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 $\varphi ,\psi$ 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.