Let's fix some notations: $M$ will denote a real smooth, $m$-dimensional, manifold, $F^k(M)$ is the k-th order frame bundle on $M$ and $Gl^k(m)$ is the space of $k$-jets of diffeomorphisms of $\mathbb R^m$ that fix the origin.

Now, it is well known that on the case of (pseudo) riemannian metrics, killing fields are characterized by $\mathcal{L}_Xg=0$. In particular, this means that the space of killing fields is a lie algebra.

But on the Gromov case, we don't have a metric tensor anymore, but a $Gl^k$-equivariant map $\sigma:F^k(M)\to Z$, where $Z$ is a smooth real algebraic variety where $Gl^k$ acts algebraically. It is not difficult to see that a killing field is characterized by $(d\sigma)_\alpha (X_{(k)})=0$ for every $\alpha\in F^k(M)$, where $X_{(k)}$ is a natural lifting of $X$ obtained from lifting the flow $\varphi_t$ of $X$ to $F^k(M)$.

Does it remains true that the killing fields are a lie subalgebra? How to prove that the bracket $[X,Y]$ of two killing fields also integrates to isometries?


1 Answer 1


Doesn't this follow immediately from the statement that $X$ is a Killing field if and only if the prolonged (local) flow $\phi^{k}_t:F^k(M)\to F^k(M)$ of the induced (local) flow $\phi_t:M\to M$ of $X$ preserves $\sigma:F^k\to Z$? After all, one has the identity of Lie brackets $[X_{(k)},Y_{(k)}] = [X,Y]_{(k)}$, and I believe that this implies what you want.


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