16.3 Frobenius theorem: dual form

中文

Let $D^k$ be a $k$-dimensional distribution on a manifold $M$, assigning a $k$-dimensional subspace $D^k(p)$ of the tangent space at each point $p\in M$ . Its annihilator subspace $(D^k{)}^{\perp }(p)$ (see Problem 3.16) consists of the set of covectors at $p$ that vanish on $D^k(p)$ ,

$$(D^k{)}^{\perp }(p)=\{{\omega }_p\mid ⟨{\omega }_p,X_p⟩=0\text{ for all }{X}_p\in D^k(p)\}.$$

Since the distribution $D^k$ is required to be $C^{\infty }$ , it follows from Theorem 3.7 that every point $p$ has a neighbourhood $U$ and a basis $e_i$ of smooth vector fields on $U,$ , such that $e_{r+1},\dots ,e_n$ span $D^k(q)$ at every point $q\in U$ , where $r=n-k$ . The dual basis of 1-forms ${\omega }^i$ defined by $⟨{\omega }^i,e_j⟩={\delta }_j^i$ has the property that the first $r$ 1-forms ${\omega }^1,{\omega }^2,\dots ,{\omega }^r$ are linearly independent and span the annihilator subspace $(D^k{)}^{\perp }(q)$ at each $q\in U$

The annihilator property is reciprocal: given $r$ linearly independent 1-forms ${\omega }^a(a=1,\dots ,r)$ on an open subset $U$ of $M,$ , they span the annihilator subspace $(D^k{)}^{\perp }$ of the $k=(n-r)$ -dimensional distribution

$$D^k=\{X\in \mathcal{T}(U)|⟨{\omega }^a,X⟩=0\}.$$

As shown at the end of Section 8.3, the simple differential $k$-form

$$\Omega ={\omega }^1\wedge {\omega }^2\wedge \cdots \wedge {\omega }^k$$

is uniquely defined up to a scalar field factor by the subspace $(D^k{)}^{\perp }$ , and has the property that a 1-form $\omega$ belongs to $(D^k{)}^{\perp }$ if and only if $\omega \wedge \Omega =0$

Suppose the distribution $D^k$ is involutive, so that $X$, $Y\in D^k\Rightarrow [X,Y]\in D^k$ . From Eq. (16.14)

$$\mathrm{d}{\omega }^a(X,Y)=\frac{1}{2}(X(⟨Y,{\omega }^a⟩)-Y(⟨X,{\omega }^a⟩)-⟨[X,Y],{\omega }^a⟩)=0$$

for any pair of vectors $X$, $Y\in D^k$ . Conversely, if all ${\omega }^a$ and $\mathrm{d}{\omega }^a$ vanish when restricted to the distribution $D^k$ , then $⟨[X,Y],{\omega }^a⟩=0$ for all $\zeta ,Y\in D^k$ . Thus, a necessary and sufficient condition for a distribution $D^k$ to be involutive is that for all ${\omega }^a\in (D^k{)}^{\perp }$ the exterior derivative $\mathrm{d}{\omega }^a$ vanishes on $D^k$

Let $A_{ij}^a=-A_{ji}^a$ be scalar fields such that $\mathrm{d}{\omega }^a=A_{ij}^a{\omega }^i\wedge {\omega }^j.\mathrm{If}\mathrm{d}{\omega }^a(e_{\alpha },e_{\beta })=0$ for all $\alpha$, $\beta =r+1,\dots ,n$ , then $A_{\alpha \beta }^a=0$ and

$$\mathrm{d}{\omega }^a=A_{bc}^a{\omega }^b\wedge {\omega }^c+A_{b\beta }^a{\omega }^b\wedge {\omega }^{\beta }+A_{\alpha c}^a{\omega }^{\alpha }\wedge {\omega }^c.$$

Thus, $D^k$ is involutive if and only if for the 1-forms $\mathrm{d}{\omega }^a$ there exist 1-forms ${\theta }_b^a$ such that

$$\mathrm{d}{\omega }^a={\theta }_b^a\wedge {\omega }^b.$$

On the other hand, the Frobenius theorem 15.4 asserts that $D^k$ is involutive if and only if there exist local coordinates $(U;x^i)$ at any point $p$ such that $e_{\alpha }=B_{\alpha }^{\beta }{\partial }_{x^{\beta }}$ for an invertible matrix of scalar fields $\left[B_{\alpha }^{\beta }\right]$ on $U.$ . In these coordinates, set ${\omega }^a=A_b^a\mathrm{d}{x}^b+W_{\alpha }^a\mathrm{d}{x}^{\alpha }$ , and using $⟨{\omega }^a,e_{\alpha }⟩=0.$ , we have ${\mathcal{W}}_{\alpha }^a=0$ . Hence an alternative necessary and sufficient condition for $D^k$ to be involutive is the existence of coordinates $(U;x^i)$ ) such that

$${\omega }^a=A_b^a\mathrm{d}{x}^b.$$

Theorem 16.4 · Frobenius theorem in differential-form language

Let ${\omega }^a(a=1,\dots ,r)$ be a set of 1-forms on an open set $U,$ , linearly independent at every point $p\in U$ . Thefollowing statements are all equivalent:

(i) There exist local coordinates $(U;x^i)$ at every point $p\in U$ such that ${\omega }^a=A_b^a\mathrm{d}{x}^b$

(ii) There exist 1-forms ${\theta }_b^a$ such that $\mathrm{d}{\omega }^a={\theta }_b^a\wedge {\omega }^b$

(iii) $\mathrm{d}{\omega }^a\wedge \Omega =0$ where $\Omega ={\omega }^1\wedge {\omega }^2\wedge \cdot \cdot \cdot \wedge {\omega }^r$

(iv) $\mathrm{d}\Omega \wedge {\omega }^a=0.$

(v) There exists a 1-form $\theta$ such that $\mathrm{d}\Omega =\theta \wedge \Omega$

Proof

We have seen by the above remarks that $(\mathrm{i})\iff (\mathrm{ii})$ as both statements are equivalent to the statement that the distribution $D^k$ that annihilates all ${\omega }^a$ is involutive. Condition (ii) $\iff (\mathrm{iii})$ since ${\omega }^a\wedge \Omega =0$ , while the converse follows on setting $\mathrm{d}{\omega }^a={\theta }_b^a\wedge {\omega }^b+A_{\alpha \beta }^a{\omega }^{\alpha }\wedge {\omega }^{\beta }$ , where ${\omega }^i(i=1,\dots ,n)$ is any local basis of 1-forms completing the ${\omega }^a$

The implication $(\mathrm{iii})\iff (\mathrm{iv})$ follows at once from Eq. (16.3), and $(\mathrm{v})⟹(\mathrm{iv})$ since

$$\mathrm{d}\Omega =\theta \wedge \Omega ⟹\mathrm{d}\Omega \wedge {\omega }^a=\theta \wedge \Omega \wedge {\omega }^a=0.$$

Finally, $(\ddot{1}\dot{1})\Rightarrow (\mathbf{v})$ , for if $\mathrm{d}{\omega }^a={\theta }_b^a\wedge {\omega }^b$ then

$$\begin{array}{l}\mathrm{d}\Omega =\mathrm{d}{\omega }^1\wedge {\omega }^2\wedge \cdots \wedge {\omega }^r-{\omega }^1\wedge \mathrm{d}{\omega }^2\wedge \cdots \wedge {\omega }^r+\dots \\ ={\theta }_1^1\wedge {\omega }^1\wedge {\omega }^2\wedge \cdots \wedge {\omega }^r-{\omega }^1\wedge {\theta }_2^2\wedge {\omega }^2\wedge \cdots \wedge {\omega }^r+\dots \\ =({\theta }_1^1+{\theta }_2^2+\cdots +{\theta }_r^r)\wedge {\omega }^1\wedge {\omega }^2\wedge \cdots \wedge {\omega }^r\\ =\theta \wedge \Omega \end{array}$$

where $\theta ={\theta }_a^a$ . Hence $(\mathrm{iv})\Rightarrow (\mathrm{iii})\Rightarrow (\mathrm{ii})\Rightarrow (\mathrm{v})$ and the proof is completed.

A system of linearly independent 1-forms ${\omega }^1,\dots ,{\omega }^r$ on an open set $U_{\text{ : }}$ , satisfying any of the conditions $(\mathrm{i})-(\mathrm{v})$ of this theorem is said to be completely integrable . The equations defining the distribution $D^k\left(k=n-r\right)$ that annihilates these ${\omega }^a$ is given by the equations $⟨{\omega }^a,X⟩=0$ , often written as a Pfaffian system of equations

$${\omega }^a=0(a=1,\dots ,r).$$

Condition (i) says that locally there exist $r$ functions $g^a(x^1,\dots ,x^n)$ on $U$ such that

$${\omega }^a=f_b^a\mathrm{d}{g}^b$$

where the functions $f_b^a$ form a non-singular $r\times r$ matrix at every point of $U.$ . The functions $g^a$ are known as a first integral of the system. The $r$-dimensional submanifolds $(N_{\mathbf{c}},{\psi }_{\mathbf{c}})$ defined by $g^a(x^1,\dots ,x^n)=c^a=$ const. have the property

$${\psi }_{\mathbf{c}}^{\ast }{\omega }^a=f_b^a\circ {\psi }_{\mathbf{c}}\mathrm{d}{c}^b=0,$$

and are known as integral submanifolds of the system.

Example 16.2

Consider a single Pfaffian equation in three dimensions,

$$\omega =P(x,y,z)\mathrm{d}x+Q(x,y,z)\mathrm{d}y+R(x,y,z)\mathrm{d}z=0.$$

If $\omega =f\mathrm{d}g$ where $f(0,0,0)\ne 0$ , the function $f(x,y,z)$ is said to be an integrating factor. It is immediate then tha

$$\mathrm{d}\omega =\mathrm{d}f\wedge \mathrm{d}g=\mathrm{d}f\wedge \frac{1}{f}\omega =\theta \wedge \omega$$

where $\theta =d(\ln f)$ . This is equivalent to conditions (ii) and (v) ofTheorem 16.4. Conditions (iii) and (iv) are identical since $\Omega =\omega ,$ , and follow at once from

$$\mathrm{d}\omega \wedge \omega =\theta \wedge \omega \wedge \omega =0,$$

which reduces to Euler’s famous integrability condition for the existence of an integrating factor,

$$P\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)+Q\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)+R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)=0.$$

For example, if $\omega =\mathrm{d}x+z\mathrm{d}y+\mathrm{d}z$ there is no integrating factor, $\omega =f\mathrm{d}g$ , since

$$\mathrm{d}\omega \wedge \omega =\mathrm{d}z\wedge \mathrm{d}y\wedge \mathrm{d}x=-\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z\ne 0.$$

On the other hand, if $\omega =2xz\mathrm{d}x+2yz\mathrm{d}y+\mathrm{d}z$ , then

$$\mathrm{d}\omega \wedge \omega =(2x\mathrm{d}z\wedge \mathrm{d}x+2y\mathrm{d}z\wedge \mathrm{d}y)\wedge \omega =4xyz(\mathrm{d}z\wedge \mathrm{d}x\wedge \mathrm{d}y+\mathrm{d}z\wedge \mathrm{d}y\wedge \mathrm{d}x)=0.$$

It should therefore be possible locally to express $\omega$ in the form $f$ $\mathrm{d}g$. The functions $f$ and $g$ are not unique, for if $G(g)$ is an arbitrary function then $\omega =F$ $\mathrm{d}G$ where $F=f/(\mathrm{d}G/\mathrm{d}g)$ To find an integrating factor $f$ we solve a system of three differential equations

$$f\frac{\partial g}{\partial x}=2xz,\text{\hspace{1em}(a)}$$ $$f\frac{\partial g}{\partial y}=2yz,\text{\hspace{1em}(b)}$$ $$f\frac{\partial g}{\partial z}=1.\text{\hspace{1em}(c)}$$

Eliminating $f$ from (a) and (b) we have

$$\frac{1}{x}\frac{\partial g}{\partial x}=\frac{1}{y}\frac{\partial g}{\partial y},$$

which can be expressed as

$$\frac{\partial g}{\partial x^2}=\frac{\partial g}{\partial y^2}.$$

This equation has a general solution $g(z,u)$ where $u=x^2+y^2$ , and eliminating $f$ from (b) and (c) results in

$$\frac{\partial g}{\partial z}=2yz\frac{\partial g}{\partial y}⟹\frac{\partial g}{\partial \ln z}=\frac{\partial g}{\partial y^2}=\frac{\partial g}{\partial u}.$$

Hence $g=G(\ln z+u)$ , and since it is possible to pick an arbitrary function $G$ we can set $g=z{\mathrm{e}}^{x^2+y^2}$ . From (c) it follows that $f={\mathbf{e}}^{-x^2-y^2}$ , and it is easy to check that

$$\omega ={\mathrm{e}}^{-x^2-y^2}\mathrm{d}\left(z{\mathrm{e}}^{x^2+y^2}\right)=2xz\mathrm{d}x+2yz\mathrm{d}y+\mathrm{d}z.$$

Problems

Problem 16.7

Let $\omega =yz\mathbf{d}x+xz+z^2$ $\mathrm{d}z$. Show that the Pfaffian system $\omega =0$ has integral sur faces $g=z^3{\mathbf{e}}^{xy}=const.$ ., and express $\omega$ in the form $f\mathrm{d}\mathbf{g}$

Problem 16.8

Given an $r\times r$ matrix of 1-forms $\Omega ,$ show that the equation

$$\mathrm{d}\mathsf{A}=\Omega \mathsf{A}-\mathsf{A}\Omega$$

is soluble for an $r\times r$ matrix of functions $A$ only if

$$\Theta \mathsf{A}=\mathsf{A}\Theta$$

where $\Theta =\mathrm{d}\Omega -\Omega \wedge \Omega$

If the equation has a solution for arbitrary initial values $\mathsf{A}={\mathsf{A}}_0$ at any point $p\in M$ , show that there exists a 2-form $\alpha$ such that ” $\alpha I$ and $\mathrm{d}\alpha =0$