# 0708-1300/Class notes for Tuesday, October 16

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## Dror's Computer Program for C+$\alpha$ C

Our handout today is a printout of a Mathematica notebook that computes the measure of the projection of $C\times C$ in a direction $t\in[0,\pi/2]$ (where $C$ is the standard Cantor set). Here's the notebook, and here's a PDF version. Also, here's the main picture on that notebook:

## Typed Class Notes - First Hour

Today's Agenda:

Proof of Sard's Theorem. That is, for $f:M^m\rightarrow N^n$ being smooth, the measure $\mu$(critical values of f) = 0.

Comments regarding last class

1) In our counterexample to Sard's Theorem for the case of $C^1$ functions it was emphasized that there are functions f from a Cantor set C' to the Cantor set C. We then let $g(x,y) = f(x) + f(y)$ and hence the critical values are $C+C = [0,2]$ as was shown last time. The sketch of such an f was the same as last class.

Furthermore, in general we can find a $C^n$ such function where we make the "bumps" in f smoother as needed and so $f(C') = C''$ where $C''$ is a "very thin" Cantor set. But now let $g(x,y,z,\ldots) = f(x) + f(y) + f(z) +\ldots$ which will have an image of $C'' + C'' + C'' + \ldots =$ an interval.

2) The code, and what the program does, for Dror's program (above) was described. It is impractical to describe it here in detail and so I will only comment that it computes the measure of $C+\alpha C$ for various alpha and that the methodology relied on the 2nd method of proof regarding C+C done last class.

Proof of Sard's Theorem

Firstly, it is enough to argue locally.

Further, the technical assumption about manifolds that until now has been largely ignored is that our M must be second countable. Recall that this means that there is a countable basis for the topology on M.

As a counterexample to Sard's Theorem when M is NOT second countable consider the real line with the discrete topology, a zero dimensional manifold, mapping via the identity onto the real line with the normal topology. Every point in the real line is thus a critical point and the real line has non zero measure.

We can restrict our neighborhoods so that we can assume $M=\mathbb{R}^m$ and $N=\mathbb{R}^n$

The general idea here is that if we consider a function g=f' that is nonzero at p but that f is zero at p, the inverse image is (in some chart) a straight line (a manifold). As such, we will inductively reduce the dimension from m down to zero. For m=0 there is nothing to prove. Hence we assume true for m-1.

Now, set $D_k = \{p |$ all partial derivities of f of order $\leq k\}$ for $k\geq 1$

Set $D_0 = \{p\ |\ df_p$ is not onto }. This is just the critical points.

Clearly $D_0\supset D_1\supset\ldots\supset D_i\supset\ldots\supset D_m$

We will show by backwards induction that $\mu(F(D_k)) = 0$

Comment:

We have not actually defined the measure $\mu$. We use it merely to denote that $F(D_k))$ has measure zero, a concept that we DID define.

Claim 1

$F(D_m)$ has measure 0.

Proof

W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size $\epsilon$ in 1D becomes a smaller set of size $\epsilon^2$ in 2D etc. More precisely, for $f = (f_1,\ldots,f_n)$, $f(D_n)\subset f_1(D_m)\times{R}^{n-1}$. Applying the proposition that if A is of measure zero in $\mathbb{R}$ then $A\times\mathbb{R}^{n-1}$ is measure zero in $\mathbb{R}^n$ we now see that assuming n=1 is justified.

Reminder

Taylor's Theorem: for smooth enough $g:\mathbb{R}\rightarrow\mathbb{R}$ and some $x_0$ then $g(x) = \sum_{j=0}^{m} \frac{g^{(j)}(x_0)}{j!}(x-x_0)^j + \frac{g^{(m+1)}(t)}{(m+1)!}(x-x_0)^{m+1}$ for some t between x and $x_0$.

For $x_0\in D_m$ all but the last term vanishes and so we can conclude that f(x) is bounded by a constant times $(x-x_0)^{m+1}$.

Now let us consider a box B in $\mathbb{R}^m$ containing a section of $D_m$. We divide B into $C_1\lambda^m$ boxes of side $1/\lambda$.

By Taylor's Theorem, $f(B_i)\subset$ of an interval of length $C_2\frac{1}{\lambda^{m+1}}$ where the constant $C_2$ is determined by Taylor's Theorem. Call this interval $I_i$

Hence,

$f(D_m)\subset\bigcup_{i:B_i\cap D_m \neq 0} f(B_i)\subset \bigcup_{i:B_i\cap D_m \neq 0}I_i$

But $\sum_{i} length(I_i)\leq C_1\lambda^m C_2\frac{1}{\lambda^{m+1}}$ which tends to zero as $\lambda$ tends to infinity.

Q.E.D for Claim 1

Claim 2

$f(D_k)$ has measure zero for $k\geq 1$. We just proved this for k=m.

Now, W.L.O.G. $D_{k+1}$ is the empty set. If not, just consider $M^m - D_{k+1}$ which is still a manifold as $D_{k+1}$ is closed (as it is determined by the "closed" condition that a determinant equals zero)

So, there is some kth derivative g of f such that $dg\neq 0$.

So $D_k\subset g^{-1}(0)$ but $g^{-1}(0)$ is at least locally a manifold of dimension 1 less. So, $f(D_k)\subset f(D_k\cap g^{-1}(0)$ which has measure zero due to our induction hypothesis.

Q.E.D for Claim 2

Claim 3

$f(D_0)$ is of measure zero.

Recall $D_0$ is defined differently from the $D_k$ and so requires a different technique to prove.

W.L.O.G. lets assume that $D_1$ is the empty set. So, some derivative of f is not zero. W.L.O.G. $\frac{\partial f_1}{\partial x_1}$ is non zero near some point p. We can simply move to a coordinate system where this is true.

The idea here is to prove that the intersection with any "slice" has measure zero where we will then invoke a theorem that will claim everything has measure zero.

So, let U be an open neighborhood of a point $p\in M$. Consider $f_1:U\rightarrow\mathbb{R}$ and let $df_1$ be onto. Using our previous theorem for the local structure of such a submersion W.L.O.G. let us assume $f_1:\mathbb{R}^m\rightarrow\mathbb{R}$ via $(x_1,\ldots,x_m)\mapsto x_1$. That is, $f_1 = x_1$.

Our differential df then is just the matrix whose first row consists of $(1,0,\ldots,0)$. Then df is onto if the submatrix consisting of all but the first row and first column is invertible.

For notational convenience let us say $f =(f_1,f_{rest})$.

Now define $U_t = \{t\times\mathbb{R}^{m-1}\}$ Also lets denote "critical points of f by CP(f) and "critical values of f" by CV(f)

Claim 4

The $CP(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CP(f_{rest}|_{U_t})$

$\Rightarrow$

$CV(f) = \bigcup_{t\in\mathbb{R}}\{t\}\times CV(f_{rest}|_{U_t})$.

But $CV(f_{rest}|_{U_t})$ has measure zero by our induction.

Lemma 1

If $A\subset I^2$ is closed and has $\mu(A\cap(\{t\}\times I)) = 0\ \forall t$ then $\mu(A)=0$.

Proof

Note: We prove this significantly differently then in Bredon

Sublemma

If $\{t\}\times U$ for an open U cover $\{t\}\times I\cap A$ for a closed A then $\exists\epsilon>0$ such that $(t-\epsilon,t+\epsilon)\times U\supset (t-\epsilon,t+\epsilon)\times I\cap A$

Indeed, let $d:A-(I\times U)\rightarrow\mathbb{R}$ be $d(x) = |x_1-t|$ then d is a continuous function of a compact set and so obtains a minimum and since d>0 then $min(d)>0 \rightarrow d>\epsilon>0$. But this $\epsilon$ works for the claim. Q.E.D

The rest of the proof of Lemma 1, and of Sard's Theorem will be left until next class