Let $\mathbb{H}^3$ be the three-dimensional hyperbolic space. Let $K$ be an imaginary quadratic number field and $\mathcal{O}_K$ its ring of integers. Then $SL_2(\mathcal{O}_K)$ acts on $\mathbb{H}^3$ and the quotient $$ M=\mathbb{H}^3 / SL_2(\mathcal{O}_K) $$ is a non-compact hyperbolic manifold. I have heard several times that the Borel-Serre compactification of $M$ consists of adding elliptic curves with complex multiplication, as much as the class number of $K$. Does anybody know a good reference for this or can provide more details? I would be grateful
1 Answer
I do not know of any specific reference, but I will try to explain the situation a bit. For most of the stuff, one does not actually need references besides the paper of Borel-Serre defining the Borel-Serre compactification. Maybe the book of Elstrodt, Grunewald, Mennicke "Groups acting on hyperbolic space" contains some useful information, I currently do not have access to it to check.
To figure out the structure of $\mathbb{H}^3/SL_2(\mathcal{O}_K)$, one can first compactify $\mathbb{H}^3$ by a $\mathbb{CP}^1$ at infinity. The action of $SL_2(\mathbb{C})$ on $\mathbb{H}^3$ can be extended to the $\mathbb{CP}^1$ at infinity; the action of $SL_2(\mathbb{C})$ on $\mathbb{CP}^1$ is given by Möbius transformations. Then there is a bijection between the cusps of $M=\mathbb{H}^3/SL_2(\mathcal{O}_K)$, the orbits of $SL_2(\mathcal{O}_K)$ acting on $\mathbb{CP}^1$, and the ideal class group of $\mathcal{O}_K$. A short explanation of this can be found in Tom Church's answer to this MO-question. A slightly longer explanation of this bijection between orbits and ideal classes is given in these notes of Keith Conrad (the link on Keith Conrad's page of notes did not seem to work). This explains why the boundary components of the Borel-Serre compactification are in bijection with the ideal class group.
Now, to explain what the right boundary components should be, look at the definition of the Borel-Serre compactification. The cusps are related to the intersection of parabolic subgroups with $SL_2(\mathcal{O}_K)$. For example, the cusp for the trivial ideal class is stabilized by the subgroup of upper triangular matrices: $$ \left\{\left(\begin{array}{cc} \pm 1&\alpha\\0&\pm 1\end{array}\right)\mid \alpha\in\mathcal{O}_K\right\}. $$ [Edit: Note that, as pointed out by Ian Agol this is not true for $K=\mathbb{Q}(\sqrt{-1})$ and $K=\mathbb{Q}(\sqrt{-3})$. These number fields have additional units, therefore the stabilizers of cusps are bigger than written above.] The Borel-Serre compactification (of $\mathbb{H}^3$) adds for this cusp a copy of $\mathbb{R}^2\cong \mathcal{O}_K\otimes_{\mathbb{Z}}\mathbb{R}$ on which the above stabilizer of the cusp acts via translations by elements of $\mathcal{O}_K$. The Borel-Serre compactification of the quotient $\mathbb{H}^3/SL_2(\mathcal{O}_K)$ then adds the quotient of $\mathbb{R}^2$ by the translation group $\mathcal{O}_K$ - but this is exactly a $2$-torus. Of course, there is a natural identification of $\mathbb{R}^2$ with $\mathbb{C}$, such that the corresponding $2$-torus has an induced complex structure which makes it an elliptic curve with complex multiplication by $\mathcal{O}_K$. The same happens at all the other cusps, but the stabilizer groups of these cusps are more difficult to write down explicitly. But up to conjugacy in $SL_2(K)$, everything is as in the case of the trivial ideal class. [Edit: As in the comment of Ian Agol, for $K=\mathbb{Q}(\sqrt{-1})$ and $K=\mathbb{Q}(\sqrt{-3})$ the boundary is the orbifold quotient of $\mathbb{C}$ modulo the corresponding stabilizer group $$ \left\{\left(\begin{array}{cc} \beta&\alpha\\0&\beta^{-1}\end{array}\right)\mid \alpha\in\mathcal{O}_K,\beta\in\mathcal{O}_K^\times\right\}.] $$
However, I would consider the elliptic curve structure as somewhat more of an accident. Just like the quotients of symmetric spaces modulo arithmetic group actions may or may not be varieties (in the case at hand, the locally symmetric space fails to be an algebraic variety simply because it is $3$-dimensional and hence can not be a complex variety), the same is also true for the boundary of the Borel-Serre compactification. I think the natural thing to say is that the Borel-Serre compactification adds $2$-torus for each ideal class. [Edit: As explained in the comment of Ben Wieland, the complex structure is natural, induced from hyperbolic space. What I wanted to say is that for other types of arithmetic groups acting on other types of symmetric spaces, the Borel-Serre boundary does not necessarily have a complex structure (sometimes just for dimension reasons).]
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$\begingroup$ The complex structure is natural. The boundary of the locally symmetric space is a subquotient of the boundary of the symmetric space, from which it inherits local properties. As you said, the boundary of 3d hyperbolic space is $\mathbb C\mathbb P^1$; in general, the boundary of hyperbolic space has a unique conformal structure preserved by the group of hyperbolic isometries. $\endgroup$ Commented Jul 23, 2014 at 21:44
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$\begingroup$ Mathias, thanks for this wonderful answer! $\endgroup$– hyp93Commented Jul 23, 2014 at 21:57
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$\begingroup$ There's a minor correction: for $K=\mathbb{Q}[\sqrt{-1}], \mathbb{Q}[\sqrt{-3}]$, the cusp is actually an orbifold quotient of the elliptic curve by the group of units. $\endgroup$– Ian AgolCommented Jul 24, 2014 at 4:25
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1$\begingroup$ @jacob: I do not really understand what you mean. What exactly would you like to know? (I thought about this a bit and I am not sure if I can say much about those neighbourhoods apart from the definition.) $\endgroup$ Commented Aug 21, 2014 at 13:52
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1$\begingroup$ @MatthiasWendt I am just confused as to what the topology looks like at the boundary, and would like to understand it better. So I'd like to understand what the open sets like. For instance, in terms of $NAM$ decomposition relative to a cusp, we add a copy of $M/(M\cap K)$ at the boundary. But then if I actually want a neighborhood of a point $m$ in there, can it be built of product sets such as $B_N\times B_M\times B_A$? If so, what do these sets look like? $\endgroup$– jacobCommented Aug 23, 2014 at 0:07