0708-1300/Class notes for Tuesday, October 30

0708-1300/Navigation Panel   Add your name / see who's in! # Week of... Links Fall Semester 1 Sep 10 About, Tue, Thu 2 Sep 17 Tue, HW1, Thu 3 Sep 24 Tue, Photo, Thu 4 Oct 1 Questionnaire, Tue, HW2, Thu 5 Oct 8 Thanksgiving, Tue, Thu 6 Oct 15 Tue, HW3, Thu 7 Oct 22 Tue, Thu 8 Oct 29 Tue, HW4, Thu, Hilbert sphere 9 Nov 5 Tue,Thu, TE1 10 Nov 12 Tue, Thu 11 Nov 19 Tue, Thu, HW5 12 Nov 26 Tue, Thu 13 Dec 3 Tue, Thu, HW6 Spring Semester 14 Jan 7 Tue, Thu, HW7 15 Jan 14 Tue, Thu 16 Jan 21 Tue, Thu, HW8 17 Jan 28 Tue, Thu 18 Feb 4 Tue 19 Feb 11 TE2, Tue, HW9, Thu, Feb 17: last chance to drop class R Feb 18 Reading week 20 Feb 25 Tue, Thu, HW10 21 Mar 3 Tue, Thu 22 Mar 10 Tue, Thu, HW11 23 Mar 17 Tue, Thu 24 Mar 24 Tue, HW12, Thu 25 Mar 31 Referendum,Tue, Thu 26 Apr 7 Tue, Thu R Apr 14 Office hours R Apr 21 Office hours F Apr 28 Office hours, Final (Fri, May 2) Register of Good Deeds Errata to Bredon's Book

Announcements go here

Today's Agenda

Debts

A bit more about proper functions on locally compact spaces.

Smooth Retracts and Smooth Brouwer

Theorem. There does not exist a smooth retract $r:D^{n+1}\to S^{n}$.

Corollary. (The Brouwer Fixed Point Theorem) Every smooth $f:D^{n}\to D^{n}$ has a fixed point.

Suggestion for a good deed. Tell Dror if he likes the Brouwer fixed point theorem, for he is honestly unsure. But first hear some drorpaganda on what he likes and what he doesn't quite.

Corollary. The sphere $S^{n}$ is not smoothly contractible.

Challenge. Remove the word "smooth" everywhere above.

Continuous Brouwer's Fixed-Point Theorem Proof: Let $f:D^{n}\to D^{n}$ be a continuous function. By Weierstrass theorem and since $D^{n}$ is compact there is a sequence of polynomials $p_{n}$ converging uniformly to f. By the smooth Brouwers' Theorem each $p_{n}$ has a fixed point $x_{n}$. By the compactness of $D^{n}$ there is cluster point of $x_{n}$ which should be a fixed point of f.

Smooth Approximation

Theorem. Let $A$ be a closed subset of a smooth manifold $M$, let $f:M\to {\mathbb {R} }$ be a continuous function whose restriction $f|_{A}$ to $A$ is smooth, and let $\epsilon$ be your favourite small number. Then there exists a smooth $g:M\to {\mathbb {R} }$ so that $f|_{A}=g|_{A}$ and $||f-g||<\epsilon$. Furthermore, $f$ and $g$ are homotopic via an $\epsilon$-small homotopy.

Theorem. The same, with the target space replaced by an arbitrary compact metrized manifold $N$.

Tubular Neighborhoods

Theorem. Every compact smooth submanifold $M^{m}$ of ${\mathbb {R} }^{n}$ has a "tubular neighborhood".

Entertainment

A student told me about this clip on YouTube (lyrics). Enjoy!

There is this one too but it is in Spanish. Romance of the Derivative and the Arctangent

Further Notes

With Brouwer's fixed point theorem you can prove amazing things

1) There are two antipodal points in the equator with the same temperature.

2) There are two antipodal points with the same temperature and the same pressure.

3) You can through three potatoes in the air and with just one swing cut all of them in half.

4) Every non-bold person has a swirl of hair or some other problem ordering their hair...

5) If you have a car with a loose antenna and you always go in your car in a trip exactly the same way every day then there is an initial position of the antenna such that it wont fall during your trip.

6) It doesn't matter how much you stir your coffee at least one point will be in the same position.

Constructive proof of Brouwer's Fixed-Point Theorem

Several proves of Brouwer's theorem had been given. See A Constructive Proof of the Brouwer Fixed-Point Theorem and Computational, R. B. Kellogg; T. Y. Li; J. Yorke

Most of them motivated by the fact that the first proof of the fixed-point theorem was a non-constructive indirect proof i.e. using reductio ad absurdum and hence using the excluded middle axiom. This axiom is rejected by Brouwer's itself paradigm of Foundations of Mathematics and the intuitionist school of which Brouwer is one of the founders.

Typed 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

Claim 1:

Consider $f:X\rightarrow Y$ to be continuous and proper and Y being locally compact. Then it is closed.

Definition 1

A set $F\subset X$ (X a topological space) is called locally closed if every $x\in X$ has a neighbourhood U such that $U\cap F$ is closed in U.

Claim 2:

Locally closed $\Leftrightarrow$ closed.

Proof of Claim 2:

($\Leftarrow$) is tautological

($\Rightarrow$) If $x\in F^{c}$ then $\exists U_{x}$ such that $U\cap F$ is closed in U so $U\cap F^{c}$ is open in U and hence also in X. Thus, x has a neighbourhood in $F^{c}$ so $F^{c}$ is open and thus F is closed.

Proof of Claim 1:

To be repeated next class.

Definition 2

If $Y\subset X$, a retract is a continuous fucntion $f:X\rightarrow Y$ such that $r|_{Y}=id_{Y}$

Theorem 1

There is no smooth retract $f:D^{n+1}\rightarrow S^{n}$ where

$D^{n+1}=\{x\in \mathbb {R} ^{n+1}\ :\ ||x||\leq 1\}\supset S^{n}=\{x\in \mathbb {R} ^{n+1}\ :\ ||x||=1\}$

Proof of Theorem 1

Assume not. Then $\exists r:D^{n+1}\rightarrow S^{n}$ such that r is a smooth retract. By Sard's theorem we can find a non critical value y of r. So $r^{-1}(y)$ is a submanifold (with boundary) of dimension 1 of $D^{n+1}$.

Previously we proved that such things were manifolds but only if they were without boundary. The proof with boundary works in exactly the same was, that is, use the theorem that immersions looks like projections in $R^{n}$ only this time allow neighborhoods to be homeomorphic to the half space $H^{n}$.

Now the classification theorem for 1 dimensional manifolds shows that $r^{-1}(y)$ must be homeomorphic to a closed interval (as it has at least one boundary point at y) and hence it must also have a second boundary point. But $\partial (r^{-1}(y))\subset \partial D^{n+1}=S^{n}$ and hence this second boundary point must be some other point y' on $S^{n}$. But such points map to themselves by assumption so $y'=r(y')=y$. This establishes the contradiction.

Theorem 2

Brouwer's Fixed Point Theorem

Consider a function $f:D^{n}\rightarrow D^{n}$, f being smooth. Then $\exists p\in D^{n}$ such that f(p)=p, i.e., there exists a fixed point of the function.

Proof of Brouwer's Theorem

Assume not. We define a function from $D^{n}$ to $S^{n-1}$ in the following way. Consider $x\in D^{n}$ and consider $f(x)\in D^{n}$. These points are different by assumption. Consider the line from f(x) towards x. Let r(x) be the point on this line intersecting the boundary (choosing the point going from f(x) to x not the reverse). While we have not (but could) written down a precise formula for this, it is apparent that it is a smooth function from $D^{n}$ to $S^{n-1}$ that is fixed on the boundary. Hence this is a retract, which is proved could not exist and thus establishing the contradiction.

Second Hour

Theorem 3

Let A be a closed subset of M and let $f:M\rightarrow \mathbb {R}$ be continuous such that $f|_{A}$ is smooth. Then, $\forall \epsilon >0\ \exists g:M\rightarrow \mathbb {R}$ such that $g|_{A}=f|_{A}$ and $||f-g||_{L^{\infty }}<\epsilon$ and g is smooth.

Corollary 1

This implies that there doesn't not exist such continuous retracts which in turn implies the continuous version of Brouwer's Theorem.

Proof of theorem 3

The idea is as follows: It is obviously locally true and hence it is globally true. Lets justify this.

First off, by "locally true" what we mean is that $\forall x\in M\ \exists U_{x}$ such that $\exists g_{x}:U_{x}\rightarrow \mathbb {R}$ such that the rest of the conditions are true.

Why is this locally true? Well, if $x\in A$, by smoothness of $f|_{A}$ $\exists$ smooth extension g of $f|_{A}$ to a neighborhood of x. Lets take this extension. Now if $x\notin A$ then let y = f(x) and set $g_{x}=y$. This works in a neighborhood of x.

We now want to extend this to being globally true. To do this we use a partition of unity to assemble the local property to a global property.

Consider our previous cover $\{U_{x}\}$. $\forall \alpha \exists x(\alpha )$ such that supp $\lambda _{\alpha }\subset U_{x(\alpha )}$ and $\sum \lambda _{\alpha }=1$.

Set $g(p)=\sum _{\alpha }\lambda _{\alpha }(p)g_{x(\alpha )}(p)$. Smoothness is obvious.

Now, if $p\in A$ then $g(p)=\sum _{\alpha }\lambda _{\alpha }(p)g_{x(\alpha )}(p)=\sum _{\alpha }\lambda _{\alpha }(p)f(p)=f(p)$

Now, if $p\in M$ then $||g(p)-f(p)||\leq \sum _{\alpha }\lambda _{\alpha }||g_{x(\alpha )}-f||=\sum _{\alpha :p\in U_{\alpha }}\lambda _{\alpha }||g_{x(\alpha )}-f||\leq \sum _{\alpha :p\in U_{\alpha }}\lambda _{\alpha }\epsilon =\epsilon$

Q.E.D

We can now generalize the previous theorem with the following theorem.

Theorem 4

Let N be a compact metrized manifold (i.e. we have actually specified a metric not just claimed one can exist which we always know for manifolds). Also let $A\subset M$ be closed in the manifold. Let f$:M\rightarrow N$ be a continuous function such that $f|_{A}$ is smooth. Let $\epsilon >0$ then $\exists$ a smooth $g:M\rightarrow N$ such that

1) $g|_{A}=f|_{A}$

2) $\forall p\in M,\ d(f(p),g(p))<\epsilon$.

Proof of Theorem 4

By the Whitney embedding theorem we let N be a submanifold of $\mathbb {R} ^{2n+1}$. We consider our function f from M to N but thought of as going into $\mathbb {R} ^{2n+1}$. By the previous theorem we can approximate this by a smooth function g' going into $\mathbb {R} ^{2n+1}$. The problem is that this may not take points actually onto N but near by it in the ambient space. We then want to construct from g' a smooth function g that actually goes into N as we would want. To do this we will use the idea of tubular neighbourhoods.

Definition 3

Let N be a (compact) submanifold of $\mathbb {R} ^{k}$. Then the normal bundle of N in $\mathbb {R} ^{k}$ is ${\mathcal {N}}(N)=\{(x,v)|x\in N\subset \mathbb {R} ^{k},\ v\in T_{x}(\mathbb {R} ^{k}),\ v\perp T_{x}(N)\}$ where $T_{x}(N)=\theta _{*}T_{\theta ^{-1}(x)}(N)$ from an immersion $\theta :N\rightarrow \mathbb {R} ^{k}$ (also an embedding as N is compact).

${\mathcal {N}}$ is a manifold of dimension n+ (k-n) = k. We of course would need to verify that this is in fact a manifold.

Theorem 5

Tubular Neighbourhood Theorem

$\theta |_{{\mathcal {N}}(N,\epsilon )}$ is a smooth homeomorphism of ${\mathcal {N}}(N,\epsilon )=\{(x,v):||v||<\epsilon \}$ with $\{p\in \mathbb {R} ^{k}:d(p,N)<\epsilon \}$ for small $\epsilon$.