# Cup products in simplicial cohomology

This post is a walkthrough through a computation I just did - and one of the main reasons I post it is for you to find and tell me what I've done wrong. I have a nagging feeling that the cup product just plain doesn't work the way I tried to make it work, and since I'm trying to understand cup products, I'd appreciate any help anyone has.

I've picked out the examples I have in order to have two spaces with the same Betti numbers, but with different cohomological ring structure.

## Sphere with two handles

I choose a triangulation of the sphere with two handles given the boundary of a tetrahedron spanned by the nodes a,b,c,d and the edges be, ef, bf and cg, ch, gh spanning two triangles.

Computing nullspaces and images, we get a one-dimensional [tex]H^0[/tex], generated by [tex]a^*+b^*+c^*+d^*+e^*+f^*+g^*+h^*[/tex], a two-dimensional [tex]H^1[/tex] with one basis given by [tex]ef^*[/tex] and [tex]gh^*[/tex], and a one-dimensional [tex]H^2[/tex] with the linear dual of any of the four 2-cells as an acceptable basis choice, for instance [tex]bcd^*[/tex]. Throughout, given a simplex [tex]\sigma[/tex] I write [tex]\sigma^*[/tex] for the functional [tex]C_*(X)\to\mathbb Z[/tex] that takes [tex]\sigma\mapsto 1[/tex] and [tex]\tau\mapsto 0[/tex] for all other simplices [tex]\tau[/tex].

So, any product of something in [tex]H^0[/tex] with anything else is already clear. And any product of a degree 1 or 2 coclass with a degree 2 coclass will have to vanish - since there isn't anything for it to possibly hit. Remains the three possible products of two coclasses both of degree 1.

So all the higher degree cohomology products have to vanish.

## Torus

Now, setting up the same computations to get the cohomology classes, we arrive at a one-dimensional class in degree zero, represented by the sum of all duals of all vertices, two classes in degree one, with representatives given, for instance, by [tex]bc^*-ce^*-ci^*-di^*+ef^*-fh^*-fi^*-gi^*[/tex] and [tex]dg^*+di^*+eg^*+eh^*+fh^*+fi^*[/tex], and one class in degree two, represented by the dual of any 2-cell - for instance [tex]fhi^*[/tex].

This, somehow, feels fishy. Could anyone check my reasoning for me, please?

*Edited to add:*indeed this was fishy. The kind of blatant reordering I did to compute with dgi isn't permissible. Also, we know that the square of a degree 1 coclass has to vanish, since the cohomology ring is graded commutative. However, we're not far from the truth: If we want to compute [tex][u]\cup[v][/tex], we should look at all pairs of coclasses in the expression for each, and look for pairs that "connect" to form a valid 2-cell. So, from u we can immediately discard bc, ci, di, fh, fi and gi, since nothing in v starts with c,h or i. Remains to check which of the cells ce and ef connect to anything at all. We get, in the end, the pairings