Frobenius Theorem

2 Frobenius Theorem

Consider a smooth manifold \(M\) of dimension \(m\). For local questions we can take \(M=\mathbb {R}^m\), which could correspond to a chart around some point \(x_0\in M\). All functions, vector fields and differential forms are presumed to be smooth (\(C^\infty \)).

Definition 1 involutivity
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Let \(L_i = \sum _{k=1}^m f_i^k(x) \partial /\partial x^j\), \(i=1,\ldots ,r\le m\), be first order differential operators, such that the vector fields \(v_i(x) = (f_i^k(x))_{k=1}^m\) are linearly independent. They are said to be in involution when there exist functions \(c^k_{ij}(x)\) such that

\[ L_i L_j - L_j L_i = \sum _{k=1}^r c^k_{ij}(x) L_k . \]
Theorem 2 local Frobenius

If the first order differential operators \(L_i\), \(i=1,\ldots ,r \le m\), are in involution, then there exist \(m-r\) smooth functions \(u^k(x)\) that satisfy the equations \(L_i u^k(x) = 0\) and such that their gradients \(\nabla u^k(x)\), \(k=1,\ldots ,m-r\) are linearly independent.

Proof

This proof consists of chaining together several intermediate results, which are proven in separate lemmas below.

The first step is to replace the \(L_i\) operators by some better behaved operators \(L'_{i'}\), namely satisfying \([L'_{i'}, L'_{j'}] = 0\) (Lem. 5(e)) and having a form adapted to a split local coordinate system \((y,z)\) around \(x_0\). The equations \(L'_{i'} u = 0\) and \(L_i u = 0\) are equivalent (Lem. 3, 4). Lem. 7 shows that there exists a new local coordinate system \((y,\bar{Z})\) on a neighborhood of \(x_0\), where \(\bar{Z}^k=\bar{Z}^k(y,z)\), which is better adapted to our differential equations. Lem. 6 actually shows that the contructed coordinates give us the desired solutions via \(u^k(x) = \bar{Z}^k(y(x),z(x))\).

Lemma 3

If the \(L_i\) as in Def. 1 are in infolution, then the \(L'_{i'} = \sum _{i=1}^r \alpha _{i'}^i L_i\), \(i'=1,\ldots ,r\), with smooth pointwise invertible \(\alpha _{i'}^i\), are also in involution.

Proof

It is sufficient to compute the commutator

\begin{align*} L’_{i'} L’_{j'} - L’_{j'} L’_{i'} & = \alpha _{i'}^i \alpha _{j'}^j (L_i L_j - L_j L_i) + \alpha _{i'}^i L_i(\alpha _{j'}^j) L_j - \alpha _{j'}^j L_j(\alpha _{i'}^i) L_i \\ & = \alpha _{i'}^i \alpha _{j'}^j \left(c_{ij}^k + (\alpha ^{-1})_j^{j'} L_i(\alpha _{j'}^j) \delta _j^k - (\alpha ^{-1})_i^{i'} L_j(\alpha _{i'}^i) \delta _i^k \right) L_k \\ & = \alpha _{i'}^i \alpha _{j'}^j \left(c_{ij}^k + (\alpha ^{-1})_j^{j'} L_i(\alpha _{j'}^j) \delta _j^k - (\alpha ^{-1})_i^{i'} L_j(\alpha _{i'}^i) \delta _i^k \right) (\alpha ^{-1})_k^{k'} L’_{k'} \\ & = \sum _{k'=1}^r c’^{k'}_{i'j'} L’_{k'} , \end{align*}

where the formula for \(c'^{k'}_{i'j'}\) can be read off from the last equality.

Lemma 4

For the operators \(L'_{i'}\) as in Lem. 3, a smooth function \(u\) solves \(L_i u = 0\), \(i=1,\ldots ,r\), iff it solves \(L'_{i'} u = 0\), \(i'=1,\ldots ,r\).

Proof

The computation

\[ L'_{i'} u = \sum _{i=1}^r \alpha _{i'}^i (L_i u) \]

shows that \(L_i u = 0\), \(i=1,\ldots ,r\), implies \(L'_{i'} u = 0\) for any \(i'=1,\ldots ,r\).

Lemma 5

For the operators \(L_i\) from Def. 1, given \(x_0 \in M\), there exists an open coordinate neighborhood \(U\ni x_0\) such that (a) there an exists invertible \(\alpha _{i'}^i\) as in Lem. 3, (b) there exists a split local chart \((y,z)\colon M \supset U' \cong \mathbb {R}^r \times \mathbb {R}^{m-r}\), with (c) \((y(x_0),z(x_0)) = (0,0)\), (d)

\[ L'_{i'} = \frac{\partial }{\partial y^{i'}} + \sum _{j=1}^{m-r} f_{i'}^j(y,z) \frac{\partial }{\partial z^j} . \]

and (e) \([L'_{i'}, L'_{j'}] = 0\), for \(i',j'=1,\ldots ,r\), which expressed in terms of \(f_{i'}^j\) means (f)

\[ \frac{\partial }{\partial y^{i'}} f_{j'}^j + \sum _{k=1}^{m-r} f_{i'}^k \frac{\partial }{\partial z^k} f_{j'}^j = \frac{\partial }{\partial y^{j'}} f_{i'}^j + \sum _{k=1}^{m-r} f_{j'}^k \frac{\partial }{\partial z^k} f_{i'}^j . \]
Proof

Start with the coordinates \((x^1,\ldots ,x^m)\) on \(U\) and consider the coordinate components \(L_i = a_i^j(x) \frac{\partial }{\partial x^j}\). The rank of the matrix \(a_i^j(x_0)\) must be \(r\), otherwise the \(L_i\) vectors do not constitute a frame for the distribution \(\mathcal{D}\). Hence, there exists a subset \(I \subseteq \{ 1,\ldots ,r\} \) such that the matrix minor \((a_i^j)_{i\in I, 1\le j\le m}\) is non-singular. Define the coordinates \(y^{i'} = x^{I(i')} - (x_0)^{I(i')}\), \(i'=1,\ldots ,r\), and \(z^{j} = x^{I^c(j)} - (x_0)^{I^c(j)}\), \(j=1,\ldots ,m-r\), where \(I(i')\) and \(I^c(j)\) is some ordering of the sets \(I\) and its complement \(I^c\). Then, restrict to a sub-neighborhood \(U'' \subseteq U\) that is split with respect to the \((y,z)\) coordinates.

The new coordinate components are

\[ L_{i} = \sum _{i'=1}^r a_{i}^{I(i')}(x(y,z)) \frac{\partial }{\partial y^{i'}} + \sum _{j=1}^{m-r} a_{i}^{I^c(j)}(x(y,z)) \frac{\partial }{\partial z^j} . \]

Let \(\beta _{i}^{i'}(y,z) = a_{i}^{I(i')}(x(y,z))\) and \(\gamma _{i}^{j}(y,z) = a_{i}^{I^c(j)}(x(y,z))\), so that by construction \(\beta _{i}^{i'}(0,0)\) is non-singular. Since \(\beta \colon U'' \to \operatorname {Mat}(r,r)\) is smooth (hence a fortiriori continuous) and the subset of non-singular matrices in \(\operatorname {Mat}(r,r)\) is open, there is a possibly smaller split sub-neighborhood \(U' \subseteq U''\) on which \(\beta \) is everywhere non-singular. So, defining \(\alpha _{i'}^j(y,z) = (\beta _{i}^{i'}(y,z))^{-1}\) on \(U''\) satisfies the desired conclusions (a), (b), (c) and (d), where \(f_{i'}^j(y,z) = \alpha _{i'}^i(y,z) \gamma _{i}^j(y,z)\).

To prove (e) and (f), consider the computation

\begin{align*} [L’_{i'}, L’_{j'}] & = L’_{i'} L’_{j'} - L’_{j'} L’_{i'} \\ = \sum _{k'=1}^r c’^{k'}_{i'j'} L’_{k'} & = \sum _{j=1}^{m-r} \left(\frac{\partial }{\partial y^{i'}} f_{j'}^j - \frac{\partial }{\partial y^{j'}} f_{i'}^j\right)\frac{\partial }{\partial z^j} + \sum _{k=1}^{m-r} \sum _{j=1}^{m-r} \left(f_{i'}^j \frac{\partial }{\partial z^j} f_{j'}^k - f_{j'}^j \frac{\partial }{\partial z^j} f_{i'}^k\right) \frac{\partial }{\partial z^k} \\ & = \sum _{j=1}^{m-r} \left( \frac{\partial }{\partial y^{i'}} f_{j'}^j + \sum _{k=1}^{m-r} f_{i'}^k \frac{\partial }{\partial z^k} f_{j'}^j - \frac{\partial }{\partial y^{j'}} f_{i'}^j - \sum _{k=1}^{m-r} f_{j'}^k \frac{\partial }{\partial z^k} f_{i'}^j . \right) \end{align*}

Hence, for each fixed \(i',j'\), the \(\frac{\partial }{\partial y^{k'}}\) components of the right-hand side vanish, while those of the left-hand side equal \(\sum _{k'=1}^{m-r} c'^{k'}_{i'j'} \frac{\partial }{\partial y^{k'}}\), meaning that all components of \(c'^{k'}_{i'j'}\) must vanish, proving (e). On the other hand, the vanishing of the right-hand side of the last equality proves (f).

Lemma 6
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Consider the operators \(L'_{i'}\) and the split neighborhood \(U\ni x_0\) as in Lem. 5. Let \(Z^k(y,z)\) and \(\bar{Z}^k(y,z)\) satisfy the inversion identity \(\bar{Z}^k(y, Z(y, z)) = z^k\), for all \(z\) on a sufficiently small neighborhood of \(z=0\), and for all \(y\) on a sufficiently small neighborhood of \(y=0\). Suppose that \(Z^j(y,z)\) satisfies (a) \(Z^j(0,z) = z^j\) and (b)

\[ \frac{\partial Z^j}{\partial y^{i'}}(y,z) = f_{i'}^j(y, Z^j(y,z)) . \]

Then

\[ L'_{i'} \bar{Z}(y,z) = 0, \quad i'=1,\ldots ,r , \]

and vice versa.

Proof

Start by differentiating the inversion identity:

\begin{align*} 0 = \frac{\partial }{\partial y^{i'}} z^k & = \frac{\partial }{\partial y^{i'}} \bar{Z}^k(y, Z(y, z)) \\ & = \left. \frac{\partial \bar{Z}^k}{\partial y^{i'}}(y, z’) \right|_{z'=Z(y,z)} + \frac{\partial Z^j}{\partial y^{i'}}(y,z) \left. \frac{\partial \bar{Z}^k}{\partial z'^j}(y,z’) \right|_{z'=Z(y,z)} \\ & = \left. \left(\frac{\partial \bar{Z}^k}{\partial y^{i'}}(y, z’) + f_{i'}^j(y,z’) \frac{\partial \bar{Z}^k}{\partial z'^j}(y,z’)\right) \right|_{z'=Z(y,z)} \\ & \quad {} + \left. \left(\frac{\partial Z^j}{\partial y^{i'}}(y,z) - f_{i'}^j(y,z’)\right) \frac{\partial \bar{Z}^k}{\partial z'^j}(y,z’) \right|_{z'=Z(y,z)} . \end{align*}

Recall that being a diffeomorphism, the Jacobian \(\frac{\partial \bar{Z}^k}{\partial z'^j}(y,z')\) non-singular on the sufficiently small split domain, with \((\frac{\partial \bar{Z}^k}{\partial z'^j}(y,z'))^{-1} = \left. \frac{\partial Z^j}{\partial z^k}(y,z) \right|_{z=\bar{Z}(y,z')}\). Hence, rearranging the last equality, we find

\[ \left. \frac{\partial Z^j}{\partial z^k}(y,z) L'_{i'} \bar{Z}^k(y,z') \right|_{z'=Z(y,z)} = -\left(\frac{\partial Z^j}{\partial y^{i'}}(y,z) - f_{i'}^j(y,Z(y,z))\right) . \]

Hence, if one side of the equality vanishes, then so does the other, which proves the desired equivalence.

Lemma 7

Let \(Z^j(y,z)\) be as in Lem. 6. Then, \(\zeta ^j(t,y,z) = Z^j(ty,z)\) satisfies \(\zeta ^j(0,y,z) = z^j\) and the equations

\[ \frac{\partial }{\partial t} \zeta ^j(t,y,z) = y^{i'} f_{i'}^j(ty,\zeta (t,y,z)) . \]

Conversely, if \(\zeta ^j(t,y,z)\) satisfies the initial value problem above, then there exists a sufficiently small neighborhood of \((y,z)=(0,0)\) for which \(Z^j(ty,z)\) exists up to at least \(t=1\). Then \(Z^j(y,z) = \zeta ^j(1,y,z)\) satisfies the conditions in the hypotheses of Lem. 6.

Proof

The easy direction is proved by the following computation:

\begin{align*} \frac{\partial }{\partial t} \zeta ^j(t,y,z) & = \frac{\partial }{\partial t} Z^j(ty,z) \\ & = y^{i'} \left.\frac{\partial Z^j}{\partial y'^{i'}}(y’,z) \right|_{y'=ty} \\ & = y^{i'} \left.f_{i'}^j(y’,Z^j(y’,z)) \right|_{y'=ty} \\ & = y^{i'} f_{i'}^j(ty,Z(ty,z)) = y^{i'} f_{i'}^j(ty, \zeta (t,y,z)) . \end{align*}

For the converse direction, consider the following computation, where we use the ODE satisfied by \(\zeta ^j(t,y,z)\) and the identity from Lem. 5(f):

\begin{align*} \frac{\partial }{\partial t} \left(\frac{\partial }{\partial y^{i'}} \zeta ^j(t,y,z) - t f_{i'}^j(ty, \zeta (t,y,z))\right) & = \frac{\partial }{\partial y^{i'}} \frac{\partial }{\partial t} \zeta ^j(t,y,z) \\ & \quad {} - f_{i'}^j(ty, \zeta (t,y,z)) - \left.t y^{j'} \frac{\partial }{\partial y'^{j'}} f_{i'}^{j}(y’, \zeta (t,y,z))\right|_{y'=ty} - t \left(\frac{\partial }{\partial t} \zeta ^k(t,y,z)\right) \left.\frac{\partial }{\partial z'^k} f_{i'}^j(ty, z’)\right|_{z'=\zeta (t,y,z)} \\ & = \frac{\partial }{\partial y^{i'}} \left(y^{j'} f_{j'}^j(ty, \zeta (t,y,z))\right) \\ & \quad {} - f_{i'}^j(ty, \zeta (t,y,z)) - t y^{j'} \left. \frac{\partial }{\partial y'^{j'}} f_{i'}^{j}(y’, z’)\right|_{y'=ty, z'=\zeta (t,y,z)} - t y^{j'} \left. f_{j'}^k(y’,z’) \frac{\partial }{\partial z'^k} f_{i'}^j(y’, z’)\right|_{y'=ty, z'=\zeta (t,y,z)} \\ & = \left.\left(f_{i'}^{j}(y’, z’) + t y^{j'} \frac{\partial }{\partial y'^{i'}} f_{j'}^j(y’, z’) + \left(\frac{\partial }{\partial y^{i'}} \zeta ^k(t,y,z)\right) y^{j'} \frac{\partial }{\partial z'^k} f_{j'}^j(y’, z’) \right)\right|_{y'=ty, z'=\zeta (t,y,z)} \\ & \quad {} - \left.\left( f_{i'}^j(y’, z’) + t y^{j'} \frac{\partial }{\partial y'^{i'}} f_{j'}^{j}(y’, z’) + t y^{j'} f_{i'}^k(y’,z’) \frac{\partial }{\partial z'^k} f_{j'}^j(y’, z’) \right)\right|_{y'=ty, z'=\zeta (t,y,z)} \\ & = \left(\frac{\partial }{\partial y^{i'}} \zeta ^k(t,y,z) - t f_{i'}^k(ty, \zeta (t,y,z))\right) \left. y^{j'} \frac{\partial }{\partial z'^k} f_{j'}^j(y’, z’)\right|_{y'=ty, z'=\zeta (t,y,z)} \end{align*}

Hence, we find that \(\eta (t,y,z) = \frac{\partial }{\partial y^{i'}} \zeta ^k(t,y,z) - t f_{i'}^k(ty, \zeta (t,y,z))\) satisfies a linear ODE. Hence, by the uniqueness of ODE solutions (Lem. 8(b)), if the initial condition \(\eta (0,y,z) = 0\) is satisfied, the solution must identically vanish, \(\eta (t,y,z) = 0\), which upon setting \(t=1\) proves that \(Z^j(y,z) = \zeta ^j(1,y,z)\) satisfies the desired differential equation, (Lem. 8(c)). It remains to check the vanishing initial condition:

\begin{align*} \eta (0,y,z) & = \frac{\partial }{\partial y^{i'}} \zeta ^j(0,y,z) - 0 \cdot f_{i'}^j(0 \zeta (0,y,z)) \\ & = \frac{\partial }{\partial y^{i'}} z^j - 0 = 0 . \end{align*}

The proof is completed by noting that the inverse function \(\bar{Z}(y,z)\) exists on a sufficiently small neighborhood of \((y,z)=(0,0)\), because the continuity of \(\frac{\partial Z^j}{\partial z^k}\) and the property that \(\left.\frac{\partial Z^j}{\partial _z^k}\right|_{z=0} = \delta ^j_k\) ensures that \(Z^j(y,z)\) is an immersion (has non-singular jacobian) on a neighborhood of \((y,z)=(0,0)\) and hence a diffeomorphism on a possibly smaller neighborhood (use inverse function theorem).

Lemma 8

An ODE initial value problem (a sufficiently general one to cover the one for \(\zeta ^j(t,y,z)\) in Lem. 7 and the one for \(\eta _{i'}^j(t,y,z)\) in the proof of Lem. 7) (a) has a solution that is jointly smooth in \((t,y,z)\), which (b) is unique, and (c) exists (at least) up to time \(t=1\) on a sufficiently small neighborhood of \((y,z) = (0,0)\).

Proof

This should follow from the Picard-Lindelöf ODE existence and uniqueness theorem with parameters.

Definition 9 differential forms
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Definition 10 differential ideal
Theorem 11 differential form Frobenius

If \(\alpha _i\), \(i=1,\ldots k\le m-k\) are 1-forms on \(M\) that generate a closed differential ideal. Then there exist smooth scalar functions \(u_i(x)\), \(i=1,\ldots ,m-k\) such that the exact 1-forms \(du_i\), \(i=1,\ldots ,m-k\) generate the same differential ideal.

Definition 12 tangent distribution

A tangent distribution on a manifold \(M\) is a vector sub-bundle \(\mathcal{D}\hookrightarrow TM\) (equivalently, an embedding of vector bundles).

Definition 13 Lie bracket

On a manifold \(M\), given two vector fields \(u\), \(v\) (sections of the tangent bundle \(TM\)), their Lie bracket \(w = [u,v]\) is the vector field that satisfies the identity \(w(f) = u(v(f)) - v(u(f))\), where vector fields act as first order differential operators on a smooth function \(f\). In coordinate form, if \(u = u^i\partial _i\), \(v = v^i\partial _i\), \(w = w^i\partial _i\), then \(w^j = u^i \partial _i v^j - v^i \partial _i u^i\). The vector fields \(u\), \(v\) commute (or are in involution in the sense of Def. 1) if \([u,v] = 0\).

Definition 14 involutive distribution

A tangent distribution \(\mathcal{D} \hookrightarrow TM\) is involutive if, for any two vector field sections \(u\), \(v\) of \(\mathcal{D}\), the Lie bracket \([u,v]\) is also a section of \(\mathcal{D}\).

Definition 15 integral submanifold

Given a manifold \(M\) with a tangent distribution \(\mathcal{D} \hookrightarrow TM\) of rank \(r\) (as a vector bundle), a submanifold \(\iota \colon N \hookrightarrow M\) passing through \(x_0 \in M\) is called an integral submanifold of the distribution \(\mathcal{D}\) if it is everywhere tangent to \(\mathcal{D}\), \(T\iota (TN) \subseteq \mathcal{D}\), where naturally \(\dim N \le r\). In the case \(\dim N = r\), the integral submanifold is called maximal (in dimension).

Definition 16 foliation
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Theorem 17 vector field Frobenius

Let \(\mathcal{D} \subseteq TM\) be an involutive tangent space distribution of rank \(r\le m = \dim M\). Then, for every \(x_0\in M\), there exists a maximal integral submanifold \(\iota \colon \mathbb {R}^n \hookrightarrow M\) of \(\mathcal{D}\) such that \(\iota (0) = x_0\). Moreover, these integral submanifolds collect into a \(r\)-dimensional foliation of \(M\) whose leaves are everywhere tangent to the distribution \(\mathcal{D}\).