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For a finite-dimensional smooth manifold $M$, let $\mathrm{Diff}(M)$ be its diffeomorphism group.

Suppose we are given a $2$-tensor $\mathcal{K}$ on $M$, and let $$\mathrm{Diff}_{~\mathcal{K}}(M) = \{~\phi \in \mathrm{Diff}(M) : \phi^*(\mathcal{K}) = \mathcal{K}~\}.$$

Some special cases are:

  • If $\mathcal{K}$ is a metric, then $\mathrm{Diff}_{~\mathcal{K}}(M)$ is just the isometry group of $\mathcal{K}$. It is well-known that this is a finite-dimensional Lie group. In some cases, this group might even be trivial, consisting of the identity map alone.
  • On the other hand, if $\mathcal{K} = 0$, then $\mathrm{Diff}_{~\mathcal{K}}(M) = \mathrm{Diff}(M)$, which is not a (finite-dimensional) Lie group.
  • The previous example might seem too extravagant, but even if $\mathcal{K}$ is a non-degenerate $2$-form, it might happen that $\mathrm{Diff}_{~\mathcal{K}}(M)$ is still too big to be a (again, finite-dimensional) Lie group. This is what happens with symplectic forms, for instance.

These examples show that this group depends quite a bit on conditions we impose on $\mathcal{K}$, and it's not clear to me exactly how this happens, or why. So, my questions are:

1) Is there a general theory that says under which conditions (on $\mathcal{K}$) can we expect this group to be finite-dimensional?

2) Is there any 'big picture' explanation for this phenomenon?


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You should modify your title because a metric is not 2-form! – Mercy King Jun 30 '12 at 1:56
@Mercy I guess you're right.. I meant 2-tensor. Thanks. – student Jun 30 '12 at 2:09
Is there any particular reason to restrict this question to $(0,2)$-tensors? – Neal Jun 30 '12 at 2:21
@Neal It seems to me that $2$-tensors deserve a special place among the others. But of course, the same question goes for any tensor. – student Jul 1 '12 at 17:20

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