(v_1,\ldots,v_p)</math>
</ol>



====Proof====

The first statement is easy to show and is left as an exercise. The second statement is more of a convenient definition. Note that <math>A^0(V)\!</math> consists of all maps that take no vectors and return a real number since the other properties are vacuous when the domain is empty. We can thus interpret an element in this space simply as a real number. The third statement is clear as the defintions of <math>A^1(V)\!</math> and <math>V^*\!</math> coincide.

As for the fourth, note that <math>0 = \omega(v_1, \ldots, v_j + v_k, \ldots, v_j+v_k, \ldots, v_p)</math> so that using linearity we obtain

<math>0= \omega(v_1, \ldots, v_j , \ldots, v_j, \ldots, v_p) +\omega(v_1, \ldots, v_k , \ldots, v_j, \ldots, v_p) + \omega(v_1, \ldots, v_j , \ldots, v_k, \ldots, v_p) +\omega(v_1, \ldots, v_k , \ldots, v_k, \ldots, v_p) </math>

 and hence <math>\omega(v_1, \ldots, v_k , \ldots, v_j, \ldots, v_p) + \omega(v_1, \ldots, v_j , \ldots, v_k, \ldots, v_p) = 0</math>.

 The fifth statement then follows from repeated application of the fourth. <math>\Box\!</math>

 

Our computation in the previous proof shows that we could equally well have defined <math>A^p(V)\!</math> to consist of all those multilinear maps from <math>V^k\!</math> to <math>\mathbb{R}\!</math> that change sign when two arguments are interchanged.

 One of the nicest things about these spaces is that we can define a sort of multiplication of elements of <math>A^p(V)\!</math> with <math>A^q(V)\!</math>. This multiplication is called the wedge product and is defined as follows. 

===Definition===



For each <math>p,q \in \mathbb{N}\!</math> the wedge product is the map <math>\wedge : A^p(V) \times A^q(V) \to A^{p+q}(V), (\omega,\lambda) \mapsto \omega \wedge \lambda</math> defined by

<div align="center"><math>(\omega \wedge \lambda) (v_1,\ldots,v_{p+q}) = \sum_{\sigma \in S_{p,q}} (-1)^\sigma\omega(v_{\sigma(1)},\ldots,v_{\sigma(p)})\lambda(v_{\sigma(p+1)},\ldots,v_{\sigma(p+q)})</math> </div>

 for every <math>v_1 ,\ldots,v_{p+q} \in V</math>, where <math>S_{p,q} = \{ \sigma \in S_{p+q} | \sigma(1) < \ldots < \sigma(p), \sigma(p+1) < \ldots < \sigma(p+q)\}</math>.<math>\Box</math>




 

The idea behind this definition is to feed vectors to <math>\omega \wedge \lambda\!</math> in as many ways as possible. We could equally well have set 

<div align="center"> <math>(\omega \wedge \lambda) (v_1,\ldots,v_{p+q}) = \frac{1}{p!q!} \sum_{\sigma \in S_{p+q}} (-1)^\sigma\omega(v_{\sigma(1)},\ldots,v_{\sigma(p)})\lambda(v_{\sigma(p+1)},\ldots,v_{\sigma(p+q)})</math>. </div>

The factor of <math>\frac{1}{p!q!}</math> compensates for the overcounting that we do by summing over all permutations, since their are <math>p!\!</math> ways of rearranging the <math>p\!</math> vectors fed to <math>\omega\!</math> if we don't care about order, but only one way if we do care. The same argument accounts for the <math>q!\!</math>.

Of course, as we have defined it, it is not immediately clear that <math>\omega \wedge \lambda\in A^{p+q}(V)\!</math>. However, multilinearity is obvious and it is fairly clear that the <math>(-1)^\sigma\!</math> takes care of the skew-symmetry.

In fact, <math>\wedge\!</math> has a number of nice properties:

===Proposition===



The following statements hold:

<ol>
<li> <math>\wedge\!</math> is a bilinear map.
<li> <math>\wedge\!</math> is associative.
<li> <math>\wedge\!</math> is supercommutative: <math>\omega \wedge \lambda = (-1)^{\mathrm{deg}(\omega)\mathrm{deg}(\lambda)} \lambda \wedge \omega\!</math>.
</ol>



====Proof====

 Bilinearity is clear. Associativity and supercommutativity follow from some combinatorial arguments. <math>\Box\!</math> 

 

 It turns out that we can use the wedge product to find bases for <math>A^p(V)\!</math>:
===Proposition===



 If <math>\{\omega_1,\ldots,\omega_n \}\subset V^* \!</math> is a basis for <math>V^*\!</math> then <math>\{\omega_{i_1}\wedge\cdots\wedge\omega_{i_p} \in A^p(V) | i_1 < \ldots < i_p \}\!</math> is a basis for <math>A^p(V)\!</math>



====Proof====

Let <math>\{v_1,\ldots,v_n \}\subset V \!</math> be the dual basis to <math>\{\omega_1,\ldots,\omega_n \}</math>, so that <math>\omega_i(v_j) = \delta_{ij}\!</math>. Let <math>\rho_p = \{(i_1,\ldots,i_p) \in \mathbb{N}^p | i_1 < \ldots < i_p\}\!</math>. For <math>I,J \in \rho_p\!</math> with <math>I = (i_1,\ldots,i_p)\!</math> and <math>J= (j_1,\ldots,j_p)\!</math>, let <math>\omega_I = \omega_{i_1} \wedge \cdots \wedge \omega_{i_p}</math>, and let <math>v_J = (v_{j_1},\ldots,v_{j_p})</math>. Then <math>\omega_I(v_J) = 1\!</math> if <math>I=J\!</math> and <math>\omega_I(v_J) = 0\!</math> otherwise.

We claim that if <math>\omega \in A^p(V)\!</math>, then <math>\omega = \sum_{I\in\rho_p} \omega(v_I) \omega_I\!</math>. But <math>\sum_{I\in\rho_p} \omega(v_I) \omega_I(v_J) = \sum_{I\in\rho_p} \omega(v_I) \delta_{IJ} = \omega(v_J)\!</math>, so equality holds for ordered sequences of basis vectors. Equality then holds for any sequence of vectors by skew-symmetry and linearity. We claim further that the <math>\omega_I\!</math> are linearly independent. But if <math>0 = \sum_{I \in \rho_p} \alpha_I \omega_I\!</math>, then <math>\alpha_J = 0 \! </math> by applying <math>\sum_{I \in \rho_p} \alpha_I \omega_I\!</math> to <math>v_J\!</math>. Hence the <math>\omega_I\!</math> form a basis.<math>\Box\!</math>

===Corollary===



 <math>\mathrm{dim}(A^p(V)) = \frac{n!}{p!(n-p)!}\!</math>, where <math>n=\mathrm{dim}(V)\!</math>. <math>\Box\!</math>



 

 We may now define differential forms. The idea is to smoothly assign to each point <math>x\!</math> in a manifold <math>M\!</math> an element of <math>A^p(T_xM)\!</math>. 

===Definition===



Let <math> M\!</math> be a smooth manifold of dimension <math>m\!</math>. For <math>0 \le p \le m\!</math>, a differential <math>p\!</math>-form on <math>M\!</math> (or simply a p-form) is an assignment to each <math>x \in M\!</math> an element <math>\omega_x \in A^p(T_x M)\!</math> that is smooth in the sense that if <math>X_1,\ldots,X_p\!</math> are smooth vector fields on <math>M\!</math> then the map <math>M \ni x \mapsto \omega_x(X_1(x),\ldots,X_p(x)) \in \mathbb{R}\!</math> is <math>C^\infty\!</math>.

The collection of <math>p\!</math>-forms on <math>M\!</math> will be denoted by <math>\Omega^p(M)\!</math>.<math>\Box\!</math>



 

If <math>\omega_1,\ldots,\omega_n \in \Omega^1(M)\!</math> are such that <math>(\omega_1)_x,\ldots,(\omega_n)_x\!</math> form a basis for <math>(T_xM)^*\!</math> for each <math>x\in U\!</math> with <math>U \subset M\!</math> open, then <math>\lambda \in \Omega^k(M)\!</math> can be written (for <math>x\in U\!</math>) as

<div align="center"> <math> \lambda_x = \sum_{I \in \rho_k} a_I(x) (\omega_I)_x </math> </div>

where the maps <math>a_I : U \to \mathbb{R}\!</math> are smooth. In fact, we could have taken this property as our definition of smoothness on <math>U\!</math>.

0708-1300/Class notes for Tuesday, October 2

2007-11-06T21:14:05Z

Jvoltz:

{{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) = D(f\circ\theta)</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) \!</math>
|<math> = D(\theta^* f)\!</math>
|-
|
|<math>=\sum_{i=1}^m a_i\frac{\partial}{\partial x_i}(f\circ\theta) </math>
|-
|
|<math>=\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. In other words, <math>\theta_*D = \sum_{j=1}^{n} \sum_{i=1}^{m}a_i \frac{\partial \theta_j}{\partial x_i} \frac{\partial }{\partial y_j} </math>

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 exist charts <math>\varphi</math> acting on <math>U\subset M</math> and <math>\psi</math> acting on <math>V\subset N</math> such that for <math>p\in U, \varphi(p) = \psi(\theta(p)) = 0</math> such that the following diagram commutes:

<math>\begin{matrix}
U&\rightarrow^{\varphi}&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''