0708-1300/Class notes for Thursday, September 27 - Revision history

Trefor: /* Class Notes */

2007-11-02T19:35:35Z

Class Notes

← Older revision		Revision as of 15:35, 2 November 2007
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	2) We only need to check this on an arbitrary <math>x_j</math> as they span all such functions.		2) We only need to check this on an arbitrary <math>x_j</math> as they span all such functions.
	So, <math>~~Dx_j =~~ \sum Dx_i \frac{\partial x_j}{\partial x_i}		So, <math>\sum Dx_i \frac{\partial x_j}{\partial x_i}
	= \sum Dx_i \delta_{i,j} = Dx_j</math>		= \sum Dx_i \delta_{i,j} = Dx_j</math>

Aptikuis: /* Class Notes */

2007-10-06T00:58:35Z

Class Notes

← Older revision		Revision as of 20:58, 5 October 2007
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	1) Consider the analogy with a (smooth) car which must stop when approaching a sharp bend. When it does stop, everything around the car, such as a tree, stops moving relative to the car as well		1) Consider the analogy with a (smooth) car which must stop when approaching a sharp bend. When it does stop, everything around the car, such as a tree, stops moving relative to the car as well

	2) There is a map h going from the restriction of <math>R^{2}</math> to our set into <math>R</math> as well as a map (f,g) going in reverse that satisfies <math>h\circ(f,g)=I_{d}</math>. We can then apply the chain rule (think about why!) to get <math>h_{x}f' + h_{y}g' = 1</math>. However, <math>f=\pm g</math> and both cases occur at adjacent points, resulting in <math>f' =\pm g'</math> at adjacent point and thus establishing the contradiction.		2) There is a map h going from the restriction of <math>\mathbb{R}^{2}</math> to our set into <math>\mathbb{R}</math> as well as a map (f,g) going in reverse that satisfies <math>h\circ(f,g)=I_{d}</math>. We can then apply the chain rule (think about why!) to get <math>h_{x}f' + h_{y}g' = 1</math>. However, <math>f=\pm g</math> and both cases occur at adjacent points, resulting in <math>f' =\pm g'</math> at adjacent point and thus establishing the contradiction.

	3) This hint uses methods from beyond page 71. It is possible to find two linearly independent directional derivatives on functions on our set A near zero. However this is a contradiction as a one dimensional space cannot have a two dimensional tangent space.		3) This hint uses methods from beyond page 71. It is possible to find two linearly independent directional derivatives on functions on our set A near zero. However this is a contradiction as a one dimensional space cannot have a two dimensional tangent space.
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	It is easily checked that the tangent space <math>T_{0}R^{n}</math> forms an n dimensional ''vector space''. This is because the D's are linear and because the D is determined by the n constants <math>Dx_{i}</math>.		It is easily checked that the tangent space <math>T_{0}\mathbb{R}^{n}</math> forms an n dimensional ''vector space''. This is because the D's are linear and because the D is determined by the n constants <math>Dx_{i}</math>.

	We wish to generalize this concept to show that <math>T_{p}M^{n}</math> is a vector space. This is easily done as there is a canonical isomorphism between <math>T_{p}M^{n}</math> and <math>T_{0}R^{n}</math> via the chart <math>\varphi</math>		We wish to generalize this concept to show that <math>T_{p}M^{n}</math> is a vector space. This is easily done as there is a canonical isomorphism between <math>T_{p}M^{n}</math> and <math>T_{0}\mathbb{R}^{n}</math> via the chart <math>\varphi</math>


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	<math>		<math>
	R^n</math> possesses canonical functions <math>(x_1,...,x_n)</math> that are merely the levels <math>x_i = const</math>.		\mathbb{R}^n</math> possesses canonical functions <math>(x_1,...,x_n)</math> that are merely the levels <math>x_i = const</math>.

	The pullback of these into the manifold under <math>\varphi^{-1}</math> yields a similar 'grid' of lines on the manifolds only these lines are curves. Formally, we equip the manifold with functions <math>x^{o}_1 = x_1\circ\varphi~</math>, <math>x^{o}_2 = x_2\circ\varphi~</math>, etc...		The pullback of these into the manifold under <math>\varphi^{-1}</math> yields a similar 'grid' of lines on the manifolds only these lines are curves. Formally, we equip the manifold with functions <math>x^{o}_1 = x_1\circ\varphi~</math>, <math>x^{o}_2 = x_2\circ\varphi~</math>, etc...

	Now, <math>\forall f:M\rightarrow R\ \exists g:R^n\rightarrow R</math> such that <math>f=g\circ\varphi</math> and <math>f(p) = g(\varphi(p)) = g((x_1(\varphi(p)),...,x_n(\varphi(p))) = g(x_1^{o},...,x_m^{o})</math>		Now, <math>\forall f:M\rightarrow \mathbb{R}\ \exists g:\mathbb{R}^n\rightarrow \mathbb{R}</math> such that <math>f=g\circ\varphi</math> and <math>f(p) = g(\varphi(p)) = g((x_1(\varphi(p)),...,x_n(\varphi(p))) = g(x_1^{o},...,x_m^{o})</math>

	Conventionally the distinction between x and <math>x^{0}</math> is not made.		Conventionally the distinction between x and <math>x^{0}</math> is not made.

Trefor: /* Class Notes */

2007-09-28T02:07:00Z

Class Notes

← Older revision		Revision as of 22:07, 27 September 2007
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			'''General comments regarding the wiki page'''
	coming soon, just testing the page creation

			1) Use the history/recent changes to track your own work

			2) Never post/upload without linking


			'''Comments on Problem 4, page 71, Assignment 1'''

			Dror gave three hints towards a solution to this this problem:

			1) Consider the analogy with a (smooth) car which must stop when approaching a sharp bend. When it does stop, everything around the car, such as a tree, stops moving relative to the car as well

			2) There is a map h going from the restriction of <math>R^{2}</math> to our set into <math>R</math> as well as a map (f,g) going in reverse that satisfies <math>h\circ(f,g)=I_{d}</math>. We can then apply the chain rule (think about why!) to get <math>h_{x}f' + h_{y}g' = 1</math>. However, <math>f=\pm g</math> and both cases occur at adjacent points, resulting in <math>f' =\pm g'</math> at adjacent point and thus establishing the contradiction.

			3) This hint uses methods from beyond page 71. It is possible to find two linearly independent directional derivatives on functions on our set A near zero. However this is a contradiction as a one dimensional space cannot have a two dimensional tangent space.


			At this point, the discussion returned to the previous days class regarding the theorem of the equivalence of our two definitions of a tangent vector. It was reiterated that a major point in proving the bijection between the two types of vectors was indeed onto is that it was possible, as a result of Hadamard's Lemma, to determine D by the n constants <math>Dx_{i}</math>


			It is easily checked that the tangent space <math>T_{0}R^{n}</math> forms an n dimensional ''vector space''. This is because the D's are linear and because the D is determined by the n constants <math>Dx_{i}</math>.

			We wish to generalize this concept to show that <math>T_{p}M^{n}</math> is a vector space. This is easily done as there is a canonical isomorphism between <math>T_{p}M^{n}</math> and <math>T_{0}R^{n}</math> via the chart <math>\varphi</math>


			'''Proof of Hadamard's Lemma'''

			<math>f(p)-f(0)=\int_0^1 \frac{d}{dt}f(tp)\, dt </math>
			<math>=\int_0^1 \sum_{i=1}^{n} \frac{\partial f}{\partial x_{i}}(tp)x_{i}dt</math>
			<math>=\sum_{i=1}^{n}x_{i}\int_0^1\frac{\partial f}{\partial x_{i}}(tp)dt</math>
			<math>=\sum_{i=1}^{n}x_{i}g_i (p)</math>

			where <math>g_i (p)=\int_0^1\frac{\partial f}{\partial x_{i}}(tp)dt</math>

			f is smooth with respect to p and so <math>g_i</math> is, as derivatives with respect to p can pass through the integral which is with respect to t.

			''QED''

			'''Corollary''':

			<math>g_i(0)=\frac{\partial f}{\partial x_{i}}(0)</math>


			'''Local Coordinates'''

			<math>
			R^n</math> possesses canonical functions <math>(x_1,...,x_n)</math> that are merely the levels <math>x_i = const</math>.

			The pullback of these into the manifold under <math>\varphi^{-1}</math> yields a similar 'grid' of lines on the manifolds only these lines are curves. Formally, we equip the manifold with functions <math>x^{o}_1 = x_1\circ\varphi~</math>, <math>x^{o}_2 = x_2\circ\varphi~</math>, etc...

			Now, <math>\forall f:M\rightarrow R\ \exists g:R^n\rightarrow R</math> such that <math>f=g\circ\varphi</math> and <math>f(p) = g(\varphi(p)) = g((x_1(\varphi(p)),...,x_n(\varphi(p))) = g(x_1^{o},...,x_m^{o})</math>

			Conventionally the distinction between x and <math>x^{0}</math> is not made.


			''Question:''
			How do you express <math>D\in T_p(M)</math> using the local coordinates?

			'''Claim'''

			1) <math>\frac{\partial}{\partial x_i}</math> is a tangent vector; <math>\frac{\partial}{\partial x_i}(f):=\frac{\partial}{\partial x_i}(g)</math> where <math>g=f\circ\varphi^{-1}</math>

			2) <math>D = \sum (Dx_i)\frac{\partial}{\partial x_i}</math>

			'''Proof'''

			1) We need to check linearity and liebnitz's rule (easy)

			2) We only need to check this on an arbitrary <math>x_j</math> as they span all such functions.
			So, <math>Dx_j = \sum Dx_i \frac{\partial x_j}{\partial x_i}
			= \sum Dx_i \delta_{i,j} = Dx_j</math>

Trefor at 01:12, 28 September 2007

2007-09-28T01:12:25Z

← Older revision		Revision as of 21:12, 27 September 2007
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			{{0708-1300/Navigation}}

	==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>

Trefor at 01:10, 28 September 2007

2007-09-28T01:10:56Z

New page

==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>

coming soon, just testing the page creation