Cohomology of lattice with coefficients in field of rational functions

Question

In my research, I came across a 1-cocycle in the following group cohomology complex:

Let $\Lambda_\mathbb{Z}$ be a lattice (i.e. isomorphic to $\mathbb{Z}^n)$; let $\Lambda_\mathbb{C} = \Lambda_\mathbb{Z} \otimes_\mathbb{Z} \mathbb{C}$; let $\mathbb{C}(\Lambda_\mathbb{C})$ be the rational functions on $\Lambda_\mathbb{C}$, and let the $\times$ denote the multiplicative group.

Let $\Lambda_\mathbb{Z}$ act on $\mathbb{C}(\Lambda_\mathbb{C})$ by translation. Then the bar resolution takes the form $\mathbb{C}(\Lambda_\mathbb{C})^\times \rightarrow Fun(\Lambda_\mathbb{Z}, \mathbb{C}(\Lambda_\mathbb{C})^\times) \rightarrow Fun(\Lambda_\mathbb{Z}^2, \mathbb{C}(\Lambda_\mathbb{C})^\times) \rightarrow \ldots$

where $Fun$ refers to arbitrary functions.

Is anything known about the cohomology of this complex? I think I've figured out the first cohomology group (the group is not easy to explain, so I'd rather not type it out if there is an immediate answer), and the zeroth cohomology group is clearly $\mathbb{C}$ (as the 0-cocycles are constant functions).

For a more general question, is anything known about cohomology with coefficients in the multiplicative group of a field?

Edited to add: I didn't know Hilbert's theorem 90, which says that if we have a Galois extension $L/K$ with Galois group $G$, then $H^1(G, L^\times)$ is trivial. That doesn't seem to apply to my original question, but answers the more general one.

Edit 2: In the case $n = 1$, I have a proof that the 2nd cohomology group is trivial, which I think generalizes to show that all higher cohomology groups are trivial. In that case, the 1st cohomology group is $\mathbb{C}(\mathbb{C}^\times)^\times$.

Edit 3: One thing that helps here is decomposing $\mathbb{C}(\Lambda_\mathbb{C})^\times)$ into $\mathbb{Z}[\Lambda_\mathbb{Z}]$-modules; unique factorization gives us that it decomposes as $\oplus_{O} \oplus_{p \in O} \mathbb{Z}p$ where $p$ ranges over irreducible polynomials, and $O$ are the orbits of irreducibles; the summands of the decomposition are $\oplus_{p \in O} \mathbb{Z}p$.

Let $C_p = \{c \in \Lambda_\mathbb{C}|T_c(p) = p\}$; because $p$ is a polynomial and polynomials aren't periodic, $C_p$ is a subspace. Therefore $C_p \cap \Lambda_\mathbb{Z}$ is a summand of $\Lambda_\mathbb{Z}$. We can set $\Lambda_\mathbb{Z} = \mathbb{Z}^n \supset C_p \cap \Lambda_\mathbb{Z} = \mathbb{Z}^m \oplus \{0\}$; then $\oplus_{p \in O} \mathbb{Z}p \simeq \mathbb{Z}^n/(\mathbb{Z}^m \oplus \{0\})$

I think I have a finite-step free resolution, which should then allow determining the group cohomology on each of the individual representations easily.

Edit 4: I think the group homology may be informative; is there a universal coefficient theorem for group cohomology that works with an action of the group on the coefficients? I've found one where the action is trivial, but not generally.

Answer 1

I am not aware of any computations of the cohomology of $\mathbb{C}(t_1,\dots,t_n)^\times$ with the $\mathbb{Z}^n$-action by translation. However, there are some direct consequences of basic group cohomology statements that you might like to know.

First of all, the group $\mathbb{Z}^n$ has cohomological dimension $n$ (because $\mathbb{R}^n/\mathbb{Z}^n$ is a classifying space of dimension $n$). This means, in particular, that $H^i(\mathbb{Z}^n,M)=0$ for any $i>n$ and any coefficient module $M$. So your computation in the case $n=1$ follows from this general fact.

The other thing I wanted to point out is that low-dimensional cohomology groups have well-known explicit interpretations (and you seemed to be interested in 1-cocycles). $H^0(G,M)$ for a $G$-module $M$ is the invariants, and as you noticed these are the constants $\mathbb{C}^\times$ since there are no other periodic rational functions.

$H^1(G,M)$ can be described in terms of "crossed homomorphisms". In the case $n=1$, a crossed homomorphism $\mathbb{Z}\to\mathbb{C}(t)^\times$ is determined by its value at $1\in\mathbb{Z}$ and every value is realizable because $\mathbb{Z}$ is free on one generator. The trivial crossed homomorphisms are those which map $1$ to the function $f(T+1)/f(T)$, where $f\in\mathbb{C}(T)^\times$. So $H^1(\mathbb{Z},\mathbb{C}(T)^\times)$ is the quotient of $\mathbb{C}(T)^\times$ modulo functions of the form $f(T+1)/f(T)$, $f(T)\in\mathbb{C}(T)^\times$, which somehow does not seem to agree with your computation.

In the case $n>1$, the description of $H^1$ via crossed homomorphisms might still be useful. Generally, you can easily see that the crossed homomorphisms $\mathbb{Z}^n\to\mathbb{C}(t_1,\dots,t_n)^\times$ will be determined by their values on the standard vectors $e_i\in\mathbb{Z}^n$, but for $n>2$ there will be non-trivial conditions coming from commutativity of $\mathbb{Z}^n$.

Finally, $H^2(G,M)$ with $G$ a group and $M$ a $G$-module can be interpreted to classify extensions. In the case at hand, these would be extensions of the form $$ 1\to\mathbb{C}(t_1,\dots,t_n)^\times\to E\to\mathbb{Z}^n\to 0 $$ such that the induced conjugaction action of $\mathbb{Z}^n$ on $\mathbb{C}(t_1,\dots,t_n)^\times$ is translation.

All these low-dimensional statements come from writing out the bar-complex, but they will only get you so far. The best bet is to find other interpretations of the cohomology groups. It seems that there should be some interesting "geometric" interpretation of these cohomology groups, but I do not have to offer anything in this direction at the moment. Maybe something in the direction of translational scissors congruence? By the way, in what context did this question appear?