\classheader{2013-01-23}

\begin{theorem}[Whitney embedding theorem]
Let $X$ and $Y$ be $\Cinfty$ manifolds, and assume that $X$ is compact. If $2\dim(X)+1\leq\dim(Y)$, then every $f_0:X\to Y$ can be approximated by an embedding.
\end{theorem}

\begin{proof}
We will give a proof in the case when $Y$ is $\R^m$, and using tubular neighborhoods it will work even when $Y$ is compact.

Let's write $Y=V$ where $V$ is a vector space of dimension $m$ (a notational aid for when we are using the vector space structure). Let $g_1,\ldots,g_r$ be $\Cinfty$ functions from $X$ to $\R$, and let $v_1,\ldots,v_m$ be a basis for $V$. Define a deformation $F:X\times S\to Y$ of $f_0$, where $S=\R^{rm}$, by
\[F(p,t)=f_0(p)+\sum_{i,j}t_{ij}g_i(p)v_j,\]
where $t=(t_{ij})\in S$.

Define $G:(X\times X\setminus\Delta X)\times S\to V\times V$ by $G(p,q,t)=(F(p,t),F(q,t))$. We want $G$ to be transverse to $\Delta V\subset V\times V$.

Let $H:V\times V\to V$ be defined by $H(v_1,v_2)=v_1-v_2$. Then $H'(v_1,v_2)$ is onto, 0 is a regular value of $H$, and $H^{-1}(0)=\Delta V$.

We see that we want to demonstrate the surjectivity of $(H\circ G)'(p,q,t)$ at all points in
\[\{(p,q,t)\mid p\neq q\text{ and }F(p,t)=F(q,t)\}.\]
Now note that
\[(H\circ G)(p,q,t)=f_0(p)-f_0(q)+\sum_{i,j}t_{ij}(g_i(p)-g_i(q))v_j.\]
Let's look at the derivative evaluated at points of the form $(0,0,?)$, as it is easiest to compute in this case, and it will suffice for surjectivity anyway.

Because the $v_j$ form a basis for $V$, a necessary and sufficient condition that
\begin{center}
\begin{tikzcd}[row sep=0in]
TS \ar{r} & V\\
(0,0,?) \ar[mapsto]{r} & (H\circ G)'(p,q,t)(0,0,?)
\end{tikzcd}
\end{center}
is surjective is that there is some $i$ such that $g_i(p)\neq g_i(q)$.

For instance, you could consider an embedding
\begin{center}
\begin{tikzcd}
X \ar[hook]{r} \ar{rd}[swap]{g_i} & \R^N \ar{d}{p_i}\\
& \R
\end{tikzcd}
\end{center}
where $N=r$.

Now, $G^{-1}(\Delta V)$ is a $\Cinfty$ submanifold of $(X\times X\setminus \Delta X)\times S$.

Let $Q:G^{-1}(\Delta V)\to S$ be the projection.

Consider the regular values of $Q$ (which are a subset of $S$). For any regular value $s$, we have
\[Q^{-1}(s)=\{(p,q)\in X\times X\mid p\neq q\text{ and }f_s(p)=f_s(q)\}\]
is a $\Cinfty$ submanifold of $X\times X\setminus\Delta X$ of dimension $2\dim(X)-\dim(Y)$. Thus, if $2\dim(X)<Y$, this set is empty.

For $f:X\to Y$ to be an embedding, we need that
\begin{itemize}
\item[(i)] $f$ is injective,
\item[(ii)] $f'(p)$ is injective for all $p\in X$,
\item[(iii)] $f|_X:X\to f(X)$ is a homeomorphism and $f(X)$ is a closed subset of $Y$,
\end{itemize}
though if $X$ is compact then (iii) is not required.

\subsection*{The tangent bundle}

The tangent bundle of $X$ is denoted by $TX$. As a set, it is equal to the disjoint union
\[TX=\coprod_{x\in X}T_xX.\]
Given an open $\Omega \subseteq\R^n$, there is a natural identification $T\Omega\cong \Omega\times\R^n$ (this is the topology and $\Cinfty$ structure on $T\Omega$).

In general, the topology on $TX$ is defined as follows. There is a natural projection $\pi:TX\to X$ defined by $\pi(T_xX)=x$ for all $x\in X$.

If $f:P\to Q$ is $\Cinfty$, then there is an induced commutative diagram
\begin{center}
\begin{tikzcd}
TP \ar{d}[swap]{\pi} \ar{r}{F} & TQ\ar{d}{\pi}\\
P \ar{r}[swap]{f} & Q
\end{tikzcd}
\end{center}
where $F$ is the map defined by $F|_{T_xP}=f'(x)$ for all $x\in P$.


Now let $X$ be a $\Cinfty$ manifold. For all charts $(\Omega,g:\Omega\to U)$ on $X$, i.e. with $g:\Omega\to U$ a diffeomorphism from an open $\Omega\subset\R^n$ to an open $U\subseteq X$, we get a commutative diagram
\begin{center}
\begin{tikzcd}
T\Omega\ar{d} \ar{r}{\text{homeo}} \ar[bend left=45]{rr}{h} & TU \ar[hook]{r} \ar{d} & TX \ar{d}\\
\Omega \ar{r}[swap]{g} & U \ar[hook]{r} & X
\end{tikzcd}
\end{center}
The $\Cinfty$ structure is transported via $h$.

Returning to our argument, we have $F:X\times S\to V$, a deformation of $f_0$. This gives rise to $\widetilde{F}:(TX)\times S\to TV$, a $\Cinfty$ map.

Let $w\in T_pX$ be such that $\widetilde{F}(w,s)=F'(p,s)(w,0)$.

Now define $T'X$ to be the complement of the zero section of $TX$; in other words,
\[T'X=\coprod_{x\in X}(T_xX\setminus\{0\}).\]
Clearly, $T'X$ is an open subset of $TX$.

We claim that $\widetilde{F}:(T'X)\times S\to V$ has $0\in V$ as a regular value.

Let's assume the claim for now. Then $M:=\widetilde{F}^{-1}(0)$ is a $\Cinfty$ submanifold with
\[2\dim(X)-\dim(S)-\dim(\widetilde{F}^{-1}(0))=\dim(V)=\dim(Y).\]
Consider the map $Q':M\to S$, and let $s$ be a regular value of $Q'$. Then
\[\dim((Q')^{-1}(s))=\dim(M)-\dim(S)=2\dim(X)-\dim(Y).\]
Under the assumption that $2\dim(X)\leq \dim(Y)$, the fiber is forced to be empty,
\[(Q')^{-1}(s)=\{(p,w)\in TX\mid 0\neq w\in T_pX \text{ for }p\in X,\text{ and }f_s'(p)w=0\}.\]
\fbox{If $2\dim(X)\leq \dim(Y)$, then $f_s'(p):T_pX\to T_{f_s(p)}V$ is injective for all $p\in X$.}

\begin{proof}[Proof of claim]
Fix $p\in X$ and $0\neq w\in T_pX$. We have the map $\widetilde{F}|(p,w)\times S\to V$. We need to find an $i$ such that $g_i'(p)w\neq 0$.

Taking the derivative of $H\circ G$, we get
\[f_0'(p)w+\sum_{i,j}t_{ij}(g_i'(p)w)v_j.\]
In our case, $j$ is a $\Cinfty$ embedding
\begin{center}
\begin{tikzcd}
X \ar[hook]{r}{j} & \R^N \ar{d}{p_i}\\
& \R
\end{tikzcd}
\end{center}
$0\neq w\in T_pX$ implies that $j'(p)w\neq 0$, which implies that there is some $i$ such that $(p_i\circ j)'(p)w\neq 0$, i.e. $g_i'(p)\neq 0$.
\end{proof}
Thus, we are done.
\end{proof}


Our proof shows a form of the Whitney immersion theorem (at least when $Y=\R^n$), then every $f_0:X\to Y$ can be approximated by $f:X\to Y$ which is an immersion (i.e. $f'(p):T_pX\to T_{f(p)}Y$ is injective for all $p\in X$) if $2\dim(X)\leq \dim(Y)$.

Next time we'll do Sard's theorem and maybe a bit more about vector bundles.

\begin{corollary}[of Sard's theorem]
Let $M$ and $N$ be $\Cinfty$ smooth manifolds (we also assume second countable and $T_2$), let $f:M\to N$ be a $\Cinfty$ map, and assume that $\dim(M)<\dim(N)$. Then $f$ is \textbf{not} onto. If $a\in N$ is a regular value of $f$, then $f^{-1}(a)$ manifold of dimension $\dim(M)-\dim(N)$.
\end{corollary}