Lecture Notes

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====== 2020-01-31, Friday ======

$$\gdef\d\partial$$

Today is our first encounter with a non-trivial theorem, the Sard theorem.\\
**Theorem(Sard)**Given a smooth map $F: M \to N$, the set of critical values of $F$ is a measure zero set in $N$. 

We recall the definitions
  * A point $p \in M$ is a //critical point// of $F$, if $dF: T_p M \to T_{F(p)} N$ is not surjective. The set of critical point is denoted as $Cr(F)$. 
  * A point $q \in N$ is a //critical value// of $F$, if $q$ is the image of a critical point. The set of critical value is denoted as $\Delta_F$, or called the discriminant locus. 
  * A subset $A \subset N^n$ is //of measure zero// (or negligible), if for any coordinate chart $(U, \varphi)$ on $N$, $\varphi(A \cap N)$ is of measure zero in $\R^n$. 

Note: we don't have a canonical measure on $N$, but we have a class smooth density on $N$, which can be written as a smooth positive function times the Lebesgue measure induced by the coordinates. 

It suffices to prove a local version of the Sard theorem, in a coordinate chart. 

**Theorem (Sard on $\R^n$) ** Let $U \subset \R^n$ be an open set, $F: U \to \R^m$ be a smooth map. Assume $n \geq m$. Then the discriminant set $\Delta_F$ is of measure zero. 

**Remark:** //We don't say anything about the measure of $Cr(F)$, in fact, it can be quite large. For example, consider a non-negative function $f: \R \to \R$, where $f$ vanishes on $[-1,1]$, then $[-1,1] \subset Cr(f)$, but $f([-1,1])$ is just a single point, $0$. //

(I am copying the proof from [[https://www3.nd.edu/~lnicolae/Lectures.pdf|Nicholascu's note, page 30]], which was originally due to Milnor and Pontryagin.)

We denote $Cr^k(F) \subset Cr(F)$ denote the subset of points in $U$ such that all partial derivatives of $F$ up to order $k$ vanishes (check: this notion does not depends on the choice of coordinates). We obtain a descreasing filtration of closed sets 
$$ Cr(F) \supset Cr^1(F) \supset Cr^2(F) \supset Cr^3(F) \supset \cdots $$

**Note:** A point $p \in Cr(F)$ means $dF(p)$ is not surjective, a point $p \in Cr^1(F)$ means $dF(p)=0$. 

We prove by induction. The case $n=0$ is trivial. We assume the case is true for any $n' < n$ and any $m \leq n'$. The inductive step is divided into 3 steps

   - Step 1: Show that $F(Cr_F - Cr^1_F)$ is neligible .
   - Step 2: Show that $F(Cr^k_F - Cr^{k+1}_F)$ is neglible for all $k \geq 1$
   - Step 3: Show that the set $F(Cr^k_F)$ is neglible for some sufficiently large $k$. 

Step 1: Set $Cr'_F = Cr_F - Cr_F^1$. We will show that there exists a countable open cover $\gdef\cO{\mathcal O} \{\cO_j\}_{j=1}^\infty$ of $Cr'_F$, such that $F(\cO_j \cap Cr'_F)$ is neglibile for all $j$. Since $Cr'_F$ is contained in a second countable space $\R^n$, every open cover has a countable refinement, hence suffice to prove that for each $u \in Cr'_F$, there is a neighborhood $\gdef\cN{\mathcal N} \cN$, such that $F(\cN \cap Cr'_F)$ is negligible. 

Suppose $p \in Cr'_F$, since $dF(p)$ is not identifically zero, we may choose coordinate chart $(U, (x_i))$ 
centered around $p$ and $(V, (y_j))$ centered around $F(p)$, such that in this coordinate $F(x_1,\cdots,x_n) = (F_1(x), \cdots, F_m(x))$, and 
$$ F_1(x) = x_1. $$
( Why we can have such coordinates? We can choose coordinate $(V, (y_j))$ first, and consider $y_j \circ F$ on $F^{-1}(V)$ for all $j$, there exists at least one $j$, such that $d(y_j\circ F)(p) \neq 0$, otherwise $dF(p)=0$. Then, wlog, assume $j=1$, and define $x_1 = y_1 \circ F$. )

Next, for every $t \in \R$, set 
$$\cN_t = \{ x \in \cN \mid x_1 = t\} $$
and define 
$$ G_t: \cN_t \to \R^{m-1}, p \mapsto (F_2(p), \cdots, F_m(p)) $$
Observe that
$$ F(\cN \cap Cr'_F) = \bigcup_{t} \{t\} \times G_t(Cr_{G_t}) $$
By the induction hypothesis, we have the statement when the source dimension is $n-1$. Hence the (m-1) Lebesgue measure of $G_t(Cr_{G_t})$ is zero. By Fubini theorem
$$ \mu_{m} (F(\cN \cap Cr'_F)) = \int \mu_{m-1}( G_t(Cr_{G_t})) dt = 0. $$

Step 2: Set $Cr^{(k)}_F := Cr^k_F - Cr^{k+1}_F$, and suppose $p \in Cr^{(k)}_F$. We may choose local coordinate $s_1, \cdots, s_n$ around $p$ and $y_1, \cdots, y_m$ around $F(p)$, such that $s_i(p)=0$ for all $i$ and $y_j(F(p))=0$ for all $j$. And furthermore, we assume that
$$ \frac{\d^{k+1} y_1}{\d s_1^{k+1}}(p) \neq 0 $$
Then, we define $x_1 = \frac{\d^{k} y_1}{\d s_1^{k}}$ and $x_2=s_2,\cdots,x_n=s_n$. We choose $\cN$ to be a small enough neighborhood around $p$, such that $(x_i)$ forms a coordinate. Then $Cr^{k}_F \cap \cN$ is contained in the hyperplane $x_1=0$. (indeed, if $x_1 \neq 0$, then one $k$-th derivative of $F$ is non-zero, hence the point is not in $Cr^k_F$ by definition).

Define 
$$ G: \cN \cap \{x_1=0\} \to \R^m, \quad G(p) = F(p) \forall p\cN \cap \{x_1=0\} $$
Then 
$$ Cr^k_F \cap \cN = Cr_G^k, \quad F(Cr^k_F \cap \cN ) = G(Cr_G^k) $$
By induction hypothesis, $G(Cr_G^k)$ is negligible in $\R^m$, hence $F(Cr^k_F \cap \cN)$ is neglibile. By covering $Cr^{(k)}_F$ by such open cover, and take countable refinement, we can conclude that $F(Cr^{(k)}_F)$ is negligible. 

Step 3 (the key step): Suppose $k > n / m - 1$. We will show that $F(Cr^k_F)$ is neglibile. More precisely, for every compact subset $S \subset U$, we will show that $F(S \cap Cr^k_F)$ is negligible. 

From Taylor expansion around points in $S \cap Cr^k_F$, we know there exists $0< r_0 < 1$ and $\lambda_0>0$, depending only on $S$, such that if $C$ is a cube with sides $r < r_0$ and intersects $Cr^k_F \cap S$. Then 
$$ diam(F(C)) < \lambda_0 r^{k+1} $$
where for any set $A \subset \R^m$,  the diameter is defined as 
$$diam(A) = sup \{|a_1 - a_2|, a_1, a_2 \in A \}.$$ 

( Recall the Taylor expansion formula, if $f: \R^n \to \R$ is a smooth function, then for any $k \geq 0$, there exists $r>0$, and $C > 0$, such that for any $|x|<r$, we have 
$$ f(x_1, \cdots, x_n) = f(0) + \sum_{1 \leq |\alpha| \leq k } f^{(\alpha)}\frac{x_1^{\alpha_1}\cdots x_n^{\alpha_n}}{\alpha_1! \cdots \alpha_n!} + R_k(x)$$
where the remainder
$| R_k(x) | < C |x|^{k+1}$. Apply this remainder estimate to each function $F_i$, one can get the diameter)

Hence, the Lebesgue measure of the image is 
$$ \mu_m(F(C)) < C_1 r^{m(k+1)} = C_1 \mu_n(C)^{m(k+1)/n} $$

Now, we cover $Cr^k_F \cap S$ by finitely many cubes $\{C_l\}_{l=1}^N$, of edges $r< r_0$, with disjoint interiors. For each positive integer $P$, we may subdivide each $C_l$ into $P^n$ many subcubes of equal sizes. For every sub-cube $C_l^\sigma$ ** that intersects $Cr^k_F$**, we have 
$$ \mu_m(F(C_l^\sigma)) \leq C_1 \mu_n(C_l^\sigma)^{m(k+1)/n} = \frac{C_1}{P^{m(k+1)}}\mu_n(C_l) $$
Hence
$$\mu_m(F(C_l \cap Cr^k_F)) = \sum_{\sigma} \mu_m(F(C_l^\sigma \cap Cr^k_F)) \leq P^{n-m(k+1)} \mu_n(C). $$
Now, we may send $P$ to $\infty$, and conclude that $\mu_m(F(C_l \cap Cr^k_F)) = 0$.
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