0708-1300/Class notes for Thursday, October 4 - Revision history

Bpym: /* Theorem */

2007-10-08T17:58:40Z

Theorem

← Older revision		Revision as of 13:58, 8 October 2007
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	<p>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.</p>		<p>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.</p>
	<br>		<br>
	<p>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>. </p>		<p>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>. </p>
	<br>		<br>
	<p>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.</p>		<p>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.</p>

Bpym: /* Proof */

2007-10-05T22:33:07Z

Proof

← Older revision		Revision as of 18:33, 5 October 2007
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	<p>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.</p>		<p>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.</p>
	<br>		<br>
	<p> 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></p>		<p> 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></p>

	===Theorem===		===Theorem===

Bpym at 22:31, 5 October 2007

2007-10-05T22:31:23Z

← Older revision		Revision as of 18:31, 5 October 2007
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	===Remarks===		===Remarks===
	<p>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.</p>		<p>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.</p>

	===Lemma===		===Lemma===
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	====Proof====		====Proof====
	<p>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.</p>		<p>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.</p>
	<br>		<br>
	<p> 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></p>		<p> 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></p>
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	====Proof====		====Proof====

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

Bpym at 05:35, 5 October 2007

2007-10-05T05:35:06Z

← Older revision		Revision as of 01:35, 5 October 2007
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	==Class Notes==		==Class Notes==
	<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>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:</p>

	===Definition===		===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 <b>submersion</b>.		<p>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 <b>submersion</b>. </p>

			===Remarks===
			<p>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.</p>

			===Lemma===
			<p> 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>.</p>

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

	===Theorem===		===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		<p>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</p>

	<ol>		<ol>
Line 22:		Line 36:
	====Proof====		====Proof====

	<p>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 so that we may assume <math>U_0 \subset \theta^{-1}\left(V\right)\!</math> without loss of generality.</p>		<p>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.</p>
	<br>		<br>
	<p>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 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>. </p>		<p>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>. </p>
	<br>		<br>
	<p>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~~ 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.</p>		<p>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.</p>
	<br>		<br>
	<p>It remains to check that <math>\pi \circ \phi = \psi \circ \theta\|_U\!</math>, but this is clear: if <math>(x,y~~) = \phi_1(q~~)\!</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>. <math>\Box\!</math>		<p>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>

Bpym at 04:35, 5 October 2007

2007-10-05T04:35:47Z

← Older revision		Revision as of 00:35, 5 October 2007
Line 5:		Line 5:

	==Class Notes==		==Class Notes==
			<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>

	===Definition===		===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 <b>submersion</b>.		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 <b>submersion</b>.

Bpym: /* Proof */

2007-10-05T04:35:05Z

Proof

← Older revision		Revision as of 00:35, 5 October 2007
Line 21:		Line 21:

	====Proof====		====Proof====

			<p>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 so that we may assume <math>U_0 \subset \theta^{-1}\left(V\right)\!</math> without loss of generality.</p>
			<br>
			<p>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 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>. </p>
			<br>
			<p>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 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.</p>
			<br>
			<p>It remains to check that <math>\pi \circ \phi = \psi \circ \theta\|_U\!</math>, but this is clear: if <math>(x,y) = \phi_1(q)\!</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>. <math>\Box\!</math>

Bpym at 03:40, 5 October 2007

2007-10-05T03:40:25Z

← Older revision		Revision as of 23:40, 4 October 2007
Line 3:		Line 3:
	==Movie Time==		==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.		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==

			===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 <b>submersion</b>.

			===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
			<p align="center">[[Image:07-10-04-submersion-diagram.png]]</p>
			<p> commutes, where <math>\pi : \mathbb{R}^m = \mathbb{R}^n \times \mathbb{R}^{m-n} \rightarrow \mathbb{R}^n\!</math> is the canonical projection. </p>
			</ol>

			====Proof====

Drorbn at 10:37, 4 October 2007

2007-10-04T10:37:37Z

← Older revision		Revision as of 06:37, 4 October 2007
Line 2:		Line 2:

	==Movie Time==		==Movie Time==
	With the word "immersion" in our minds, we watch the movie "Outside In". Also see [http://www.geom.uiuc.edu/docs/outreach/oi/], ~~[http://www.math.toronto.edu/~drorbn/~~Talks/UofT-040205/index.html] and [http://video.google.com/videoplay?docid=-6626464599825291409].		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.

Drorbn at 23:33, 3 October 2007

2007-10-03T23:33:37Z

New page

{{0708-1300/Navigation}}

==Movie Time==
With the word "immersion" in our minds, we watch the movie "Outside In". Also see [http://www.geom.uiuc.edu/docs/outreach/oi/], [http://www.math.toronto.edu/~drorbn/Talks/UofT-040205/index.html] and [http://video.google.com/videoplay?docid=-6626464599825291409].