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Is it possible to have a faithful action of a connected nontrivial Lie group on a hyperbolic surface $ \Sigma_g $?

Some background:

It is impossible to have a transitive action of a connected nontrivial Lie group on a hyperbolic surfaces $ \Sigma_g $.

It is impossible to have a faithful action of a nontrivial connected Lie group by isometries or by conformal transformations on $ \Sigma_g $.

I think any action by a compact group is in some sense an action by isometries so every compact connected group must act trivially on $ \Sigma_g $.

This is perhaps related to the fact that $ \Sigma_g $ are spin manifolds and a theorem of Atiyah and Hirzebruch that for certain spin manifolds of dimension $4k$ (obviously doesn't directly apply since dimension here is $2$ but maybe it's a similar idea) every action by a compact connected group is trivial. See https://mathoverflow.net/a/89358/387190

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    $\begingroup$ What does nontrivial mean? The group $\mathbb{R}$ acts on the closed unit disk in the complex plane by the formula $\phi_t(z) = e^{i(1-|z|)t}z$. Since this action is the identity on the boundary of the disk, this can be done on any closed disk contained in any surface. $\endgroup$
    – Jim Belk
    Commented Apr 20, 2022 at 21:57
  • $\begingroup$ @JimBelk that's a good point. Do you want to just post that as an answer? $\endgroup$
    – Ian Gershon Teixeira
    Commented Sep 29, 2022 at 16:53

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Yes. The group $\mathbb{R}$ acts on the closed unit disk in the complex plane by the formula $\phi_t(z) = e^{(1-|z|)t}z$. Since this action is the identity on the boundary of the disk, this can be done on any closed disk contained on any surface.

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