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First recall how the cup product is defined for the cohomology of a group $G$:

Fix a projective resolution $P \to \mathbb{Z}$ over $\mathbb{Z}G$. Then $P \otimes P \to \mathbb{Z} \otimes \mathbb{Z} = \mathbb{Z}$ is a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G \otimes \mathbb{Z}G=\mathbb{Z}[G \times G]$. Since the diagonal $$D: G \to G \times G,\;g \mapsto (g,g)$$ is a group homomorphism, $P\otimes P$ can be considered as (acyclic) complex of $\mathbb{Z}G$-modules via $D$. By standard homological algebra there is a $\mathbb{Z}G$-linear map $\Delta: P \to P \otimes P$ (called a diagonal approximation) that extends $id: \mathbb{Z} \to \mathbb{Z}$. Finally, if $M,N$ are $\mathbb{Z}G$-modules, the cup product is defined on cochain level by the morphism

$$\begin{array}{lll} Hom_{\mathbb{Z}G}(P,M) \otimes Hom_{\mathbb{Z}G}(P,N) & \xrightarrow{} & Hom_{\mathbb{Z}(G\times G)}(P\otimes P,M\otimes N) \newline & \xrightarrow{\Delta^\ast} & Hom_{\mathbb{Z}G}(P,M\otimes N) \end{array}$$


Obviously, the same construction can be made with any group homomorphism $G \to G \times G$ in place of $D$.

Question 1: What is the motivation to choose the diagonal $D$ for the definition of the cup product ?

Or, to put it the other way round:

Question 2: What "cup product" do be get if we choose one of the group homomorphisms
$$G \to G \times G,\;g \mapsto (g,1) \quad\text{ or }\quad G \to G \times G,\; g \mapsto (1,g)\;\; ? $$

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1  
Answer to 1: Because it always exists. If you use the others than... Answer to 2: You just get the map $H^*G \otimes H^*G \rightarrow H^*G$ that quotients out one of the factors in the tensor product. –  Dylan Wilson Dec 15 '12 at 23:09
    
"...use others than..." should be "use others then" –  Dylan Wilson Dec 15 '12 at 23:09
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Nice observations, TJ. (1) The diagonal map is a gift. It makes any set (any space) into a coalgebra. So any symm. monoidal contravariant functor from sets (spaces) gives you an algebra for every set (space). This is why, morally, cohomology of a space has the structure of an algebra. So when you have a functor like "cochains on a group," it's natural to examine the diagonal, motivated (for me) by this standard fact from topology, to yield an algebraic structure like the cup product. –  Hiro Lee Tanaka Dec 16 '12 at 1:03
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It is interesting to note that Cartan and Eilenberg, in the chapter on products of their book, consider cup products induced by arbitrary "diagonal" maps. They have to assume coassociativity to get associativity of the product and so on, of course (there is more leeway, anyways, as one needs only a diagonal map on resolutions of $\mathbb Z$, and those can be taken non-coassociative and worse :-) ) –  Mariano Suárez-Alvarez Dec 16 '12 at 2:52
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@Mariano: The diagonal map is unique in the following sense. In any cartesian category (i.e. category with finite products, in which the symmetric monoidal structure is chosen to be categorical product), on any object the diagonal map is the unique comultiplication making that object into a counital coassociative coalgebra. So the answer to @TJ's question is that the other maps do not give a unital cup product, as can be seen explicitly from @John's answer, in which "cup product with $\beta$" is sometimes the zero map. –  Theo Johnson-Freyd Dec 16 '12 at 5:10

1 Answer 1

up vote 1 down vote accepted

The diagonal map $\Delta$ is "coassociative": the two maps $(\Delta \otimes 1) \circ \Delta$ and $(1 \otimes \Delta) \circ \Delta$ from $\mathbb{Z}G$ to $(\mathbb{Z}G)^{\otimes 3}$ are equal. Therefore $\Delta$ induces an associative product on cohomology. Similarly, the map $\varepsilon: \mathbb{Z}G \rightarrow \mathbb{Z}$ defined by $\varepsilon(g)=1$ for all $g \in G$ induces the unit map $\mathbb{Z} \to H^*(G)$ in cohomology for the product induced by $\Delta$, because of how $\Delta$ and $\varepsilon$ interact. Other choices for "diagonals" won't do this, in general.

(The fancy thing to say is that $\mathbb{Z}G$ is a Hopf algebra, with structure maps given by the usual product and unit, along with $\Delta$ and $\varepsilon$, and also a map $\chi: \mathbb{Z}G \to \mathbb{Z}G$ defined by $\chi(g) = g^{-1}$ for $g \in G$.)

For question 2, Dylan is saying that the map $g \mapsto (g,1)$ will induce the "product" map $\alpha \otimes \beta \mapsto \alpha \beta$ if $\beta \in H^0(G)$ (i.e. $\beta$ is a scalar), $\alpha \otimes \beta \mapsto 0$ otherwise.

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Actually, the map $g \mapsto (g, 1)$ induces a product which carries $\alpha \otimes \beta$ to $\alpha$ only when $\beta \in H^0(G)$; otherwise the "product" is $0$. –  Craig Westerland Dec 16 '12 at 1:26
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@Craig Westerland: Oh, you're right. fixed. –  John Palmieri Dec 16 '12 at 1:46

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