15.6 Distributions and Frobenius theorem

中文

A $k$-dimensional distribution $D^k$ on a manifold $M$ is an assignment of a $k$-dimensional subspace $D^k(p)$ of the tangent space $T_p(M)$ at every point $p\in M$ . The distribution is said to be $C^{\infty }$ or smooth if for all $p\in M$ there is an open neighbourhood $U$ and $k$ smooth vector fields $X_1,\dots ,X_k$ on $U$ that span $D^k(q)$ at each point $q\in U$ . A vector field $X$ on an open domain $A$ is said to lie in or belong to the distribution $D^k$ if $X_p\in D^k(p)$ ) at each point $p\in A$ . A one-dimensional distribution is equivalent to a vector field up to an arbitrary scalar factor at every point, and is sometimes called a direction field.

An integral manifold of a distribution $D^k$ is a $k$-dimensional submanifold $(K,\psi )$ of $M$ such that all vector fields tangent to the submanifold belong to $D^k$

$${\psi }_{\ast }(T_p(K))=D^k(\psi (p)).$$

Every one-dimensional distribution has integral manifolds, for if $X$ is any vector field that spans a distribution $D^1$ then any family ofintegral curves of $X$ act as integral manifolds ofthe distribution $D^1$ . We will see, however, that not every distribution ofhigher dimension has integral manifolds.

A distribution $D^k$ is said to be involutive if for any pair of vector fields $X$, $Y$ lying in $D^k$ , their Lie bracket $[X,Y]$ also belongs to $D^k$ . If $\{e_1,\dots ,e_k\}$ is any local basis of vecto fields spanning an involutive $D^k$ on an open neighbourhood $U$, then

$$[e_{\alpha },e_{\beta }]=\sum _{\gamma =1}^k{C}_{\alpha \beta }^{\gamma }{e}_{\gamma }(\alpha ,\beta =1,\dots ,k)\text{\hspace{1em}(15.40)}$$

where $C_{\alpha \beta }^{\gamma }=-C_{\beta \alpha }^{\gamma }$ are $C^{\infty }$ functions on $U.$ . Conversely, if there exists a local basis $\{e_{\alpha }\}$ satisfying Eq. (15.40) for some scalar structure fields $C_{\alpha \beta }^{\gamma }$ , the distribution is involutive, for if $X=X^{\alpha }{e}_{\alpha }$ and $Y=Y^{\beta }{e}_{\beta }$ then

$$[X,Y]=[X^{\alpha }{e}_{\alpha },Y^{\beta }{e}_{\beta }]=(X(Y^{\gamma })-Y(X^{\gamma })+X^{\alpha }{Y}^{\beta }{C}_{\alpha \beta }^{\gamma })e_{\gamma },$$

which belongs to $D^k$ as required. For example, ifthere exists a coordinate chart $(U;x^i)$ such that the distribution $D^k$ is spanned by the first $k$ coordinate basis vector fields

$$e_1={\partial }_{x^1},e_2={\partial }_{x^2},\dots ,e_k={\partial }_{x^k}$$

then $D^k$ is involutive on $U$ since all $[e_{\alpha },e_{\beta }]=0$ , a trivial instance of the relation Eq. (15.40). In this case we can restrict the chart to a cubical neighbourhood $U^{\prime }=\{p|-a<x^i(p){<}^a\}$ , and the ‘slices’ $x^a=\mathrm{const}(a=k+1,\dots ,n)$ are local integral manifolds of the distribution $D^k$ . The key result is the Frobenius theorem:

Theorem 15.4 · Frobenius theorem

A smooth $k$-dimensional distribution $D^k$ on a manifold $M$ is involutive if and only ifevery point $p\in M$ lies in a coordinate chart $(U;x^i)$ ) such that the coordinate vector fields $\partial /\partial x^{\alpha }for\alpha =1,\dots ,k$ span $D^k$ at each point of $U$. $D^k$

Proof

The ifpart follows from the above remarks. The converse will be shown by induc tion on the dimension $k$. The case $k=1$ follows immediately from Theorem 15.3. Suppose now that the statement is true for all ($k$ 1)-dimensional distributions, and let $D^k$ be a $k$-dimensional involutive distribution spanned at all points of an open set $A$ by vector fields $\{X_1,\dots ,X_k\}$ . At any point $p\in A$ there exist coordinates $(V;y^i)$ such that $X_k={\partial }_{y^k}$ $k$ . Set

$$Y_{\alpha }=X_{\alpha }-(X_{\alpha }{y}^k)X_k,Y_k=X_k$$

where Greek indices $\alpha ,\beta ,\dots$ . range from 1 to $k-1$ . The vector fields $Y_1,Y_2,\dots ,Y_k$ clearly span $D^k$ on $V_{;}$ , and

$$Y_{\alpha }{y}^k=0,Y_k{y}^k=1.\text{\hspace{1em}(15.41)}$$

Since $D^k$ is an involutive we can write

$$\begin{array}{r}[Y_{\alpha },Y_{\beta }]=C_{\alpha \beta }^{\gamma }{Y}_{\gamma }+a_{\alpha \beta }{Y}_k,\\ [Y_{\alpha },Y_k]=C_{\alpha }^{\gamma }{Y}_{\gamma }+a_{\alpha }{Y}_k.\end{array}$$

Applying both sides of these equations to the coordinate function $y^k$ and using Eq. (15.41), we find $a_{\alpha \beta }=a_{\alpha }=0$ , whence

$$[Y_{\alpha },Y_{\beta }]=C_{\alpha \beta }^{\gamma }{Y}_{\gamma },\text{\hspace{1em}(15.42)}$$

$$[Y_{\alpha },Y_k]=C_{\alpha }^{\gamma }{Y}_{\gamma }.\text{\hspace{1em}(15.43)}$$

The distribution $D^{k-1}$ spanned by $Y_1,Y_2,\dots ,Y_{k-1}$ is therefore involutive on $V,$ and by the induction hypothesis there exists a coordinate chart $(W;z^i)$ such that $D^{k-1}$ is spanned by $\left\{{\partial }_{z^1},\dots ,{\partial }_{z^{k-1}}\right\}$ . Set

$$\frac{\partial }{\partial z^{\alpha }}=A_{\alpha }^{\beta }{Y}_{\beta }$$

where $[A_{\alpha }^{\beta }]$ is a non-singular matrix of functions on $W$ . The original distribution $D^k$ is spanned on $W$ by the set of vector fields

$$\{{\partial }_{z^1},{\partial }_{z^2},\dots ,{\partial }_{z^{k-1}},Y_k\}.$$

It follows then from Eq. (15.43) that

$$[{\partial }_{z^{\alpha }},Y_k]=K_{\alpha }^{\beta }{\partial }_{z^{\beta }}\text{\hspace{1em}(15.44)}$$

for some functions $K_{\alpha }^{\beta }$ . If we write

$$Y_k=\sum _{\alpha =1}^{k-1}{\xi }^{\alpha }{\partial }_{z^{\alpha }}+\sum _{a=k}^n{\xi }^a{\partial }_{z^a}$$

and apply Eq. (15.44) to the coordinate functions $z^a\left(a=k,\dots ,n\right)$ , we find

$$\frac{\partial {\xi }^a}{\partial z^{\alpha }}=0.$$

Hence ${\xi }^a={\xi }^a(z^k,\dots ,z^n)$ for all $a\ge k$ . Since $Y_k$ is linearly independent of the vectors ${\partial }_Z\alpha$ , the distribution $D^k$ is spanned by the set of vectors $\left\{{\partial }_{z^1},{\partial }_{z^2},\dots ,{\partial }_{z^{k-1}},Z\right\}$ , where

$$Z=Y_k-{\xi }^{\alpha }{\partial }_{z^{\alpha }}={\xi }^a(z^k,\dots ,z^n){\partial }_{z^a}.$$

By Theorem 15.3 there exists a coordinate transformation not involving the first $(k-1)$ coordinates,

$$x^k=x^k(z^k,\dots ,z^n),x^{k+1}=x^{k+1}(z^k,\dots ,z^n),\dots ,x^n=x^n(z^k,\dots ,z^n)$$

such that $Z={\partial }_{x^k}$ . Setting $x^1=z^1,\dots ,x^{k-1}=z^{k-1}$ , we have coordinates $(U;x^i)$ in which $D^k$ is spanned by $\left\{{\partial }_{x^1},\dots ,{\partial }_{x^{k-1}},{\partial }_{x^k}\right\}$ -

Theorem 15.5 · Commuting vector fields and coordinate bases

A set ofvector fields $\{X_1,X_2,\dots ,X_k\}$ is equal to thefirst $k$ basisfields of a local coordinate system, $X_1={\partial }_{x^1},\dots ,X_k={\partial }_{x^k}$ ifand only ifthey commute with each other, $[X_{\alpha },X_{\beta }]=0$

Proof

The vanishing of all commutators is clearly a necessary condition for the vector fields to be local basis fields of a coordinate system, for if $X_{\alpha }={\partial }_{x^{\alpha }}(\alpha =1,\dots ,r)$ then $[X_{\alpha },X_{\beta }]=[{\partial }_{x^{\alpha }},{\partial }_{x^{\beta }}]=0.$

To prove sufficiency, we again use induction on $k$. The case $k=1$ is essentially Theorem 15.3. By the induction hypothesis, there exists local coordinates $(U;x^i)$ such that $X_{\alpha }={\partial }_{x^{\alpha }}$ for $\alpha =1,\dots ,k-1$ . Set $Y=X_k=Y^i(x^1,\dots ,x^n){\partial }_{x^i}$ , and by Example 15.15 ${Y^i}_{,\alpha }=0.$ so that we may write

$$Y=\sum _{\alpha =1}^{k-1}{Y}^{\alpha }(x^k,\dots ,x^n){\partial }_{x^{\alpha }}+\sum _{a=k}^n{Y}^a(x^k,\dots ,x^n){\partial }_{x^a}.$$

Using Theorem 15.3 we may perform a coordinate transformation on the last $n-k+1$ coordinates such that

$$Y=\sum _{\alpha =1}^{k-1}{Y}^{\alpha }(x^k,\dots ,x^n){\partial }_{x^{\alpha }}+{\partial }_{x^k}.$$

A coordinate transformation

$$\begin{array}{ll}{x}^{\prime \alpha }=x^{\alpha }+f^{\alpha }(x^k,\dots ,x^n)& (\alpha =1,\dots ,k-1)\\ x^{\prime a}=x^a& (a=k,\dots ,n)\end{array}$$

has the effect

$$Y=\sum _{\beta =1}^{k-1}\left(Y^{\beta }+\frac{\partial f^{\beta }}{\partial x^k}\right)\frac{\partial }{\partial x^{\prime \beta }}+\frac{\partial }{\partial x^{\prime k}}.$$

Solving the differential equations

$$\frac{\partial f^{\beta }}{\partial x^k}=-Y^{\beta }(x^k,\dots ,x^n)$$

by a straightforward integration leads to $Y={\partial }_{x^{\prime k}}$ as required.

Example 15.16

On ${\dot{\mathbb{R}}}^3={\mathbb{R}}^3-\{(0,0,0)\}$ let $X_1,X_2,X_3$ be the three vector field

$$X_1=y{\partial }_z-z{\partial }_y,X_2=z{\partial }_x-x{\partial }_z,X_3=x{\partial }_y-y{\partial }_x.$$

These three vector fields generate a two-dimensional distribution $D^2$ , as they are not linearly independent

$$x{X}_1+y{Y}_2+z{X}_3=0.$$

The Lie bracket of any pair of these vector fields is easily calculated,

$$\begin{array}{rl}& [X_1,X_2]f=[y{\partial }_z-z{\partial }_y,z{\partial }_x-x{\partial }_z]f\\ & =yz[{\partial }_z,{\partial }_x]f+y{f}_{,x}-yx[{\partial }_z,{\partial }_z]f-z^2[{\partial }_y,{\partial }_x]f+zx[{\partial }_y,{\partial }_z]f-x{f}_{,y}\\ & =(-x{\partial }_y+y{\partial }_x)f=-X_{3}f.\end{array}$$

There are similar identities for the other commutators,

$$[X_1,X_2]=-X_3,[X_2,X_3]=-X_1,[X_3,X_1]=-X_2.\text{\hspace{1em}(15.45)}$$

Hence the distribution $D^2$ is involutive and by the Frobenius theorem it is possible to find a local transformation to coordinates $y^1,y^2,y^3$ such that ${\partial }_{y^1}$ 1 and ${\partial }_{y^2}$ span all three vector fields $X_1,X_2$ and $X_3$

The vector field $X=x{\partial }_x+y{\partial }_y+z{\partial }_z$ commutes with all $X_i$ : for example,

$$\begin{array}{rl}& [X_3,X]f=[x{\partial }_y-y{\partial }_x,x{f}_{,x}+y{f}_{,y}+z{f}_{,z}]f\\ & =x^2[{\partial }_y,{\partial }_x]f-x{\partial }_{y}f+x{\partial }_{y}f+xy[{\partial }_y,{\partial }_y]f-y^2[{\partial }_x,{\partial }_y]f\\ & +xz[{\partial }_y,{\partial }_z]f-yz[{\partial }_x,{\partial }_z]f\\ & =0.\end{array}$$

Hence the distribution $E^2$ generated by the pair of vector fields $\{X_3,X\}$ is also involutive. Let us consider spherical polar coordinates, Eq. (15.2), having inverse transformations

$$r=\sqrt{x^2+y^2+z^2},\theta ={\cos }^{-1}\left(\frac{z}{r}\right),\varphi ={\tan }^{-1}\left(\frac{y}{x}\right).$$

Express the basis vector fields in terms of these coordinates

$${\partial }_x=\frac{\partial r}{\partial x}{\partial }_r+\frac{\partial \theta }{\partial x}{\partial }_{\theta }+\frac{\partial \varphi }{\partial x}{\partial }_{\varphi }=\sin \theta \cos \varphi {\partial }_r+\frac{\cos \theta \cos \varphi }{r}{\partial }_{\theta }-\frac{\sin \varphi }{r\sin \theta }{\partial }_{\varphi },$$ $${\partial }_y=\frac{\partial r}{\partial y}{\partial }_r+\frac{\partial \theta }{\partial y}{\partial }_{\theta }+\frac{\partial \varphi }{\partial y}{\partial }_{\varphi }=\sin \theta \sin \varphi {\partial }_r+\frac{\cos \theta \sin \varphi }{r}{\partial }_{\theta }+\frac{\cos \varphi }{r\sin \theta }{\partial }_{\varphi },$$ $${\partial }_z=\frac{\partial r}{\partial z}{\partial }_r+\frac{\partial \theta }{\partial z}{\partial }_{\theta }+\frac{\partial \varphi }{\partial z}{\partial }_{\varphi }=\cos \theta {\partial }_r-\frac{\sin \theta }{r}{\partial }_{\theta },$$

and a simple calculation gives

$$X_1=y{\partial }_z-z{\partial }_y=-\sin \varphi {\partial }_{\theta }-\cot \theta \cos \varphi {\partial }_{\varphi },$$ $$X_2=z{\partial }_x-x{\partial }_z=-\cos \varphi {\partial }_{\theta }-\cot \theta \sin \varphi {\partial }_{\varphi },$$ $$X_3=x{\partial }_y-y{\partial }_x={\partial }_{\varphi },$$ $$X=x{\partial }_x+y{\partial }_y+z{\partial }_z=r{\partial }_r={\partial }_{r^{\prime }}\text{ where }{r}^{\prime }=\ln r.$$

The distribution $D^2$ is spanned by the basis vector fields ${\partial }_{\theta }$ and ${\partial }_{\varphi }$ , while the distribution $E^2$ is spanned by the vector fields ${\partial }_r$ and ${\partial }_{\varphi }$ in spherical polars.

Exercise

Find a chart, two of whose basis vector fields span the distribution generated by $X_1$ and $X.$ Do the same for the distribution generated by $X_2$ and $X.$

Problems

Problem 15.26

Let $D_k$ be an involutive distribution spanned locally by coordinate vector fields $e_{\alpha }=\partial /\partial x^{\alpha }$ , where Greek indices $\alpha ,\beta ,$ , etc, all range from 1 to $k$. If $X_{\alpha }=A_{\alpha }^{\beta }{e}_{\beta }$ is any local basis spanning a distribution $D^k$ , show that the matrix of functions $[A_{\alpha }^{\beta }]$ is non-singular everywhere on its region of definition, and that $[X_{\alpha },X_{\beta }]=C_{\alpha \beta }^{\gamma }{X}_{\gamma }$ where

$$C_{\alpha \beta }^{\gamma }=(A_{\alpha }^{\delta }{A}_{\beta ,\delta }^{\eta }-A_{\beta }^{\delta }{A}_{\alpha ,\delta }^{\eta })(A^{-1}{)}_{\eta }^{\gamma }.$$

Problem 15.27

There is a classical version of the Frobenius theorem stating that a system of partia differential equations of the form

$$\frac{\partial f^{\beta }}{\partial x^j}=A_j^{\beta }\left(x^1,\dots ,x^k,f^1(x),\dots ,f^r(x)\right)$$

where $i,j=1,\dots ,k$ and $\alpha ,\beta =1,\dots ,r$ has a unique local solution through any point $(a^1,\dots ,a^k,b^1,\dots ,b^r)$ if and only if

$$\frac{\partial A_j^{\beta }}{\partial x^i}-\frac{\partial A_i^{\beta }}{\partial x^j}+A_i^{\alpha }\frac{\partial A_j^{\beta }}{\partial y^{\alpha }}-A_j^{\alpha }\frac{\partial A_i^{\beta }}{\partial y^{\alpha }}=0$$

where $A_j^{\beta }=A_j^{\beta }(x^1,\dots ,x^k,y^1,\dots ,y^r)$ . Show that this statement is equivalent to the version given in Theorem 15.4. [Hint: On ${\mathbb{R}}^n$ where $n=r+k$ consider the distribution spanned by vectors

$$Y_i=\frac{\partial }{\partial x^i}+A_i^{\beta }\frac{\partial }{\partial y^{\beta }}(i=1,\dots ,k)$$

and show that the integrability condition is precisely the involutive condition $[Y_i,Y_j]=0$ , while the condition for an integral submanifold of the form $y^{\beta }=f^{\beta }(x^1,...,x^k)$ is $A_j^{\beta }=f_{,j}^{\beta }.]$