Another group cohomology cup product question

Question

I am wondering if there is a way to see the cup product, in some cases, without using cochain complexes. The situation I am interested in is the following:

Let $G=F/R$ be a finitely presented group and $k$ a finite field. Then $H_1(G;k)$ is easy to find as a finitely generated abelian group. Since I'm taking field coefficients, $H^1$ is isomorphic to this abelian group. If I keep track of normalizing the relations matrix I can even get a set of generators in terms of $G$. Using Hopf's formula for $H_2(G;k)$, I can get generators for $H^2$. Is there a way to see what the cup product of terms from $H^1$ is in $H^2$? Can I get elements of $H^2, H^3, H^4$ this way?

Answer 1

Is there a way to see what the cup product of terms from $H^1$ is in $H^2$?

By the universal coefficient theorem there is an isomorphism $$ H^2(G;k) \cong Hom_k(H_2(G;k),k) $$ and an embedding $$ H_2(G;\mathbb{Z}) \otimes k \hookrightarrow H_2(G;k). $$ Identify $H_2(G;\mathbb{Z})=[F,F] \cap R / [F,R]$ and $H^2(G;k)=Hom_k(H_2(G;k),k)$.

Let $\alpha, \beta \in H^1(G;k)=Hom(G,k)$. Then the restriction of $\alpha \cup \beta$ to $H_2(G;\mathbb{Z}) \otimes k$ is given as follows:

For $x=\Pi_j [x_j,y_j] \in [F,F] \cap R$ and $\hat{x}=x$ mod $[F,R]$ we have

$$ (\alpha \cup \beta)(\hat{x} \otimes 1) = \sum_j [\alpha(\overline x_j)\beta(\overline y_j)-\alpha(\overline y_j)\beta(\overline x_j)], $$

where $\overline x_j \in G$ denotes the image of $x_j \in F$ under the map $F \to G$.

This follows from the correspondence between elements of $[F,F] \cap R$ and 2-cycles of the bar resolution of $G$ that is described in exercise 4c) in II.5 of "Brown: Cohomology of Groups".

Can I get elements of $H^2,H^3,H^4$ this way?

Of course. But the description depends on the representation of the cohomology group. For example, if $H^n$ is represented by cocycles of the bar resolution, then the cup product $\alpha_1 \cup ... \cup \alpha_n $ of $\alpha_1, ..., \alpha_n \in Hom(G,k))$is represented by the cocyle $[g_1|...|g_n] \to \alpha_1(g_1)...\alpha_n(g_n).$

Answer 2

It's OK to "relate" $H_1$ and $H^1$, but you should say that you view $H^1$ as the dual of $H_1$, namely as linear maps $\alpha:H_1\to\Bbbk$.

The cup product in degree 1 is then very simple: it's $\alpha\cdot\beta:H_2\to\Bbbk$ given by $(\alpha\cdot\beta)([f_1,f_2])=\alpha(f_1)\beta(f_2)-\alpha(f_2)\beta(f_1)$, extended by linearity. Recall that $H_2 = ([F,F]\cap R)/[F,R]$, by Hopf's formula. It takes a little bit of work to show that the above formula is independent of the expression of an element of $H_2$ as a product of commutators.

The Massey (or $A_\infty$) products in degree 1 are almost as simple, using iterated commutators if I recall well. However, I'm not aware of a slick way of computing the cup product in higher degrees.

Answer 3

I'm not sure what kind of answer you're hoping for: how will you describe an element of $H^4$, for example, without using cocycles?

Having said that, perhaps syzygies would help. Let $I$ denote the augmentation ideal of $kG$; then every element of $H^1$ gives a $kG$-module map $I \rightarrow k$. More generally, each element of $H^n$ gives a $kG$-module map $I^{\otimes n} \rightarrow k$. If you have two such maps, say $\alpha: I^{\otimes n} \rightarrow k$ and $\beta: I^{\otimes m} \rightarrow k$, then the cup product of the corresponding cohomology classes is represented by their "composition": $$ I^{\otimes m+n} \xrightarrow{I^{\otimes m} \otimes \alpha} I^{\otimes m} \xrightarrow{\beta} k. $$ (To be more precise, every element of $H^n$ corresponds to an equivalence class of maps $I^{\otimes n} \rightarrow k$...)

Does this count as "mucking around with cochains"?