\classheader{2013-02-18}

Though it'll seem like we're leaving integral curves, we'll return to them in the middle of the lecture.

Recall that given a $\Cinfty$ manifold $M$, a point $p\in M$, and a tangent vector $v\in T_pM$, there is an $\R$-linear functional $v:\Cinfty(M)\to\R$, sending a $\Cinfty$ function $f:M\to\R$ to $v(f)\in\R$. It satisfies the Leibniz rule,
\[v(fg)=f(p)v(g)+g(p)v(f).\]
This is a generalization of the notion of directional derivative in Euclidean space.

Now let $v$ be a vector field on $M$. Let $R=\Cinfty(M)$. Now we have an $\R$-linear map $v:R\to R$, defined by $v(f)(p)=v(p)(f)$ for all $p\in M$. For example, if $M=\R^n$ and $v=(a_1,\ldots,a_n)=\sum_{i=1}^n a_i\frac{\partial}{\partial x_i}$, we have that
\[v(f)=\sum_{i=1}^na_i\frac{\partial f}{\partial x_i}.\]
For any ring $S$, a function $D:S\to S$ is a derivation when $D(fg)=D(f)\cdot g+f\cdot D(g)$. Very often, we are given a subring $T\subset S$ contained in the center of $S$, that we require to satisfy $D(t)=0$ for all $t\in T$. Note that the map $v:R\to R$ sending $f\mapsto v(f)$ is a derivation, and $v(\text{any constant function})=0$ (observe that we can $\R$ is a subring of $R$).


The following is an easy lemma.

\begin{lemma}
If $D_1,D_2:R\to R$ are derivations, then $(D_1\circ D_2)-(D_2\circ D_1)$ is also a derivation.
\end{lemma}

In particular, if $v,w$ are $\Cinfty$ vector fields on $M$, $U$ is an open subset of $M$, and $\Cinfty(U)$ is the ring of $\Cinfty$ functions on $U$, the map $f\mapsto v(w(f))-w(v(f))$ is a derivation of $\Cinfty(U)$. If we fix a point $p\in M$, we can consider neighborhoods $U$ of $p\in M$, and the map
\[f\mapsto (v(w(f))-w(v(f)))(p)\]
induces an $\R$-linear map on germs $\Cinfty_{M,p}\to\R$. Being a derivation, this is equal to $h(p)(f)$ for a unique $h(p)\in T_pM$. It is true (though we won't check) that $p\mapsto h(p)$ is a $\Cinfty$ vector field on $M$, and we define the Lie bracket of $v$ and $w$ to be this $h$. We write $h=[v,w]$. Thus,
\[[v,w](f)=v(w(f))-w(v(f))\]
for all $\Cinfty$ maps $f:U\to\R$.

\begin{lemma}
Let $\Omega$ be an open subset of $\R^n$, and let $v,w$ be $\Cinfty$ vector fields on $\Omega$. Then
\[[v,w]=D_vw-D_wv,\]
where
\[(D_vw)(x)=\frac{d}{dt}\,w(x+tv)\bigg|_{t=0}.\]
\end{lemma}

\begin{proposition}
The $\R$-vector space of $\Cinfty$ vector fields on $M$, together with the bracket, satisfies the axioms of a Lie algebra:
\begin{enumerate}
\item $[v,w]=-[w,v]$ for all $\Cinfty$ vector fields $v$ and $w$.
\item $[v_1,[v_2,v_3]]+[v_2,[v_3,v_1]]+[v_3,[v_1,v_2]]=0$ for all $\Cinfty$ vector fields $v_1,v_2,v_3$.
\item $[tv,w]=t[v,w]$ for all $t\in\R$.
\end{enumerate}
\end{proposition}

\begin{definition}
Let $M$ and $N$ be $\Cinfty$ manifolds, and let $\phi:M\to N$ be a $\Cinfty$ map. Given vector fields $v$ on $M$ and $w$ on $N$, we say that $v$ and $w$ are $\phi$-related if for all $x\in M$,
\[\phi'(x)v(x)=w(\phi(x)).\]
\end{definition}
\begin{lemma-N}
Given vector fields $v$ on $M$ and $w$ on $N$, they are $\phi$-related if and only if $\phi(\gamma)$ is an integral curve of $w$ for any integral curve $\gamma$ of $v$.
\end{lemma-N}
\begin{proof}
Assume that $v$ and $w$ are $\phi$-relted. Let $\gamma:(a,b)\to M$ be an integral curve for $v$, so that for all $t\in (a,b)$, we have
\[\gamma'(t)=v(\gamma(t)).\]
Let $\delta=\phi\circ \gamma$. Then
\[\delta'(t)=\phi'(\gamma(t))\gamma'(t)=w(\delta(t)).\]
Everything is reversible, so we are done.
\end{proof}

Last time, I mentioned that if a vector field is non-zero at a point, then in some neighborhood it looks like $\frac{\partial}{\partial x_1}$. There is a proof of this in Warner's book on page 40.

\begin{example}
Let $w$ be a vector field on $N$ and suppose that $w(p)\neq 0$. Then there is a chart centered at $p$ such that $w$ is transformed to $\frac{\partial}{\partial x_n}$.
\end{example}
\begin{proof}
Let $Z$be a codimension 1 closed submanifold of $N$ containing $p$, and suppose that it is transverse, i.e. that $T_pZ\oplus\R w(p)=T_pN$. Let $\delta_y(t)$ be an integral curve of $w$ with initial value $y$, i.e. $\delta_y(0)=y$. Let $M=Z\times(-c,c)$, and let $\phi:M\to N$ be the map defined by
\[\phi(z,t)=\delta_z(t).\]
This is a diffeomorphism in a neighborhood of $Z\times\{0\}$ by the inverse function theorem, and the curves $t\mapsto (z,t)$ on $M$ are sent by $\phi$ to the curves $\delta_z(t)$ on $N$, which are integral curves of $w$. Thus, $t\mapsto (z,t)$ is an integral curve for $\frac{\partial}{\partial x_n}$.
\end{proof}
\begin{lemma-N}
Let $M$ and $N$ be $\Cinfty$ manifolds, and let $\phi:M\to N$ be $\Cinfty$.
\begin{itemize}
\item[(a)] If $v$ on $M$ and $w$ on $N$ are $\phi$-related, then $v(\phi^*f)=\phi^*w(f))$ for any $\Cinfty$ map $f:N\to\R$; this is just a restatement of the definition.
\item[(b)] If $v_1$ is $\phi$-related to $w_1$ and $v_2$ is $\phi$-related to $w_2$, then $[v_1,v_2]$ and $[w_1,w_2]$ are $\phi$-related.
\end{itemize}
\end{lemma-N}
\begin{proof}[Proof of (b)]
We have
\[v_1(v_2(\phi^*(f)))=v_1(\phi^*(w_2(f)))=\phi^*(w_1(w_2(f))).\]
Now interchange and subtract.
\end{proof}
\begin{remark}
This has an important consequence. If $M$ is a locally closed submanifold of $N$, $\phi:M\to N$ is the inclusion, and $w$ is a vector field on $N$, then to say that there is some $v$ on $M$ that is $\phi$-related to $w$ is equivalent to saying that $w(x)\in T_xM$ for all $x\in M$ (because $w(x)=v(x)$). Thus, Lemma 2 is saying something about vector fields that are tangent to submanifolds; if $w_1$ and $w_2$ are vector fields on $N$ such that $w_1(x),w_2(x)$ belong to $T_xM$ for all $x\in M$, then $[w_1,w_2]$ has the same property.
\end{remark}
\begin{definition}
Let $M$ be a $\Cinfty$ manifold, and let $W$ be a $\Cinfty$ subbundle of $TM$ of rank $r$. A locally closed submanifold $A$ of $M$ is a leaf if for all $x\in A$, $T_xA=W(x)$.
\end{definition}

(The notion of leaf can defined in more generality than what is given here.)

Suppose that there is a leaf of $W$ through every point of $M$. If $w_1,w_2$ are $\Cinfty$ sections of $W$, then $[w_1,w_2]$ is necessarily also a section of $W$; we can see this easily as follows. Let $p\in M$ and let $Z$ be a leaf through $p$. Because $Z$ is a leaf, $w_1$ and $w_2$ are tangential to $Z$, so $[w_1,w_2]$ is tangential to $Z$, i.e. $[w_1,w_2](p)\in T_pZ=W(p)$ for all $p\in M$.

\begin{definition}
A $\Cinfty$ subbundle $W$ of $TM$ is said to be involutive (alternatively, integrable) if for all $\Cinfty$ sections $w_1,w_2$ of $W$, $[w_1,w_2]$ is also a section of $W$.
\end{definition}
We have already proven one piece of the following theorem:
\begin{theorem}[Frobenius]
Let $W$ be a subbundle of $TM$. The following are equivalent:
\begin{enumerate}
\item $W$ is involutive.
\item There is a leaf of $W$ through every point.
\item For all $p\in M$, there is a diffeomorphism $h$ from a neighborhood of $p$ to $U_1\times U_2$, where $U_i$ is an open subset of $\R^{n_i}$ for $i=1,2$, such that $h(W)$ is the constant $\R^{n_1}\times\{0\}$ bundle on $U_1\times U_2$.
\end{enumerate}
\end{theorem}
\begin{proof}
It is clear that 3 $\implies$ 2, and we have already proven that 2 $\implies$ 1, so it remains to prove that 1 $\implies$ 3. This proof is taken from Narasimhan (the proof is originally due to Volterra).

\textit{Step 1.} Let $W$ be an involutive subbundle of rank $r$. Then in a neighborhood of any $p\in M$, we can find vector fields $w_1,\ldots,w_r$ which are a frame for $W$, i.e. $w_1(x),\ldots,w_r(x)$ are a basis for $W(x)$ for all $x$ in the neighborhood, and such that $[w_i,w_j]=0$ for all $i,j$.

Let me make a linear algebra observation: given a vector space $V=V_1\oplus V_2$, subspaces $W\subset V$ such that the projection to $V_1$ is an isomorphism, i.e.
\begin{center}
\begin{tikzcd}
W \ar[hook]{r} \ar[bend right]{rr}[swap]{\cong} & V \ar{r}{p_1} & V_1
\end{tikzcd}
\end{center}
can be identified with graphs of linear transformations $S:V_1\to V_2$.

Now write $\R^N=V_1\times V_2$, where $N=\dim(M)$, where $V_1$ and $V_2$ have been chosen such that $p|_{W(p)}:W(p)\to V_1$ is an isomorphism ($p$ is the projection $\R^N\to V_1$), so that $W(x)\cong V_1$ for all $x$ in some neighborhood of $p$. Thus, for each $x$, we get $S(x):V_1\to V_2$, and
\[W(x)=\{(v_1,S(x)v_1)\mid v_1\in V_1\}.\]
Let $\Omega\subset V_1\times V_2=\R^N$ be open. WLOG we have $V_1=\R^r$, where $e_1,\ldots,r_r$ are the standard basis of $\R^r$. We have $S(x)e_i=u_i(x)$, where $u_i:\Omega\to V_2$ is some $\Cinfty$ function. Thus $W(x)$ is the linear space of the $e_i+u_i$. For any $i,j$, we have that $[e_i+u_i,e_j+u_j]$ is a section of $W$, and using the formula
\[[\alpha,\beta]=D_\alpha\beta-D_\beta\alpha\]
on Euclidean space, we have that $[e_i+u_i+e_j+u_j]$ is a section of $V_2$ (i.e. a function $\Omega\to V_2$); but it also has to be a section of $W$, so it has to be 0 since $V_2\cap W(x)=0$ for all $x\in\Omega$.
\end{proof}

We'll finish the proof of this with Step 2 next time.