\classheader{2013-02-13}

\begin{theorem}[Ehresmann fibration theorem]
Let $f:X\to Y$ be a $\Cinfty$ proper submersion. Then $f$ is a $\Cinfty$ fiber bundle.
\end{theorem}
\begin{proof}
Let $y_0\in Y$, and let $F=f^{-1}(y_0)$, which is compact because $f$ is proper. We may regard $X$ as a closed submanifold of $\R^k$ for some $k$. Thus, we have $F\subset X\subset\R^k$, and we apply the tubular neighborhood theorem to $i:F\hookrightarrow\R^k$. Thus, we get an open neighborhood $U$ of $F$ in $\R^k$ and a $\Cinfty$ retraction $r:U\to F$. Now replace $(U,r)$ by $(U\cap X,r|_{U\cap X})$.

We now have that $U$ is a neighborhood of $F$ in $X$, and $r:U\to F$ is still $\Cinfty$. Consider the map $(f|_U\times r):U\to Y\times F$. We claim that this induces an isomorphism on tangent space at all $x\in F\subset U$. We see that the sequence
\begin{center}
\begin{tikzcd}
0 \ar{r} & T_x(f^{-1}(f(x))) \ar{r} & T_x(U) \ar{r}{f'(x)} & T_{f(x)}(Y) \ar{r} & 0
\end{tikzcd}
\end{center}
is exact for all $x\in X$ because $f$ is a submersion, and in particular, for $x\in F=f^{-1}(y_0)$, we have that the sequence
\begin{center}
\begin{tikzcd}
0 \ar{r} & T_xF \ar{r} & T_xU \ar{r} & T_{y_0}Y\ar{r} & 0
\end{tikzcd}
\end{center}
is exact. Thus, the claim is now equivalent to the claim that \begin{center}
\begin{tikzcd}[column sep=0.9in]
T_xF\ar{r}{r'(x)|_{T_xF}} & T_x F
\end{tikzcd}
\end{center}
is an isomorphism. Why? If we have a map of vector spaces
\begin{center}
\begin{tikzcd}
V \ar{r}{(\alpha,\beta)} & A\oplus B
\end{tikzcd}
\end{center}
and we know that $\alpha$ is onto, then $(\alpha,\beta)$ is an isomorphism if and only if $\beta|_{\ker(\alpha)}:\ker(\alpha)\to B$ is an isomorphism. But $r'(x)|_{T_xF}$ is the identity, so this is true in our case.

Note that the set
\[V=\{z\in U\mid f|_U\times r\text{ is an isomorphism of tangent spaces }T_zX\to (\cdots)\}\]
is open, so $X\setminus V$ is closed. Thus $f(X\setminus V)$ is closed, because $f$ is proper. Its complement necessarily contains $f^{-1}(U_{y_0})$, where $U_{y_0}$ is a neighborhood of $y_0$ in $Y$. In other words,
\begin{center}
\begin{tikzcd}
f^{-1}(U_{y_0}) \ar{r}{f|\times r} & U_{y_0}\times F
\end{tikzcd}
\end{center}
induces an isomorphism of tangent space and is injective on $F\subset f^{-1}(U_{y_0})$. It follows that $f\times r$ is injective on $f^{-1}(U_{y_0}')$ for some neighborhood $U_{y_0}'\subset U_{y_0}$ of $y_0$.
\end{proof}

Here is a variant of the theorem.

\begin{theorem}
Let $(X,\boundary X)$ be a $\Cinfty$ manifold with boundary. Suppose that $f:X\to Y$ is a $\Cinfty$ proper submersion such that $f|_{\boundary B}$ is also a submersion. Then $f$ is a $\Cinfty$ fiber bundle.
\end{theorem}


\subsection*{Existence and uniqueness of solutions of ODEs}

A reference for this is Warner's \textit{Foundations of Differentiable Manifolds and Lie Groups}.

\begin{definition}
Let $M$ be a $\Cinfty$ manifold. We say that $A$ is a continuous vector field on $M$ when $A$ is a continuous section $A:M\to TM$ the tangent bundle map $p:TM\to M$. A $\Cinfty$ vector field is just a $\Cinfty$ section.
\end{definition}

\begin{definition}
An integral curve of a vector field $v$ on $M$ is a $C^1$ curve $\gamma:(a,b)\to M$ such that $\gamma'(t)=v(\gamma(t))$ for all $t\in (a,b)$.
\end{definition}

\begin{theorem}[Existence and uniqueness]
Let $v$ be a $C^1$ vector field on $M$. Let $x_0\in M$. Then there exists an integral curve $\gamma:U\to M$ for some connected open neighborhood $U\subseteq\R$ of $0$ such that $\gamma(0)=x_0$. If $\gamma_i:U_i\to M$ for $i=1,2$ are integral curves of $v$ such that $\gamma_1(0)=\gamma_2(0)=x_0$, then there is some open neighborhood $U\subset U_1\cap U_2$ of 0 such that $\gamma_1|_U=\gamma_2|_U$.
\end{theorem}
\begin{corollary}
In fact, $\gamma_1|_{U_1\cap U_2}=\gamma_2|_{U_1\cap U_2}$.
\end{corollary}
\begin{proof}[Proof of corollary]
Both $U_1$ and $U_2$ are connected open neighborhoods of $0$ so it suffices to check the claim on the positive side and on the negative side. Let $U_i\cap [0,\infty)=[0,b_i)$ for $i=1,2$. We may assume WLOG that $b_1\leq b_2$. Let
\[c=\sup\{t\geq 0\mid \gamma_1(t)=\gamma_2(t)\}.\]
If $c=b_1$, then we are done. If $c<b_1$, then continuity implies that $\gamma_1(c)=\gamma_2(c)$. Consider the curves $\gamma_1(t+c)$ and $\gamma_2(t+c)$, which are both integral curves for $v$ with the same initial value, i.e. they agree when $t=0$. But the theorem implies that, for any integral curve $\delta$ of $v$ with initial value $\gamma_1(c)=\gamma_2(c)$, we have that $\delta(t)=\gamma_1(t+c)$ for all $t$ in a neighborhood of 0, and $\delta(t)=\gamma_2(t+c)$ for all $t$ in a neighborhood of 0, and therefore $\gamma_1(t)=\gamma_2(t)$ for all $t\in (c-\epsilon,c+\epsilon)$ for some $\epsilon>0$. But this contradicts the construction of $c$; thus $c=b_1$ and $\gamma_1|_{U_1\cap U_2\cap [0,\infty)}=\gamma_2|_{U_1\cap U_2\cap [0,\infty)}$. Now do the same for $(-\infty,0]$.

Now consider the set of all pairs
\[Z=\left\{(U,\gamma)\;\middle\vert\;\begin{array}{c}
U\text{ is an open connected neighborhood of }0\text{ in }\R,\\
\gamma \text{ is an integral curve of }v\text{ and }\gamma(0)=x_0
\end{array}\right\}.\]
This partially ordered set has a greatest element; let $V=\bigcup\{U\mid \text{there is some }(U,\gamma)\in Z\}$, we have $\gamma:V\to M$, apply Zorn's lemma as usual, etc.
\end{proof}

\begin{examples}
$\text{}$

\begin{itemize}
\item Let $M=\R^n$, and let $v$ be a constant vector field. Then for any $x\in M$, the curve $\gamma:\R\to M$ defined by $\gamma(t)=x+tv$ is an integral curve.
\item Let $M=\R^n$, and let $S:\R^n\to\R^n$ be a linear transformation. Define the vector field $v(x)=Sx$ for all $x\in\R^n$. As you've seen in Victor Ginzburg's class, for any $A\in\M_n(\R)$ we can define
\[\exp(A)=\sum_{k=0}^\infty\frac{A^k}{k!},\]
which satisfies $\exp(A+B)=\exp(A)\exp(B)$ when $AB=BA$, and $\|\exp(A)\|\leq \exp(\|A\|)$. In particular, $\exp(A)\in \GL_n(\R)$, and $\exp((t_1+t_2)A)=\exp(t_1A)\exp(t_2A)$.

The integral curve of $v(x)=Sx$ with $\gamma(0)=x$ is simply the curve $\gamma(t)=(\exp(tS))x$.
\end{itemize}
\end{examples}

\begin{proof}[Proof of existence and uniqueness]
We use the contraction principle, which states that if $X$ is a complete metric space and $0\leq\lambda<1$, and $T:X\to X$ is a continuous function such that $d(Tx_1,Tx_2)\leq\lambda d(x_1,x_2)$ for all $x_1,x_2\in X$, then there is a unique fixed point of $T$ (this follows simply by observing that $\{T^nx\}_{n\in\N}$ is a Cauchy sequence for any $x\in X$).

WLOG, let $x_0=0$, and let $M=$ an open set $\Omega\subseteq\R^n$ which is a neighborhood of 0. We have our $C^1$ vector field $v$ on $M$.

We want to construct a $\gamma:(-c,c)\to\Omega$ such that $\gamma(0)=0$ and such that $\gamma'(t)=v(\gamma(t))$ for all $t\in(-c,c)$. Thus, we want
\[\gamma(t)=\int_0^tv(\gamma(t))\,dt.\]
Let $r>0$ such that $\overline{B_r(0)}=\{x\in\R^n\mid \|x\|\leq r\}\subset\Omega$.

Let $X$ be the space of continuous functions $[-c,c]\to \overline{B_r(0)}$ such that $\gamma(0)=0$. We give $X$ the metric induced by the sup norm.

For any $\gamma\in X$, define
\[(T\gamma)(t)=\int_0^tv(\gamma(t))\,dt.\]
We want $T$ to be a well-defined map $X\to X$, and to be a contraction.

Because $\overline{B_r(0)}$ is compact and $v$ is continuous, we have that $v(\overline{B_r(0)})\subseteq B_C(0)$ for some $C$.

\begin{itemize}
\item[(i)] By taking $c$ to be sufficiently small, we have $cC\leq r$. This implies that $T$ is a well-defined map $T:X\to X$, because $t\in[-c,c]$ and $v(\gamma(t))\in\overline{B_r(0)}$ implies that
\[|(T\gamma)(t)|=\left|\int_0^t v(\gamma(t))\,dt\right|\leq cC\leq r,\]
so that $T\gamma\in X$.
\item[(ii)] Now we want to show that $T$ is a contraction. For any $\gamma_1,\gamma_2\in X$, we have
\[|(T\gamma_1)(t)-(T\gamma_2)(t)|=\left|\int_0^t v(\gamma_1(h))-v(\gamma_2(h))\,dh\right|\leq\int_0^t\|v'\|_\infty\|\gamma_1(h)-\gamma_2(h)\|\,dh\leq cM\|\gamma_1-\gamma_2\|. \]
By shrinking $c$ further if necessary, we can make $cM\leq \lambda<1$, where $M$ is defined by
\[M=\sup_{\substack{x\in\R^n\\ \|x\|\leq r}}\{\|v'(x)\|\},\]
where $\|v'(x)\|$ denotes the operator norm of $v'(x):\R^n\to\R^n$.\qedhere
\end{itemize}

\end{proof}












