\classheader{2013-02-27}

Let $M$ be a $\Cinfty$ manifold and $v$ a $\Cinfty$ vector field on $M$. Let $\phi_t(x)=\gamma_x(t)$ be the integral curve for $v$ with $\gamma_x(0)=x$. Let $\omega$ be any object attached to the manifold, such as for example a section of $TM^{\otimes m}\otimes T^*M^{\otimes n}$. Then the Lie derivative of $\omega$ with respect to $v$ makes sense:
\[L_v\omega=\frac{d}{dt}\phi_t^*\omega\bigg|_{t=0}\]
In particular, $L_vw$ is defined when $w$ is a vector field.

\begin{proposition}
For all vector fields $v,w$ on $M$, we have $L_vw=[v,w]$.
\end{proposition}
\begin{lemma}[Leibniz rule for sections of bundles]
Let $v$ be a vector field.
\begin{enumerate}
\item $L_v(\omega\wedge \eta)=(L_v\omega)\wedge \eta+\omega\wedge L_v(\eta)$, where $\omega$ is a $k$-forma and $\eta$ is an $\ell$-form
\item $L_vi_w\theta=i_{L_v(w)}\theta+i_wL_v\theta$< where $w$ is a vector field, and $\theta$ is a $k$-form
\item $v(\theta(w))=\theta(L_v(w)+(L_v\theta)(w)$, where $\theta$ is a 1-form (this is just a special case of 2)
\end{enumerate}
\end{lemma}
\begin{proof}
Let $V_1,V_2,V_3$ be vector bundles on $M$, and let $B$ be a bilinear map
\begin{center}
\begin{tikzcd}
V_1\times_M V_2 \ar{rr}{B} \ar{rd} & & V_3 \ar{ld}\\
 &M
\end{tikzcd}
\end{center}
i.e. $B(x):V_1(x)\times V_2(x)\to V_3(x)$ is bilinear for all $x\in M$. Let $s_t^1$, $s_t^2$ be families of $\Cinfty$ sections of $V_1$ and $V_2$ respectively, indexed by $t\in(-\epsilon,\epsilon)$. Let $p_1$ be the projection $p_1:M\times(-\epsilon,\epsilon)\to M$, so that each $s_i$ is a secton of $p_1^* V_i$. Then
\[\frac{d}{dt} B(s_t^1,s_t^2)=B\left(\frac{d}{dt}\,s_t^1,s_t^2\right)+B\left(s_t^1,\frac{d}{dt}\,s_t^2\right).\]
How will we apply this - we want to choose $s_i=\phi_t^*(?)$.

Let $V_1=TM$, $V_2=\Lambda^kT^*M$, $V_3=\Lambda^{k-1}T^*M$, and let $B(x):T_xM\times \Lambda^kT_x^*M\to \Lambda^{k1-}T_x^*M$ be defined by $B(x)(\omega,\theta)=i_{\omega}(\theta)$.

Part 2 is then an application of the Leibniz rule
\[i_{v^*}(\omega\wedge\eta)=i_{v^*}(\omega)\wedge\eta+(-1)^{\deg(\omega)}\omega\wedge i_{v^*}(\eta)\]
where $v^*\in V^*$ and $\omega\in\Lambda^kV$, and 3 is just 2 for $k=1$.

Given $\theta=df$, where $f:M\to\R$ is a $\Cinfty$ map, then
\[\theta(w)=(df)(w)=w(f)\]
That $v(\theta(w))=v(w(f))$ is just the left side of 3. But
\[\theta(L_vw)=(L_vw)(f),\]
hence
\[L_v(\theta)=L_v(df)=dL_vf=d(v(f)),\]
hence
\[(L_v\theta)(w)=w(v(f)).\]
Now 3 reads as
\[v(w(f))=w(v(f))+(L_vw)(f),\]
i.e.
\[(L_vw)(f)=v(w(f))-w(v(f))=[v,w](f)\]
for all $\Cinfty$ maps $f:M\to\R$.
\end{proof}


\begin{corollary}[Special case of Cartan's formula]
Let $\omega$ be a 1-form, and let $v_1$ and $v_2$ be vector fields. Then
\[d\omega(v_1,v_2)=v_1(\omega(v_2))-v_2(\omega(v_1))-\omega([v_1,v_2])\]
\end{corollary}
\begin{remark}
Note that we can identify $\Lambda^k T_x^*M$ with $(\Lambda^k T_xM)^*$ as follows: given $\omega\in \Lambda^kT_x^*M$, we define
\[\omega(v_1,v_2,\ldots,v_k)=i_{v_k}i_{v_{k-1}}\cdots i_{v_1}\omega\in \Lambda^k T_x^*M\in\R\]
for $v_1,\ldots,v_k\in T_xM$. 
\end{remark}
\begin{proof}
We have that $L_v=i_vd+di_v$. Thus,
\[i_{v_1}d\omega=L_{v_1}\omega-d(i_{v_1}\omega),\]
so that
\begin{align*}
d\omega(v_1,v_2)&=(i_{v_1}d\omega)v_2=(L_{v_1}\omega)v_2-v_2(\omega(v_1))\\
&=L_{v_1}(\omega(v_2))-\omega(L_{v_1}v_2)-v_2(\omega(v_1))\\
&=v_1(\omega(v_2))-\omega([v_1,v_2])-v_2(\omega(v_1)).\qedhere
\end{align*}
\end{proof}

\begin{remark}
We defined
\[L_v\omega=\frac{d}{dt}\,\phi_t^*\omega\bigg|_{t=0}.\]
It is more generally true that
\[\frac{d}{dt}\,\phi_t^*(\omega)\bigg|_{t=t_0}=\phi_{t_0}^*(L_v\omega).\]
Note that we haven't said what kind of thing $\omega$ is; it only makes sense for certain natural bundles. But this works in particular when $\omega$ is some vector field $w$. Then $\phi_t$ is the flow associated to $v$; also, let $\psi_s$ be the flow associated to $w$. Then
\[[v,w]=0\iff L_v(w)=0\iff \frac{d}{dt}(\phi_t^*w)=0\text{ for all }t\iff \phi_t^*w=w\text{ for all }t.\]
Assume that $[v,w]=0$, so that $\phi_t^*w=w$, and thus $\phi_t(\text{integral curve of }w)$ is an integral curve of $w$. This is equivalent to saying that $\phi_t\circ\psi_s=\psi_s\circ\phi_t$. Thus, we have established the following:
\begin{corollary}
$[v,w]=0\iff \phi_t\circ\psi_s=\psi_s\circ \phi_t$
\end{corollary}
\end{remark}

\begin{theorem}[Ehresmann's theorem] Let $f:X\to Y$ be a proper submersion. Then $f$ is a $\Cinfty$ fiber bundle. 
\end{theorem}


We will give a second proof of this using flows.

\begin{proof}
Let $X\hookrightarrow\R^m$ be an embedding of $X$ in Euclidean space. Thus, for any $x\in X$, $T_xX$ gets an inner product. We have a short exact sequence
\begin{center}
\begin{tikzcd}
0 \ar{r} & T_xf^{-1}(f(x)) \ar{r} & T_xX\ar{r}{f'(x)} & T_{f(x)}Y \ar{r} & 0
\end{tikzcd}
\end{center}
where we have used that $f$ is a submersion. Let $W(x)=T_xf^{-1}(f(x))^\perp$, so that we get a subbundle $W$ of $TX$ such that $f'(x):W(x)\xrightarrow{\;\cong\;} T_{f(x)}Y$.

Assume that $Y=(-1,1)^n\subset\R^n$. Thne $\frac{\partial}{\partial y_1},\ldots,\frac{\partial}{\partial y_n}$ are vector fields on $Y$, i.e. sections of $TY$, and so we get corresponding sections $w_1,\ldots,w_n$ of $W$ such that $f'(x)(w_i)=\frac{\partial}{\partial y_i}$ for all $i$. Note that even though the $\frac{\partial}{\partial y_i}$ all commute with each other, we need not have that the $w_i$ all commute with each other.

Let $\phi_t^i$ denote the flow associated to $w_i$. One sees that for any compact $K\subseteq X$, there is an $\epsilon>0$ such that $\phi_{t_1}^1\cdots\phi_{t_n}^n(x)$ are defined for all $x\in K$ and $t_i\in(-\epsilon,\epsilon)$.

Let $K=f^{-1}(0)$, which is compact because $f$ is proper. Then we have a commutative diagram
\begin{center}
\begin{tikzcd}
K\times(-\epsilon,\epsilon)^n \ar{r}{h} \ar{d}[swap]{p_2} & X\ar{d}{f} \\
(-\epsilon,\epsilon)^n \ar[hook]{r} & (-1,1)^n
\end{tikzcd}
\end{center}
and $h$ induces isomorphisms on tangent spaces at $K\times 0$, so it must do so in a neighborhood of $K\times 0$. Because $p_2$ is proper, it follows that by shrinking $\epsilon$ if necessary, we may assume that $h$ induces isomorphisms on tangent spaces everywhere, and that $h$ is one-to-one. Then $h$ is then a diffeomorphism onto its image $U$, which is open in $X$. We want to show that $U=X$; thus, let $F=X\setminus U$. Then $F$ is closed in $X$, and because $f$ is proper, we have that $f(F)$ is a closed set (we're using Hausdorffness here). Then $F\cap f^{-1}(0)=\emptyset$, because $0\notin f(F)$, and now replace $(-1,1)^n$ by the complement of $f(F)$.
\end{proof}

We can now state a refinement of Ehresmann's theorem.

\begin{theorem}
Let $f:X\to Y$ be a proper submersion, and let $A\subseteq X$ be a closed $\Cinfty$ submanifold. Assume also that $f|_A:A\to Y$ is a submersion. Then $f:(X,A)\to Y$ is a fiber-bundle pair.
\end{theorem}
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