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0708-1300/Class notes for Tuesday, November 6

2008-02-10T21:21:51Z

Bpym: /* 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 there 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

2007-11-19T18:13:15Z

Bpym: /* 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 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, November 6

2007-11-17T20:08:17Z

Bpym: /* 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 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, November 6

2007-11-17T19:59:35Z

Bpym: /* 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 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 

<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>

===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. 

===Remarks===

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 this is clear. 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

2007-11-15T18:54:50Z

Bpym:

Template:0708-1300/Navigation

2007-11-15T17:55:36Z

Bpym:

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
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|colspan=3 align=center|'''Fall Semester'''
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|Sep 10
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|Sep 17
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|Sep 24
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|Oct 1
|[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]
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|Oct 8
|Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]
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|Oct 22
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|Nov 5
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0708-1300/Class notes for Thursday, October 18

2007-10-21T21:35:26Z

Bpym: /* Remarks */

{{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.

===Outline===

Today we stated the Whitney Embedding Theorem and began to discuss its proof. Along the way, we also encountered some related notions that will serve us well in the future. We begin by stating the theorem:

===Theorem (Whitney Embedding)===
Let <math>M\!</math> be a smooth <math>m\!</math>-manifold. Then there exists an embedding <math>\Phi : M \to \mathbb{R}^{2m+1}</math>.

====Proof====
=====Outline=====

We will break the proof of the theorem into three parts:

<ol>
<li> Find an embedding of a compact <math>M\!</math> into <math>\mathbb{R}^N\!</math> for some <math>N\!</math>.
<li> Use Sard's Theorem to reduce <math>N\!</math> to <math>2m+1\!</math>.
<li> Use the "Zebra Trick" to prove the theorem for non-compact <math>M\!</math>.
</ol>

Parts two and three shall be left to the next lecture.

=====Part 1=====

 Suppose that <math>M\!</math> is compact. Let <math>\{\phi_\alpha : U_\alpha \to \mathbb{R}^m \}_{\alpha \in A}</math> be an atlas for <math>M\!</math>, and note that <math>\{U_\alpha\}_{\alpha \in A}\!</math> is an open cover of <math>M\!</math>. Hence it possesses a finite subcover <math>\{U_j\}_{1 \le j \le J}\!</math>, and the corresponding collection <math>\{\phi_j\}_{1 \le j \le J}</math> of charts is an atlas. 

Choose smooth functions <math>\{\lambda_j : M \to \mathbb{R}_{\ge 0} \}_{1 \le j \le J}</math> with the following properties:

<ol>
<li> <math>\lambda_j |_{M \setminus U_j} = 0 </math>
<li> <math>\bigcup_{j=1}^{J} \lambda_j^{-1}\left((0,\infty)\right) = M</math>
<li> <math>\mathrm{supp}\left(\lambda_j\right) \subset U_j</math> for <math>1 \le j \le J </math>
</ol>

where <math>\mathrm{supp}(f)\!</math> for <math>f : M \to \mathbb{R}</math> is the support of <math>f\!</math>, ie. the closure of <math>f^{-1}(\mathbb{R} \setminus \{0\}).</math> The existence of such functions follows from the existence of smooth partitions of unity for manifolds---a concept that will be discussed later on.

Now define <math>\Phi : M \to \mathbb{R}^{J + mJ}</math> by <math>\Phi(p) = \left(\lambda_1(p), \ldots, \lambda_J(p), \lambda_1 \tilde \phi_1 (p), \ldots, \lambda_J \tilde \phi_J(p)\right)</math>, where <math>\tilde \phi_j : M \to \mathbb{R}^m</math> is defined by <math>\tilde \phi_j |_{U_j} = \phi_j</math> and <math>\tilde \phi_j |_{M \setminus U_j} = 0</math>.

We claim that <math>\Phi\!</math> is an embedding. That <math>\Phi\!</math> is smooth follows immediately from its construction (the <math>\lambda_j\!</math>s have been used to smear out the <math>\phi_j\!</math> to smooth functions on all of <math>M\!</math>). That <math>\Phi\!</math> is injective is also clear. It takes a bit of work to show that <math>\Phi\!</math> is an immersion, but this is left as an exercise. It remains to see that <math>\Phi\!</math> is a homeomorphism, but this fact follows from the following topological lemma. <math>\Box\!</math>

===Lemma===

Let <math>(X,\tau)\!</math> and <math>(Y,\sigma)\!</math> be topological spaces. Suppose that <math>X\!</math> is compact, <math>Y\!</math> is Hausdorff, and that <math> f : X \to Y</math> is continuous and injective. Then <math>f : X \to f(X)</math> is a homeomorphism.

====Proof====

 Since <math>f\!</math> is an injection onto its image, it is a bijection. Since <math>f\!</math> is continuous, it remains to show that <math>f^{-1}\!</math> is continuous. Thus, it suffices to see that <math>f\!</math> takes closed sets to closed sets. Let <math>A \subset X</math> be closed. Since <math>X\!</math> is compact, so is <math>A\!</math>. Hence <math>f(A)\!</math> is compact since <math>f\!</math> is continuous. But <math>Y\!</math> is Hausdorff, and every compact subset of a Hausdorff space is closed. Hence <math>f(A)\!</math> is closed. <math>\Box</math>

===Remarks===

 The smearing functions we used in Part 1 of the proof of the Whitney Embedding Theorem are very similar to partitions of unity---collections of functions that break the constant function <math>p \mapsto 1</math> into a bunch of bump functions. We will now formalize this notion and show that such collections of functions exist for smooth manifolds.

===Definition===

Let <math>U = \{U_\alpha\}_{\alpha \in A}</math> be an open cover of a topological space (manifold) <math>M\!</math>. A partition of unity subordinated to <math>U\!</math> is a collection <math>\{\lambda_\beta : M \to \mathbb{R}_{\ge 0} \}_{\beta \in B}</math> of continuous (smooth) functions such that

<ol>
<li> For every <math>\beta \in B</math> there is an <math>\alpha \in A</math> such that <math>\mathrm{supp}(\lambda_\beta) \subset U_\alpha</math>
<li> <math>\{\mathrm{supp}(\lambda_\beta)\}_{\beta \in B}</math> is locally finite, ie. for every <math>p \in M</math>, <math>p \in \mathrm{supp}(\lambda_\beta)</math> for only finitely many <math>\beta\!</math>.
<li> <math>\sum_{\beta \in B} \lambda_\beta = 1</math>
</ol>

 A local refinement of <math>U\!</math> is an open cover <math>\{V_\gamma\}_{\gamma \in C}</math> of <math>M\!</math> such that for every <math>\gamma \in C</math>, <math>V_\gamma \subset U_\alpha</math> for some <math>\alpha \in A</math>. <math>M\!</math> is called paracompact if every open cover of <math>M\!</math> has a locally finite refinement.

===Remarks===

For further information on paracompactness, we refer the reader to the corresponding [http://en.wikipedia.org/wiki/Paracompact Wikipedia entry]. Note, in particular, that locally compact, second-countable topological spaces---such as manifolds---are paracompact, and that paracompact spaces are shrinking spaces. The following result (which follows immediately from these facts) will be useful for constructing partitions of unity on manifolds:

===Theorem===

Manifolds are paracompact. In particular, if <math>\{U_\alpha\}_{\alpha \in A}</math> is locally finite then there is an open cover <math>\{V_\alpha\}_{\alpha \in A}</math> such that <math>\bar{V}_\alpha \subset U_\alpha</math> for every <math>\alpha \in A</math>.

0708-1300/Class notes for Thursday, October 18

2007-10-19T03:27:55Z

Bpym: /* 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.

===Outline===

Today we stated the Whitney Embedding Theorem and began to discuss its proof. Along the way, we also encountered some related notions that will serve us well in the future. We begin by stating the theorem:

===Theorem (Whitney Embedding)===
Let <math>M\!</math> be a smooth <math>m\!</math>-manifold. Then there exists an embedding <math>\Phi : M \to \mathbb{R}^{2m+1}</math>.

====Proof====
=====Outline=====

We will break the proof of the theorem into three parts:

<ol>
<li> Find an embedding of a compact <math>M\!</math> into <math>\mathbb{R}^N\!</math> for some <math>N\!</math>.
<li> Use Sard's Theorem to reduce <math>N\!</math> to <math>2m+1\!</math>.
<li> Use the "Zebra Trick" to prove the theorem for non-compact <math>M\!</math>.
</ol>

Parts two and three shall be left to the next lecture.

=====Part 1=====

 Suppose that <math>M\!</math> is compact. Let <math>\{\phi_\alpha : U_\alpha \to \mathbb{R}^m \}_{\alpha \in A}</math> be an atlas for <math>M\!</math>, and note that <math>\{U_\alpha\}_{\alpha \in A}\!</math> is an open cover of <math>M\!</math>. Hence it possesses a finite subcover <math>\{U_j\}_{1 \le j \le J}\!</math>, and the corresponding collection <math>\{\phi_j\}_{1 \le j \le J}</math> of charts is an atlas. 

Choose smooth functions <math>\{\lambda_j : M \to \mathbb{R}_{\ge 0} \}_{1 \le j \le J}</math> with the following properties:

<ol>
<li> <math>\lambda_j |_{M \setminus U_j} = 0 </math>
<li> <math>\bigcup_{j=1}^{J} \lambda_j^{-1}\left((0,\infty)\right) = M</math>
<li> <math>\mathrm{supp}\left(\lambda_j\right) \subset U_j</math> for <math>1 \le j \le J </math>
</ol>

where <math>\mathrm{supp}(f)\!</math> for <math>f : M \to \mathbb{R}</math> is the support of <math>f\!</math>, ie. the closure of <math>f^{-1}(\mathbb{R} \setminus \{0\}).</math> The existence of such functions follows from the existence of smooth partitions of unity for manifolds---a concept that will be discussed later on.

Now define <math>\Phi : M \to \mathbb{R}^{J + mJ}</math> by <math>\Phi(p) = \left(\lambda_1(p), \ldots, \lambda_J(p), \lambda_1 \tilde \phi_1 (p), \ldots, \lambda_J \tilde \phi_J(p)\right)</math>, where <math>\tilde \phi_j : M \to \mathbb{R}^m</math> is defined by <math>\tilde \phi_j |_{U_j} = \phi_j</math> and <math>\tilde \phi_j |_{M \setminus U_j} = 0</math>.

We claim that <math>\Phi\!</math> is an embedding. That <math>\Phi\!</math> is smooth follows immediately from its construction (the <math>\lambda_j\!</math>s have been used to smear out the <math>\phi_j\!</math> to smooth functions on all of <math>M\!</math>). That <math>\Phi\!</math> is injective is also clear. It takes a bit of work to show that <math>\Phi\!</math> is an immersion, but this is left as an exercise. It remains to see that <math>\Phi\!</math> is a homeomorphism, but this fact follows from the following topological lemma. <math>\Box\!</math>

===Lemma===

Let <math>(X,\tau)\!</math> and <math>(Y,\sigma)\!</math> be topological spaces. Suppose that <math>X\!</math> is compact, <math>Y\!</math> is Hausdorff, and that <math> f : X \to Y</math> is continuous and injective. Then <math>f : X \to f(X)</math> is a homeomorphism.

====Proof====

 Since <math>f\!</math> is an injection onto its image, it is a bijection. Since <math>f\!</math> is continuous, it remains to show that <math>f^{-1}\!</math> is continuous. Thus, it suffices to see that <math>f\!</math> takes closed sets to closed sets. Let <math>A \subset X</math> be closed. Since <math>X\!</math> is compact, so is <math>A\!</math>. Hence <math>f(A)\!</math> is compact since <math>f\!</math> is continuous. But <math>Y\!</math> is Hausdorff, and every compact subset of a Hausdorff space is closed. Hence <math>f(A)\!</math> is closed. <math>\Box</math>

===Remarks===

 The smearing functions we used in Part 1 of the proof of the Whitney Embedding Theorem are very similar to partitions of unity---collections of functions that break the constant function <math>p \mapsto 1</math> into a bunch of bump functions. We will now formalize this notion and show that such collections of functions exist for smooth manifolds.

===Definition===

Let <math>U = \{U_\alpha\}_{\alpha \in A}</math> be an open cover of a topological space (manifold) <math>M\!</math>. A partition of unity subordinated to <math>U\!</math> is a collection <math>\{\lambda_\beta : M \to \mathbb{R}_{\ge 0} \}_{\beta \in B}</math> of continuous (smooth) functions such that

<ol>
<li> For every <math>\beta \in B</math> there is an <math>\alpha \in A</math> such that <math>\mathrm{supp}(\lambda_\beta) \subset U_\alpha</math>
<li> <math>\{\mathrm{supp}(\lambda_\beta)\}_{\beta \in B}</math> is locally finite, ie. for every <math>p \in M</math>, <math>p \in \mathrm{supp}(\lambda_\beta)</math> for only finitely many <math>\beta\!</math>.
<li> <math>\sum_{\beta \in B} \lambda_\beta = 1</math>
</ol>

 A local refinement of <math>U\!</math> is an open cover <math>\{V_\gamma\}_{\gamma \in C}</math> of <math>M\!</math> such that for every <math>\gamma \in C</math>, <math>V_\gamma \subset U_\alpha</math> for some <math>\alpha \in A</math>. <math>M\!</math> is called paracompact if every open cover of <math>M\!</math> has a locally finite refinement.

===Remarks===

For further information on paracompactness, we refer the reader to the corresponding [http://en.wikipedia.org/wiki/Paracompact Wikipedia entry]. Note, in particular, that locally compact, Hausdorff spaces---such as manifolds---are paracompact, and that paracompact spaces are shrinking spaces. The following result (which follows immediately from these facts) will be useful for constructing partitions of unity on manifolds:

===Theorem===

Manifolds are paracompact. In particular, if <math>\{U_\alpha\}_{\alpha \in A}</math> is locally finite then there is an open cover <math>\{V_\alpha\}_{\alpha \in A}</math> such that <math>\bar{V}_\alpha \subset U_\alpha</math> for every <math>\alpha \in A</math>.

0708-1300/Class notes for Thursday, October 18

2007-10-19T03:19:55Z

Bpym: /* 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.

===Outline===

Today we stated the Whitney Embedding Theorem and began to discuss its proof. Along the way, we also encountered some related notions that will serve us well in the future. We begin by stating the theorem:

===Theorem (Whitney Embedding)===
Let <math>M\!</math> be a smooth <math>m\!</math>-manifold. Then there exists an embedding <math>\Phi : M \to \mathbb{R}^{2m+1}</math>.

====Proof====
=====Outline=====

We will break the proof of the theorem into three parts:

<ol>
<li> Find an embedding of a compact <math>M\!</math> into <math>\mathbb{R}^N\!</math> for some <math>N\!</math>.
<li> Use Sard's Theorem to reduce <math>N\!</math> to <math>2m+1\!</math>.
<li> Use the "Zebra Trick" to prove the theorem for non-compact <math>M\!</math>.
</ol>

Parts two and three shall be left to the next lecture.

=====Part 1=====

 Suppose that <math>M\!</math> is compact. Let <math>\{\phi_\alpha : U_\alpha \to \mathbb{R}^m \}_{\alpha \in A}</math> be an atlas for <math>M\!</math>, and note that <math>\{U_\alpha\}_{\alpha \in A}\!</math> is an open cover of <math>M\!</math>. Hence it possesses a finite subcover <math>\{U_j\}_{1 \le j \le J}\!</math>, and the corresponding collection <math>\{\phi_j\}_{1 \le j \le J}</math> of charts is an atlas. 

Choose smooth functions <math>\{\lambda_j : M \to \mathbb{R}_{\ge 0} \}_{1 \le j \le J}</math> with the following properties:

<ol>
<li> <math>\lambda_j |_{M \setminus U_j} = 0 </math>
<li> <math>\bigcup_{j=1}^{J} \lambda_j^{-1}\left((0,\infty)\right) = M</math>
<li> <math>\mathrm{supp}\left(\lambda_j\right) \subset U_j</math> for <math>1 \le j \le J </math>
</ol>

where <math>\mathrm{supp}(f)\!</math> for <math>f : M \to \mathbb{R}</math> is the support of <math>f\!</math>, ie. the closure of <math>f^{-1}(\mathbb{R} \setminus \{0\}).</math> The existence of such functions follows from the existence of smooth partitions of unity for manifolds---a concept that will be discussed later on.

Now define <math>\Phi : M \to \mathbb{R}^{J + mJ}</math> by <math>\Phi(p) = \left(\lambda_1(p), \ldots, \lambda_J(p), \lambda_1 \tilde \phi_1 (p), \ldots, \lambda_J \tilde \phi_J(p)\right)</math>, where <math>\tilde \phi_j : M \to \mathbb{R}^m</math> is defined by <math>\tilde \phi_j |_{U_j} = \phi_j</math> and <math>\tilde \phi_j |_{M \setminus U_j} = 0</math>.

We claim that <math>\Phi\!</math> is an embedding. That <math>\Phi\!</math> is smooth follows immediately from its construction (the <math>\lambda_j\!</math>s have been used to smear out the <math>\phi_j\!</math> to smooth functions on all of <math>M\!</math>). That <math>\Phi\!</math> is injective is also clear. It takes a bit of work to show that <math>\Phi\!</math> is an immersion, but this is left as an exercise. It remains to see that <math>\Phi\!</math> is a homeomorphism, but this fact follows from the following topological lemma. <math>\Box\!</math>

===Lemma===

Let <math>(X,\tau)\!</math> and <math>(Y,\sigma)\!</math> be topological spaces. Suppose that <math>X\!</math> is compact, <math>Y\!</math> is Hausdorff, and that <math> f : X \to Y</math> is continuous and injective. Then <math>f : X \to f(X)</math> is a homeomorphism.

====Proof====

 Since <math>f\!</math> is an injection onto its image, it is a bijection. Since <math>f\!</math> is continuous, it remains to show that <math>f^{-1}\!</math> is continuous. Thus, it suffices to see that <math>f\!</math> takes closed sets to closed sets. Let <math>A \subset X</math> be closed. Since <math>X\!</math> is compact, so is <math>A\!</math>. Hence <math>f(A)\!</math> is compact since <math>f\!</math> is continuous. But <math>Y\!</math> is Hausdorff, and every compact subset of a Hausdorff space is closed. Hence <math>f(A)\!</math> is closed. <math>\Box</math>

===Remarks===

 The smearing functions we used in Part 1 of the proof of the Whitney Embedding Theorem are very similar to partitions of unity---collections of functions that break the constant function <math>p \mapsto 1</math> into a bunch of bump functions. We will now formalize this notion and show that such collections of functions exist for smooth manifolds.

===Definition===

Let <math>U = \{U_\alpha\}_{\alpha \in A}</math> be an open cover of a topological space (manifold) <math>M\!</math>. A partition of unity subordinated to <math>U\!</math> is a collection <math>\{\lambda_\beta : M \to \mathbb{R}_{\ge 0} \}_{\beta \in B}</math> of continuous (smooth) functions such that

<ol>
<li> For every <math>\beta \in B</math> there is an <math>\alpha \in A</math> such that <math>\mathrm{supp}(\lambda_\beta) \subset U_\alpha</math>
<li> <math>\{\mathrm{supp}(\lambda_\beta)\}_{\beta \in B}</math> is locally finite, ie. for every <math>p \in M</math>, <math>p \in \mathrm{supp}(\lambda_\beta)</math> for only finitely many <math>\beta\!</math>.
<li> <math>\sum_{\beta \in B} \lambda_\beta = 1</math>
</ol>

 A local refinement of <math>U\!</math> is an open cover <math>\{V_\gamma\}_{\gamma \in C}</math> of <math>M\!</math> such that for every <math>\gamma \in C</math>, <math>V_\gamma \subset U_\alpha</math> for some <math>\alpha \in A</math>. <math>M\!</math> is called paracompact if every open cover of <math>M\!</math> has a locally finite refinement.

The following result will be useful for constructing partitions of unity on manifolds:
===Theorem===

Manifolds are paracompact. In particular, if <math>\{U_\alpha\}_{\alpha \in A}</math> is locally finite then there is an open cover <math>\{V_\alpha\}_{\alpha \in A}</math> such that <math>\bar{V}_\alpha \subset U_\alpha</math> for every <math>\alpha \in A</math>.

0708-1300/Class notes for Thursday, October 18

2007-10-19T01:54:01Z

Bpym:

==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.

Template:0708-1300/Navigation

2007-10-19T01:53:00Z

Bpym:

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
|- align=right
|<div class="NavFrame"><div class="NavHead">[[0708-1300]]/[[Template:0708-1300/Navigation|Navigation Panel]]  </div>
<div class="NavContent">
{| border="1px" cellpadding="1" cellspacing="0" width="220" style="margin: 0 0 1em 0.5em; font-size: small"
|-
|colspan=3 align=center|[[Image:0708-1300-ClassPhoto.jpg|215px]] [[0708-1300/Class Photo|Add your name / see who's in!]]
|- align=center
!#
!Week of...
!Links
|-
|colspan=3 align=center|'''Fall Semester'''
|- align=left
|align=center|1
|Sep 10
|[[0708-1300/About This Class|About]], [[0708-1300/Class notes for Tuesday, September 11|Tue]], [[0708-1300/Class notes for Thursday, September 13|Thu]]
|- align=left
|align=center|2
|Sep 17
|[[0708-1300/Class notes for Tuesday, September 18|Tue]], [[0708-1300/Homework Assignment 1|HW1]], [[0708-1300/Class notes for Thursday, September 20|Thu]]
|- align=left
|align=center|3
|Sep 24
|[[0708-1300/Class notes for Tuesday, September 25|Tue]], [[0708-1300/Class Photo|Photo]], [[0708-1300/Class notes for Thursday, September 27|Thu]]
|- align=left
|align=center|4
|Oct 1
|[[0708-1300/Questionnaire|Questionnaire]], [[0708-1300/Class notes for Tuesday, October 2|Tue]], [[0708-1300/Homework Assignment 2|HW2]], [[0708-1300/Class notes for Thursday, October 4|Thu]]
|- align=left
|align=center|5
|Oct 8
|Thanksgiving, [[0708-1300/Class notes for Tuesday, October_9|Tue]], [[0708-1300/Class notes for Thursday, October 11|Thu]]
|- align=left
|align=center|6
|Oct 15
|[[0708-1300/Class notes for Tuesday, October 16|Tue]], [[0708-1300/Homework Assignment 3|HW3]], [[0708-1300/Class notes for Thursday, October 18|Thu]]
|- align=left
|align=center|7
|Oct 22
|
|- align=left
|align=center|8
|Oct 29
|HW4
|- align=left
|align=center|9
|Nov 5
|[[0708-1300/Term Exam 1|TE1 on Thu]]
|- align=left
|align=center|10
|Nov 12
|HW5
|- align=left
|align=center|11
|Nov 19
|
|- align=left
|align=center|12
|Nov 26
|HW6
|- align=left
|align=center|13
|Dec 3
|
|-
|colspan=3 align=center|'''Spring Semester'''
|- align=left
|align=center|14
|Jan 7
|HW7
|- align=left
|align=center|15
|Jan 14
|
|- align=left
|align=center|16
|Jan 21
|HW8
|- align=left
|align=center|17
|Jan 28
|
|- align=left
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|Feb 4
|HW9
|- align=left
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|Feb 11
|TE2; Feb 17: last chance to drop class
|- align=left
|align=center|R
|Feb 18
|
|- align=left
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|Feb 25
|HW10
|- align=left
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|Mar 3
|
|- align=left
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|Mar 10
|HW11
|- align=left
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|Mar 17
|
|- align=left
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|Mar 24
|HW12
|- align=left
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|Mar 31
|
|- align=left
|align=center|26
|Apr 7
|
|- align=center
|colspan=3|[[0708-1300/Errata to Bredon's Book|Errata to Bredon's Book]]
|}
</div></div>
|}

0708-1300/Class notes for Tuesday, October 2

2007-10-09T01:26:59Z

Bpym: /* 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) \!</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_j \frac{\partial \theta_i}{\partial x_j} \frac{\partial }{\partial y_i} </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 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 Thursday, October 4

2007-10-08T17:58:40Z

Bpym: /* Theorem */

{{0708-1300/Navigation}}

==Movie Time==
With the word "immersion" in our minds, we watch the movie "Outside In". Also see the movie's [http://www.geom.uiuc.edu/docs/outreach/oi/ home], a {{Home Link|Talks/UofT-040205/index.html|talk}} I once gave, and the [http://video.google.com/videoplay?docid=-6626464599825291409 movie] itself, on google video.

==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.

During the previous class, we discussed immersions---smooth maps whose differentials are injective. This class deals with the dual notion of submersions, defined as follows:

===Definition===
Let <math>\theta : M^m \rightarrow N^n\!</math> be a smooth map between manifolds. If for each <math>p \in M\!</math> the differential <math>d\theta_p : T_p M \rightarrow T_{\theta(p)} N\!</math> is surjective, <math>\theta\!</math> is called a submersion. 

===Remarks===
We had previously seen that immersions induce "nice" coordinate charts---ones where the immersion looks like the canonical inclusion <math>\iota:\mathbb{R}^m\rightarrow\mathbb{R}^n</math> (where <math>n \ge m\!</math>). The proof of this theorem made use of the Inverse Function Theorem on a function defined on a chart of <math>N\!</math>. In the case of submersions, there is a similar theorem. Submersions locally look like the canonical projection <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math>, and the proof of this fact makes use of the Inverse Function Theorem for a function define on a chart of <math>M\!</math> (duality!). However, before we can prove this theorem, we will need the following lemma.

===Lemma===
 Let <math>V\!</math> and <math>W\!</math> be finite-dimensional vector spaces over <math>\mathbb{R}\!</math> and let <math>T:V \rightarrow W \!</math> be a surjective linear map. Then there exist bases <math>v=\{v_1,\ldots,v_m\}\!</math> for <math>V\!</math> and <math>w=\{w_1,\ldots,w_n\}\!</math> for <math>W\!</math> such that the matrix representative of <math>T\!</math> with respect to <math>v\!</math> and <math>w\!</math> is that of the canonical projection <math>\pi : \mathbb{R}^m \rightarrow \mathbb{R}^n\!</math>.

====Proof====
Let <math>w=\{w_1,\ldots,w_n\}\!</math> be any basis for <math>W\!</math> and choose <math>\{v_1,\ldots,v_n\} \subset V\!</math> such that <math>T(v_i)=w_i\!</math> for each <math>i\in\{1,\ldots,n\}\!</math> (this may be done since <math>T\!</math> is surjective). We claim that the set <math>v' = \{v_1,\ldots,v_n\} \subset V\!</math> is linearly independent. Suppose it were not. Then there would exist <math>\{c_1,\ldots,c_n\} \subset \mathbb{R}\!</math> with <math>\sum_{i=1}^n c_i v_i = 0\!</math> and <math>c_i \ne 0\!</math> for some <math>i\in\{1,\ldots,n\}\!</math>. But then <math>0 = T\left(\sum_{i=1}^n c_i v_i\right) = \sum_{i=1}^n c_i T\left( v_i\right) = \sum_{i=1}^n c_i w_i\!</math> by linearity of <math>T\!</math>, contradicting the assumption that <math>w\!</math> is a basis.
 
 Note that <math>\mathrm{dim}(\mathrm{ker}(T)) = m-n\!</math>. Hence we may find a basis <math>v'' = \{v_{n+1},\ldots,v_m\} \subset V\!</math> for <math>\mathrm{ker}(T)\!</math>. Since <math>\mathrm{span}(v') \cap \mathrm{ker}(T) = \{0 \in V\}\!</math>, the set <math>v = v' \cup v''</math> must be linearly independent and hence form a basis for <math>V\!</math>. We then have <math>w_i = \sum_{i=j}^m \delta_{ij} T(v_j)\!</math>, so that the matrix representative of <math>T\!</math> is <math>\left( I_{n \times n} | 0_{n \times (m-n)} \right)\!</math>, which is the matrix representative of <math>\pi\!</math>. <math>\Box\!</math>

===Theorem===
If <math>\theta : M^m \rightarrow N^n\!</math> is a smooth map between manifolds and for some <math>p \in M\!</math> the differential <math>d\theta_p : T_p M \rightarrow T_{\theta(p)} N\!</math> is surjective then there exist charts <math>\phi : U \rightarrow U' \subset \mathbb{R}^m\!</math> and <math>\psi : V \rightarrow V' \subset \mathbb{R}^n\!</math> on <math>M\!</math> and <math>N\!</math>, respectively, such that

<ol>
<li> <math>\phi(p) = 0\!</math>
<li> <math>\psi\left(\theta(p)\right) = 0\!</math>
<li> The diagram
[[Image:07-10-04-submersion-diagram.png]]
 commutes, where <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math> is the canonical projection. 
</ol>

====Proof====

Since translations are diffeomorphisms of <math>\mathbb{R}^k\!</math> for every <math>k \in \mathbb{N}\!</math>, it is trivial to find charts <math>\phi_0 : U_0 \rightarrow U_0' \subset \mathbb{R}^m\!</math> and <math>\psi : V \rightarrow V' \subset \mathbb{R}^m\!</math> such that <math>\phi(p) = 0 \in \mathbb{R}^m\!</math> and <math>\psi(\theta(p)) = 0 \in \mathbb{R}^n\!</math>. Furthermore, <math>V\!</math> is open, <math>U_0\!</math> is open and <math>\theta\!</math> is continuous, so <math>U_0 \cap \theta^{-1}\left(V\right) \subset M\!</math> is open and contains <math>p\!</math>. Hence, we may assume <math>U_0 \subset \theta^{-1}\left(V\right)\!</math> without loss of generality.
 
Let <math>\theta_0 = \psi \circ \theta \circ \phi_0^{-1} : U_0' \rightarrow V'\!</math> be the local representative of <math>\theta\!</math> and let <math>D_0 : \mathbb{R}^m \rightarrow \mathbb{R}^n \!</math> be the local representative of <math>d\theta_p\!</math>. Since <math>d\theta_p\!</math> is onto, we may apply the previous lemma to obtain a change of coordinates <math>T : \mathbb{R}^m \rightarrow \mathbb{R}^m</math> such that <math>D_0 = \pi \circ T</math>. 
 
Let <math>\phi_1 = T \circ \phi_0\! : U_0 \rightarrow U_1' \subset \mathbb{R}^m</math>. Then <math>\phi_1\!</math> is a chart because <math>T\!</math> is a diffeomorphism. Let <math>\theta_1 = \psi \circ \theta \circ \phi_1^{-1} : U_1' \rightarrow V'\!</math> be the corresponding local representative, define <math>\zeta : U_1' \rightarrow \mathbb{R}^m\!</math> by <math>\zeta(x,y) = (\theta_1(x,y),y)\!</math>, and let <math>D_1 : \mathbb{R}^m \rightarrow \mathbb{R}^n\!</math> be the differential of <math>\theta_1\!</math> at <math>0\!</math>. Then, by construction of <math>T\!</math>, we have that <math>D_1 = \pi\!</math> and hence <math>d \zeta_0 = \mathrm{id}_{\mathbb{R}^m}\!</math>, which is invertible. Hence, the Inverse Function Theorem gives the existence of non-empty open set <math>U' \subset \zeta(U_1')\!</math> such that <math>\zeta|_{\zeta^{-1}(U')}\!</math> is a diffeomorphism. Put <math>U = \phi_1^{-1}(\zeta^{-1}(U'))\!</math> and <math>\phi = \zeta \circ \phi_1|_U\!</math>. Then <math>\phi\!</math> is a chart.
 
It remains to check that <math>\pi \circ \phi = \psi \circ \theta|_U\!</math>, but this is clear: if <math>\phi_1(q)=(x,y)\!</math> for some <math>q \in U\!</math>, then <math>\pi\circ\phi(q) = \pi\circ\zeta(x,y) = \theta_1(x,y) = \psi(\theta(q))\!</math> by definition of <math>\theta_1\!</math>. Hence, <math>\pi \circ \phi = \psi \circ \theta|_U\!</math> and the proof is complete.<math>\Box\!</math>

0708-1300/Class notes for Thursday, October 4

2007-10-05T22:33:07Z

Bpym: /* Proof */

{{0708-1300/Navigation}}

==Movie Time==
With the word "immersion" in our minds, we watch the movie "Outside In". Also see the movie's [http://www.geom.uiuc.edu/docs/outreach/oi/ home], a {{Home Link|Talks/UofT-040205/index.html|talk}} I once gave, and the [http://video.google.com/videoplay?docid=-6626464599825291409 movie] itself, on google video.

==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.

During the previous class, we discussed immersions---smooth maps whose differentials are injective. This class deals with the dual notion of submersions, defined as follows:

===Definition===
Let <math>\theta : M^m \rightarrow N^n\!</math> be a smooth map between manifolds. If for each <math>p \in M\!</math> the differential <math>d\theta_p : T_p M \rightarrow T_{\theta(p)} N\!</math> is surjective, <math>\theta\!</math> is called a submersion. 

===Remarks===
We had previously seen that immersions induce "nice" coordinate charts---ones where the immersion looks like the canonical inclusion <math>\iota:\mathbb{R}^m\rightarrow\mathbb{R}^n</math> (where <math>n \ge m\!</math>). The proof of this theorem made use of the Inverse Function Theorem on a function defined on a chart of <math>N\!</math>. In the case of submersions, there is a similar theorem. Submersions locally look like the canonical projection <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math>, and the proof of this fact makes use of the Inverse Function Theorem for a function define on a chart of <math>M\!</math> (duality!). However, before we can prove this theorem, we will need the following lemma.

===Lemma===
 Let <math>V\!</math> and <math>W\!</math> be finite-dimensional vector spaces over <math>\mathbb{R}\!</math> and let <math>T:V \rightarrow W \!</math> be a surjective linear map. Then there exist bases <math>v=\{v_1,\ldots,v_m\}\!</math> for <math>V\!</math> and <math>w=\{w_1,\ldots,w_n\}\!</math> for <math>W\!</math> such that the matrix representative of <math>T\!</math> with respect to <math>v\!</math> and <math>w\!</math> is that of the canonical projection <math>\pi : \mathbb{R}^m \rightarrow \mathbb{R}^n\!</math>.

====Proof====
Let <math>w=\{w_1,\ldots,w_n\}\!</math> be any basis for <math>W\!</math> and choose <math>\{v_1,\ldots,v_n\} \subset V\!</math> such that <math>T(v_i)=w_i\!</math> for each <math>i\in\{1,\ldots,n\}\!</math> (this may be done since <math>T\!</math> is surjective). We claim that the set <math>v' = \{v_1,\ldots,v_n\} \subset V\!</math> is linearly independent. Suppose it were not. Then there would exist <math>\{c_1,\ldots,c_n\} \subset \mathbb{R}\!</math> with <math>\sum_{i=1}^n c_i v_i = 0\!</math> and <math>c_i \ne 0\!</math> for some <math>i\in\{1,\ldots,n\}\!</math>. But then <math>0 = T\left(\sum_{i=1}^n c_i v_i\right) = \sum_{i=1}^n c_i T\left( v_i\right) = \sum_{i=1}^n c_i w_i\!</math> by linearity of <math>T\!</math>, contradicting the assumption that <math>w\!</math> is a basis.
 
 Note that <math>\mathrm{dim}(\mathrm{ker}(T)) = m-n\!</math>. Hence we may find a basis <math>v'' = \{v_{n+1},\ldots,v_m\} \subset V\!</math> for <math>\mathrm{ker}(T)\!</math>. Since <math>\mathrm{span}(v') \cap \mathrm{ker}(T) = \{0 \in V\}\!</math>, the set <math>v = v' \cup v''</math> must be linearly independent and hence form a basis for <math>V\!</math>. We then have <math>w_i = \sum_{i=j}^m \delta_{ij} T(v_j)\!</math>, so that the matrix representative of <math>T\!</math> is <math>\left( I_{n \times n} | 0_{n \times (m-n)} \right)\!</math>, which is the matrix representative of <math>\pi\!</math>. <math>\Box\!</math>

===Theorem===
If <math>\theta : M^m \rightarrow N^n\!</math> is a smooth map between manifolds and for some <math>p \in M\!</math> the differential <math>d\theta_p : T_p M \rightarrow T_{\theta(p)} N\!</math> is surjective then there exist charts <math>\phi : U \rightarrow U' \subset \mathbb{R}^m\!</math> and <math>\psi : V \rightarrow V' \subset \mathbb{R}^n\!</math> on <math>M\!</math> and <math>N\!</math>, respectively, such that

<ol>
<li> <math>\phi(p) = 0\!</math>
<li> <math>\psi\left(\theta(p)\right) = 0\!</math>
<li> The diagram
[[Image:07-10-04-submersion-diagram.png]]
 commutes, where <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math> is the canonical projection. 
</ol>

====Proof====

Since translations are diffeomorphisms of <math>\mathbb{R}^k\!</math> for every <math>k \in \mathbb{N}\!</math>, it is trivial to find charts <math>\phi_0 : U_0 \rightarrow U_0' \subset \mathbb{R}^m\!</math> and <math>\psi : V \rightarrow V' \subset \mathbb{R}^m\!</math> such that <math>\phi(p) = 0 \in \mathbb{R}^m\!</math> and <math>\psi(\theta(p)) = 0 \in \mathbb{R}^n\!</math>. Furthermore, <math>V\!</math> is open, <math>U_0\!</math> is open and <math>\theta\!</math> is continuous, so <math>U_0 \cap \theta^{-1}\left(V\right) \subset M\!</math> is open and contains <math>p\!</math>. Hence, we may assume <math>U_0 \subset \theta^{-1}\left(V\right)\!</math> without loss of generality.
 
Let <math>\theta_0 = \psi \circ \theta \circ \phi_0^{-1} : U_0' \rightarrow V'\!</math> be the local representative of <math>\theta\!</math> and let <math>D_0 : \mathbb{R}^m \rightarrow \mathbb{R}^n \!</math> be the local representative of <math>d\theta_p\!</math>. Since <math>d\theta_p\!</math> is onto, we may apply the previous lemma to obtain a change of basis <math>T : \mathbb{R}^m \rightarrow \mathbb{R}^m</math> such that <math>D_0 = T^{-1} \circ \pi \circ T</math>. 
 
Let <math>\phi_1 = T \circ \phi_0\! : U_0 \rightarrow U_1' \subset \mathbb{R}^m</math>. Then <math>\phi_1\!</math> is a chart because <math>T\!</math> is a diffeomorphism. Let <math>\theta_1 = \psi \circ \theta \circ \phi_1^{-1} : U_1' \rightarrow V'\!</math> be the corresponding local representative, define <math>\zeta : U_1' \rightarrow \mathbb{R}^m\!</math> by <math>\zeta(x,y) = (\theta_1(x,y),y)\!</math>, and let <math>D_1 : \mathbb{R}^m \rightarrow \mathbb{R}^n\!</math> be the differential of <math>\theta_1\!</math> at <math>0\!</math>. Then, by construction of <math>T\!</math>, we have that <math>D_1 = \pi\!</math> and hence <math>d \zeta_0 = \mathrm{id}_{\mathbb{R}^m}\!</math>, which is invertible. Hence, the Inverse Function Theorem gives the existence of non-empty open set <math>U' \subset \zeta(U_1')\!</math> such that <math>\zeta|_{\zeta^{-1}(U')}\!</math> is a diffeomorphism. Put <math>U = \phi_1^{-1}(\zeta^{-1}(U'))\!</math> and <math>\phi = \zeta \circ \phi_1|_U\!</math>. Then <math>\phi\!</math> is a chart.
 
It remains to check that <math>\pi \circ \phi = \psi \circ \theta|_U\!</math>, but this is clear: if <math>\phi_1(q)=(x,y)\!</math> for some <math>q \in U\!</math>, then <math>\pi\circ\phi(q) = \pi\circ\zeta(x,y) = \theta_1(x,y) = \psi(\theta(q))\!</math> by definition of <math>\theta_1\!</math>. Hence, <math>\pi \circ \phi = \psi \circ \theta|_U\!</math> and the proof is complete.<math>\Box\!</math>

0708-1300/Class notes for Thursday, October 4

2007-10-05T22:31:23Z

Bpym:

{{0708-1300/Navigation}}

==Movie Time==
With the word "immersion" in our minds, we watch the movie "Outside In". Also see the movie's [http://www.geom.uiuc.edu/docs/outreach/oi/ home], a {{Home Link|Talks/UofT-040205/index.html|talk}} I once gave, and the [http://video.google.com/videoplay?docid=-6626464599825291409 movie] itself, on google video.

==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.

During the previous class, we discussed immersions---smooth maps whose differentials are injective. This class deals with the dual notion of submersions, defined as follows:

===Definition===
Let <math>\theta : M^m \rightarrow N^n\!</math> be a smooth map between manifolds. If for each <math>p \in M\!</math> the differential <math>d\theta_p : T_p M \rightarrow T_{\theta(p)} N\!</math> is surjective, <math>\theta\!</math> is called a submersion. 

===Remarks===
We had previously seen that immersions induce "nice" coordinate charts---ones where the immersion looks like the canonical inclusion <math>\iota:\mathbb{R}^m\rightarrow\mathbb{R}^n</math> (where <math>n \ge m\!</math>). The proof of this theorem made use of the Inverse Function Theorem on a function defined on a chart of <math>N\!</math>. In the case of submersions, there is a similar theorem. Submersions locally look like the canonical projection <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math>, and the proof of this fact makes use of the Inverse Function Theorem for a function define on a chart of <math>M\!</math> (duality!). However, before we can prove this theorem, we will need the following lemma.

===Lemma===
 Let <math>V\!</math> and <math>W\!</math> be finite-dimensional vector spaces over <math>\mathbb{R}\!</math> and let <math>T:V \rightarrow W \!</math> be a surjective linear map. Then there exist bases <math>v=\{v_1,\ldots,v_m\}\!</math> for <math>V\!</math> and <math>w=\{w_1,\ldots,w_n\}\!</math> for <math>W\!</math> such that the matrix representative of <math>T\!</math> with respect to <math>v\!</math> and <math>w\!</math> is that of the canonical projection <math>\pi : \mathbb{R}^m \rightarrow \mathbb{R}^n\!</math>.

====Proof====
Let <math>w=\{w_1,\ldots,w_n\}\!</math> be any basis for <math>W\!</math> and choose <math>\{v_1,\ldots,v_n\} \subset V\!</math> such that <math>T(v_i)=w_i\!</math> for each <math>i\in\{1,\ldots,n\}\!</math> (this may be done since <math>T\!</math> is surjective). We claim that the set <math>v' = \{v_1,\ldots,v_n\} \subset V\!</math> is linearly independent. Suppose it were not. Then there would exist <math>\{c_1,\ldots,c_n\} \subset \mathbb{R}\!</math> with <math>\sum_{i=1}^n c_i v_i = 0\!</math> and <math>c_i \ne 0\!</math> for some <math>i\in\{1,\ldots,n\}\!</math>. But then <math>0 = T\left(\sum_{i=1}^n c_i v_i\right) = \sum_{i=1}^n c_i T\left( v_i\right) = \sum_{i=1}^n c_i w_i\!</math> by linearity of <math>T\!</math>, contradicting the assumption that <math>w\!</math> is a basis.
 
 Note that <math>\mathrm{dim}(\mathrm{ker}(T)) = m-n\!</math>. Hence we may find a basis <math>v'' = \{v_{n+1},\ldots,v_m\} \subset V\!</math> for <math>\mathrm{ker}(T)\!</math>. Since <math>\mathrm{span}(v') \cap \mathrm{ker}(T) = \{0 \in V\}\!</math>, the set <math>v = v' \cup v''</math> must be linearly independent and hence form a basis for <math>V\!</math>. We then have <math>w_i = \sum_{i=j}^m \delta_{ij} T(v_j)\!</math>, so that the matrix representative of <math>T\!</math> is <math>\left( I_{n \times n} | 0_{n \times (n-m)} \right)\!</math>, which is the matrix representative of <math>\pi\!</math>. <math>\Box\!</math>

===Theorem===
If <math>\theta : M^m \rightarrow N^n\!</math> is a smooth map between manifolds and for some <math>p \in M\!</math> the differential <math>d\theta_p : T_p M \rightarrow T_{\theta(p)} N\!</math> is surjective then there exist charts <math>\phi : U \rightarrow U' \subset \mathbb{R}^m\!</math> and <math>\psi : V \rightarrow V' \subset \mathbb{R}^n\!</math> on <math>M\!</math> and <math>N\!</math>, respectively, such that

<ol>
<li> <math>\phi(p) = 0\!</math>
<li> <math>\psi\left(\theta(p)\right) = 0\!</math>
<li> The diagram
[[Image:07-10-04-submersion-diagram.png]]
 commutes, where <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math> is the canonical projection. 
</ol>

====Proof====

Since translations are diffeomorphisms of <math>\mathbb{R}^k\!</math> for every <math>k \in \mathbb{N}\!</math>, it is trivial to find charts <math>\phi_0 : U_0 \rightarrow U_0' \subset \mathbb{R}^m\!</math> and <math>\psi : V \rightarrow V' \subset \mathbb{R}^m\!</math> such that <math>\phi(p) = 0 \in \mathbb{R}^m\!</math> and <math>\psi(\theta(p)) = 0 \in \mathbb{R}^n\!</math>. Furthermore, <math>V\!</math> is open, <math>U_0\!</math> is open and <math>\theta\!</math> is continuous, so <math>U_0 \cap \theta^{-1}\left(V\right) \subset M\!</math> is open and contains <math>p\!</math>. Hence, we may assume <math>U_0 \subset \theta^{-1}\left(V\right)\!</math> without loss of generality.
 
Let <math>\theta_0 = \psi \circ \theta \circ \phi_0^{-1} : U_0' \rightarrow V'\!</math> be the local representative of <math>\theta\!</math> and let <math>D_0 : \mathbb{R}^m \rightarrow \mathbb{R}^n \!</math> be the local representative of <math>d\theta_p\!</math>. Since <math>d\theta_p\!</math> is onto, we may apply the previous lemma to obtain a change of basis <math>T : \mathbb{R}^m \rightarrow \mathbb{R}^m</math> such that <math>D_0 = T^{-1} \circ \pi \circ T</math>. 
 
Let <math>\phi_1 = T \circ \phi_0\! : U_0 \rightarrow U_1' \subset \mathbb{R}^m</math>. Then <math>\phi_1\!</math> is a chart because <math>T\!</math> is a diffeomorphism. Let <math>\theta_1 = \psi \circ \theta \circ \phi_1^{-1} : U_1' \rightarrow V'\!</math> be the corresponding local representative, define <math>\zeta : U_1' \rightarrow \mathbb{R}^m\!</math> by <math>\zeta(x,y) = (\theta_1(x,y),y)\!</math>, and let <math>D_1 : \mathbb{R}^m \rightarrow \mathbb{R}^n\!</math> be the differential of <math>\theta_1\!</math> at <math>0\!</math>. Then, by construction of <math>T\!</math>, we have that <math>D_1 = \pi\!</math> and hence <math>d \zeta_0 = \mathrm{id}_{\mathbb{R}^m}\!</math>, which is invertible. Hence, the Inverse Function Theorem gives the existence of non-empty open set <math>U' \subset \zeta(U_1')\!</math> such that <math>\zeta|_{\zeta^{-1}(U')}\!</math> is a diffeomorphism. Put <math>U = \phi_1^{-1}(\zeta^{-1}(U'))\!</math> and <math>\phi = \zeta \circ \phi_1|_U\!</math>. Then <math>\phi\!</math> is a chart.
 
It remains to check that <math>\pi \circ \phi = \psi \circ \theta|_U\!</math>, but this is clear: if <math>\phi_1(q)=(x,y)\!</math> for some <math>q \in U\!</math>, then <math>\pi\circ\phi(q) = \pi\circ\zeta(x,y) = \theta_1(x,y) = \psi(\theta(q))\!</math> by definition of <math>\theta_1\!</math>. Hence, <math>\pi \circ \phi = \psi \circ \theta|_U\!</math> and the proof is complete.<math>\Box\!</math>

0708-1300/Class notes for Thursday, October 4

2007-10-05T05:35:06Z

Bpym:

{{0708-1300/Navigation}}

==Movie Time==
With the word "immersion" in our minds, we watch the movie "Outside In". Also see the movie's [http://www.geom.uiuc.edu/docs/outreach/oi/ home], a {{Home Link|Talks/UofT-040205/index.html|talk}} I once gave, and the [http://video.google.com/videoplay?docid=-6626464599825291409 movie] itself, on google video.

==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.

During the previous class, we discussed immersions---smooth maps whose differentials are injective. This class deals with the dual notion of submersions, defined as follows:

===Definition===
Let <math>\theta : M^m \rightarrow N^n\!</math> be a smooth map between manifolds. If for each <math>p \in M\!</math> the differential <math>d\theta_p : T_p M \rightarrow T_{\theta(p)} N\!</math> is surjective, <math>\theta\!</math> is called a submersion. 

===Remarks===
We had previously seen that immersions induce "nice" coordinate charts---ones where the immersion looks like the inclusion of <math>\iota:\mathbb{R}^m\rightarrow\mathbb{R}^n</math> (where <math>n \ge m\!</math>). The proof of this theorem made use of the Inverse Function Theorem on a function defined on a chart of <math>N\!</math>. In the case of submersions, there is a similar theorem. Submersions locally look like the canonical projection <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math>, and the proof of this fact makes use of the Inverse Function Theorem for a function define on a chart of <math>M\!</math> (duality!). However, before we can prove this theorem, we will need the following lemma.

===Lemma===
 Let <math>V\!</math> and <math>W\!</math> be finite-dimensional vector spaces over <math>\mathbb{R}\!</math> and let <math>T:V \rightarrow W \!</math> be a surjective linear map. Then there exist bases <math>v=\{v_1,\ldots,v_m\}\!</math> for <math>V\!</math> and <math>w=\{w_1,\ldots,w_n\}\!</math> for <math>W\!</math> such that the matrix representative of <math>T\!</math> with respect to <math>v\!</math> and <math>w\!</math> is that of the canonical projection <math>\pi : \mathbb{R}^m \rightarrow \mathbb{R}^n\!</math>.

====Proof====
Let <math>w=\{w_1,\ldots,w_n\}\!</math> be any basis for <math>W\!</math> and choose <math>\{v_1,\ldots,v_n\} \subset V\!</math> such that <math>T(v_i)=w_i\!</math> for each <math>i\in\{1,\ldots,n\}\!</math>. We claim that the set <math>v' = \{v_1,\ldots,v_n\} \subset V\!</math> is linearly independent. Suppose it were not. Then there would exist <math>\{c_1,\ldots,c_n\} \subset \mathbb{R}\!</math> with <math>\sum_{i=1}^n c_i v_i = 0\!</math> and <math>c_i \ne 0\!</math> for some <math>i\in\{1,\ldots,n\}\!</math>. But then <math>0 = T\left(\sum_{i=1}^n c_i v_i\right) = \sum_{i=1}^n c_i T\left( v_i\right) = \sum_{i=1}^n c_i w_i\!</math> by linearity of <math>T\!</math>. But this contradicts the assumption that <math>w\!</math> is a basis.
 
 Note that <math>\mathrm{dim}(\mathrm{ker}(T)) = m-n\!</math>. Hence we may find a basis <math>v'' = \{v_{n+1},\ldots,v_m\} \subset V\!</math> for <math>\mathrm{ker}(T)\!</math>. Since <math>\mathrm{span}(v') \cap \mathrm{ker}(T) = \{0 \in V\}\!</math>, the set <math>v = v' \cup v''</math> must be linearly independent and hence form a basis for <math>V\!</math>. We then have <math>w_i = \sum_{i=j}^m \delta_{ij} T(v_j)\!</math>, so that the matrix representative of <math>T\!</math> is <math>\left( I_{n \times n} | 0_{n \times (n-m)} \right)\!</math>, which is the matrix representative of <math>\pi\!</math>. <math>\Box\!</math>

===Theorem===
If <math>\theta : M^m \rightarrow N^n\!</math> is a smooth map between manifolds and for some <math>p \in M\!</math> the differential <math>d\theta_p : T_p M \rightarrow T_{\theta(p)} N\!</math> is surjective then there exist charts <math>\phi : U \rightarrow U' \subset \mathbb{R}^m\!</math> and <math>\psi : V \rightarrow V' \subset \mathbb{R}^n\!</math> on <math>M\!</math> and <math>N\!</math>, respectively, such that

<ol>
<li> <math>\phi(p) = 0\!</math>
<li> <math>\psi\left(\theta(p)\right) = 0\!</math>
<li> The diagram
[[Image:07-10-04-submersion-diagram.png]]
 commutes, where <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math> is the canonical projection. 
</ol>

====Proof====

Since translations are diffeomorphisms of <math>\mathbb{R}^k\!</math> for every <math>k \in \mathbb{N}\!</math>, it is trivial to find charts <math>\phi_0 : U_0 \rightarrow U_0' \subset \mathbb{R}^m\!</math> and <math>\psi : V \rightarrow V' \subset \mathbb{R}^m\!</math> such that <math>\phi(p) = 0 \in \mathbb{R}^m\!</math> and <math>\psi(\theta(p)) = 0 \in \mathbb{R}^n\!</math>. Furthermore, since <math>V\!</math> is open, <math>U_0\!</math> is open and <math>\theta\!</math> is continuous, <math>U_0 \cap \theta^{-1}\left(V\right) \subset M\!</math> is open and contains <math>p\!</math> so that we may assume <math>U_0 \subset \theta^{-1}\left(V\right)\!</math> without loss of generality.
 
Let <math>\theta_0 = \psi \circ \theta \circ \phi_0^{-1} : U_0' \rightarrow V'\!</math> be the local representative of <math>\theta\!</math> and let <math>D_0 : \mathbb{R}^m \rightarrow \mathbb{R}^n \!</math> be the local representative of <math>d\theta_p\!</math>. Since <math>d\theta_p\!</math> is onto, we may apply the previous lemma to obtain a change of basis <math>T : \mathbb{R}^m \rightarrow \mathbb{R}^m</math> such that <math>D_0 = T^{-1} \circ \pi \circ T</math>. 
 
Let <math>\phi_1 = T \circ \phi_0\!</math>. Then <math>\phi_1\!</math> is a chart because <math>T\!</math> is a diffeomorphism. Let <math>\theta_1 = \psi \circ \theta \circ \phi_1^{-1} : U_0' \rightarrow V'\!</math> be the corresponding local representative, define <math>\zeta : U_0' \rightarrow \mathbb{R}^m\!</math> by <math>\zeta(x,y) = (\theta_1(x,y),y)\!</math>, and let <math>D_1 : \mathbb{R}^m \rightarrow \mathbb{R}^n\!</math> be the differential of <math>\theta_1\!</math> at <math>0\!</math>. Then, by construction of <math>T\!</math>, we have that <math>D_1 = \pi\!</math> and hence <math>d \zeta_0 = \mathrm{id}_{\mathbb{R}^m}\!</math>, which is invertible. Hence, the Inverse Function Theorem gives the existence of non-empty open set <math>U' \subset \zeta(U_1')\!</math> such that <math>\zeta|_{\zeta^{-1}(U')}\!</math> is a diffeomorphism. Put <math>U = \phi_1^{-1}(\zeta^{-1}(U'))\!</math> and <math>\phi = \zeta \circ \phi_1|_U\!</math>. Then <math>\phi\!</math> is a chart.
 
It remains to check that <math>\pi \circ \phi = \psi \circ \theta|_U\!</math>, but this is clear: if <math>\phi_1(q)=(x,y)\!</math> for some <math>q \in U\!</math>, then <math>\pi\circ\phi(q) = \pi\circ\zeta(x,y) = \theta_1(x,y) = \psi(\theta(q))\!</math> by definition of <math>\theta_1\!</math>. Hence, <math>\pi \circ \phi = \psi \circ \theta|_U\!</math> and the proof is complete.<math>\Box\!</math>

0708-1300/Class notes for Thursday, October 4

2007-10-05T04:35:47Z

Bpym:

0708-1300/Class notes for Thursday, October 4

2007-10-05T04:35:05Z

Bpym: /* Proof */

0708-1300/Class notes for Thursday, October 4

2007-10-05T03:40:25Z

Bpym:

File:07-10-04-submersion-diagram.png

2007-10-05T03:31:19Z

Bpym: A commutative diagram for the statement of a theorem from class on 2007-10-04

A commutative diagram for the statement of a theorem from class on 2007-10-04

0708-1300/Class notes for Thursday, September 20

2007-09-29T06:23:19Z

Bpym: Some discussion of configurations spaces

{{0708-1300/Navigation}}
==Dror's Note==
Come to {{Home Link|Talks/Fields-0709/|my talk}} today at 4:30PM at the Fields Institute!

{{0708-1300/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==
{{07008-1300/Solution Warning}}

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>.

===Dror's Evaluation===
The solution for <math>n=2</math> seems right but hard to understand. For <math>n=3</math> the solution is absolutely right. Unfortunately for <math>n>3</math> the solution is wrong. (Why?)

--[[User:Drorbn|Drorbn]] 19:46, 24 September 2007 (EDT)

==Enrichment: Configuration Spaces of Mechanical Systems==
As Prof. Bar-Natan mentioned in class, manifolds are quite useful for studying the physics of mechanical systems like robots and steam engines. There is actually a very general mathematical theory of rigid body mechanics that makes plenty of use of differential geometry. See, for example,
{{ref|BulloLewis_04}} for a discussion of one of these models and its application to nonlinear control theory for mechanical systems. The following is a brief introduction to the idea of a mechanical system's configuration space, and makes use of the notation of {{ref|BulloLewis_04}}.

Imagine that you have some rigid body <math>B\!</math> floating around in space. The first thing to do is to set up an orthonormal frame <math>(O_{spatial},(s_1,s_2,s_3))\!</math> with respect to which we will describe <math>B\!</math>'s position. Here <math>O_{spatial}\!</math> is some point in Euclidean (affine) space which serves as the origin of our coordinate system and <math>(s_1,s_2,s_3)\!</math> is an orthonormal basis for Euclidean space centred at <math>O_{spatial}\!</math>.

Now suppose that we rigidly attach a frame <math>(O_{body},(b_1,b_2,b_3))\!</math> to <math>B\!</math> in the sense that the frame's origin is fixed to a specific point on the body (such as the centre of mass) and the basis <math>(b_1,b_2,b_3)\!</math> rotates around with <math>B\!</math>. We can now specify <math>B\!</math>'s configuration with the vector

::<math>r = O_{body} - O_{spatial} \in \mathbb{R}^3\!</math>

and the matrix <math> R \in SO(3;\mathbb{R})\!</math> which gives the orientation of the basis <math>(b_1,b_2,b_3)\!</math> relative to <math>(s_1,s_2,s_3)\!</math>. Here <math>SO(3;\mathbb{R})\!</math> is the group of <math>3\times 3\!</math> real matrices with unit determinant---the group of rotations of <math>\mathbb{R}^3\!</math>. Concisely, the configuration space of a rigid body is the space <math>Q_{free}=SO(3;\mathbb{R}) \times \mathbb{R}^3\!</math>.

Lucky for us, <math>SO(3;\mathbb{R})\!</math> is actually a manifold. It's an example of something called a Lie group, which is a group that is also a manifold and whose group operations are smooth according to the differentiable structure of the manifold. Hence <math>Q_{free}\!</math> is also a manifold.

Now imagine we have <math>k\!</math> rigid bodies. If the bodies are free to move around independently, and if we briefly suspend disbelief and allow them to pass right through each other instead of colliding, our configuration space is <math>Q_{free}^k= \left(SO(3;\mathbb{R}) \times \mathbb{R}^3\right)^k</math>. If, on the other hand, the bodies are interconnected---ie., their motions are not independent---or if the motion of the bodies is constrained in some other way, then the configuration space for the system will be some subset <math>Q \subset Q_{free}^k\!</math>. In fact, it is (usually) a submanifold.

This situation is precisely what's happening with the roach's configuration. In this case the system consists of 12 rigid bodies (six legs composed of two segments each) that are constrained to move in a plane, connected together at their ends, and pinned down.

Hopefully, it is starting to become clear how we can model some of the physics now: we can describe the motion of a mechanical system with a curve <math>\gamma : I \rightarrow Q\!</math> where <math>I \subset \mathbb{R}\!</math> is an interval. The velocities of all the bodies (both translational and rotational) at time <math>t\!</math>are encapsulated in the tangent vector <math>\gamma'(t) \in T_{\gamma(t)}Q\!</math>. Similarly, there are nice differential-geometric ways of describing the kinetic and potential energies, forces and so on.

==References==
{{note|BulloLewis_04}}Francesco Bullo and Andrew D. Lewis, Geometric Control of Mechanical Systems: Modeling, Analysis, and Design for Simple Mechanical Control Systems, Springer-Verlag, New York-Heidelberg-Berlin, 2004. Number 49 in Texts in Applied Mathematics.

0708-1300/Class Photo

2007-09-29T04:17:20Z

Bpym:

{{0708-1300/Navigation}}

Our class on September 27, 2007:

[[Image:0708-1300-ClassPhoto.jpg|thumb|centre|600px|Class Photo: click to enlarge]]

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
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The first option is more fun but less private.

===Who We Are...===

{| align=center border=1
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn @ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
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0708-1300/Class Photo

2007-09-29T04:09:16Z

Bpym:

{{0708-1300/Navigation}}

Our class on September 27, 2007:

[[Image:0708-1300-ClassPhoto.jpg|thumb|centre|600px|Class Photo: click to enlarge]]

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn @ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
{{Photo Entry|last=Bazett|first=Trefor|userid=Trefor|email=trefor.bazett @ toronto.ca|location=tallest person a little right of center in a beige shirt|comments=}}
{{Photo Entry|last=Bjorndahl|first=Adam|userid=ABjorndahl|email=adam.bjorndahl @ utoronto.ca|location=back row, fifth from the left, under the "f(tp)dt"|comments=Looking forward to a great year!}}
{{Photo Entry|last=Chow|first=Aaron|userid=aaron.chow|email=aaron @ utoronto.ca|location=Third from right, in a black shirt.|comments=Hope we have a good year together!}}
{{Photo Entry|last=Isgur|first=Abraham|userid=Abisgu|email=abraham.isgur@ math.toronto.edu|location=2nd person in the back row, from the right, the one with the beard and long hair|comments=}}
{{Photo Entry|last=Pym|first=Brent|userid=Bpym|email=bpym @ math.toronto.edu|location=10th from the right (cumulatively)---under the <math>T_P(M)</math>|comments=Manifolds are fun!}}
{{Photo Entry|last=Vera Pacheco|first=Franklin|userid=Franklin|email=franklin.vp @ gmail.com|location=Xth from left to right|comments=To find me you must first go to [[http://www.deathball.net/notpron/]] solve the first 4 pages. Once this done you will know how to find me. Once this done go back to NOTPRON an solve the rest of the puzzle}}
{{Photo Entry|last=Wong|first=Silian|userid=kuramay|email=kurama_y @ hotmail.com|location=One of the Asian-looking girls...with sparkling teeth(??)|comments=I'll write up some comments after their existences}}
|}

0708-1300/Class Photo

2007-09-29T04:08:23Z

Bpym: I added myself to the photo index

{{0708-1300/Navigation}}

Our class on September 27, 2007:

[[Image:0708-1300-ClassPhoto.jpg|thumb|centre|600px|Class Photo: click to enlarge]]

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn @ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
{{Photo Entry|last=Bazett|first=Trefor|userid=Trefor|email=trefor.bazett @ toronto.ca|location=tallest person a little right of center in a beige shirt|comments=}}
{{Photo Entry|last=Bjorndahl|first=Adam|userid=ABjorndahl|email=adam.bjorndahl @ utoronto.ca|location=back row, fifth from the left, under the "f(tp)dt"|comments=Looking forward to a great year!}}
{{Photo Entry|last=Chow|first=Aaron|userid=aaron.chow|email=aaron @ utoronto.ca|location=Third from right, in a black shirt.|comments=Hope we have a good year together!}}
{{Photo Entry|last=Isgur|first=Abraham|userid=Abisgu|email=abraham.isgur@ math.toronto.edu|location=2nd person in the back row, from the right, the one with the beard and long hair|comments=}}
{{Photo Entry|last=Pym|first=Brent|userid=Bpym|email=bpym @ math.toronto.edu|location=10<math>^{th}</math> from the right (cumulatively)---under the <math>T_P(M)</math>|comments=Manifolds are fun!}}
{{Photo Entry|last=Vera Pacheco|first=Franklin|userid=Franklin|email=franklin.vp @ gmail.com|location=Xth from left to right|comments=To find me you must first go to [[http://www.deathball.net/notpron/]] solve the first 4 pages. Once this done you will know how to find me. Once this done go back to NOTPRON an solve the rest of the puzzle}}
{{Photo Entry|last=Wong|first=Silian|userid=kuramay|email=kurama_y @ hotmail.com|location=One of the Asian-looking girls...with sparkling teeth(??)|comments=I'll write up some comments after their existences}}
|}