0708-1300/Class notes for Tuesday, September 25: Difference between revisions
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<math>D_{\gamma}(f) = (f\circ\gamma)'(0)</math> |
<math>D_{\gamma}(f) = (f\circ\gamma)'(0)</math> |
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It is clear that |
It is clear that <math>D_{\gamma}</math> is linear and satisfies the leibnitz rule. |
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Claim: <math>\Phi</math> is a bijection. |
Claim: <math>\Phi</math> is a bijection. |
Latest revision as of 14:47, 2 November 2007
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Dror's Notes
- Class photo is on Thursday, show up and be at your best! More seriously -
- The class photo is of course not mandatory, and if you are afraid of google learning about you, you should not be in it.
- If you want to be in the photo but can't make it on Thursday, I'll take a picture of you some other time and add it as an inset to the main picture.
- I just got the following email message, which some of you may find interesting:
NSERC - CMS Math in Moscow Scholarships The Natural Sciences and Engineering Research Council (NSERC) and the Canadian Mathematical Society (CMS) support scholarships at $9,000 each. Canadian students registered in a mathematics or computer science program are eligible. The scholarships are to attend a semester at the small elite Moscow Independent University. Math in Moscow program www.mccme.ru/mathinmoscow/ Application details www.cms.math.ca/bulletins/Moscow_web/ For additional information please see your department or call the CMS at 613-562-5702. Deadline September 30, 2007 to attend the Winter 2008 semester.
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.
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 that these charts would satisfy the property of a manifold, defined in the atlas sense, that is smooth where defined.
Proof
is smooth is smooth i is smooth where is the coordinate projection map.
Now, since is always smooth,
But then we have and so, by a property of functional structures, and hence where it is defined and thus is smooth. QED
Definition 1 (induced structure)
Suppose and suppose Y is equipped with a functional structure then the "induced functional structure" on X is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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\}}
Claim: this does in fact define a functional structure on X
Definition 2
This is the reverse definition of that given directly above. Let and let X be equipped with a functional structure . Then we get a functional structure on Y by
Claim: this does in fact define a functional structure on X
Example 1
Let ~
where the equivalence relation ~ is given by x~x for >0
We thus get a canonical projection map
and hence, there is an induced functional structure on .
Claim:
1) This induced functional structure makes into a manifold
2) This resulting manifold is the same manifold as from the atlas definition given previously
Example 2
Consider the torus thought of as , i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in and (n,m) in
As in the previous example, the torus inherits a functional structure from the real plane we must again check that 1) We get a manifold 2) This is the same manifold as we had previously with the atlas definition
Example 3
Let denote the n dimensional complex projective space, that is,
~ where
~
where
Again, this space inherits a functional structure from and we again need to claim that this yields a manifold.
Proof of Claim
We consider the subsets for
Clearly
Now, for each there is a unique representative for its equivalence class of the form where the 1 is at the ith location.
We thus can get a map from by Hence we have shown (loosely) that our functional structure is locally isormorphic to
Definition 3 Product Manifolds
Suppose and are manifolds. Then the product manifold, on the set MxN has an atlas given by are charts in resp. manifolds}
Claim: This does in fact yield a manifold
Example 4
It can be checked that gives the torus a manifold structure, by the product manifold, that is indeed the same as the normal structure given previously.
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 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 and be manifolds. We say that a function is smooth if is smooth, where it makes sense, charts
Definition 2 Smooth Function, Functional Structure Sense
Let and be manifolds. We say that a function is smooth if such that h is smooth, 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() (these would be functions for sets, continuous functions for topological spaces, smooth functions for manifolds, etc.) along with
1)Composition maps :mor() mor() mor ()
2) For any x there exists an element mor()
such that several axioms are satisfied, including
a) Associativity of the composition map
b) 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 and are smooth maps between manifolds then is smooth
2) 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
We define an equivalence relation between paths by ~ if is tangent to as paths in .
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 .
such that,
1) D(af +bg) = aDf + bDg for a,b constants in 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, = the set of all tangent vectors at a point p
Claim 3
We will show later that in fact is a vector space
Consider a map . We are interested in defining an associated map between the tangent spaces, namely,
We get two different definitions from the two different definitions of a tangent vector
Definition 7 (atlas sense)
Definition 8 (functional structure sense)
Proof of Theorem 1
We consider a curve and denote its equivalence class of curves by
We consider the map defined by
It is clear that is linear and satisfies the leibnitz rule.
Claim: is a bijection.
We will need to use the following lemma, which will not be proved now:
Hadamard's Lemma
If is smooth near 0 then there exists smooth such that
Now, let D be a tangent vector. One can quickly see that D(const) = 0.
So, by Hadamard's lemma, = as
and hence, Df is a linear comb of fixed quantities.
Now, let be such that
but
We can now claim that is onto because given a D take
is trivially 1:1.
Smoot
In this class It was riced the question of what a "Smoot" is. Here it is 0708-1300/Smoot