0708-1300/Class notes for Tuesday, November 6 - Revision history

Bpym: /* Definition */

2008-02-10T21:21:51Z

Definition

← Older revision		Revision as of 17:21, 10 February 2008
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	<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>		<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>

	<p>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>.</p>		<p>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>.</p>

	<p>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.</p>		<p>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.</p>

Jvoltz: /* Definition */

2007-11-20T14:18:03Z

Definition

← Older revision		Revision as of 10:18, 20 November 2007
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	<i>		<i>

	<p>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 </p>		<p>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 </p>

	<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>		<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>

Jvoltz: /* Class Notes */

2007-11-20T14:17:06Z

Class Notes

← Older revision		Revision as of 10:17, 20 November 2007
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	<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>		<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>

	<p>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:</p>		<p>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:</p>

	<ol>		<ol>

Bpym: /* Definition */

2007-11-19T18:13:15Z

Definition

← Older revision		Revision as of 14:13, 19 November 2007
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	<p>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>.</p>		<p>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>.</p>

	<p>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.		<p>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.</p>

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

Bpym: /* Class Notes */

2007-11-17T20:08:17Z

Class Notes

← Older revision		Revision as of 16:08, 17 November 2007
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	===Definition===		===Definition===

			<i>

	<p>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 </p>		<p>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 </p>

	<p><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></p>		<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>

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

			</i>

	===Proposition===		===Proposition===

			<i>

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

			</i>

	====Proof====		====Proof====
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	<p> 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>.</p>		<p> 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>.</p>

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

			<br>

	===Remarks===

	<p>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.</p>		<p>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.</p>
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	</i>		</i>

			<br>

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

	<p>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></p>		<p>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></p>

	===Corollary===		===Corollary===
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	</i>		</i>

			<br>

	<p> 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>. </p>		<p> 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>. </p>
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	</i>		</i>

			<br>

	<p>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</p>		<p>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</p>

Bpym: /* Definition */

2007-11-17T19:59:35Z

Definition

@@ Line 50: / Line 50: @@
 ===Definition===
 <p>For each <math>p,q \in \mathbb{N}\!</math> the <b>wedge product</b> 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</p>
-<p> <math>(\omega \wedge \lambda) (v_1,\ldots,v_{p+q}) = \sum_{\sigma \in S_{p+q}} \frac{(-1)^\sigma}{p!q!} \omega(v_{\sigma(1)},\ldots,v_{\sigma(p)})\lambda(v_{\sigma(p+1)},\ldots,v_{\sigma(p+q)})</math> </p>
+<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>

Bpym at 18:54, 15 November 2007

2007-11-15T18:54:50Z

New page

{{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

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

 for every <math>v_1 ,\ldots,v_{p+q} \in V</math>.