# Difference between revisions of "0708-1300/Class notes for Tuesday, October 23"

Announcements go here

## Dror's Notes

• You're all invited to my talk today at 12, "Non-Commutative Gaussian Elimination and Rubik's Cube".
• Today's office hours will go 1-2.
• Our handout today is a printout of a Mathematica notebook that demonstrates a "space-filling" Peano curve. Here's the notebook, and here's a PDF version. Also, here's the main picture on that notebook:

## Typed Notes

### First Hour

Diversion

It is noted that there are no smooth curves that cover the plane. However, there ARE continuous curves that cover the plane.

As an example we consider the continuous function from the unit interval to the unit square that is defined iteratively that looks like the function above.

The construction is done as follows:

$f_0$ draws a diagonal line from the bottom left corner to the top right corner of the unit square. For $f_1$ one breaks the unit interval into 7 sections. The map $f_1$ takes the 1st, 3rd, 5th, 7th sections respectively to a diagonal line in the bottom left, bottom right, top left and top right subboxes respectively where the diagonal line goes from the bottom left to the top right corner of each subbox. The 2nd, 4th and 6th "filler" sections of the unit interval simply draw the lines that map the end of one diagonal to the start of the other. The $f_i$ is defined iteratively. See the diagram above to see what this looks like.

We note that this is obviously continuous and we get uniform convergence to a continuous function into the unit square. As every point in the square get approached arbitrary close to a point in the image of one of the iterates of the function, compactness tells us the entire square is covered.

Definition

A general definition of "locally something" is typically that every point has a neighborhood in which this something property holds. Or perhaps a neighborhood basis where this property holds.

Hence, precisely:

1) A cover $\{U_{\alpha}\}$ is locally finite if every point has a neighborhood V such that V intersects only finitely many $U_{\alpha}$'s.

2) A space is paracompact if:

a) Every open cover has a locally finite refinement

b) If a cover $\{U_{\alpha}\}$ is locally finite then $\exists\ \{V_{\alpha}\}$ so that $\{V_{\alpha}\}$ still covers the space but that $\bar{V_{\alpha}}\subset U_{\alpha}$

Note: Manifolds are paracompact

Whitney Embedding Theorem

We recall the 3 steps to prove this theorem mentioned last class. It is noted that step three will actually be broken up into two steps:

3a: For an arbitrary $M^m$ we can embed $M^m$ in $\mathbb{R}^{4m+3}$

3b: We can then embed it in $\mathbb{R}^{2m+1}$

Proof of Theorem

We had previously proved most of part 1, but what we still had to show was that such partitions of unity actually exist.

Reminder of Definition:

A partition of unity subordinate to $\{U_{\alpha}\}$ is a collection of functions $\lambda_{\alpha}:M\rightarrow\mathbb{R}$ such that:

1) Supp $\lambda_{\alpha}\subset U_{\alpha}$

2) $\sum_{\alpha} \lambda_{\alpha} = 1$

Claim: If $\phi:U\rightarrow\mathbb{R}^n$ is a chart and $K\subset U$ is compact then we can find a function $\lambda:M\rightarrow\mathbb{R}$ that is compact and $K\subset$ supp $\lambda \subset U$

Proof of Claim

Because we are inside a chart, it is enough to just do this in $\mathbb{R}^n$.

For every $p\in K$ we can find a radius r such that $B_{r(p)}(p)\subset U$. By compactness we can take only finitely many such p's. Hence, $\{B_{r(p_i)}(p_i)\}$ cover K.

We want to put a bump function of each ball and sum them up to give us our $\lambda$.

Let $f_r(x) = e^{\frac{1}{x^2-r^2}}$ for x<r and 0 otherwise.

Now let $\lambda(p):=\sum_i f_{r(p_i)}(d^2(p,p_i))$

Q.E.D

Theorem

On a manifold, given an open cover, you can find a partition of unity subordinate to a locally finite refinement of it.

Proof

WLOG, the cover is by charts and each one is bounded and the cover is locally finite $\{U_{\alpha}\}$

By paracompactness, find $V_{\alpha}\subset U_{\alpha}$ such that $\bar{V_{\alpha}}\subset U_{\alpha}$ and $\cup V_{\alpha} = M$. By the previous claim can find $\bar{V_{\alpha}}\subset$ supp $\lambda^t_{\alpha}\subset U_{\alpha}$

Now consider $\sum \lambda_{alpha}^t =: \lambda^t$. This is a finite sum.

By local finiteness, it is smooth and so we define $\lambda_{\alpha} :=\frac{\lambda^{t}_{\alpha}}{\lambda}$

Q.E.D. for part 1 of Whitney

### Second Hour

coming soon to a wikipedia near you.