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I have a question related to fundamental vector fields. For that I first setup the notations and properties etc.

Let $G$ be a lie group acting smoothly on the manifold $M$. Let $\mathfrak{g}$ be its lie algebra and let $\xi\in\mathfrak{g}$. Then the fundamental vector field generated by $\xi$ which I denote as $\#(\xi)$ is given by

$$\#(\xi)_p = \left.\frac{d}{dt}p\exp(-t\xi)\right|_{t=0}.$$

Now define the map $\phi_p : G \to M$ as $\phi_p(g) = g\cdot p$. Then this is a smooth map (if I am not mistaken about the definition of a smooth action).

Now I assume that $p$ is free (i.e it is moved by all elements) so that $\phi_p$ is invertible because it is 1-1. I now prove that its differential is injective. Let $\gamma(t)$ be any curve in $G$ then

$$\#(\xi)_p = (\phi_{p})_{*} \left(\left.\frac{d}{dt}(\gamma(t))\right|_{t=0}\right) = \left.\frac{d}{dt}\phi_p(\gamma(t))\right|_{t=0}.$$

But since $p$ is free (i.e fundamental vector field generated by every element is non-zero in some nbd) $\phi_p(\gamma(t))$ is a smooth curve with non-zero tangent. Therefore $(\phi_{p})_{*}$ has zero kernel at this point. Then does this mean that dimension of the lie algebra and hence the lie group should be smaller than or equal to the dimension of the manifold?

Note: This particulary holds if the critical points of the action with respect to each $\xi$ is isolated. Let ${e_{\alpha_j}}$ be a basis Indeed let $p$ be a point where $\#(e_{\alpha_1})_p =0$ Then we can find a nbd s.t $\#(e_{\alpha_1})$ is never zero except at $p$. Take such a point and continue on by shrinking the nbd until you find a nbd on which non of $\#(e_{\alpha_j})$ is zero.

Thanks

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You should specify properties of $\gamma$ since you use it to rewrite the fundamental vector field $\#(\xi)$. –  gofvonx Nov 26 '13 at 17:55
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