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Let $G$ be a subgroup of the group of Möbius transformations $$ z \mapsto \frac{az+b}{cz+d}.$$ What is the relationship between the two conditions:

(1) $G$ being discrete.

(2) $G$ acting properly discontinuously.

Partial answer: Clearly (2) implies (1), since if $G$ is non-discrete, there is a convergent sequence in $G$, say $g_n \to g\in G$. Now take the sequence $g^{-1}\circ g_n$ which converges to the identity, and the orbit of any $z\in \widehat{\mathbb{C}}$ has an accumulation point, so $G$ does not act properly discontinuously.

What can be said about the other implication ? In case the two conditions are not equivalent, can we specify additional requirements on the elements of $G$ so that (1) implies (2).

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1 Answer 1

It is a theorem in the theory of Fuchsian groups that $\Gamma$ is a Fuchsian group (discrete subgroup of the Möbius group) if and only if $\Gamma$ acts as a properly disconituous group of isometries on the upper half plane $\mathbb{H}$.

I believe there is a proof of this Theorem in Katok's book Fuchsian Groups.

You can also find a proof starting on page 129 of these notes on hyperbolic geometry kindly hosted online by Charles Walkden.

It's important to note however, that it is not true in general that for a metric space $M$, a subgroup $\Gamma$ of the isometry group $\operatorname{Isom}(M)$ is discrete if and only if it is properly discontinuous.

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If $M$ is a complete Riemannian manifold, however, (1) and (2) are equivalent. –  user641 Jan 21 '13 at 13:38

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