Lie group action on manifold and the vector field generated

Suppose $$G$$ is a Lie group and acts smoothly on a compact oriented manifold $$M$$, and $$X\in\mathfrak{g}$$ is an element in Lie algebra of $$G$$ such that the vector field $$\tilde X$$ generated by $$X$$

$$\tilde X(p)=\frac{d}{dt}\exp(tX)\big\vert^{t=0}.p$$

has isolated zeros. Then $$\tilde X$$ act on space of vector field on $$M$$ by Lie bracket: $$F\mapsto [\tilde X,F]$$. This action induces a well-defined action on $$T_pM$$ for any point $$p$$ with $$\tilde X(p)=0$$. This action is defined by

$$L(X,p):T_pM\to T_pM$$

$$~~~~~~~~~~~~~~~~~~~~~~v\mapsto [\tilde X,F_v](p)~~~~\text{where } F_v\text{ is any vector field with }F_v(p)=v$$

The question is how to show that $$L(X,p)$$ is invertible.

I tried to show that if there is a non-zero vector $$\xi\in T_pM$$ s.t. $$L(X,p)(\xi)=0$$ then $$\exp(tX)$$ fixes $$\exp_p(sv)$$ for $$t,s$$ small. This would then contradict the fact that $$p$$ is isolated. But I don't know how to show this.

• First, your vector field $\tilde{X} \in \mathfrak{X}(M)$ is ill-defined. Try this: Suppose $G$ acts simply transitively on $M$. If we set $\mathcal{L}_g: G \to G$ to be left-multiplication by $g \in G$, then $T\mathcal{L}_g: T_I(G) \to T_g(G)$, and there is a left-invariant vector field $\overline{X}$ in $\mathfrak{X}(G)$ given by $\overline{X}(g) = T\mathcal{L}_g(X)$. Fix a basepoint $* \in M$ and a "base vector" $v_0 \in T_*(M)$. Let $\Phi: G \to \text{Diffeo}(M)$ be the action of $G$ on $M$. (con't) Commented Oct 22, 2021 at 2:12
• Then $T\Phi: TG \to \text{VB}(TM)$, where $\text{VB}(TM)$,is the vector bundle isomophisms of $TM$. Finally we can define $\displaystyle \tilde{X}(p)= T\Phi[\overline{X}(g)](*,v_0) \in T_pM$, where $\Phi(g)(*) = p$. Commented Oct 22, 2021 at 2:17
• Second, I think the part about the Lie group is a red herring. Let $\tilde{X} \in \mathfrak{X}(M)$ be any vector field with isolated zeros. Then this induces a well-defined action of $\tilde{X}$ on $T_p(M)$ for any $p \in M$ with $\tilde{X}(p) = 0$ by $L(\tilde{X},p): T_p(M) \to T_p(M): v \mapsto [\tilde{X},F_v](p)$, where $F_v$ is any vector field in $\mathfrak{X}(M)$ with $F_v(p) = v$. Commented Oct 22, 2021 at 2:36

For reasons of laziness/neatness/misuse/abuse of notation, write not only $$X$$ for the element of the lie algebra $$\mathfrak g$$ of $$G$$, but also for the vector field on $$M$$.

First of all, what exactly is meant by the zeros are 'isolated'? (I am sure there is an accepted definition, but...)

Namely:

If 'isolated' means that there is an open set $$U\subset M$$ around $$p$$, on which $$X$$ vanishes only at $$p$$, I think the statement is false.

Counter-example (with this definition): on (the non-compact) $$M=\mathbb R$$, if $$(X f) (x) = x^2 f'(x)$$, then $$0$$ is an 'isolated' zero of $$X$$. On the other hand, $$[X,{d\over dx}]$$ vanishes at $$0$$. The one-parameter group action on $$M$$ is $$t\cdot x = {x \over 1 - xt}.$$ To get a counter-example in the compact case, take $$M = S^1$$ (circle), and set $$(X f) (\theta) = \sin^2 (\theta) f'(\theta).$$ The action of the one-parameter group is a bit more of a pain to write out ( but without singularities, as there was for the non-compact real-line, which is why I am bothering to state it), but the two examples are basically identical...

On the other hand:

Choose a trivialization chart around $$p$$ such that we can write
$$X = y_1(x) {\partial \over \partial x_1} +\cdots + y_n(x) {\partial \over \partial x_n}.$$

If the statement

$$\text{0 is an isolated zero of X}$$

is equivalent to

$$\text{y_k vanish at 0, and span (give rise to a basis of) \mathfrak m/\mathfrak m^2},$$ where $$\mathfrak m$$ is the germ of functions vanishing at $$0$$, (in other words, the 'hypersurfaces' $$y_k = 0$$ meet transversally at $$0$$), then the result holds even without compactness of $$M$$. We can identify $$L(X,0)$$ with the linear map:

$$v \mapsto - \big( v(y_1), \cdots, v(y_n) \big) .$$ Since [under my (!) hypothesis] $$y_k \pmod {\mathfrak m^2}$$ form a basis for the dual to the tangent space at $$0$$, the (original) map is an isomorphism.

• Gee, I hadn't noticed that the question is so old. Sorry, Mathilda! I saw it because of Jeffrey's much more recent comments, but no doubt you have 'moved on' to other things in the last year. I hope so! But how sad, sad, sad for me, though. Commented Nov 7, 2021 at 17:12