0708-1300/Class notes for Tuesday, September 25 - Revision history

Trefor: /* Second Hour */

2007-11-02T18:47:32Z

Second Hour

← Older revision		Revision as of 14:47, 2 November 2007
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	<math>D_{\gamma}(f) = (f\circ\gamma)'(0)</math>		<math>D_{\gamma}(f) = (f\circ\gamma)'(0)</math>

	It is clear that ~~this map~~ is linear ~~is f~~ and satisfies the leibnitz rule.		It is clear that <math>D_{\gamma}</math> is linear and satisfies the leibnitz rule.

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

Aptikuis: /* Second Hour */

2007-10-06T00:40:27Z

Second Hour

← Older revision		Revision as of 20:40, 5 October 2007
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	Now, let <math>\gamma:\mathbb{R}\rightarrow \mathbb{R}^{n}</math> be such that <math>\gamma(0) = 0</math>		Now, let <math>\gamma:\mathbb{R}\rightarrow \mathbb{R}^{n}</math> be such that <math>\gamma(0) = 0</math>

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

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

Aptikuis: /* Second Hour */

2007-10-05T02:25:48Z

Second Hour

← Older revision		Revision as of 22:25, 4 October 2007
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	===Second Hour===		===Second Hour===
	Aim: We consider two manifolds, M and N, and a function f between them. We aim for the analogous idea of the tangent in the reals, namely that every smooth f:M -> N has a good linear approximation. Of course, we will need to define what is meant by such a ''smooth'' function, as well as what this linear approximation is.		Aim: We consider two manifolds, M and N, and a function f between them. We aim for the analogous idea of the tangent in the reals, namely that every smooth <math>f:M \rightarrow N</math> has a good linear approximation. Of course, we will need to define what is meant by such a ''smooth'' function, as well as what this linear approximation is.


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

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


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

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

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

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

	such that several axioms are satisfied, including		such that several axioms are satisfied, including
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	This (loosely) amounts to (easily) checking the two claims that		This (loosely) amounts to (easily) checking the two claims that

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

	2) <math>1_{M}</math> is smooth		2) <math>1_{M}</math> is smooth
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	Consider the set of curves on the manifold, that is		Consider the set of curves on the manifold, that is
	<math>\{\gamma:R\rightarrow M\ \|\ s.t.\ \gamma\ smooth,\ \gamma(0) = p\}</math>		<math>\{\gamma:\mathbb{R}\rightarrow M\ \|\ s.t.\ \gamma\ smooth,\ \gamma(0) = p\}</math>

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

	Then, a ''tangent vector'' is an equivalence class of curves.		Then, a ''tangent vector'' is an equivalence class of curves.
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	'''Definition 5''' Tangent Vector, Functional Structure Sense		'''Definition 5''' Tangent Vector, Functional Structure Sense

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

	such that,		such that,

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

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

	If <math>f:R^{m}\rightarrow R</math> is smooth near 0 then there exists smooth <math>g_{i}:R^{m}\rightarrow R</math> such that		If <math>f:\mathbb{R}^{m}\rightarrow \mathbb{R}</math> is smooth near 0 then there exists smooth <math>g_{i}:\mathbb{R}^{m}\rightarrow \mathbb{R}</math> such that
	<math>		<math>
	f(x) = f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)</math>		f(x) = f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)</math>
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	Now, let D be a tangent vector. One can quickly see that D(const) = 0.		Now, let D be a tangent vector. One can quickly see that D(const) = 0.

	So, by ~~hadamard~~'s lemma,		So, by Hadamard's lemma,
	<math>Df = D(f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)) = D(\Sigma_{i=1}^{n}x_{i}g_{i}(x)) = \Sigma_{i=1}^{n}(Dx_{i})g_{i}(0) + \Sigma_{i=1}^{n}x_{i}(0)Dg_{i}</math> = <math>\Sigma_{i=1}^{n}(Dx_{i})g_{i}(0)</math>		<math>Df = D(f(0) + \Sigma_{i=1}^{n}x_{i}g_{i}(x)) = D(\Sigma_{i=1}^{n}x_{i}g_{i}(x)) = \Sigma_{i=1}^{n}(Dx_{i})g_{i}(0) + \Sigma_{i=1}^{n}x_{i}(0)Dg_{i}</math> = <math>\Sigma_{i=1}^{n}(Dx_{i})g_{i}(0)</math>
	as <math>x_{i}(0) = 0</math>		as <math>x_{i}(0) = 0</math>
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	and hence, Df is a linear comb of fixed quantities.		and hence, Df is a linear comb of fixed quantities.

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

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

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

Aptikuis: /* First hour */

2007-10-04T23:56:25Z

First hour

← Older revision		Revision as of 19:56, 4 October 2007
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	''Proof''		''Proof''

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

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

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

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

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

	Claim: this does in fact define a functional structure on X		Claim: this does in fact define a functional structure on X
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	'''Definition 2'''		'''Definition 2'''
	This is the reverse definition of that given directly above. Let <math>\pi:X\rightarrow Y</math> and let X be equipped with a functional structure <math>F_{X}</math>. Then we get a functional structure on Y by		This is the reverse definition of that given directly above. Let <math>\pi:X\rightarrow Y</math> and let X be equipped with a functional structure <math>F_{X}</math>. Then we get a functional structure on Y by
	<math>F_{Y}(V) = \{g:V\rightarrow R\ \|\ g\circ\pi\in F_{X}(\pi^{-1}(V)\}		<math>F_{Y}(V) = \{g:V\rightarrow \mathbb{R}\ \|\ g\circ\pi\in F_{X}(\pi^{-1}(V)\}
	</math>		</math>
	Claim: this does in fact define a functional structure on X		Claim: this does in fact define a functional structure on X
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	'''Example 1'''		'''Example 1'''
	Let <math>S^{2} = R^{3}-\{0\}/</math>~		Let <math>S^{2} = \mathbb{R}^{3}-\{0\}/</math>~
	where the equivalence relation ~ is given by x~<math>\alpha</math>x for <math>\alpha</math>>0		where the equivalence relation ~ is given by x~<math>\alpha</math>x for <math>\alpha</math>>0
	We thus get a canonical projection map <math>\pi:R^{3}-\{0\}\rightarrow S^{2}</math>		We thus get a canonical projection map <math>\pi:\mathbb{R}^{3}-\{0\}\rightarrow S^{2}</math>
	and hence, there is an induced functional structure on <math>S^{2}</math>.		and hence, there is an induced functional structure on <math>S^{2}</math>.
	Claim:		Claim:
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	'''Example 2'''		'''Example 2'''
	Consider the torus thought of as <math>T^{2} = R^{2}/Z^{2}</math>, i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in <math>R^{2}</math>and (n,m) in <math>Z^{2}</math>		Consider the torus thought of as <math>T^{2} = \mathbb{R}^{2}/\mathbb{Z}^{2}</math>, i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in <math>\mathbb{R}^{2}</math>and (n,m) in <math>\mathbb{Z}^{2}</math>

	As in the previous example, the torus inherits a functional structure from the real plane we must again check that		As in the previous example, the torus inherits a functional structure from the real plane we must again check that
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	'''Example 3'''		'''Example 3'''
	Let <math>CP^{n}</math> denote the n dimensional complex projective space, that is,		Let <math>CP^{n}</math> denote the n dimensional complex projective space, that is,
	<math>CP^{n} = C^{n+1}-\{0\}/</math>~ where <math>[z_{0},...,z_{n}]</math>		<math>CP^{n} = \mathbb{C}^{n+1}-\{0\}/</math>~ where <math>[z_{0},...,z_{n}]</math>
	~ <math>[\alpha z_{0},...,\alpha z_{n}]</math>		~ <math>[\alpha z_{0},...,\alpha z_{n}]</math>
	where <math>\alpha\in C</math>		where <math>\alpha\in \mathbb{C}</math>

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

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

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


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

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


	===Second Hour===		===Second Hour===

Franklin: /* Smoot */

2007-09-27T20:49:09Z

Smoot

← Older revision		Revision as of 16:49, 27 September 2007
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	<math>\Phi</math> is trivially 1:1.		<math>\Phi</math> is trivially 1:1.
	===Smoot===		==Smoot==
	In this class It was riced the question of what a "Smoot" is. Here it is [[0708-1300/Smoot]]		In this class It was riced the question of what a "Smoot" is. Here it is [[0708-1300/Smoot]]

Franklin: /* Smoot */

2007-09-27T20:48:09Z

Smoot

← Older revision		Revision as of 16:48, 27 September 2007
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	<math>\Phi</math> is trivially 1:1.		<math>\Phi</math> is trivially 1:1.
	===Smoot===		===Smoot===
	In this class It was riced the question of what a "Smoot" is. Here it is [0708-1300/Smoot]		In this class It was riced the question of what a "Smoot" is. Here it is [[0708-1300/Smoot]]

Franklin at 20:46, 27 September 2007

2007-09-27T20:46:54Z

← Older revision		Revision as of 16:46, 27 September 2007
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	<math>\Phi</math> is trivially 1:1.		<math>\Phi</math> is trivially 1:1.
			===Smoot===
			In this class It was riced the question of what a "Smoot" is. Here it is [0708-1300/Smoot]

Drorbn: /* Class Notes - Second Hour */

2007-09-26T00:29:48Z

Class Notes - Second Hour

← Older revision		Revision as of 20:29, 25 September 2007
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	==~~Class Notes -~~ Second Hour==		===Second Hour===
	Aim: We consider two manifolds, M and N, and a function f between them. We aim for the analogous idea of the tangent in the reals, namely that every smooth f:M -> N has a good linear approximation. Of course, we will need to define what is meant by such a ''smooth'' function, as well as what this linear approximation is.		Aim: We consider two manifolds, M and N, and a function f between them. We aim for the analogous idea of the tangent in the reals, namely that every smooth f:M -> N has a good linear approximation. Of course, we will need to define what is meant by such a ''smooth'' function, as well as what this linear approximation is.

Trefor: /* Class Notes - First hour */

2007-09-25T23:37:07Z

Class Notes - First hour

← Older revision		Revision as of 19:37, 25 September 2007
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	'''Example 4'''		'''Example 4'''
	It can be checked that <math>T^{2} = S^{1}\times S^{1}</math> gives the torus a manifold structure, by the product manifold, that is indeed the same as the normal structure given previously.		It can be checked that <math>T^{2} = S^{1}\times S^{1}</math> gives the torus a manifold structure, by the product manifold, that is indeed the same as the normal structure given previously.


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


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

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


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

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


			'''Claim 1'''

			The two definitions are equivalent



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

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

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

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

			such that several axioms are satisfied, including

			a) Associativity of the composition map

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


			'''Claim 2'''

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

			1) if f:M-> N and g:N->L are smooth maps between manifolds then gof:M->L is smooth

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


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

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

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

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

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


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

			A tangent vector D at p, often called a directional derivative in this sense, is an operator that takes the smooth real valued functions near p into '''R'''.

			such that,

			1) D(af +bg) = aDf + bDg for a,b constants in '''R''' and f,g such functions

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


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


			'''Definition 6''' Tangent Space

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

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


			'''Claim 3'''

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

			</math> is a vector space


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

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

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

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

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

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


			'''Proof of Theorem 1'''

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

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

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

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

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

			''Hadamard's Lemma''

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


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

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

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

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

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

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

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

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

Drorbn at 22:30, 25 September 2007

2007-09-25T22:30:12Z

← Older revision		Revision as of 18:30, 25 September 2007
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	Deadline September 30, 2007 to attend the Winter 2008 semester.		Deadline September 30, 2007 to attend the Winter 2008 semester.

			{{0708-1300/Class Notes}}

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