What is this cochain complex about, whose $H^1 = \mathbb{R}$?

Question

$\DeclareMathOperator\QEnd{QEnd}$Let $C^n$ be the set of functions $\mathbb{Z}^n \to \mathbb{Z}$, and $B^n$ the set of bounded such functions. For $a_1,...,a_{n+1} \in \mathbb{Z}$, the differential of this complex $d^n : C^n \to C^{n+1}$ is defined by :

$$d^n f(a_1,...,a_{n+1}) = f(a_2,...,a_{n+1}) + \sum_{k=1}^n (-1)^n f(a_1,...,a_k +a_{k+1},... a_{n+1})$$ $$+ \;(-1)^{n+1} f(a_1,...,a_{n})$$

This differential is null for additive functions. Let's call $\QEnd(Z)\subset C^1$ the set of functions whose differential is bounded.

Let's consider the complex $\bar{C}= C^n/B^n$ with the induced differential $\bar{d}$. It can be shown that the first cohomology group of this complex is the field of real numbers $\mathbb{R}$ (Eudoxe-Schanuel numbers) : $$ H^1(\bar{C^n}) = \QEnd(Z)/B^1 = \mathbb{R}$$ But what is this complex about ? The formula for the differential looks like its about group cohomology, but for what group and with what coefficients ? It is not $\mathbb{Z}$ whose cohomology is the one for $S^1$. Or is it related to any space ?

And how to compute $H^2$,... $H^n$ ?

Answer 1

As you have a short exact sequence of cochain complexes $0 \to B \to C \to \overline C \to 0$, you have a long exact sequence of their cohomology groups: $$\cdots \to H^{i-1}(\overline C) \to H^i(B) \to H^i(C) \to H^i(\overline C) \to \cdots$$

Now the group cohomology of the integers, $H^i(C)$, is $\Bbb Z$ in degrees $0$ and $1$ and is zero otherwise.

Now $H^0(B) = \Bbb Z$ and by the proof of the Lemma on page 11 here we have $H^i(B) = 0$ for $i > 2$.

Thus for $i > 1$ we have an exact sequence $0 = H^i(C) \to H^i(\overline C) \to H^{i+1}(B) = 0$, and thus $H^i(\overline C) = 0$ for $i > 1$. Your direct computations show that $H^0(\overline C) = 0$ and $H^1(\overline C) = \Bbb R$.

In general your complex $C$ interpolates the usual group cohomology and the bounded cohomology. If $G$ is, say, amenable and the fundamental group of some aspherical manifold of dimension $n$, then $H^i(\overline C_G) = 0$ for $i > n$ by the same argument.