# Finite homology of a homogeneous space

Let $$\Gamma$$ be a cocompact lattice in $$\operatorname{SL}(2,\mathbb R)$$ and $$X=\operatorname{SL}(2,\mathbb R)/\Gamma$$ be the underlying homogeneous space. Can the homology group $$H_1(X,\mathbb Z)$$ be finite?

• Absolutely. Any hyperbolic triangle group $\Gamma_0=\langle a,b,c\mid a^p=b^q=c^r=abc=1\rangle$ (here, hyperbolic means $1/p+1/q+1/r<1$) is a subgroup of $PSL_2(\mathbb{R})$ and the quotient $M$ is the unit tangent bundle of the corresponding triangle orbifold. This $M$ is a kind of Seifert fibred 3-manifold called a Brieskorn homology sphere, which in particular has trivial $H_1$. Taking $\Gamma$ to be the preimage of $\Gamma_0$ in $SL_2(\mathbb{R})$ obviously gives the same quotient.
– HJRW
Commented Feb 6, 2023 at 10:50
• Many thanks for your insight! Could you perhaps suggest to me a related reference? Commented Feb 6, 2023 at 10:51
• I think this will be enough for me to move forwards. Thanks a lot once more. Commented Feb 6, 2023 at 11:00
• I like the reference "The geometries of 3-manifolds" by Peter Scott, available from his webpage here: math.lsa.umich.edu/~pscott/8geoms.pdf . I'm not sure he describes the Brieskorn spheres explicitly, but he does introduce the basic notions like orbifold and Seifert-fibred manifold that make it easy to think about these things.
– HJRW
Commented Feb 6, 2023 at 11:01
• The $\pi_1$ of $G/\Gamma$ is isomorphic to $\tilde{G}/\tilde{\Gamma}$ where $\tilde{G}$ is the universal covering and $\tilde{\Gamma}$ is the inverse image of $\Gamma$ therein. So one has to find $\Gamma$ for which this has finite abelianization and actually every cocompact Fuchsian group $\Gamma$ with finite abelianization itself turns out to work (because otherwise the corresponding surface group extension would virtually split). In @HJRW's triangle examples it probably it can be checked directly by making the presentation of $\tilde{\Gamma}$ explicit.
– YCor
Commented Feb 6, 2023 at 11:08