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Of course, I assume the manifold is connected. I feel like this is probably true but I have no idea how to prove it. Any hints?

R Mary
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2 Answers2

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No. Take the one-parameter group action generated by a not-identically-vanishing vector field that vanishes in an open set.

Tim kinsella
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  • I guess you also need to make sure the vector field is "complete", so just kill it outside of a compact set. – Tim kinsella Jan 17 '18 at 13:57
  • Incidentally, the combination of our two answers shows the following: If you have a vector field which vanishes on an open set, then either at least one flow line doesn't close up, or they all close up but with no common period. Weird! – Jason DeVito - on hiatus Jan 18 '18 at 15:09
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Tim's answer is spot on, but I wanted to mention that the answer switches if you assume the Lie group is compact.

Suppose $G$ is a compact Lie group acting on a connected manifold $M$. If $G$ fixes a non-empty open set pointwise, then the $G$ action on $M$ is trivial.

Proof: Pick any background Riemannian metric on $M$. By averaging this action over the $G$ action, we obtain a $G$-invariant Riemannian metric. In particular, we may assume $G$ acts isometrically.

Now, let $U$ be a non-empty open subset of $M$ on which $G$ acts trivially and let $p\in U$. Then, for any $g\in G$, we of course have $g\ast p = p$, but more is true: $d_p g:T_p M\rightarrow T_p M$ is the identity function. To see this, pick $v\in T_p M$ and let $\gamma$ be a curve with image entirely in $U$ for which $\gamma'(0) = v$. Then $g \gamma(t) = \gamma(t)$ and differentiation both sides at $t = 0$ gives $d_p g (v) = v$.

But an isometry of a connected Riemannian manifold is determined by its action at a single point. Since multiplication by $g$ and the identity map do the same thing at $p$, $g = Id$. In particular, $G$ acts trivially.

Jason DeVito - on hiatus
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