0708-1300/Class notes for Tuesday, September 25
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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 - 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]\displaystyle{ \varphi,\psi }[/math] that these charts would satisfy the property of a manifold, defined in the atlas sense, that [math]\displaystyle{ \psi\circ\varphi^{-1} }[/math] is smooth where defined.
Proof
[math]\displaystyle{ \psi\circ\varphi^{-1} }[/math]:Rn[math]\displaystyle{ \rightarrow }[/math]Rn is smooth [math]\displaystyle{ \leftrightarrow }[/math] [math]\displaystyle{ (\psi\circ\varphi^{-1})_i }[/math]:Rn[math]\displaystyle{ \rightarrow }[/math]R is smooth [math]\displaystyle{ \forall }[/math] i [math]\displaystyle{ \leftrightarrow }[/math] [math]\displaystyle{ \pi_i\circ\psi\circ\varphi^{-1} }[/math] is smooth where [math]\displaystyle{ \pi_i }[/math] is the [math]\displaystyle{ i^{th} }[/math] coordinate projection map.
Now, since [math]\displaystyle{ \pi_i }[/math] is always smooth, [math]\displaystyle{ \pi_i\in F_{R^{n}}(U^{'}_{\psi}) }[/math]
But then we have [math]\displaystyle{ \pi_{i}\circ\psi\in F_{M}(U_{\psi}) }[/math] and so, by a property of functional structures, [math]\displaystyle{ \pi_{i}\circ\psi |_{U_{\varphi}\bigcap U_{\psi}}\in F_{M}(U_{\varphi}\bigcap U_{\psi}) }[/math] and hence [math]\displaystyle{ \pi_{i}\circ\psi\circ\varphi^{-1}\in F_{R^{n}} }[/math] where it is defined and thus is smooth. QED
Definition 1 (induced structure)
Suppose [math]\displaystyle{ \pi:X\rightarrow Y }[/math] and suppose Y is equipped with a functional structure [math]\displaystyle{ F_Y }[/math] then the "induced functional structure" on X is
[math]\displaystyle{ 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]
Claim: this does in fact define a functional structure on X
Definition 2
This is the reverse definition of that given directly above. Let [math]\displaystyle{ \pi:X\rightarrow Y }[/math] and let X be equipped with a functional structure [math]\displaystyle{ F_{X} }[/math]. Then we get a functional structure on Y by
[math]\displaystyle{ F_{Y}(V) = \{g:V\rightarrow R\ |\ g\circ\pi\in F_{X}(\pi^{-1}(V)\}
}[/math]
Claim: this does in fact define a functional structure on X
Example 1
Let [math]\displaystyle{ S^{2} = R^{3}-\{0\}/ }[/math]~
where the equivalence relation ~ is given by x~[math]\displaystyle{ \alpha }[/math]x for [math]\displaystyle{ \alpha }[/math]>0
We thus get a canonical projection map [math]\displaystyle{ \pi:R^{3}-\{0\}\rightarrow S^{2} }[/math]
and hence, there is an induced functional structure on [math]\displaystyle{ S^{2} }[/math].
Claim:
1) This induced functional structure makes [math]\displaystyle{ S^{2} }[/math] 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 [math]\displaystyle{ 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]\displaystyle{ R^{2} }[/math]and (n,m) in [math]\displaystyle{ Z^{2} }[/math]
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 [math]\displaystyle{ CP^{n} }[/math] denote the n dimensional complex projective space, that is,
[math]\displaystyle{ CP^{n} = C^{n+1}-\{0\}/ }[/math]~ where [math]\displaystyle{ [z_{0},...,z_{n}] }[/math]
~ [math]\displaystyle{ [\alpha z_{0},...,\alpha z_{n}] }[/math]
where [math]\displaystyle{ \alpha\in C }[/math]
Again, this space inherits a functional structure from [math]\displaystyle{ C^{n+1} }[/math] and we again need to claim that this yields a manifold.
Proof of Claim
We consider the subsets [math]\displaystyle{ CP^{n}\supset U_{i} = \{[z_{0},...,z_{n}]\ |\ z_{i}\neq 0\} }[/math] for [math]\displaystyle{ 0\leq i \leq n }[/math]
Clearly [math]\displaystyle{ \bigcup U_{i} = CP^{n} }[/math]
Now, for each [math]\displaystyle{ p\in U_{i} }[/math] there is a unique representative for its equivalence class of the form [math]\displaystyle{ [z_{0},...,1,...,z_{n}] }[/math] where the 1 is at the ith location.
We thus can get a map from [math]\displaystyle{ \varphi_{i}:U_{i}\rightarrow C^{n} = R^{2n} }[/math] by [math]\displaystyle{ 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]\displaystyle{ (R,C^\infty) }[/math]
Definition 3 Product Manifolds
Suppose [math]\displaystyle{ M^{m} }[/math] and [math]\displaystyle{ N^{n} }[/math] are manifolds. Then the product manifold, on the set MxN has an atlas given by [math]\displaystyle{ \{\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}
Claim: This does in fact yield a manifold
Example 4
It can be checked that [math]\displaystyle{ 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.
