9. The Frobenius Theorem

9.1Integral Submanifold

If $M^n$ is decomposed into $k$-dimensional manifolds $L^k\subset M$ which are the image of injective immersions, the $i:L\hookrightarrow M$ and $(D_{p}i)(T_{p}L)$ is a $k$-dimensional subspace of $T_{i(p)}M$.

Suppose that $E\subset TM$ is a rank $k$ subbundle.

定义 9.1.1. A submanifold $S\stackrel{i}{\subset }M$ is called an integral submanifold for $E$ if $\forall p\in S$,$$(D_{p}i)(T_{p}S)\subset E_p$$

定义 9.1.2. $E$ is called integrable if through every point $p\in M$, there is a $k$-dimensional integral submanifold for $E$.

9.2The Frobenius Theorem

定理 9.2.1 (Frobenius Theorem). For a rank $k$ subbundle $E\subset TM$, the following are equivalent:

(1)	$E$ is integrable.
(2)	$\Gamma (E)$ is closed under $[,]$.
(3)	there is an atlas $(U_i,{\phi }_i)$ for $M$, $i\in I$, such that $\forall p\in U_i$, $$(D_p{\phi }_i)(E_p)\ni \frac{\partial }{\partial x_j}\text{for}j\in \{1,\dots ,k\}$$

证明. (3)$\Rightarrow$(1): Let $(U,\phi )$ be a chart as in (3). In $\phi (U)$, the slices given by$$x_{k+1}=c_{k+1},\dots ,x_n=c_n$$are $k$-dimensional integral submanifold of $D\phi (E)$. Applying ${\phi }^{-1}$, we obtain $k$-dimensional integral submanifold for $E{|}_U$.
(1)$\Rightarrow$(2): Let $L\subset M$ be a $k$-dimensional integral submanifold for $E$ through $p\in M$. If $X,Y\in \Gamma (E)$, then there exist unique $\tilde{X},\tilde{Y}\in \mathfrak{X}(L)$, s.t.$$\begin{array}{rl}{D}_i(\tilde{X})& =X{|}_{i(L)}\\ D_i(\tilde{Y})& =Y{|}_{i(L)}\end{array}$$$$[X,Y](p)=[D_i(\tilde{X}),D_i(\tilde{Y})](p)=(D_i[\tilde{X},\tilde{Y}])(p)\in E_p$$The second equlity is by the following claim:

Claim 9.2.2. $f:M\to N$ is a smooth map, $X,Y\in \mathfrak{X}(M)$.$$D_{p}f([X,Y](p))=[Df(X),Df(Y)](f(p))$$

证明. Let $h\in {\mathcal{C}}^{\infty }(N)$.$$\begin{array}{rl}(L_{Df(X)}h)(q)& =D_{q}h(D_{p}f(X(p)))\\ & =D_p(h\circ f)(X(p))\\ & =(L_X(h\circ f))(p)\end{array}$$Note that $q=f(p)$. So$$(L_{Df(X)}h)\circ f=L_X(h\circ f)$$Then$$\begin{array}{rl}{L}_{[Df(X),Df(Y)]}h& =L_{Df(X)}{L}_{Df(Y)}h-L_{Df(Y)}{L}_{Df(X)}h\\ & =L_X((L_{Df(Y)}h)\circ f)-L_Y((L_{Df(X)}h)\circ f)\\ & =L_X{L}_Y(h\circ f)-L_Y{L}_X(h\circ f)\\ & =L_{[X,Y]}(h\circ f)\\ & =L_{Df[X,Y]}h\end{array}$$Thus,$$[Df(X),Df(Y)]=Df[X,Y]$$

$\square$

(2)$\Rightarrow$(3): Proving (3) is a local problem, so we may work on an open neighborhood $U$ of $0$ in ${\mathbb{R}}^n$.

Step 1: Consider the projection $\begin{array}{rl}[t]\pi :U& \to {\mathbb{R}}^k\\ (x_1,\dots ,x_n)& \mapsto (x_1,\dots ,x_k)\end{array}$
Suppose that $D_0\pi {|}_{E_0}$ is an isomorphism. Then we may assume $D_p\pi {|}_{E_p}$ is an isomorphism for all $p\in U$.

Step 2: After a linear change of coordinates on ${\mathbb{R}}^n$, we may assume that $D_0\pi {|}_{E_0}$ is an isomorphism. By Step 1, the same is then true for all $p\in U$.

Step 3: Let $U$ and $\pi$ be as above. Fix $z_i\in \Gamma (E{|}_U)$, so that$$D\pi (z_i)=\frac{\partial }{\partial x_i}\text{for}i\in \{1,\dots ,k\}$$Then $z_1(p),\dots ,z_k(p)$ are a basis of $E_p$ for every $p\in U$. By (2), we have $[z_i,z_j]\in \Gamma (E)$. Then$$D\pi [z_i,z_j]=[D\pi (z_i),D\pi (z_j)]=\left[\frac{\partial }{\partial x_i},\frac{\partial }{\partial x_j}\right]=0$$By injectivity of $D\pi {|}_E$, we conclude $[z_i,z_j]=0$.

Step 4: Since $z_i$ pairwise commute, there are local coordinates, s.t. $z_i=\frac{\partial }{\partial x_i}$.

$\square$

9.3Foliation

定义 9.3.1. Let $M$ be a smooth $n$-dimensional manifold, $0⩽k⩽n$.
A $k$-dimensional foliation $\mathcal{F}$ of $M$ is a decomposition of $M$ into $k$-dimensional injectively immersed manifolds which is locally trivial in the following sense: $\forall p\in M$, $\exists$ open neighborhood $U$ and a diffeomorphism $\phi :U\to {\mathbb{R}}^n$, s.t. the intersections of injective immersed manifolds making up $\mathcal{F}$ with $U$ are mapped by $\phi$ to the slices$$x_{k+1}=c_{k+1},\dots ,x_n=c_n$$

A subbundle $E\subset TM$ is integrable if and only if $E$ consists of vectors tangent to the leaves of a foliation, this is true if and only if $\Gamma (E)$ is closed under $[,]$.

例 9.3.2. Every rank $1$ subbundle $E\subset TM$ is integrable to a $1$-dimensional foliation.

例 9.3.3. $k=2$, locally $E=\mathrm{span}\{x,y\}$. Then

Integrate $x$ to get $1$-dimensional integral submanifold for $E$.

Integrate $y$ to get $1$-dimensional integral submanifold for $E$.

例 9.3.4. $M=T^2=S^1\times S^1=([0,1]\times [0,1])/\sim$
$E\subset T{T}^2$ spanned by $a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}=X$
If $a/b\in \mathbb{Q}$, then all flowlines of $X$ are periodic, so $=S^1$.
If $a/b\notin \mathbb{Q}$, then all flowlines of $X$ are $\cong \mathbb{R}$, and are dense in $T^2$.
Let $T_i=S^1\times D^2$, then $S^3={\mathbb{R}}^2\cup \{\infty \}=T_1\cup T_2$.

This is the Reeb Foliation of $S^3$