Difference between revisions of "07081300/Class notes for Thursday, September 27"
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−  +  '''General comments regarding the wiki page'''  
+  
+  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> 
Revision as of 21:07, 27 September 2007

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 comments regarding the wiki page
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 to our set into as well as a map (f,g) going in reverse that satisfies . We can then apply the chain rule (think about why!) to get . However, and both cases occur at adjacent points, resulting in 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
It is easily checked that the tangent space forms an n dimensional vector space. This is because the D's are linear and because the D is determined by the n constants .
We wish to generalize this concept to show that is a vector space. This is easily done as there is a canonical isomorphism between and via the chart
Proof of Hadamard's Lemma
where
f is smooth with respect to p and so is, as derivatives with respect to p can pass through the integral which is with respect to t.
QED
Corollary:
Local Coordinates
possesses canonical functions that are merely the levels .
The pullback of these into the manifold under yields a similar 'grid' of lines on the manifolds only these lines are curves. Formally, we equip the manifold with functions , , etc...
Now, such that and
Conventionally the distinction between x and is not made.
Question:
How do you express using the local coordinates?
Claim
1) is a tangent vector; where
2)
Proof
1) We need to check linearity and liebnitz's rule (easy)
2) We only need to check this on an arbitrary as they span all such functions. So,