# Wanted: A purely algebraic proof of the Frobenius theorem on distributions

Is there a purely algebraic proof of the Frobenius theorem? Here's a rough sketch of what i'm looking for:

Let $Der(R)$ denote the $R$-module of ($R$-valued) derivations of the algebra $R$ endowed with the lie bracket given by the commutator.

Definiton: A distribution $D$ is a submodule of $Der(R)$.

"Frobenius" Theorem - Under certain restriction on the base algebra $R$ (and on the algebra $S$ that will be introduced) the following holds:

A distribution $D \subset Der(R)$ is closed under the lie bracket of $Der(R)$ iff for every maximal ideal $m \subset R$ there exists an epimorphism $f: R \to S$, such that after localizing $R$ by $m$ and $S$ by $f(m)$ we have: $v \in D_m \iff \exists u \in Der(S)_{f(m)}$ satisfying $f_m \circ v = u \circ f_m$.

I'm sure there is a "nicer" algebraic formulation of this problem but that's the best i could do with my current knowledge - any improvement suggestions would be very welcome. Is there such a general theorem? Does it even make sense?

Denoting the exterior algebra of $Der(R)$ by $\mathcal{A}^*$. Am i right that the following equivalence is purely algebraic and no geometric input is neaded? (i did prove it, i think... need to be sure):

A distribution $D \subset Der(R)$ is closed under the lie bracket $\iff$ $I(D) = \bigcup_k \{\omega \in \mathcal{A}^k : \omega(m_1,...,m_k)=0 \text{ for every tuple of elements } \{m_i\}_{i \le k} \subset D \} \subset \mathcal{A}^*$ is a differential ideal. ($d I(D) \subset I(D)$).

• I see two problems with a formulation in the style you suggest. Your formulation looks like the dual version of the inclusion of one integral submanifold for the distribution to me. But you need an integral submanifold through each point. Second, the Frobenius theorem is local in nature (even in an analyitc category). Think about the example of a torus with the foliation coming from a line with irrational slope. Then each leaf of the distribution is dense, so restricting functions to the leaf certainly is not a surjection globally. – Andreas Cap Sep 4 '15 at 8:09
• One could also try to phrase Frobenius via projections to local leaf spaces rather than inclusions of integral submanifolds. In the language you propose this would mean imposing conditions on the joint kernel of the elements of $D$. However, the local-global problem is at least as bad in such a formulation. – Andreas Cap Sep 4 '15 at 8:12
• @AndreasCap As for the first issue, I was thinking there must be a reasonable algebraic condition on $f$ that would ensure that it arises as a dual of an embedding. I didn't quite get the second part, I'm not well versed in foliations I must say... – Saal Hardali Sep 4 '15 at 8:15
• The trouble is that the inclusion of a leaf of a foliation is only locally an embedding. Globally, one gets what is called an initial submanifold, but things can easily get pathological, for example with each leaf being dense. (And still there is the issue that you need many integral submanifolds and not just one.) The second comment refers a different interpretation of the Frobenius theorem, which roughly is via functions which are constant along each leaf rather than functions on the individual leafs. – Andreas Cap Sep 4 '15 at 10:11
• @AndreasCap got it! I completely forgot the local stuff when passing to the algebra. Fixed it so that it's local now. Thanks for the help. My hope is to have someone who's familiar with commutative algebra formulate the nicest possible version of this whereas i'm sure my formulation is the worst possible. – Saal Hardali Sep 4 '15 at 10:39

Here's a (reasonably) algebraic proof of the Frobenius theorem (in the end there's a comment about the non-algebraic inputs of the proof):

Frobenius theorem: Let $M$ be an $n$-dimensional smooth manifold and let $D \subset TM$ be an involutive sub-bundle $[D,D] \subset D$ of rank $r$. Then $D$ is integrable

We first prove a slightly surprising lemma:

Lemma 1: Let $D \subset TM$ be as in the situation above. Then locally around any point $p\in M$ there's a local frame $\{X_1,...,X_r\} \subset \mathfrak{X}(U)$ for $D|_U$ with vanishing lie bracket $[X_i,X_j]=0$.

Proof:

Let $U$ be a small enough neighborhood of $p\in M$ s.t. both $\mathfrak{X}(U)$ and $\Gamma(U,D)$ are free as modules over $R:=C^{\infty}(U)$-modules.

Let $\{Y_1,...,Y_r\}$ and $\{\partial_1,...,\partial_n\}$ bases for $\Gamma(U,D)$ and $\mathfrak{X}(U)$ respectively. There's a unique matrix $A \in M_{r \times n}(R)$ satisfying:

$$\left( \begin{matrix} Y_1 \\ \vdots \\ Y_r \\ \end{matrix} \right) = A \left( \begin{matrix} \partial_1 \\ \vdots \\ \partial_n \\ \end{matrix} \right)$$

The rank of $A$ must be $r$ and so up to relabeling of indices we may assume $A = (B|C)$ for some unique $B \in GL_r(R)$ and $C \in M_{r \times n-r}(R)$.

We define the frame $\{X_1,...,X_r\}$ by:

$$\left( \begin{matrix} X_1 \\ \vdots \\ X_r \\ \end{matrix} \right) = B^{-1} \left( \begin{matrix} Y_1 \\ \vdots \\ Y_r \\ \end{matrix} \right) = B^{-1}A \left( \begin{matrix} \partial_1 \\ \vdots \\ \partial_n \\ \end{matrix} \right)$$

Since $B^{-1}A=(Id|B^{-1}C)$ we have $X_j = \partial_j+Z_j$ for some unique $Z_j \in Span \{\partial_{r+1},..., \partial_n\}$.

By the above we have $[X_i,X_j] \in Span \{\partial_{r+1},..., \partial_n\}$.

We recall now that $D$ is involutive so we also have $[X_i,X_j] \in Span \{X_1,...,X_r\}$.

But clearly the module $Span \{X_1,...,X_r\} / Span \{\partial_{r+1},..., \partial_n\} \cong Span \{\partial_1,...,\partial_r\}$ is free of rank $r$ and so we must have $Span \{X_1,...,X_r\} \cap Span \{\partial_{r+1},..., \partial_n\} = \{0\}$.

QED.

We actually proved something slightly stronger. Namely that for any involutive distribution $\mathcal{D} \subset \mathcal{T}_X$ there exists locally a retraction $\mathcal{T}_X|_U \to \mathcal{D}|_U$ and a local frame of commuting vector fields for $D|_U$ which preserve $\mathcal{I} = ker (\mathcal{T}_X|_U \to \mathcal{D}|_U)$ under the lie bracket.

Applying the theorem again to $\mathcal{I}$ we get locally a retraction $\mathcal{T}_X|_V \to \mathcal{I}|_V$ and a local frame for $\mathcal{I}|_V$ consisting of commuting vector fields preserving $ker(\mathcal{T}_X|_V \to \mathcal{I}|_V) \cong \mathcal{D}|_V$ under lie bracket. But since $\mathcal{D}|_V \cap \mathcal{I}|_V = \{0\}$ and both are involutive the frames for $\mathcal{D}|_V$ and $\mathcal{I}|_V$ must commute with each other.

We therefore have in fact proven the following very strong and surprising statement:

Theorem: Any involutive distribution $D$ is locally spanned by a frame of commuting vector fields which extends to a local frame for the entire tangent bundle also consisting of mutually commuting vector fields.

Proof of Frobenius theorem:

The dual frame in $\Omega^1_X$ of a commuting local frame for $TX$ consists entirely of closed 1-forms (this is a simple computation). By taking the dual of the frame in the theorem above we may apply Poincare's lemma to conclude that by sufficiently shrinking $U$ the basis of 1-forms can be lifted to a coordinate chart. A suitable projection will give the integrability of $D$.

QED.

The only non-algebraic statements in the proof are the Poincare lemma (a closed form is locally exact) and the existence of a locally commuting frame field (which is apparently true in the algebraic setting). A possible upshot of this is that the theorem generalizes immediately to the holomorphic case (by working everywhere with sheaves of modules over the sheaf of holomorphic functions and using the holomorphic poincare lemma).

• I did not check your proof of the lemma. Seemingly your deduction from the lemma to the theorem does not work. Consider the simplest case that $D$ is generated by a single vector field. Then the integral of an algebraic vector field need not be algebraic. The problem in your argument is that, the projection to $D$ in question is in fact a field of projections, which is not induced from a map of manifolds. – Yai0Phah May 16 at 14:59
• @Yai0Phah I'm not entirely sure what I intended with that argument. I do believe though that I've never claimed the frobenius theorem holds in the algebraic setting. It seems that my claim is that one of the non-algebraic ingredients missing is the Poincare lemma which indeed fails. Aside from that given the lemma it's clear that the frobenius theorem holds for a given algebraic distribution iff the mutually commuting collection of vector fields integrates to an algebraic action. – Saal Hardali May 16 at 17:52
• @Yai0Phah To be clear about what i mean, in the 1-dimensional case indeed every distribution is closed under lie bracket like you said. Choose a generating vector field. Algebraic integrability of this vector field is equivalent to the differential form dual to it being algebraically exact. This is of course not true even locally because Poincare lemma doesn't hold algebraically. – Saal Hardali May 16 at 17:58
• The first thing is that you need to get a closed differential form from the ODE associated to the vector field. Simply put, if your strategy works, then seemingly you reduce integrating vector fields to finite steps of algebraic manipulations and integrals and derivatives, which does not seem to be possible. My impression is that integrating vector fields is essentially transcendental. Indeed, known proofs of Picard-Lindelöf theorem involves compactness argument. – Yai0Phah May 16 at 18:35
• @Yai0Phah Where have I claimed that it's reduced to "finite steps". In the 1-dimensional case for example my proof does nothing. You're stuck with the problem of integrating a single vector field which is, like you said, a transcendental operation in general. All my lemma does is reduce the frobenius theorem to the problem of simultaneously integrating a finite collection of commuting vector fields. Nothing more nothing less. – Saal Hardali May 16 at 18:48