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

  • $\begingroup$ 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. $\endgroup$ – Andreas Cap Sep 4 '15 at 8:09
  • 1
    $\begingroup$ 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. $\endgroup$ – Andreas Cap Sep 4 '15 at 8:12
  • $\begingroup$ @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... $\endgroup$ – Saal Hardali Sep 4 '15 at 8:15
  • $\begingroup$ 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. $\endgroup$ – Andreas Cap Sep 4 '15 at 10:11
  • $\begingroup$ @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. $\endgroup$ – 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 $[X_i,X_j]=0$.


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.