What is the "higher cohomology" version of the Eudoxus reals?

Question

The "Eudoxus reals" are one way to construct $\mathbb{R}$ directly from the integers. A full account is given by Arthan; here is the short version: A function $f: \mathbb{Z} \to \mathbb{Z}$ is an "almost homomorphism" if the function $d_f: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ defined by $d_f(a,b) = f(a+b) - f(a) - f(b)$ has finite/bounded image. The set of almost homomorphisms forms a group under pointwise addition, and the quotient of this group by the subgroup of functions $f$ with bounded image is isomorphic to the real numbers.

This same construction can be applied to functions $f: G \to H$ where $G$ and $H$ are abelian groups. If they are both finitely generated, then Arthan shows that the resulting group is isomorphic to $\operatorname{Hom}_{\text{Ab}}(G, H) \otimes \mathbb{R}$. (*)

There is a "higher cohomology" version of this construction. For example, say that a function $f: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is an "almost 2-cocycle" if the function $d_f: \mathbb{Z}^3 \to \mathbb{Z}$ defined by $d_f(a,b,c) = f(b,c) - f(a+b,c) + f(a,b+c) - f(a,b)$ has finite image. These form a group under addition. The subgroup of "almost 2-coboundaries" is the subgroup generated by the true 2-coboundaries (functions of the form $f(a,b) = g(a+b) - g(a) - g(b)$ for some function $g: \mathbb{Z} \to \mathbb{Z}$) and by the functions which have finite image.

My question is: What is the quotient of the group of almost 2-cocyles by the subgroup of almost 2-coboundaries? (Call this quotient $H^2_{\text{almost}}(\mathbb{Z}, \mathbb{Z})$.) More generally, what do we get if we repeat this construction for arbitrary abelian groups $G, H$ and consider almost $n$-cocycles and almost $n$-coboundaries?

In light of (*), I would expect the answer to be some kind of higher Ext group tensored with $\mathbb{R}$. In particular, since $H^2(\mathbb{Z}, \mathbb{Z}) = 0$, I would expect that $H^2_{\text{almost}}(\mathbb{Z}, \mathbb{Z}) = 0$ as well.

Therefore, a more specific question is: Is the implication $H^2(\mathbb{Z}, \mathbb{Z}) = 0 \Rightarrow H^2_{\text{almost}}(\mathbb{Z}, \mathbb{Z}) = 0$ true, and if so, is there a concrete way to see it?

Answer 1

Let $C_b^n(G, H)$ be the group of bounded functions $G^n \to H$, for discrete $H$ this is equivalent to functions with finite image, but this definition also makes sense for $H$ a normed $\mathbb{R}$-vector space. This sequence forms a subcomplex of the usual chain complex $C^n(G, H) = \mathsf{Set}(G^n, H)$ whose cohomology is the group cohomology $H^n(G, H)$. Then we can define $$C^n_\text{almost}(G, H) = C^n(G, H)/C^n_b(G, H)$$ and $$H^n_{\text{almost}}(G, H) = H^n(C^\bullet_\text{almost}(G, H))$$ to generalize your definition at $n = 2$. Denote $H^n_b(G, H) = H^n(C_b^\bullet(G, H))$, this is often known as bounded cohomology (of $G$ with $H$ coefficients).

Associated to the definition of $H^*_\text{almost}$ is a long exact sequence: $$\dotsb \to H^n(G, H) \to H^n_{\text{almost}}(G, H) \to H^{n+1}_b(G, H) \to H^{n+1}(G, H) \to \dotsb$$ Setting $G = H = \mathbb{Z}$, we get isomorphisms $$H^n_{\text{almost}}(\mathbb{Z}, \mathbb{Z}) \cong H^{n+1}_b(\mathbb{Z}, \mathbb{Z})$$ for $n > 1$. I interpret your question about the implication as asking whether $$H^2(\mathbb{Z}, \mathbb{Z}) \to H^2_{\text{almost}}(\mathbb{Z}, \mathbb{Z})$$ is surjective. Given that $H^3(\mathbb{Z}, \mathbb{Z}) =0$, this is equivalent to asking whether $H^3_b(\mathbb{Z}, \mathbb{Z}) = 0$ as well. I will show that this is indeed the case, using facts about bounded cohomology from these notes on bounded cohomology based on lectures of Burger.

On page 22 appears the "Gersten long exact sequence" $$\dotsb \to H^k_b(\mathbb{Z}, \mathbb{R}) \to H^k(\mathbb{Z}, \mathbb{R}/\mathbb{Z}) \to H^{k+1}_b(\mathbb{Z}, \mathbb{Z}) \to H^{k+1}_b(\mathbb{Z}, \mathbb{R}) \to \dotsb$$ associated to the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$ and the observation/intuition that all $\mathbb{R}/\mathbb{Z}$-valued functions are "bounded". By the proposition on page 24, since $\mathbb{Z}$ is amenable, $H^k_b(\mathbb{Z}, \mathbb{R}) = 0$ for $k \ge 1$. Therefore $H^3_b(\mathbb{Z}, \mathbb{Z}) \cong H^2(\mathbb{Z}, \mathbb{R}/\mathbb{Z})$. Finally, using the long exact sequence in usual group cohomology for the same short exact sequence and the standard fact $H^2(\mathbb{Z}, \mathbb{R}) = 0 = H^3(\mathbb{Z}, \mathbb{Z})$, we get $H^2(\mathbb{Z}, \mathbb{R}/\mathbb{Z}) = 0$.

The same argument shows that $H^n_\text{almost}(\mathbb{Z}, \mathbb{Z}) = 0$ for all $n > 1$.