Let $ M $ be a Riemannian manifold. Let $ Iso(M) $ be the group of isometries of $ M $.
Let $ Conf(M) $ be the conformal group of $ M $ (the group of all diffeomorphisms of $ M $ that preserve the conformal structure induced by the metric).
A hyperbolic manifold is a manifold equipped with a metric of constant negative sectional curvature.
It is well known that a (complete) hyperbolic manifold with finite volume must have finite $ Iso(M) $.
What about $ Conf(M) $? That is, must $ Conf(M) $ be finite for a complete hyperbolic manifold of finite volume?
Best Answer
Let $ (M,g) $ be a finite volume hyperbolic manifold. Let $ \mathrm{Conf}(M,g) $ be the conformal group. Since $ M $ is a finite volume hyperbolic manifold (it must have dimension strictly greater than $ 1 $ and) it cannot be conformally equivalent to $ E^n $ or $ S^n $. So we can apply a theorem of Ferrand https://link.springer.com/article/10.1007/BF01446294 to conclude that there exits some other metric $ g' $ which is conformally equivalent $ g \sim g' $ and such that $ \mathrm{Conf}(M,g)=\mathrm{Iso}(M,g') $. Since $ g' $ is conformally equivalent to the finite volume hyperbolic metric $ g $ then $ (M,g') $ is also a finite volume hyperbolic manifold. But a finite volume hyperbolic manifold must have finite isometry group. Thus $ \mathrm{Iso}(M,g')=\mathrm{Conf}(M,g) $ must be finite.Although perhaps that argument is missing that point and all that really needs to be said is (according to Mike Miller Eismeier): if $ f:M \to M $ is a conformal automorphism, the lift to the universal cover is a conformal automorphism of $ \mathbb{H}^n $, hence an isometry, and the behavior on tangent spaces is the same upstairs and downstairs, so $f$ is an isometry to begin with. This shows $ \mathrm{Conf}(M)=\mathrm{Iso}(M) $ and then we use that a complete finite volume negatively curved manifold has finite isometry group Why is the isometry group of a closed negatively curved manifold finite?