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I have a few elementary questions about cup-products.


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7 Answers 7

up vote 11 down vote accepted

You can identify group cohomology with $Ext_kG(k,k)$ where $k$ is your base ring, and $kG $ is the group algebra. Here, the cup product given by Yoneda product (which there's some material on here). You should think Ext as being a generalization of Hom (it measures maps that should be there and aren't, and higher order things along the same lines), and this product as a generalization of composition.


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Is en.wikipedia.org/wiki/… under "Yoneda splices" what you had in mind? –  Reid Barton Oct 15 '09 at 22:49

The explicit formula for cup product on group cohomology is as simple as can be. For simplicity let's consider integer coefficients $H^*(G;\mathbb{Z})$, although this works for any coefficients as long as they're untwisted.

Let's define group cohomology using inhomogeneous cochains; thus we take the abelian groups $C^n(G;\mathbb{Z}) :=$ functions from $G^n$ to $\mathbb{Z}$, endowed with a differential $d: C^n \to C^{n+1}$, and then $H^n(G;\mathbb{Z})$ is the usual cohomology $\ker d_n/\operatorname{im} d_{n-1}$.

Anyway, cup product is a map from $H^k(G) \otimes H^m(G)$ to $H^{k+m}(G)$, and it comes from a map $C^k(G) \otimes C^m(G)$ to $C^{k+m}(G)$. Namely, given two cochains $f: G^n \to \mathbb{Z}$ and $g: G^m \to \mathbb{Z}$, define

$$ f \wedge g: G^{k+m} \to \mathbb{Z} $$ by

$$ f\wedge g(x_1,...x_{k+m}) = f(x_1,...x_k)g(x_{k+1},...x_{k+m}) $$


$$ d(f \wedge g) = df \wedge g + (-1)^k f \wedge dg $$

Thus this "wedge product" of cochains descends to a product on group cohomology, and this is exactly cup product. This is also how cup product is defined for de Rham cohomology; differential forms have a natural wedge product which satisfies $d(f \wedge g) = df \wedge g + (-1)^k f \wedge dg$, and so this induces the cup product on $H^*(M;R)$.

Topologically, cup product is the composition of

$$ H^k(Y) \otimes H^m(Y) \to H^{k+m}(Y \times Y) \to H^{k+m}(Y) $$

where the first map is the Kunneth map (just pullback by the two projections $Y \times Y \to Y$), and the second map is restriction to the diagonal. Applying this perspective to group cohomology, we would first define $f \times g : (G \times G)^{k+m} \to \mathbb{Z}$ by

$$ f \times g ((x_1,y_1),...(x_{k+m},y_{k+m})) = f(x_1,...x_k)g(y_{k+1},...,y_{k+m}). $$

Upon restriction to the diagonal $G < G \times G$, $f \times g$ restricts to $f \wedge g$ above.

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I don't know the algebraic side of cup products in group cohomology, but they can indeed be identified with topological cup products. The group cohomology $H^n(G;A)$ can be identified with the cohomology $H^n(BG;A)$ of the classifying space $BG$ of $G$ with coefficients in A (interpreting the action of $G$, which is the fundamental group of $BG$, on A as making A a local coefficient system on $BG$). The cup product on group cohomology is then the same as the usual cup product of singular cohomology.

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$Ext^i(A,B)$ corresponds to sequences

$0 \to A \to P_i \to P_{i-1} \to \dots \to P_1 \to B \to 0$

and the cup product

$u : Ext^i(A,B) \times Ext^j(B,C) \to Ext^{i+j}(A,C)$

is given by

$0 \to A \to P_i \to P_{i-1} \to \dots \to P_1 \to Q_j \to \dots \to Q_1 \to C \to 0$

where the connecting map $P_1 \to Q_j$ is given by the composition of the maps

$P_1 \to B\to Q_j$

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(i) these homomorphisms are functorial in A and B.




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See the discussion and links at nLab:cup product.


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