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?
Thanks.