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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.

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  • $\begingroup$ 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) $\endgroup$ Commented Oct 22, 2021 at 2:12
  • $\begingroup$ 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 $\Phi(g)(*) = p$. $\endgroup$ Commented Oct 22, 2021 at 2:17
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    $\begingroup$ 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 $F_v(p) = v$. $\endgroup$ Commented Oct 22, 2021 at 2:36

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – peter a g
    Commented Nov 7, 2021 at 17:12

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