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My friend asks me how to compute the homology group of $X=\operatorname{SL}(2,\mathbb{R})/\operatorname{SL}(2,\mathbb{Z})$. It is not hard to see that $H_0(X)=\mathbb{Z}$, and for $q\ge 4$ we have $H_q(X)=0$.

But I don't know how to compute $H_q(X)$ for $q=1,2,3$, since the action of $\operatorname{SL}(2,\mathbb{Z})$ seems 'strange' to me, and I can't make $X$ into a CW complex. Can anyone help me?

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    $\begingroup$ For $q \geq 4$ we have $H_q(X) = 0$ because $X$ is a 3-manifold. $\endgroup$ – user98602 Jan 16 at 6:32
  • $\begingroup$ @MikeMiller Thanks, my mistake. $\endgroup$ – Bonbon Jan 16 at 6:38
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    $\begingroup$ It's a well-known fact that $SL(2,\mathbb Z)$ is not cocompact in $SL(2,\mathbb R)$, so the quotient space is a non-compact manifold. It's also connected, so that $H_3(X)=0$. Now you just have to figure out $H_1$ and $H_2$. :) $\endgroup$ – Cheerful Parsnip Jan 16 at 6:51
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The space of unimodular lattices in $\mathbb{R}^2$ is acted upon transitively by $\mathrm{SL}(2,\mathbb{R})$, using the underlying action of $\mathrm{SL}(2,\mathbb{R})$ on $\mathbb{R}^2$. One can view such a matrix as containing a basis for a unimodular lattice. The stabilizer of the $\{(1,0),(0,1)\}$ lattice is $\mathrm{SL}(2,\mathbb{Z})$, and so $\mathrm{SL}(2,\mathbb{R})/\mathrm{SL}(2,\mathbb{Z})$ is in bijective correspondence with the space of unimodular lattices.

(Warning: what follows is my attempt to expand out a story someone once told me. I don't know the theory of elliptic functions well enough to know I haven't made any mistakes.)

The Weierstrass elliptic function $\wp(z;\Lambda)$ for a complex number $z$ and a lattice $\Lambda$ for $\mathbb{C}$ gives a point on the Riemann sphere $\mathbb{C}_\infty$. The quotient of $\mathbb{C}$ by a lattice is a flat torus, and $\wp(-;\Lambda)$ can be thought of as a function from a particular flat torus to the Riemann sphere with four branch points of order two, via a hyperelliptic involution. The four branch points are the images of $0,\omega_1/2,\omega_2/2,(\omega_1+\omega_2)/2$, where $\omega_1,\omega_2\in\mathbb{C}$ generate $\Lambda$. Let \begin{align*} e_1&=\wp(\omega_1/2;\Lambda)\\ e_2&=\wp(\omega_2/2;\Lambda)\\ e_3&=\wp((\omega_1+\omega_2)/2;\Lambda) \end{align*} which are distinct elements of $\mathbb{C}$ satisfying $e_1+e_2+e_3=0$. These three numbers determine $\Lambda$ in that $\wp$ is the unique solution to the differential equation $(\frac{dy}{dz})^2=4(y-e_1)(y-e_2)(y-e_3)$ with a pole at $0$. Expanded, the right-hand-side polynomial is $4y^3-g_2y-g_3$. Hence, polynomials of this form parameterize lattices, if we exclude those that have double roots. The discriminant of this polynomial is $(g_2^3-27 g_3^2)/16$, and $g_2^3-27 g_3^2$ is called the modular discriminant.

Every lattice has a unimodular representative by scaling, and in particular there is a deformation retract onto unimodular lattices. So, if we take all lattices parameterized using $(g_2,g_3)$ in $\mathbb{C}^2$, remove those for which the discriminant is zero, then intersect with the unit $3$-sphere $\{(z,w):\lvert z\rvert^2+\lvert w\rvert ^2=1\}$, we have a parameterization of unimodular lattices after isotopy through the deformation retract. It is known that the link of the singularity of $g_2^3-27 g_3^2$ (i.e., the intersection between the zero-set and $S^3$) is a trefoil knot. (There should be a way to say the condition on $g_2$ and $g_3$ so that the corresponding lattice is unimodular, but I couldn't find it.)

So: $\mathrm{SL}(2,\mathbb{R})/\mathrm{SL}(2,\mathbb{Z})$ is homeomorphic to $S^3$ minus a trefoil knot, which can concretely be written as $$\{(z,w)\in\mathbb{C}^2:\lvert z\rvert^2+\lvert w\rvert ^2=1\text{ and }z^3\neq 27w^3\}.$$ Knot complements have trivial reduced homology except $\tilde{H}_1(S^3-K)\cong \mathbb{Z}$ due to Alexander duality.

Stepping back a bit: if all we wanted was the fundamental group, then consider the following. Recall that each lattice is parameterized by three distinct points in $\mathbb{C}$ which sum to $0$. Unimodularity can be achieved after some positive real scaling factor of these points. Loops in $\mathrm{SL}(2,\mathbb{R})/\mathrm{SL}(2,\mathbb{Z})$ correspond to loops in the configuration space of triples of distinct points. Except for the small complication of rescaling which doesn't change anything homotopically, this is Artin's definition of the braid group.

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  • $\begingroup$ Very nice - this makes the picture eminently clear. $\endgroup$ – user98602 Feb 17 at 6:17
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We may write just as well $X = \widetilde{SL_2 \Bbb R}/\widetilde{SL_2 \Bbb Z}$, where the latter term is the preimage of $SL_2 \Bbb Z$ under the covering map $\widetilde{SL_2 \Bbb R} \to SL_2 \Bbb R$. This identifes $X = B(\widetilde{SL_2 \Bbb Z}, 1)$, because its universal cover is the contractible space $\widetilde{SL_2 \Bbb R}$.

You are thus asking for a group homology computation. This is especially fortuitous because as a particularly emotive parsnip observed, $H_k X = 0$ for $k \geq 3$. So all we need to know is $H_k X = H_k(\widetilde{SL_2 \Bbb Z}; \Bbb Z)$ for $k = 1, 2$, where this last term means group homology (or equivalently, the homology of any space with contractible universal cover and $\pi_1 X = \widetilde{SL_2 \Bbb Z}$).

To facilitate things, observe that as here this is the fundamental group of the complement of the trefoil knot, with presentation $\langle x, y \mid x^2 = y^3 \rangle.$ Knot complements always have aspherical universal cover, so we have $X \simeq S^3 \setminus T_{2,3}$, where $T_{2,3}$ is the trefoil knot. Knot complements always have $H_1 X = \Bbb Z$ and $H_k X = 0$ for $k > 1$ by Alexander duality or a Mayer-Vietoris calculation.

Likely there is an explicit homeomorphism, which I do not know but the clever 3-dimensional geometers do.

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    $\begingroup$ (I asked one of those clever 3-dimensional geometers and tried writing it up.) $\endgroup$ – Kyle Miller Feb 17 at 4:32

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