Assume that a Lie group G acts on a manifold M, effectively. So the Lie algebra g of G is embedded in $\chi^{\infty}(M)$ in a natural way. (effective action: if x.g=x for all x then g=e)
Under what dynamical conditions this embedding is an "Ideal embedding"?
That is : The image of g is an ideal in the Lie algebra of smooth vector fields on M.
By dynamical conditions I mean the dynamical properties arising from the action of G on M.
Are you sure that there is any example in which this is the case? I find it rather hard to imagine what a finite dimensional ideal in a Lie algebra of vector fields may look like.
The Lie algebra of smooth vector fields on a manifold has no finite dimensional ideals.
Proof. Assume that g is a finite dimensional linear subspace in the Lie algebra of vector fields over the manifold M. Then M has a finite number of points p_1,\dots,p_k such that for any X\in g, X\neq 0, at least one of the vectors X(p_1),\dots,X(p_k) is not zero. (This follows from a compactness argument on the unit sphere of g with respect to an arbitrary positive definite inner product.) Choose a vector field X\in g, X\neq 0 and a point q, such that q is different form p_1,...,p_k and X(q)\neq 0. Then there is a vector field Y on M such that [Y,X](q)\neq 0. Choose a smooth function h such that h vanishes on an open subset containing the points p_1,...,p_k and h is constant 1 on an open neighborhood of q. Put Z=hY. Then the Lie bracket [Z,X] is not zero, since
[Z,X](q)=[Y,X](q) \neq 0, on the other hand, [Z,X](p_i)=0 for i=1,...,k, thus, [Z,X] cannot be in g by the choice of the points p_1,...,p_k. This means that g cannot be an ideal.
Dear Balázs Csikós Thank you for your interesting argument.
Is the real analytic version of your statement, true? That is assume that I is an ideal in the Lie algebra of all analytic vector field on M. Can we say that I is infinite dimensional?
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