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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?

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(The ismorphism of $H_1$ with $H^1$ is quite unnatural...) What you want to do is doable: write down explicitely all the isomorphisms involed, and compose them. – Mariano Suárez-Alvarez Nov 1 '10 at 18:46
A maybe not so wild guess; the coproduct map $H_2(G)\to \Lambda^2 H_1(G)$ is given by $R\cap[F,F]/[F,R] \to [F,F]/([F,R],[F,[F,F]]) = \Lambda^2(F/R)$. – Torsten Ekedahl Nov 1 '10 at 19:01
@Mariano: But the problem is, I don't know how to multiply elements of H^1 or H^2 without knowing what they are as cocycles. – jd.r Nov 1 '10 at 19:30
The usual description of $H^1$ can be obtained at once from computing it using the Gruenberg resolution. To multiply, you probably want to compare that resolution with the bar resolution (this is one way to do this...) and induce corresponding isomorphisms on cohomology. Everything can be made explicit. – Mariano Suárez-Alvarez Nov 1 '10 at 22:39

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 $[g_1|...|g_n] \to \alpha_1(g_1)...\alpha_n(g_n).$

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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.

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+1 for saying that H^1 is the dual of H_1 in this setting. – David Roberts Nov 2 '10 at 3:47

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"?

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