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. ($d I(D) \subset I(D)$).

Saal Hardali
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  • 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 04 '15 at 08:09
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    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 04 '15 at 08:12
  • @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 04 '15 at 08:15
  • 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 04 '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 04 '15 at 10:39
  • Sorry, but I couldn't understand what is your definition of an integrable distribution (is it your second condition after the "iff" in the statement of the Frobenius theorem?). In this case, wouldn't the correct statement be an immersion $f: R \rightarrow S$ that does not depend on the maximal ideal $m$? – user40276 Sep 07 '15 at 17:41
  • @user40276 What does it mean for $f: R \to S$ to be an immersion when $R$ and $S$ are rings? – Saal Hardali Sep 07 '15 at 17:43
  • I mean after applying $Spec$. My point is that I couldn't understand what's definition and what's proposition in your question. – user40276 Sep 07 '15 at 17:45
  • @user40276 Integrability is the wierd part at the end of the proposition. You want that locally every derivation of $S$ is (-locally) "$f$-related" to some element in the (locallization of the) distribution (and vice versa). Would have been much clearer if I knew how to draw diagrams in tex. – Saal Hardali Sep 07 '15 at 17:48
  • I don't know anything about Frobenius theorem, but it seems like smoothness is necessary for it. If that's the case maybe you want your rings to be regular or normal? – user347489 Apr 23 '17 at 02:16

1 Answers1


This answer is not correct as stated! See the comments below. The correct statement is more subtle. I will correct this eventually i hope.

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 $[X_i,X_j]=0$.


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 $Span \{X_1,...,X_r\} \cap Span \{\partial_{r+1},..., \partial_n\} = \{0\}$.


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 $D$.


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).

Saal Hardali
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  • 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 '20 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 '20 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 '20 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](https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem) involves compactness argument. – Yai0Phah May 16 '20 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 '20 at 18:48
  • No, after reducing to commuting vector fields, you claim that then the only non-algebraic operation that you need is Poincaré's lemma, which I am questioning: I am claiming that you need more, nontrivial transcendental inputs. And one should be careful about the terminology "1-dim case" because it is that $D$ is generated by one, but not that the manifold is 1-dim. I would like to add that in the proof of Poincaré's lemma, one only need to integrate functions in finite steps. – Yai0Phah May 16 '20 at 18:53
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/108090/discussion-between-saal-hardali-and-yai0phah). – Saal Hardali May 16 '20 at 18:57
  • Sorry, I did not click that "continue in chat" and now the room is deleted. There are two steps: first reduction to commuting vector fields, which is indeed algebraic (I was not clear about that when I wrote the comment, but afterwards I read Lee's book and find that it is the same). The problem is about the second step: you integrate the commuting vector fields, which requires a transcendental input from the theory of ODE, not just Poincaré's lemma. – Yai0Phah Jun 03 '20 at 17:20
  • @Yai0Phah I agree with that assesment. The reason rhe second argument fails is like you pointed out although any distribution can indeed be given locally a commuting frame there's no guarantee that this frame can be completed to a commuting basis in general. When it can be Poincare lemma suffices to prove integrability. However the problem of completing this commuting frame to a locally commuting basis itself requires us to solve some PDE's so it is not algebraic. – Saal Hardali Jun 03 '20 at 17:37
  • So I guess the second step can also be broken to two steps. 1) Completing the local commuting frame to a commuting basis. 2) Using poincare lemma to integrate to a chart. The first step in general requires us to simultaneously solve a system of first order PDE's. – Saal Hardali Jun 03 '20 at 17:42