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Let $M$ be a Riemannian manifold and let $n$ a positive integer. Denote by $F_n(M) \subset M^n$ the space of all $n$-tuples of pairwise distinct points from $M$. The isometries of $M$ act co-ordinate wise on $M^n$, and this action restricts to an action on $F_n(M)$. What is known about the quotient of $F_n(M)$ by this action?

I am particularly interested in the case where $M$ is the hyperbolic plane and $n > 3$, but any starting points or references would be a great help.

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  • $\begingroup$ From a physicist's point of view, it should give you information on the symmetries of your configuration of points. $\endgroup$ Commented Dec 23, 2018 at 19:27

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Let us consider the case when $M$ is a hyperbolic plane, $M=\mathbb H^2$ and restrict to orientation preserving isometries of $\mathbb H^2$. Let's identify $\mathbb H^2$ with the open radius $1$ disk $D\subset\mathbb C$, centred at $0$. Then we have

$$D^n/PSL(2,\mathbb R)\cong (D)^{n-1}/S^1,\,\,\,\bf *$$ where $S^1$ acts on $D^{n-1}$ diagonally by rotating each $D$-factor around its centre.

To answer the original question, note that the quotient of $\mathbb C^{n-1}$ by the standard diagonal $S^1$-action is a cone over $\mathbb CP^{n-2}$. So, the quotient space $(D)^{n-1}/S^1$ is an open star-shape subset of the cone over $\mathbb CP^{n-2}$.

To see, that $\bf *$ holds, recall that one can choose in $PSL(2,\mathbb R)$ a parabolic subgroup $P$ acting transitively on $\mathbb H^2$ ($P$ preserves a point at infinity of $\mathbb H^2$). Now, we can obtain the quotient $D^n/PSL(2,\mathbb R)$ in two steps. First, for each $n$-tuple $(x_1,\ldots, x_n)\subset D^n$ we find a unique isometry $p\in P$ that sends $x_1$ to the centre of the disk $D$. This reduces the action of $PSL(2,\mathbb R)$ on $D^n$ to that of $S^1$ on $D^{n-1}$.

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  • $\begingroup$ Thank you for your answer. This gives a useful topological description, but can anything be said about its Riemannian structure? $\endgroup$
    – bjw
    Commented Dec 27, 2018 at 11:48
  • $\begingroup$ Indeed, the question was rather vague. I think that quite a lot can be said about the quotient metric. Basically, we take a quotient of $(\mathbb H^2)^{n-2}$ by a diagonal $S^1$-action. If it were not $\mathbb H^2$ but $\mathbb R^2$, we would get a standard metric cone over $\mathbb CP^{n-2}$ with the standard Fubini-Studi metric. In the case of $\mathbb H^2$ one can take Kahler reduction. It should be possible to write down an explicit metric (since the Kahler metric on $\mathbb H^2$ is explicit). Though I am not sure it this what you are looking for $\endgroup$ Commented Dec 27, 2018 at 20:33
  • $\begingroup$ Indeed my question was a bit vague. Thanks again for the answer for the extra comments! $\endgroup$
    – bjw
    Commented Dec 27, 2018 at 21:03

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