Cohomology of $SL_2(\mathbb Z)$ acting on rational functions

Let $\Gamma = SL_2(\mathbb Z)$ acts on the abelian group $\mathbb C(z)^\times$ (the multiplicative group of the field of rational function in one variable $z$ over $\mathbb C$) by $\gamma \cdot f = f\left( \frac{az+b}{cz+d}\right)$ if $\gamma=\left(\begin{matrix} a&b \\ c &d \end{matrix}\right)$. (This is a right action, but you can make it left if you prefer by letting $\gamma$ act through $\gamma^{-1}$).

Can we compute the cohomology group $H^1(\Gamma,\mathbb C(z)^\times)$?

The only 1-cocycle I know is $\gamma \mapsto cz+d$ and its powers. But what else is it in that group? Same question if $\mathbb C(z)$ is replaced by the multiplicative group of non-vanishing holomorphic functions on the Poincaré Upper Half plane...

Solution 1:

A partial answer.

Write $$\Gamma=\mathrm{SL}_2(\mathbf{Z})=\langle u,v\mid u^2=v^3, u^4=1\rangle;$$ and consider a $\Gamma$-module $(V,\pi)$ (written additively for the moment). So by definition $$Z^1(\Gamma,V)=\{b:\Gamma\to V:b(gh)=\pi(g)b(h)+b(g),\forall g,h\in\Gamma\};$$ $$B^1(\Gamma,V)=\{b:\exists\xi\in V:\forall g\in\Gamma,b(g)=\xi-\pi(g)\xi\}.$$

Let me restrict to 1-cocycles that are cohomologous to zero in restriction to both cyclic subgroups (of order 6 and 4) $\langle v\rangle$ and $\langle u\rangle$. This is automatic if $V$ is a $\mathbf{Z}[1/6]$-module, but this is not the case in the setting. Anyway, this will provide many 1-cocycles.

Hence such a cocycle $b$ is cohomologous to a cocycle that vanishes on the subgroup $\langle u\rangle$, and we can write $b(u)=0$, $b(v)=m-\pi(v)m$ for some $m\in V$ (which is uniquely determined modulo the subspace of $v$-invariant vectors); then $m$ completely determines the cocycle (because $u,v$. Conversely any such choice defines a 1-cocycle: indeed, it defines an affine action where $u$ acts by $\xi\mapsto\pi(u)\xi$, and $v$ by $\xi\mapsto\pi(v)\xi+m-\pi(m)$, and the latter has order 2, so this extends to an affine action of the free product $\mathrm{PSL}_2(\mathbf{Z})$. This 1-cocycle is a 1-coboundary iff this action has a fixed point. Namely if there exists $\xi\in V$ such that $\pi(u)\xi=\xi$ and $\pi(v)(\xi-m)=\xi-m$. This amounts to the condition $m\in V^{\pi(u)}+V^{\pi(v)}$. Hence this subspace of cohomology classes of 1-cocycles is identified with $V/(V^{\pi(u)}+V^{\pi(v)})$.

Now let's specialize this to your context (so we need to switch to multiplicative notation). We have $\pi(v).f(z)=f(-z^{-1})$ and $\pi(u).f(z)=f(-1-z^{-1})$ (I'm using the left action to be consistent).

The $\pi(v)$-invariants form, if I'm correct, the multiplicative subgroup generated by constants and guys of the form $z+s-z^{-1}$ for $s\in\mathbf{C}$. The $\pi(u)$-invariants should not be far from the subgroup generated by constants and things of the form $(z+t)(-1/(z+1)+t)(-1-z^{-1}+t)$ for $t\in\mathbf{C}$. I don't have a clean way of writing this... Anyway we should understand the subgroup $W\subset\mathbf{C}(z)^\times$ of rational functions that have the form $fg$ with $f$ $\pi(u)$-invariant and $g$ $\pi(v)$-invariant. I expect that there are in many reasonable senses ways to say that many rational functions are not of this form, e.g., any nonconstant rational function which does not have two (zeros or poles) in the same $\mathrm{PSL}_2(\mathbf{Z})$-orbit, as well as their powers. (I guess the case of $f(z)=(az+b)^n$ is more or less the example you had in mind?)

The other question is whether the 1-cohomology is larger, i.e., whether there are 1-cocycles that are cohomologically nontrivial in one of these two cyclic subgroups; I have no idea.

A little more inspiration. The quotient of $\mathbf{C}(z)^\times$ by $\mathbf{C}^\times$ can be $\mathrm{PGL}_2(\mathbf{C})$-equivariantly identified with the group $V$ of finitely supported functions $\mathbb{P}^1(\mathbf{C})\to\mathbf{Z}$ with sum zero (map $f$ to multiplicity of zeros). Understanding the 1-cohomology of the latter is probably a little more amenable, and, better that of of the group $W=\mathbf{Z}^{(\mathbb{P}^1(\mathbf{C}))}$ of all finitely supported functions $\mathbb{P}^1(\mathbf{C})\to\mathbf{Z}$. Indeed the latter splits as $\Gamma$-module as direct sum of $\mathbf{Z}^{(X)}$ where $X$ ranges over all $\Gamma$-orbits in $\mathbb{P}^1(\mathbf{C})$. Since $\Gamma$ is finitely generated, we get a decomposition $H^1(\Gamma,W)=\bigoplus_XH^1(\Gamma,\mathbf{Z}^{(X)})$.

Then there are three types of orbits: the generic ones (free $\mathrm{PSL}_2(\mathbf{Z})$-orbits, and two exceptional ones, with stabilizer of order 4 and 6. But in any case the cohomology group is well-known and well-understood: viewing $X$ as a Schreier graph, it is essentially a tree and we can identify $H^1(\Gamma,\mathbf{Z}^{(X)})$ to the set of locally constant functions $\partial X\to\mathbf{Z}$, where $\partial X$ is the boundary of $X$ (which is also its space of ends; this is a Cantor set).

Although this is not a description, this should reflect the essential features of the cohomology group you want to describe. I don't have a homological algebra textbook at hand, which should help comparing the various cohomology groups above (I'd just need something saying how $H^1$ behaves under $\Gamma$-module extensions).