Difference between revisions of "07081300/Class notes for Tuesday, October 2"
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3) Homework one is due on thursday  3) Homework one is due on thursday  
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==First Hour==  ==First Hour==  
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We can check the functorality, <math>(\lambda\circ\theta)_* = \lambda_*\circ\theta_*</math>, then <math>d(\lambda\circ\theta) = d\lambda\circ d\theta</math>  We can check the functorality, <math>(\lambda\circ\theta)_* = \lambda_*\circ\theta_*</math>, then <math>d(\lambda\circ\theta) = d\lambda\circ d\theta</math>  
This is just the chain rule.  This is just the chain rule.  
+  
+  ==Second Hour==  
+  
+  '''Defintion 1'''  
+  
+  An ''immersion'' is a (smooth) map <math>\theta:M^m\rightarrow N^n</math> such that <math>\theta_*</math> of tangent vectors is 1:1. More precisely, <math>d\theta_p: T_pM\rightarrow T_{\theta(p)}N</math> is 1:1 <math>\forall p\in M</math>  
+  
+  
+  '''Example 1'''  
+  
+  Consider the canonical immersion, for m<n given by <math>\iota:(x_1,...,x_m)\mapsto (x_1,...,x_m,0,...,0)</math> with nm zeros.  
+  
+  
+  '''Example 2'''  
+  
+  This is the map from R to <math>R^2</math> that looks like a loopdeloop on a roller coaster (but squashed into the plane of course!) The map <math>\theta</math> itself is NOT 1:1 (consider the crossover point) however <math>\theta_*</math> IS 1:1, hence an immersion.  
+  
+  
+  '''Example 3'''  
+  
+  Consider the map from R to <math>R^2</math> that looks like a check mark. While this map itself is 1:1, <math>\theta_*</math> is NOT 1:1 (at the cusp in the check mark) and hence is not an immersion.  
+  
+  
+  '''Example 4'''  
+  
+  Can there be objects, such as the graph of x that are NOT an immersion, but are constructed from a smooth function?  
+  
+  Consider the function <math>\lambda(x) = e^{1/x^2}</math> for x>0 and 0 otherwise.  
+  
+  Then the map <math>x\mapsto \begin{bmatrix}  
+  (\lambda(x),\lambda(x))& x>0\\  
+  (0,0)& x=0\\  
+  (\lambda(x),\lambda(x)) & x<0\\  
+  \end{bmatrix}</math>  
+  
+  
+  is a smooth mapping with the graph of x as its image, but is NOT an immersion.  
+  
+  
+  '''Example 5'''  
+  
+  The torus, as a subset of <math>R^3</math> is an immersion  
+  
+  
+  Now, consider the 1:1 linear map <math>T:V\rightarrow W</math> where V,W are vector spaces that takes <math>(v_1,...,v_m)\mapsto (Tv_1,...,Tv_m) = (w_1,..,w_m,w_{m+1},...,w_n)</math>  
+  
+  From linear algebra we know that we can choose a basis such that T is represented by a matrix with 1's along the first m diagonal locations and zeros elsewhere.  
+  
+  
+  '''Theorem 2'''  
+  
+  Locally, every immersion looks like the inclusion <math>\iota:R^m\rightarrow R^n</math>.  
+  
+  More precisely, if <math>\theta:M^m\rightarrow N^n</math> and <math>d\theta_p</math> is 1:1 then there exists charts <math>\varphi</math> acting on <math>U\subset M</math> and <math>\phi</math> acting on <math>V\subset N</math> such that for <math>p\in U, \phi(p) = \psi(\theta(p)) = 0</math> such that the following diagram commutes:  
+  
+  
+  <math>\begin{matrix}  
+  U&\rightarrow^{\phi}&U'\subset R^n\\  
+  \downarrow_{\theta} &&\downarrow_{\iota} \\  
+  V& \rightarrow^{\psi}& V'\subset R^n\\  
+  \end{matrix}</math>  
+  
+  
+  that is, <math>\iota\circ\varphi = \psi\circ\theta</math>  
+  
+  
+  '''Definition 2'''  
+  
+  M is a ''submanifold'' of N if there exists a mapping <math>\theta:M\rightarrow N</math> such that <math>\theta</math> is a 1:1 immersion.  
+  
+  '''Example 6'''  
+  
+  Our previous example of the graph of a "loopdeloop", while an immersion, the function is not 1:1 and hence the graph is not a sub manifold.  
+  
+  
+  '''Example 6'''  
+  
+  The torus is a submanifold as the natural immersion into <math>R^3</math> is 1:1  
+  
+  
+  '''Definition 3'''  
+  
+  The map <math>\theta:M\rightarrow N</math> is an embedding if the subset topology on <math>\theta(M)</math> coincides with the topology induced from the original topology of M.  
+  
+  
+  '''Example 7'''  
+  
+  Consider the map from <math>R\rightarrow R^2</math> whose graph looks like the open interval whose two ends have been wrapped around until they just touch (would intersect at one point if they were closed) the points 1/3 and 2/3rds of the way along the interval respectively.  
+  The map is both 1:1 and an immersion. However, any neighborhood about the endpoints of the interval will ALSO include points near the 1/3rd and 2/3rd spots on the line, i.e., the topology is different and hence this is not an embedding.  
+  
+  '''Corollary 1 to Theorem 2'''  
+  
+  The functional structure on an embedded manifold induced by the functional structure on the containing manifold is equal to its original functional structure.  
+  
+  Indeed, for all smooth <math>f:M\rightarrow R</math> and <math>\forall p\in M</math> there exists a neighborhood V of <math>\theta(p)</math> and a smooth <math>g:N\rightarrow R</math> such that <math>g_{\theta(M)\bigcap U} = f_U</math>  
+  
+  
+  ''Proof of Corollary 1''  
+  
+  Loosely (and a sketch is most useful to see this!) we consider the embedded submanifold M in N and consider its image, under the appropriate charts, to a subset of <math>R^m\subset R^n</math>. We then consider some function defined on M, and hence on the subset in R^n which we can extend canonically as a constant function in the "vertical" directions. Now simply pullback into N to get the extended member of the functional structure on N.  
+  
+  
+  ''Proof of Theorem 2''  
+  
+  We start with the normal situation of <math>\theta:M\rightarrow N</math> with M,N manifolds with atlases containing <math>(\varphi_0,U_)0)</math> and <math>(\psi_0, V_0)</math> respectively. We also expect that for <math>p\in U_0, \varphi_0(p) = \psi_0(\theta(p)) = 0</math>. I will first draw the diagram and will subsequently justify the relevant parts. The proof reduces to showing a certain part of the diagram commutes appropriately.  
+  
+  
+  <math>  
+  \begin{matrix}  
+  M\supset U_0 & \rightarrow^{\varphi_0} & U_1\subset R^m & \rightarrow^{Id} & U_2 = U_1 \\  
+  \downarrow_{\theta} & &\downarrow_{\theta_1} & &\downarrow_{\iota}\\  
+  N\supset V_0 & \rightarrow^{\psi_0} & V_1\subset R^n & \leftarrow^{\xi} & V_2\\  
+  \end{matrix}  
+  </math>  
+  
+  
+  It is very important to note that the <math>\varphi_0</math> and <math>\psi_0</math> are NOT the charts we are looking for , they are merely one of the ones that happen to act about the point p.  
+  
+  In the diagram above, <math>\theta_1 = \psi_0\circ\theta\circ\varphi^{1}</math>. So, <math>\theta_1(0) = 0</math> and <math>(d\theta_1)_0 = i</math>. Note the <math>\theta_1</math>, being merely the normal composition with the appropriate charts, does not fundamentally add anything. What makes this theorem work is the function <math>\xi</math>  
+  
+  We consider the map <math>\xi:V_2\rightarrow V_1</math> given <math>(x,y)\mapsto \theta_1(x) + (0,y)</math>. We note that <math>\xi</math> corresponds with the idea of "lifting" a flattened image back to its original height.  
+  
+  
+  Claims:  
+  
+  1) <math>\xi</math> is invertible near zero. Indeed, computing <math>d\xi_0 = I</math> which is invertible as a matrix and hence <math>\xi</math> is invertible as a function near zero.  
+  
+  2) Take an <math>x\in U_2</math>. There are two routes to get to <math>V_1</math> and upon computing both ways yields the same result. Hence, the diagram commutes.  
+  
+  
+  Hence, our immersion looks (locally) like the standard immersion between real spaces given by <math>\iota</math> and the charts are the compositions going between <math>U_0</math> to <math>U_2</math> and <math>V_0</math> to <math>V_2</math>  
+  
+  ''Q.E.D'' 
Revision as of 19:26, 2 October 2007

Contents 
English Spelling
Many interesting rules about 07081300/English Spelling
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 class comments
1) The class photo is up, please add yourself
2) A questionnaire was passed out in class
3) Homework one is due on thursday
First Hour
Today's Theme: Locally a function looks like its differential
Pushforward/Pullback
Let be a smooth map.
We consider various objects, defined with respect to X or Y, and see in which direction it makes sense to consider corresponding objects on the other space. In general will denote the push forward, and will denote the pullback.
1) points pushforward
2) Paths , ie a bunch of points, pushforward,
3) Sets pullback via Note that if one tried to pushforward sets A in X, the set operations compliment and intersection would not commute appropriately with the map
4) A measures pushforward via
5)In some sense, we consider functions, "dual" to points and thus should go in the opposite direction of points, namely
6) Tangent vectors, defined in the sense of equivalence classes of paths, [] pushforward as we would expect since each path pushes forward.
CHECK: This definition is well defined, that is, independent of the representative choice of
7) We can consider operators on functions to be in a sense dual to the functions and hence should go in the opposite direction. Hence, tangent vectors, defined in the sense of derivations, pushforward via
CHECK: This definition satisfies linearity and Liebnitz property.
Theorem 1
The two definitions for the pushforward of a tangent vector coincide.
Proof:
Given a we can construct as above. However from both and we can also construct and because we have previously shown our two definitions for the tangent vector are equivalent. We can then pushforward to get . The theorem is reduced to the claim that:
for functions
Now,
Q.E.D
Functorality
let
Consider some "object" s defined with respect to X and some "object u" defined with respect to Z. Something has the property of functorality if
and
Claim: All the classes we considered previously have the functorality property; in particular, the pushforward of tangent vectors does.
Let us consider on given a
We can arrange for charts on a subset of M into (with coordinates denoted )and on a subset of N into (with coordinates denoted )such that and
Define
Now, for a we can write
So,
Now, we want to write
and so,
where the i is at the kth location.
So, , i.e., is the differential of at p
We can check the functorality, , then
This is just the chain rule.
Second Hour
Defintion 1
An immersion is a (smooth) map such that of tangent vectors is 1:1. More precisely, is 1:1
Example 1
Consider the canonical immersion, for m<n given by with nm zeros.
Example 2
This is the map from R to that looks like a loopdeloop on a roller coaster (but squashed into the plane of course!) The map itself is NOT 1:1 (consider the crossover point) however IS 1:1, hence an immersion.
Example 3
Consider the map from R to that looks like a check mark. While this map itself is 1:1, is NOT 1:1 (at the cusp in the check mark) and hence is not an immersion.
Example 4
Can there be objects, such as the graph of x that are NOT an immersion, but are constructed from a smooth function?
Consider the function for x>0 and 0 otherwise.
Then the map
is a smooth mapping with the graph of x as its image, but is NOT an immersion.
Example 5
The torus, as a subset of is an immersion
Now, consider the 1:1 linear map where V,W are vector spaces that takes
From linear algebra we know that we can choose a basis such that T is represented by a matrix with 1's along the first m diagonal locations and zeros elsewhere.
Theorem 2
Locally, every immersion looks like the inclusion .
More precisely, if and is 1:1 then there exists charts acting on and acting on such that for such that the following diagram commutes:
that is,
Definition 2
M is a submanifold of N if there exists a mapping such that is a 1:1 immersion.
Example 6
Our previous example of the graph of a "loopdeloop", while an immersion, the function is not 1:1 and hence the graph is not a sub manifold.
Example 6
The torus is a submanifold as the natural immersion into is 1:1
Definition 3
The map is an embedding if the subset topology on coincides with the topology induced from the original topology of M.
Example 7
Consider the map from whose graph looks like the open interval whose two ends have been wrapped around until they just touch (would intersect at one point if they were closed) the points 1/3 and 2/3rds of the way along the interval respectively. The map is both 1:1 and an immersion. However, any neighborhood about the endpoints of the interval will ALSO include points near the 1/3rd and 2/3rd spots on the line, i.e., the topology is different and hence this is not an embedding.
Corollary 1 to Theorem 2
The functional structure on an embedded manifold induced by the functional structure on the containing manifold is equal to its original functional structure.
Indeed, for all smooth and there exists a neighborhood V of and a smooth such that
Proof of Corollary 1
Loosely (and a sketch is most useful to see this!) we consider the embedded submanifold M in N and consider its image, under the appropriate charts, to a subset of . We then consider some function defined on M, and hence on the subset in R^n which we can extend canonically as a constant function in the "vertical" directions. Now simply pullback into N to get the extended member of the functional structure on N.
Proof of Theorem 2
We start with the normal situation of with M,N manifolds with atlases containing and respectively. We also expect that for . I will first draw the diagram and will subsequently justify the relevant parts. The proof reduces to showing a certain part of the diagram commutes appropriately.
It is very important to note that the and are NOT the charts we are looking for , they are merely one of the ones that happen to act about the point p.
In the diagram above, . So, and . Note the , being merely the normal composition with the appropriate charts, does not fundamentally add anything. What makes this theorem work is the function
We consider the map given . We note that corresponds with the idea of "lifting" a flattened image back to its original height.
Claims:
1) is invertible near zero. Indeed, computing which is invertible as a matrix and hence is invertible as a function near zero.
2) Take an . There are two routes to get to and upon computing both ways yields the same result. Hence, the diagram commutes.
Hence, our immersion looks (locally) like the standard immersion between real spaces given by and the charts are the compositions going between to and to
Q.E.D