In a previous question, I asked an utterly trivial question, which Deane Yang correctly pointed out was utterly trivial. I will now ask a similar question, which is the one I meant to ask last time; I hope it's not similarly trivial.

I am working on $\mathbb R^n$, although in fact any manifold with volume form is good enough. And my question is local: I have an open neighborhood $U \ni 0$, and you are allowed to make it smaller if you want.

Recall that a constant-rank distribution $D$ on $U$ is a vector subbundle of the tangent bundle ${\rm T}U$. Let's fix the rank to be $k\leq n$, and suppose that everything is smooth: $\Gamma(D)$ is a $C^\infty$-submodule of $\Gamma({\rm T}U)$. The distribution is involutive if $\Gamma(D)$ is a Lie subalgebra of $\Gamma({\rm T}U)$ (over $\mathbb R$ — the Lie bracket on $\Gamma({\rm T}U)$ is not $C^\infty$-linear). The distribution $D$ is smooth if $\Gamma(D)$ is a free rank-$k$ module over $C^\infty$, i.e. if $\Gamma(D)$ has a basis $\{v_1,\dots,v_k\}$ so that $\Gamma(D) = \operatorname{Span}_{C^\infty}\{v_1,\dots,v_k\}$.

So, suppose that on $U \subseteq \mathbb R^n$ I have a smooth involutive rank-$k$ distribution. Given a basis $v_1,\dots,v_k \in \Gamma(D)$ (and I will use coordinates $x^1,\dots,x^n$ on $\mathbb R^n$, so I will write $v_a = \sum_i v_a^i(x) \frac{\partial}{\partial x^i}$), then I can define the structure coefficients $f_{a,b}^c(x)$, $a,b,c = 1,\dots,k$, via $[v_a,v_b] = \sum_c f_{a,b}^c v_c$, or, in coordinates: $$ \sum_i \left( v_a^i(x)\,\frac{\partial v_b^j}{\partial x^i} - v_b^i(x)\,\frac{\partial v_a^j}{\partial x^i}\right) = \sum_c f_{a,b}^c(x)\,v_c^j(x) $$

Question: Does there exist a basis $\{v_a\}$ for a given smooth involutive constant-rank distribution so that for each $a=1,\dots,k$ and each $x\in U$, we have $\displaystyle \sum_b f^b_{a,b}(x) = \sum_i \frac{\partial v^i_a(x)}{\partial x^i}$?

For example, by Frobenius theorem (my utterly trivial question), I can find a basis so that the LHS vanishes for each $a$. Or, by another of my trivial questions, I could make the basis entirely divergence-free. But I don't think I can simultaneously make the basis consist of divergence-free vector fields.

Question: If so, how many choices of such a basis do I have? Clearly ${\rm GL}(k,\mathbb R)$ acts on the set of choices; are there more?

  • 2
    $\begingroup$ Your title seems to be missing a verb. $\endgroup$ – S. Carnahan May 25 '10 at 0:15
  • $\begingroup$ Involutive distribution is integrable - it is a form of Frobenius theorem. Therefore it admits "flat" standard basis of vector fields. $\endgroup$ – Petya May 25 '10 at 2:06
  • $\begingroup$ I have no feeling for what you are looking for. Can you provide some motivation for why this condition is desirable or natural? $\endgroup$ – Deane Yang May 25 '10 at 2:45
  • $\begingroup$ @Scott: Oops, fixed. @Petya: yes, which was the comment that Deane made in my more trivial question. $\endgroup$ – Theo Johnson-Freyd May 25 '10 at 6:12
  • $\begingroup$ @Deane: I've been spending a while now thinking about formal integrals (Feynman diagrams). They were first defined for Morse functions. But in most interesting examples the function is invariant under some "gauge action". The most general type of gauge action is given by an integrable distribution. In the finite-dimensional case, my adviser asked me for a "purely combinatorial" proof that the formal integral does not depend on extra choices, and to generalize the proof from the physics literature I think I need the above condition. $\endgroup$ – Theo Johnson-Freyd May 25 '10 at 6:13

The answer to the question in the title is yes. More precisely I claim one may always find a divergence free basis where also all structure constants vanish. The following is a hopefully valid proof by induction on the rank k of the distribution: if k=1 then by an answer to your previous question one may always find a divergence free vector field spanning the distribution. In the case k>1 start by choosing a nowhere zero vector field $v_1$ which is in the distribution and divergence free. Also pick a n-1 dimensional submanifold $A_0\subset M$ transversal to $v_1$ (everything is local near a (fixed) point $p\in M$). On $A_0$ we have a rank $k-1$ involutive distribution (obtained by restricting the big one) and a volume form obtained by restricting $i_{v_1}\omega$ (where $\omega$ denotes the original volume form on $M$). By induction there is a base $\tilde{v}_2,\ldots,\tilde{v}_k$ of this distribution on $A_0$ which is in involution and divergence free for $i_{v_1}\omega$. With the flow of $v_1$ we extend these vector fields to obtain a full basis $v_1,\ldots,v_k$ of the rank $k$ distribution on $M$. The obtained basis will be in involution by construction. It remains to check that the $v_2,\ldots,v_k$ are divergence free for $\omega$. This follows from the fact that $\omega=dt\wedge i_{v_1}\omega$ where $t\in C^\infty(M)$ denotes the function which has value $0$ on $A_0$ and is otherwise determined by the flow of $v_1$. (Apply the usual rules for calculus with vector fields and forms.)

But you should really make sure I made no mistake.

If the proof is right it should also allow you to conclude that all manifolds with an integrable distribution and volume form look locally the same. (i.e. like the standard coordinate model). Hence the symmetry group is quite big (infinite dimensional I think).

  • $\begingroup$ Oh, well, so my guesses were all wrong. Thanks! $\endgroup$ – Theo Johnson-Freyd May 25 '10 at 16:49

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