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

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 $\varphi ,\psi$ that these charts would satisfy the property of a manifold, defined in the atlas sense, that $\psi \circ \varphi ^{-1}$ is smooth where defined.

Proof

$\psi \circ \varphi ^{-1}$:Rⁿ$\rightarrow$Rⁿ is smooth $\leftrightarrow$ $(\psi \circ \varphi ^{-1})_{i}$:Rⁿ$\rightarrow$R is smooth $\forall$ i $\leftrightarrow$ $\pi _{i}\circ \psi \circ \varphi ^{-1}$ is smooth where $\pi _{i}$ is the $i^{th}$ coordinate projection map.

Now, since $\pi _{i}$ is always smooth, $\pi _{i}\in F_{R^{n}}(U_{\psi }^{'})$

But then we have $\pi _{i}\circ \psi \in F_{M}(U_{\psi })$ and so, by a property of functional structures, $\pi _{i}\circ \psi |_{U_{\varphi }\bigcap U_{\psi }}\in F_{M}(U_{\varphi }\bigcap U_{\psi })$ and hence $\pi _{i}\circ \psi \circ \varphi ^{-1}\in F_{R^{n}}$ where it is defined and thus is smooth. QED

Definition 1 (induced structure) Suppose $\pi :X\rightarrow Y$ and suppose Y is equipped with a functional structure $F_{Y}$ then the "induced functional structure" on X is

$F_{X}(U)=\{f:U\rightarrow R\ |\ \exists g\in F_{Y}(V)\ such\ that\ V\supset \pi (U)\ and\ f=g\circ \pi \}$

Claim: this does in fact define a functional structure on X

Definition 2 This is the reverse definition of that given directly above. Let $\pi :X\rightarrow Y$ and let X be equipped with a functional structure $F_{X}$. Then we get a functional structure on Y by $F_{Y}(V)=\{g:V\rightarrow R\ |\ g\circ \pi \in F_{X}(\pi ^{-1}(V)\}$ Claim: this does in fact define a functional structure on X

Example 1 Let $S^{2}=R^{3}-\{0\}/$~ where the equivalence relation ~ is given by x~$\alpha$x for $\alpha$>0 We thus get a canonical projection map $\pi :R^{3}-\{0\}\rightarrow S^{2}$ and hence, there is an induced functional structure on $S^{2}$. Claim: 1) This induced functional structure makes $S^{2}$ 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 $T^{2}=R^{2}/Z^{2}$, i.e., the real plane with the equivalence relation that (x,y)~(x+n,y+m) for (x,y) in $R^{2}$and (n,m) in $Z^{2}$

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 $CP^{n}$ denote the n dimensional complex projective space, that is, $CP^{n}=C^{n+1}-\{0\}/$~ where $[z_{0},...,z_{n}]$ ~ $[\alpha z_{0},...,\alpha z_{n}]$ where $\alpha \in C$

Again, this space inherits a functional structure from $C^{n+1}$ and we again need to claim that this yields a manifold.

Proof of Claim

We consider the subsets $CP^{n}\supset U_{i}=\{[z_{0},...,z_{n}]\ |\ z_{i}\neq 0\}$ for $0\leq i\leq n$

Clearly $\bigcup U_{i}=CP^{n}$

Now, for each $p\in U_{i}$ there is a unique representative for its equivalence class of the form $[z_{0},...,1,...,z_{n}]$ where the 1 is at the ith location.

We thus can get a map from $\varphi _{i}:U_{i}\rightarrow C^{n}=R^{2n}$ by $p\mapsto [z_{0}/z_{i},...,z_{i-1}/z_{i},z_{i+1}/z_{i},...,z_{n}/z_{i}]$ Hence we have shown (loosely) that our functional structure is locally isormophic to $(R,C^{\infty })$

Definition 3 Product Manifolds

Suppose $M^{m}$ and $N^{n}$ are manifolds. Then the product manifold, on the set MxN has an atlas given by $\{\varphi \times \psi :U\times V\rightarrow U'\times V'\in R^{m}\times R^{n}\ |\varphi :U\rightarrow U'\subset R^{m}\ and\ \psi :V\rightarrow V'\subset R^{n}$ are charts in resp. manifolds}

Claim: This does in fact yield a manifold

Example 4 It can be checked that $T^{2}=S^{1}\times S^{1}$ gives the torus a manifold structure, by the product manifold, that is indeed the same as the normal structure given previously.