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Let $(M, \nabla)$ be a manifold together with a connection on $TM$ and let $T$ be a tensor field on $M$. Suppose the pseudogroup $\Gamma$ of locally defined smooth maps, that simultanously preserve the connection $\nabla$ and the tensor field $T$ acts transitivily on $M$. Denote by $G$ the 1-jets of maps in $\Gamma$: $G := \lbrace j^1_p f \, \vert \, f \in \Gamma, p \in \mathrm{dom}(f) \rbrace$.

Is $G \subset J^1(M,M)$ a submanifold of the first jet manifold? In other words: Are the maps in $\Gamma$ given as solutions of a PDE?

(Note, that a PDE of order $k$ on a fibre bundle $E \to M$ (in this case: $E = M \times M$ with the projection on the first factor) is defined to be a closed submanifold of the jet bundle $J^k(E)$.)

This question is a spin-off of a much wider question I asked on MSE: https://math.stackexchange.com/questions/1761598/why-is-a-pde-a-submanifold-and-not-just-a-subset. There I basically wanted to understand why defining a PDE to be submanifold is a good choice and used this one here as an example. I.e. I asked, why it is not too restrictive, as for example $\frac{\partial u}{\partial x} \frac{\partial u}{\partial y} = 0$ is not a PDE for functins $u: \Bbb R^2 \to \Bbb R$ in this sense.

Background:
My interest in this specific example comes from the following: Kostant showed, that a connected, simply connected affine manifold is a reductiv homogeneous space of a Lie group acting on $M$ by affine transformations, iff there exists a connection $D$ on $M$, that is rigid w.r.t. $\nabla$, invariant under parallelism and geodesically comlete. I'm looking for an analogous local result. I already know, that if there exists a Lie Pseudogroup acting reductiv, transitiv and affine on $M$, than there exists a connection $D$, which is invariant under parallelism and rigid wrt to $\nabla$. For the converse, I need to know, that the maps, that preserve simultenously such a connection $D$ and the difference tensor $S = D - \nabla$ are given by a PDE.

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  • $\begingroup$ Diffeomorphisms which preserve a tensor field are given by a PDE: write the tensor field in local coordinates, and then the Jacobian of the diffeomorphism plugs into it, and that is your PDE. Try a 1-form or a vector field to see what I mean. So the answer is yes. $\endgroup$
    – Ben McKay
    Commented May 12, 2016 at 17:00
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    $\begingroup$ @BenMcKay: Yes, it's easy to see, that in local coordinates they are given by a PDE. The problem is, that this depends on the chart. So the usual definition of a PDE (on a fibre bundle, here $M \times M \to M$), is that it is a closed submanifold of a convenient jet bundle. That's what I'm asking for. Maybe I should clarify this in my question. $\endgroup$
    – hase_olaf
    Commented May 12, 2016 at 17:28

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