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?
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2 Answers
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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.
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$\begingroup$ I guess you also need to make sure the vector field is "complete", so just kill it outside of a compact set. $\endgroup$ Jan 17, 2018 at 13:57
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$\begingroup$ 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! $\endgroup$ Jan 18, 2018 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.
answered Jan 16, 2018 at 17:50
Jason DeVito - on hiatusJason DeVito - on hiatus
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$\begingroup$ I am sure I have proved the statement in the last paragraph somewhere on MSE a long time ago, but I couldn't find it after a bit of googling. If someone can link to any proof of that statement, I would be grateful! $\endgroup$ Jan 16, 2018 at 17:51
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$\begingroup$ I will accept Tim's answer as it is technically the answer to my question, but thank you for this! It was exactly what I wanted! $\endgroup$– R MaryJan 16, 2018 at 18:16
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$\begingroup$ very cool. here's a link for the claim. Its a standard closed-open argument. $\endgroup$ Jan 16, 2018 at 18:29
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2$\begingroup$ It may be useful to remark that an isometry of a connected Riemannian manifold is determined by its differential at a single point (from the last paragraph I thought one just needed to know the value of the isometry at a single point). $\endgroup$ Jan 17, 2018 at 13:14