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?


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

  • $\begingroup$ I guess you also need to make sure the vector field is "complete", so just kill it outside of a compact set. $\endgroup$ – Tim kinsella Jan 17 '18 at 13:57
  • $\begingroup$ 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! $\endgroup$ – Jason DeVito Jan 18 '18 at 15:09

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.

  • $\begingroup$ 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! $\endgroup$ – Jason DeVito Jan 16 '18 at 17:51
  • $\begingroup$ 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 Mary Jan 16 '18 at 18:16
  • $\begingroup$ very cool. here's a link for the claim. Its a standard closed-open argument. $\endgroup$ – Tim kinsella Jan 16 '18 at 18:29
  • $\begingroup$ tinyurl.com/y9frhj79 $\endgroup$ – Tim kinsella Jan 16 '18 at 18:30
  • 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). $\endgroup$ – F M Jan 17 '18 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.