Frobenius Theorem

mathdifferential geometry

Let \((M,g)\) be a smooth Riemannian manifold. Write \(\nabla\) for its Levi–Civita connection and \(T_pM\) for the tangent space at \(p\). The coordinate conventions are those in tangent spaces and derivations.

The definitions of flows and vector fields used below are developed in Appendix Vector Fields and Their Flows. Suppose we prescribe a tangent subspace of fixed dimension at every point. When are these subspaces the tangent spaces of local submanifolds? Frobenius’ theorem gives a differential criterion: vector fields taking values in the prescribed subspaces must remain in them after taking a Lie bracket. Much of the terminology is introduced in Appendix Vector Fields and Their Flows.

Definition 1. With \(0 \leq n \leq m \leq k\), let \(M \subset \mathbb R^k\) be an \(m\)-dimensional manifold. Let \(E \subset TM\) be a subbundle of rank \(n\). We say \(E\) is involutive if for any two vector fields \(X,Y \in \operatorname{Vec}(M)\) taking values in \(E\) (i.e., \(X_p, Y_p \in E_p\) for all \(p \in M\)), we have \([X,Y]_p \in E_p\) for all \(p \in M\).

It is equivalent to test involutivity on local sections. Indeed, near any chosen point, multiply a local section by a bump function equal to one on a smaller neighbourhood and supported in its domain, then extend by zero. The extension is a global smooth section of \(E\), and its bracket near the chosen point agrees with the original local bracket. This justifies using local frames in the proof.

Definition 2. With the qualifiers as in Definition 1, we say \(E\) is integrable if for all \(p \in M\), there exists a submanifold \(N \subset M\) such that \(p \in N\), and \(T_q N = E_q\) for all \(q \in N\).

Definition 3. With the qualifiers as in Definition 1, we say a foliation box for \(E\) is a coordinate chart \(\varphi : U \to \Omega\) defined on an open subset \(U \subset M\) that takes values in an open set \(\Omega \subset \mathbb R^n \times \mathbb R^{m-n}\), where \(\Omega \cap (\mathbb R^n \times \{y\})\) is connected for all \(y \in \mathbb R^{m-n}\) and for all \(p \in U\) and \(v \in T_p M\), we have \(v \in E_p \iff \dif \varphi_p(v) \in \mathbb R^n \times \{0\}\).

These definitions describe the same local geometry in three forms: closure under Lie brackets, existence of integral submanifolds, and coordinates in which those submanifolds are slices.

Theorem 4. Suppose \(M \subset \mathbb R^k\) is an \(m\)-dimensional manifold, and \(E \subset TM\) is a subbundle of rank \(n\). Then, the following statements are equivalent:

  1. \(E\) is involutive;

  2. \(E\) is integrable;

  3. For every \(p_0 \in M\), there exists a foliation box \(\varphi : U \to \Omega\) with \(p_0 \in U\).

The result is local and does not use the embedding into \(\mathbb R^k\) or a Riemannian metric. Smoothness and constant rank of the subbundle are the essential hypotheses. When \(n=0\), the integral submanifolds are single points and every small chart is a foliation box. When \(n=m\), \(E=TM\), and a connected coordinate neighbourhood itself is an integral submanifold. The following construction covers the intermediate ranks.

The main implication constructs a foliation box from involutivity. We will compose flows to form a coordinate map. The lemma below ensures that the tangent directions already in the subbundle remain there under these flows.

Lemma 5. Let \(E \subset TM\) be an involutive subbundle of the tangent bundle. Let \(X \in \operatorname{Vec}(M)\) be a complete vector field such that \(X_p \in E_p\) for all \(p \in M\). Let \(\phi^t : \mathbb R \to \operatorname{Diff}(M)\) denote the flow of \(X\). Then for all \(t \in \mathbb R\) and \(p\in M\),

\[\begin{aligned} \dif \phi^t_p (E_p) = E_{\phi^t (p)} ~. \end{aligned}\]

Two details in the proof below make the local-to-global step precise. If a local frame of \(E\) has block component matrix \(\binom{B(z)}{C(z)}\), choose coordinates so that \(B(p_0)\) is invertible. On a smaller neighbourhood \(B(z)\) stays invertible, and the graph matrix is \(A(z)=C(z)B(z)^{-1}\), which is smooth. In this chart, membership in \(E\) is the closed equation \(\eta-A(z)\zeta=0\), so \(E\) is a closed subset of \(TM\).

A compactly supported smooth vector field is complete: an integral curve outside its compact support is stationary; while it moves it stays in that support. A finite family of local ODE existence neighbourhoods covering the compact support gives a common short continuation time. A finite maximal endpoint would therefore allow extension and contradict maximality. This is why the bump-function extension used in both proofs supplies global flows without assuming completeness of the Riemannian metric.

Proof of Theorem 4. We begin with the two implications that follow directly from coordinate slices and the intrinsic Lie bracket. First, suppose 4 holds. Then, let \(\varphi: U \to \Omega\) denote such a foliation box with \(p_0 \in U\). We can show that there is a submanifold \(N \subset M\) such that \(T_q N = E_q\) for all \(q \in N\) and \(p_0 \in N\). In fact, we can simply take

\[\begin{aligned} N = \varphi^{-1} \big( \Omega \cap (\mathbb R^n \times \{y_0\}) \big) ~, \end{aligned}\]

with \((x_0, y_0) = \varphi(p_0) \in \Omega\). Clearly, \(N\) is a submanifold as the preimage of a coordinate slice under the chart \(\varphi\) (in particular, it inherits the subspace topology and smooth structure), and by definition of the foliation box, its tangent space \(T_q N\) is exactly the fibre \(E_q = \{v \in T_q M : \dif \varphi_q(v) \in \mathbb R^n \times \{0\}\}\). This shows 4 implies 4.

Now suppose 4 holds. We want to show that \([X,Y]_p \in E_p\) if \(X_p,Y_p \in E_p\) (for all \(p \in M\)). To see this, fix \(p \in M\). Since \(X_q, Y_q \in E_q = T_q N\) for some integral submanifold \(N \subset M\) where \(p \in N\) and \(E_q = T_q N\) for all \(q \in N\), the usual rules apply, so \(X|_N\) and \(Y|_N\) are both well-defined smooth vector fields tangent to \(N\). Therefore, the restriction of \([X,Y]\) to \(N\) is identically the intrinsic Lie bracket on \(N\). As such, \([X,Y]_p \in T_p N = E_p\), as desired.

Finally, suppose 4 holds. Fix a point \(p_0 \in M\). By standard local extension arguments, there must exist vector fields \(X_1, \cdots, X_n \in \operatorname{Vec}(M)\) such that \(X_i (p) \in E_p\) for all \(i\) and \(p\), and \(X_1(p_0), \cdots, X_n(p_0)\) together form a basis of \(E_{p_0}\). Repeating the argument, there must exist vector fields \(Z_1, \cdots, Z_{m-n} \in \operatorname{Vec}(M)\) such that \(\{X_1(p_0), \cdots, X_n(p_0), Z_1(p_0), \cdots, Z_{m-n}(p_0)\}\) forms a basis of \(T_{p_0} M\). It is without loss of generality to assume that all vector fields \(X_i\) and \(Z_j\) have compact support (for instance, by multiplying local extensions by a bump function), implying they are complete vector fields on \(M\). Let \(\{\phi^t_i\}_{1 \leq i \leq n}\) and \(\{\psi^t_j\}_{1 \leq j \leq m-n}\) denote the flows of vector fields \(\{X_i\}_{1 \leq i \leq n}\) and \(\{Z_j\}_{1 \leq j \leq m-n}\), respectively. Define

\[\begin{aligned} \Phi : \mathbb R^n \times \mathbb R^{m-n} \to M : (x,y) \mapsto \phi_1^{x_1} \circ \cdots \circ \phi_n^{x_n} \circ \psi_1^{y_1} \circ \cdots \circ \psi_{m-n}^{y_{m-n}} (p_0) ~. \end{aligned}\]

By Lemma 5, the flows \(\phi_k\) of the vector fields \(X_k\) preserve \(E\), meaning \(\dif \phi_k^t (p) E_p = E_{\phi_k^t (p)}\). By the chain rule, pushing forward the vector field \(X_i\) by these intermediate flows keeps it taking values in \(E\). Thus, in local coordinates, we have

\[\begin{aligned} \frac{\partial \Phi}{\partial x_i} (x,y) = \dif \left( \phi_1^{x_1} \circ \cdots \circ \phi_{i-1}^{x_{i-1}} \right)_{p'} X_i(p') \in E_{\Phi(x,y)} ~, \end{aligned}\]

where \(p' = \phi_i^{x_i} \circ \cdots \circ \psi_{m-n}^{y_{m-n}}(p_0)\), for all \((x,y) \in \mathbb R^n \times \mathbb R^{m-n}\). Because \(\frac{\partial \Phi}{\partial x_i} (0,0) = X_i(p_0)\) and \(\frac{\partial \Phi}{\partial y_j} (0,0) = Z_j(p_0)\), the differential \(\dif \Phi(0,0) : \mathbb R^n \times \mathbb R^{m-n} \to T_{p_0} M\) is bijective. The inverse function theorem then implies there exists an open neighbourhood \(\Omega \subset \mathbb R^n \times \mathbb R^{m-n}\) of the origin (which we can safely restrict to be a product of open balls to smoothly satisfy connectedness of slices) such that \(U \triangleq \Phi(\Omega) \subset M\) is an open neighbourhood of \(p_0 \in M\), and \(\Phi|_{\Omega} : \Omega \to U\) is a diffeomorphism. Therefore, \(\left\{ \frac{\partial \Phi}{\partial x_i} (x,y) \right\}_{1 \leq i \leq n}\) are linearly independent for every \((x,y) \in \Omega\), so these form a basis of \(E_{\Phi(x,y)}\). As such, \(\varphi \triangleq (\Phi|_\Omega)^{-1} : U \to \Omega\) is a foliation box. ◻

As argued above, the role of Lemma 5 is to complete the equivalence statement in Frobenius’ theorem. As such, it remains to prove Lemma 5. We prove it by expressing the subbundle locally as the graph of a matrix. Involutivity becomes an identity for that matrix, and uniqueness for a linear differential equation gives the required invariance.

Proof of Lemma 5.

  1. Both this step and the next step will be devoted to proving the following fact: if \(E \subset TM\) is an involutive subbundle, and \(\beta : \mathbb R^2 \to M\) is a smooth map such that

    \[\begin{aligned} \frac{\partial \beta}{\partial s} (s,0) \in E_{\beta(s,0)} \quad,\quad \frac{\partial \beta}{\partial t} (s,t) \in E_{\beta(s,t)} \quad \forall s,t ~, \end{aligned}\tag{1}\]

    then \(\frac{\partial \beta}{\partial s} (s,t) \in E_{\beta(s,t)}\) for all \(s,t\). To see that this statement is intuitively true, note that the conditions in (1) essentially imply that \(\beta\) acts as a smooth integral curve along the “\(t\)-axis”, so we can reasonably expect that \(\beta\) admits a smooth first variation along the “\(s\)-axis” by calculus of variations.

    To see that this is formally true, pick any \(p_0 \in M\), and we choose a coordinate chart \(\phi: U \to \Omega\) for an open set \(U \subset M\) with values in \(\Omega \subset \mathbb R^n \times \mathbb R^{m-n}\), where \(p_0 \in U\) and \(\dif \phi_{p_0}(E_{p_0}) = \mathbb R^n \times \{0\}\). Then for every \(p \in U\), the linear subspace \(\dif \phi_p(E_p) \subset \mathbb R^n \times \mathbb R^{m-n}\) is the graph of a matrix \(A \in \mathbb R^{(m-n) \times n}\). This can be done by shrinking \(U\) to a small enough neighborhood in \(M\) if necessary. Because this is possible, there must exist a smooth map \(A : \Omega \to \mathbb R^{(m-n) \times n}\) (which evaluates to a matrix, not necessarily linear in its arguments) such that

    \[\begin{aligned} \dif \phi_p(E_p) = \{(\zeta, A(x,y) \zeta) : \zeta \in \mathbb R^n\} ~,~\quad (x,y) = \phi(p) \in \Omega \quad \forall p \in U ~. \end{aligned}\]

    We can now define the linear maps

    \[\begin{aligned} \frac{\partial A}{\partial x} (x,y) \cdot \zeta &\triangleq \sum_{1 \leq i \leq n} \zeta_i \frac{\partial A}{\partial x_i} (x,y) \in \mathbb R^{(m-n) \times n} ~, \\ \frac{\partial A}{\partial y} (x,y) \cdot \eta &\triangleq \sum_{1 \leq j \leq m-n} \eta_j \frac{\partial A}{\partial y_j} (x,y) \in \mathbb R^{(m-n) \times n} ~. \end{aligned}\]

    The remainder of this step will be devoted to proving two claims. First, we claim that for \((x,y) \in \Omega\), \(\zeta,\widetilde \zeta \in \mathbb R^n\), and \(\eta,\widetilde \eta \in \mathbb R^{m-n}\) defined by \(\eta \triangleq A(x,y) \zeta\) and \(\widetilde \eta \triangleq A(x,y) \widetilde \zeta\), then

    \[\begin{aligned} \left( \frac{\partial A}{\partial x}(x,y) \cdot \zeta + \frac{\partial A}{\partial y}(x,y) \cdot \eta \right) \widetilde \zeta = \left( \frac{\partial A}{\partial x}(x,y) \cdot \widetilde \zeta + \frac{\partial A}{\partial y}(x,y) \cdot \widetilde \eta \right) \zeta ~. \end{aligned}\]

    To see this, the graphs of the matrices \(A(z)\) for \(z = (x,y) \in \Omega\) determine a subbundle \(\widetilde E \subset \Omega \times \mathbb R^m\) with the fibres

    \[\begin{aligned} \widetilde E_z \triangleq \{(\zeta,\eta) \in \mathbb R^n \times \mathbb R^{m-n} : \eta = A(x,y) \zeta \} ~. \end{aligned}\]

    This subbundle is involutive, because under the bundle isomorphism \(\rho : TM|_U \to \Omega \times \mathbb R^m : (p,v) \mapsto (\phi(p), \dif \phi_p(v))\), \(\widetilde E\) is exactly the image of \(E|_U\). Furthermore, fix \(\zeta, \widetilde \zeta \in \mathbb R^n\); define vector fields \(\mathscr Z, \widetilde{\mathscr Z} : \Omega \to \mathbb R^n \times \mathbb R^{m-n}\) evaluated by \(\mathscr Z(z) \triangleq (\zeta,A(z) \zeta)\) and \(\widetilde{\mathscr Z}(z) \triangleq (\widetilde\zeta, A(z)\widetilde\zeta)\). In other words, \(\mathscr Z\) and \(\widetilde{\mathscr Z}\) are sections of \(\widetilde E\), so their Lie bracket in local coordinates \([X,Y] = D Y \cdot X - D X \cdot Y\) is given by

    \[\begin{aligned} [\mathscr Z,\widetilde{\mathscr Z}](z) = \left(0, \left(\dif A(z) \mathscr Z(z)\right)\widetilde\zeta - \left(\dif A(z) \widetilde{\mathscr Z}(z)\right) \zeta \right) ~. \end{aligned}\]

    Because \(\widetilde E\) is involutive, it follows that the Lie bracket above must evaluate to a vector taking values in the graph of \(A\). Since its first component is zero, it must be that \([\mathscr Z,\widetilde{\mathscr Z}](z)\) vanishes entirely to lie in \(A\)’s span. The claim is therefore established.

    Another claim is that if \(I,J \subset \mathbb R\) are open intervals, \(z = (x,y) : I \times J \to \Omega\) is a smooth map, and \(s_0 \in I\) and \(t_0 \in J\) are fixed points such that

    \[\begin{aligned} \frac{\partial y}{\partial s} (s_0,t_0) &= A(x(s_0,t_0), y(s_0,t_0)) \frac{\partial x}{\partial s} (s_0,t_0) ~, \\ \frac{\partial y}{\partial t} (s,t) &= A(x(s,t), y(s,t)) \frac{\partial x}{\partial t} (s,t) \end{aligned}\tag{2}\]

    for all \(s \in I\), \(t \in J\), then

    \[\begin{aligned} \frac{\partial y}{\partial s}(s_0,t) = A(x(s_0,t),y(s_0,t)) \frac{\partial x}{\partial s}(s_0,t) \quad \forall t \in J ~. \end{aligned}\tag{3}\]

    Clearly, the conclusion (3) holds for \(t=t_0\). Suppressing the arguments \((s,t)\) in favour of simply using \(z=(x,y)\) (while noting that this is an abuse of notation), we have

    \[\begin{aligned} \frac{\partial}{\partial t} \left( \frac{\partial y}{\partial s} - A \cdot \frac{\partial x}{\partial s} \right) &= \frac{\partial^2 y}{\partial s \partial t} - A \frac{\partial^2 x}{\partial s \partial t} - \left( \frac{\partial A}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial A}{\partial y} \cdot \frac{\partial y}{\partial t} \right) \frac{\partial x}{\partial s} \nonumber \\ &\overset{(2)}{=} \frac{\partial^2 y}{\partial s \partial t} - A \frac{\partial^2 x}{\partial s \partial t} - \left( \frac{\partial A}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial A}{\partial y} \cdot \bigg( A \frac{\partial x}{\partial t} \bigg) \right) \frac{\partial x}{\partial s} \nonumber \\ &\overset{\text{previous claim}}{=} \frac{\partial^2 y}{\partial s \partial t} - A \frac{\partial^2 x}{\partial s \partial t} - \left( \frac{\partial A}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial A}{\partial y} \cdot \bigg( A \frac{\partial x}{\partial s} \bigg) \right) \frac{\partial x}{\partial t} \nonumber \\ &= \frac{\partial^2 y}{\partial s \partial t} - A \frac{\partial^2 x}{\partial s \partial t} - \left( \frac{\partial A}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial A}{\partial y} \cdot \bigg( \frac{\partial y}{\partial s} - \left( \frac{\partial y}{\partial s} - A \frac{\partial x}{\partial s} \right) \bigg) \right) \frac{\partial x}{\partial t} ~. \end{aligned}\tag{4}\]

    Notice also that by differentiating (2) with respect to \(s\), we have

    \[\frac{\partial^2 y}{\partial s \partial t} = \bigg( \frac{\partial A}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial A}{\partial y} \cdot \frac{\partial y}{\partial s} \bigg) \frac{\partial x}{\partial t} + A \frac{\partial^2 x}{\partial s \partial t} \implies \frac{\partial^2 y}{\partial s \partial t} - A \frac{\partial^2 x}{\partial s \partial t} = \bigg( \frac{\partial A}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial A}{\partial y} \cdot \frac{\partial y}{\partial s} \bigg) \frac{\partial x}{\partial t} ~,\]

    Substitution cancels the first terms, leaving

    \[\begin{aligned} (4) = \left( \frac{\partial A}{\partial y} \cdot \bigg( \frac{\partial y}{\partial s} - A \frac{\partial x}{\partial s} \bigg) \right) \frac{\partial x}{\partial t} ~. \end{aligned}\]

    Defining a smooth curve \(\eta : J \to \mathbb R^{m-n}\) (for fixed \(s_0\)) via

    \[\begin{aligned} \eta(t) \triangleq \frac{\partial y}{\partial s} (s_0,t) - A \left( x(s_0,t), y(s_0,t) \right) \frac{\partial x}{\partial s} (s_0,t) ~. \end{aligned}\]

    Clearly, by (2) and the equality above, we have that \(\eta\) satisfies the linear differential equation

    \[\begin{aligned} \dot \eta(t) = \left( \frac{\partial A}{\partial y} \left( x(s_0,t), y(s_0,t) \right) \cdot \eta(t) \right) \frac{\partial x}{\partial t} (s_0,t) \quad \text{with} \quad \eta(t_0) = 0 ~, \end{aligned}\]

    so by the uniqueness of solutions to linear ODEs, \(\eta(t)=0\) for all \(t \in J\). This therefore implies the second claim.

  2. Suppose \(\beta : \mathbb R^2 \to M\) is as at the beginning of Step 1. Fix \(s_0 \in \mathbb R\). Consider the set

    \[\begin{aligned} W \triangleq \left\{t \in \mathbb R : \frac{\partial \beta}{\partial s}(s_0,t) \in E_{\beta(s_0,t)}\right\} ~, \end{aligned}\]

    and the second claim of Step 1 implies \(W\) is an open set. Because \(0 \in W\), \(W \neq \emptyset\) (this is true by (1)). Furthermore, \(W\) is closed in \(\mathbb R\) because \(E\) is a closed subbundle in \(TM\) and partial derivatives are continuous. By the connectedness of \(\mathbb R\), it follows that \(W = \mathbb R\). By the fact that \(s_0 \in \mathbb R\) is picked arbitrarily, we’ve established the desired conclusion \(\frac{\partial \beta}{\partial s} (s,t) \in E_{\beta(s,t)}\) for all \(s,t\).

  3. We can now use Steps 1 and 2 to show that Lemma 5 follows. To see this, let \(X \in \operatorname{Vec}(M)\) denote a complete vector field on \(M\) such that \(X_p \in E_p\) for all \(p \in M\). Let \(\phi^t\) denote the flow of \(X\). Pick a point \(p_0 \in M\) and a vector \(v_0 \in E_{p_0}\). Then, there exists a vector field \(Y\) on \(M\) with values in \(E\), such that \(Y(p_0) = v_0\); in particular, choose a vector field that has compact support, implying it is complete. Therefore, Picard-Lindelöf implies that there exists a global solution \(\gamma : \mathbb R \to M\) of the initial value problem

    \[\begin{aligned} \dot \gamma(s) = Y(\gamma(s)) \quad \text{with } \gamma(0) = p_0 ~. \end{aligned}\]

    Let \(\beta : \mathbb R^2 \to M : (s,t) \mapsto \phi^t(\gamma(s))\) for \(s,t \in \mathbb R\). Then, it clearly follows that

    \[\begin{aligned} \frac{\partial \beta}{\partial s} (s,0) &= \dot \gamma(s) = Y(\gamma(s)) \in E_{\beta(s,0)} ~, \text{and } \\ \frac{\partial \beta}{\partial t} (s,t) &= X(\beta(s,t)) \in E_{\beta(s,t)} \end{aligned}\]

    for all \(s,t \in \mathbb R\). It then follows from a direct application of Step 1’s main fact (proven globally in Step 2) that

    \[\begin{aligned} \dif \phi^t_{p_0} (v_0) = \dif \phi^t_{\gamma(0)} (\dot \gamma(0)) = \frac{\partial \beta}{\partial s}(0,t) \in E_{\beta(0,t)} = E_{\phi^t(p_0)} \quad \forall t \in \mathbb R ~. \end{aligned}\]

    Since \(v_0 \in E_{p_0}\) was arbitrary and \(\dif \phi^t_{p_0}\) is a linear map, this implies \(\dif \phi^t_{p_0} (E_{p_0}) \subset E_{\phi^t(p_0)}\). Applying identical logic backwards on the reversed complete vector field \(-X\) with its valid flow \(\phi^{-t}\) yields the reverse inclusion \(E_{\phi^t(p_0)} \subset \dif \phi^t_{p_0} (E_{p_0})\), establishing equality \(\dif \phi^t_p (E_p) = E_{\phi^t(p)}\).

The propagation argument is not assuming the desired integral submanifolds in advance. The graph-bracket calculation gives a homogeneous linear ODE for the defect \(\eta(t)=y_s-Ax_s\), with coefficient map \(u\mapsto(D_yA\cdot u)x_t\). Zero initial defect remains zero by uniqueness. Around any time where the defect vanishes, restrict the two-parameter map to a small rectangle lying in one graph chart; this proves that the zero set of times is open. Closedness follows from the subbundle equation, and connectedness propagates the identity along the entire time axis. The inverse flow then converts inclusion of fibres into equality.

This proves Lemma 5. ◻

Reading: Carmo (1992), Riemannian geometry.

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