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It is well known that a subgroup of $PSL(2,\mathbb{R})$ is discrete iff it acts properly discontinuously on the hyperbolic plane $\mathbb{H}$.

I wounder - does the same hold for a discrete group of isometries of $\mathbb {H}^n$ for $n>2$?

Also, a reference regarding hyperbolic isometries of high dimension would be of help, even not only about this specific question.

The way of life
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  • $\mathbb{H}^n$ is $\mathbb{H} \times \ldots \times \mathbb{H}$ ? The metric is the one induced by $\sum_{m=1}^n \frac{dx_m^2+dy_m^2}{y_m^2}$ ? – reuns Jun 11 '17 at 07:25
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    @user1952009 It looks like in the OP's first sentence, $\mathbb{H}$ is an abbreviation (or a typo) for $\mathbb{H}^2$. And later, $\mathbb{H}^n$ is used its standard way, namely as notation for $n$-dimensional hyperbolic space. This is consistent with the tags and with the last sentence. – Lee Mosher Jun 12 '17 at 18:08

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