This question was raised a while ago in a blog post by Terry Tao on the Euler-Arnold equation and he called it "quite tricky".

Has anyone in the meantime tried to formulate this question precisely, or to work it out?

I'm sketching here what I suppose the precise question should be (suggestions for improvement are very welcome). Let $M$ be a smooth orientable manifold and $\mu$ a volume form. Denote by

$$\mathfrak{g}=\text{SVect(M)}=\{X\in \text{Vect}(M)|\mathcal{L}_X\mu=0\}$$

the Lie algebra of divergence-free vector fields. The adjoint action of $\mathfrak{g}$ on itself is given by

$$ad: \mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}, \quad ad(X)(Y)=[X,Y]=\mathcal{L}_XY. $$

Thus, the Cartan-Killing form evaluated on $X,Y\in\mathfrak{g}$ should be related to the product of Lie derivative operators $\mathcal{L}_X\mathcal{L}_Y$.

Some basic questions are

  1. How to define the Cartan-Killing form? Suppose that $M$ is compact and $g$ is a Riemannian metric on $g$. We may regard the completion of $\mathfrak{g}$ to a Hilbert space with inner product $\langle X,Y \rangle=\int_Mg(X,Y)\mu$ (if $M$ is not compact, consider the sub-algebra of square integrable vector fields). Is $\mathcal{L}_X\mathcal{L}_Y$ a trace-class operator?
  2. What do we know about the relationships of SVect($M$) and $M$ that would help us identify the cases where the form can potentially be described more precisely? (I‘m thinking of things of the sort that are known about the Lie algebra of all vector fields, e.g. that M is uniquely determined by the algebraic structure of Vect($M$).)
  • $\begingroup$ Is it semisimple? Does it have a root system? $\endgroup$ – Ben McKay Sep 12 '18 at 19:22
  • $\begingroup$ I do not know. The first step for me would be to understand how the Cartan-Killing form could be defined for such an infinite dimensional Lie algebra. I have checked numerous books on the topic without success. $\endgroup$ – S.Surace Sep 13 '18 at 22:52
  • $\begingroup$ Why is it important to restrict to divergence-free vector fields? The inner product in the Riemannian case can be defined in the same way for general vector fields. Is it known that this approach leads to problems if one allows the vector fields to have non-zero divergence? $\endgroup$ – B K Sep 18 '18 at 11:47
  • $\begingroup$ @B K My intuition is that the problem for $\text{vect}(M)$ is probably even harder and the Cartan-Killing form more degenerate in some sense. But if we know how to tackle the divergence-free case then we might be able to generalize it. I quoted the question from Terence Tao, who asked it in the context of incompressible fluid mechanics. $\endgroup$ – S.Surace Sep 18 '18 at 13:00
  • $\begingroup$ I'm not sure of the answer at all, but the following paper may be relevant to this question. It doesn't really discuss Killing forms much but gives some good computations on the diffeomorphism group. math.toronto.edu/khesin/papers/H1-gafa.pdf $\endgroup$ – Gabe K Sep 18 '18 at 19:48

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