I have a few elementary questions about cup-products.

Can one develop them in an axiomatic approach as in group cohomology itself, and give an existence and uniqueness theorem that includes an explicitly computable map on cochains? Second, how do they relate to cup-products in algebraic topology? In general, are there connections between cup-products and other mathematical constructions that may provide more intuition into them?


7 Answers 7


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.

Almost anything anyone ever calls cohomology is Ext of something with something, and products come up very naturally this way. Cohomology of manifolds is Ext of the sheaf of locally constant k valued functions with itself, and the cup product again is just Yoneda product (all the explicit models for cohomology (simplicial, De Rham, Cech) are all just different resolutions of the this sheaf).

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    $\begingroup$ Is en.wikipedia.org/wiki/… under "Yoneda splices" what you had in mind? $\endgroup$ Oct 15, 2009 at 22:49
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    $\begingroup$ Thanks for your answer... +1... can you please say little more (1/2 lines) about "Almost anything anyone ever calls cohomology is Ext of something with something," Can you give small example if possible... How does one should see derham cohomology as Ext of something with something? $\endgroup$ Feb 24, 2019 at 14:13

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

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

You can check by hand that the differential interacts with this operation by

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

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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    $\begingroup$ Thanks! This is also what I was looking for- for some reason this doesn't seem to be emphasized in textbooks, and the analogy with de Rham cohomology makes things clearer. $\endgroup$ Oct 17, 2009 at 1:45
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    $\begingroup$ This is a great exposition. BTW (for others): cup product in Cech cohomology can be understood in the same way. $\endgroup$
    – Ravi Vakil
    Jul 22, 2011 at 18:06
  • $\begingroup$ Can this idea be made to work (easily) without using the Bar resolution for computing $Ext_R(k,k)$?, that is, can we define some other Kunneth product on $C(x)C$ for any $R$-resolution of $C$ of $k$? $\endgroup$ Aug 5, 2012 at 20:47

Tom and Ben have said most things I would have said here, but one last thing worth mentioning about the structure of cup products in group cohomology is that if you interpret the Ext groups as being equivalence classes of long exact sequences, then the cup product comes naturally from splicing the sequences together.


$Ext^i(A,B)$ 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

$P_1 \to B\to Q_j$


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.


If you can get your hands on a copy of Cassels&Frohlich, /Algebraic Number Theory/, pages 106->107 essentially answer your first question (can they be developed axiomatically, and how to construct them explicitly). Also they do it for Tate cohomology of a finite group rather than plain group cohomology, but it should be essentially the same thing, since Tate cohomology is the equal to a group cohomology in degree >=1 (and equal to group homology in negative degree: just as you get group cohomology by taking the cohomology of the classifying space, you can also get group homology by taking the homology of the classifying space). (And anyhow, if you want to learn local class field theory, you want the Tate cohomology version anyway.) There are probably nicer places to look for the explicit construction, but here's a summary of the axiomatic definition:

the cup product is a family of maps from H^p(G, A) \tensor H^p(G, B) to H^p(G, A \tensor B) for all A, B and all non-negative integers p, q (for Tate cohomology, put hats on all the H's and allow p, q to be arbitrary integers)

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

(ii) for p = q = 0 they are (induced by) the natural product A^G \otimes B^G -> (A \tensor B)^G

(iii and iv) they are compatible with the delta map in the long exact sequences associated to short exact sequences of G-modules in the following way: delta (a'' cup b) = (delta a'') cup b and a'' cup (delta b) = (-1)^(deg a'') a'' \cup delta b.

Writing out and checking the explicity construction can get annoying. In practice, though, at least when doing local class field theory, if you can come up with a map from H^p(G, A) \tensor H^p(G, B) to H^p(G, A \tensor B), it's probably cup product (okay, maybe up to sign).

  • $\begingroup$ Thanks! But is there a way to deduce straight from the axioms that the cup-product exists (I think uniqueness should follow from dimension-shifting and (ii))? $\endgroup$ Oct 15, 2009 at 23:17
  • $\begingroup$ I believe that existence also follows from dimension-shifting. $\endgroup$ Oct 16, 2009 at 0:39
  • $\begingroup$ Hmmm... actually I'm not sure. Cassels&Frohlich proves uniqueness just by constructing it. $\endgroup$ Oct 16, 2009 at 1:01

You should definitely take a look at Lang's "Topics in Cohomology of Groups", chapter 4. There a general notion of cup-products on delta-functors is introduced. This may look like a lot of abstract nonsense to most people, but I like it :)

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    $\begingroup$ This does look quite helpful- thanks! $\endgroup$ Oct 19, 2009 at 23:17

Indeed, the cup-product is a very general thing. Ben Webster already indicates that most cohomologies out there are actually Exts. But it's even better: an "Ext" is just a model for the Hom-operation in a model for suitably abelian (oo,1)-category. More generally, all cohomology that you will ever see is just Homs in some (oo,1)-catgeory: cohomology on an object X with coefficients in an object A, for X and A both objects of some (oo,1)-topos T, is just pi_0 Hom(X,A), as we are used to from T = Top.

From this perspective it is clear that whenever there is a pairing of coefficient objects A1 x A2 --> A3 this induces a corresponding pairing on these cohomologies.

Group cohomology is obtained as the special case where C = ooGrp and we restrict X to be a one.object 1-groupoid. Generally, T (if hypercomplete) is modeled by simplicial presheaves on somse site. Google the lecture notes by Jardine on simplicial presheaves and search them for the section on cup product. That gives pretty much the most general idea of what cup product on cohomology is.

See the discussion and links at nLab:cup product.

(Notice though that this entry deserves further details on actual realizations of cup products. Maybe someone feels like filling these in.)


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