# Cup products in simplicial cohomology

Published: Fri 12 September 2008

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.

We get a cochain complex on the form
$$0 \to \mathbb{Z}^8 \to \mathbb{Z}^{12} \to \mathbb{Z}^4 \to 0$$
with the codifferential given as
$$\begin{pmatrix} 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1\\ \end{pmatrix}$$
and
$$\begin{pmatrix} 1 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$

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

Now, the Encyclopedia of Topology, volume II has a paper by Viro and Fuchs on homology and cohomology. They state a direct construction of the cup product as defined by the following:
$$[c_1\cup c_2](\sigma) = c_1(\sigma_{0\ldots q_1})c_2(\sigma_{q_1\ldots q_1+q_2})$$ where the q:s are the degrees of the c:s. They also state that (at least if X is connected) any degree 0 coclass acts as identity under the cup product.

So, any product of something in $$H^0$$ 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.

Now, by the definition of the coclasses, we need something that is an equivalence class of linear duals of 2-cells to split into things that only operate on the two handles. However, these are geometrically disjoint - so any cell we could feed into such a product would vanish on the components. For instance,
$$(ef^*)\cup(ef^*)(bcd)=(ef^*)(bc)\cdot(ef^*)(cd)=0$$

So all the higher degree cohomology products have to vanish.

## Torus

We pick a triangulation of the torus with 9 vertices, 27 one-cells and 18 2-cells, given as the identification space of a square, in a usual manner. It's going to be obnoxious to write down boundary maps, and list all cells, so I'll just refer you to the following picture instead: 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 $$bc^*-ce^*-ci^*-di^*+ef^*-fh^*-fi^*-gi^*$$ and $$dg^*+di^*+eg^*+eh^*+fh^*+fi^*$$, and one class in degree two, represented by the dual of any 2-cell - for instance $$fhi^*$$.

Now is where I find things get tricky. Again, we get the 0-degree class acting as, essentially, an identity. And the only products that could possibly be nontrivial now would be products of two classes of degree 1. So let's call the classes $$[u]$$ and $$[v]$$, and lets call the degree 2 class $$[w]$$.
Using the Viro-Fuchs construction of the cup product, I should be able to say that if we consider $$[u]\cup[u](bce) = u(bc)\cdot u(ce) = 1\cdot(-1) = -1$$ and therefore $$[u]\cup[v]=[w]$$ (there is some reference to the codifferential matrices hidden in the signs here), and similarily, by considering the simplex idg (= dgi with a cyclic permutation) twice we would get $$[u]\cup[v]=[v]\cup[v]=[w]$$.

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 $$[u]\cup[v]$$, 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
ce,eg
ce,eh
ef,fh
ef,fi
corresponding to the resulting potential 2-coclasses $$ceg^*$$, $$ceh^*$$, $$efh^*$$ and $$efi^*$$. However, looking over our sketch again, we see that the cells ceg, ceh and efi are absent from our triangulation of the torus. Hence, the only potential 2-coclass representative surviving is $$efh^*$$, which represents the actual cup product, yielding $$[u]\cup[v]=[efh^*]$$