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I computed the cohomology ring of the double torus, i.e. the connected sum $X := T_2\#T_2$, where $T_2 = \mathbb S^1 \times \mathbb S^1$.

I am looking for a way to confirm that my computation looks correct, since I feel I got a weird answer.

I used the fundamental polygon shown in this post, to the left, with $I, II, III, IV$ named $a, b, c, d$ respectively. Since $H_2(X)=\mathbb Z$, generated by the whole surface, I described a triangulation $T$ of the double torus, which is a cycle, since it has empty boundary (assuming I checked correctly). Then I considered the cochain $f_a$ defined by $a \to 1$; $\quad b, c, d \to 0$; similarly for $f_b, f_c, f_d, f_T$. Computing the table for the cup product yields the following:

$$\begin{array}{c|c|c|c|c|} \cup & f_a & f_b & f_c & f_d & f_T \\ \hline f_a & 0 & -f_T & -f_T & -f_T & 0 \\ \hline f_b & f_T & 0 & -f_T & -f_T & 0 \\ \hline f_c & f_T & f_T & 0 & -f_T & 0 \\ \hline f_d & f_T & f_T & f_T & 0 & 0 \\ \hline f_T & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}$$

Is this correct?

Riccardo Orlando
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1 Answers1

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That looks right to me (up to a sign convention). The way I checked it is to use Poincare duality, which relates cup product to signed intersection number: look at the vertex $v \in X$ that is the result of gluing the eight corners of the octagon, then look at the four oriented loops $L_a,L_b,L_c,L_d \subset X$ that pass through $v$ and that come from gluing each of the four side pairs $a,b,c,d$, and then check the right-hand rule. It looks to me like you are using a sign convention where the "right hand rule" yields coefficient $-1$ and the "left hand rule" yields the coefficient $+1$. So, for example, the pair $L_a$,$L_b$ passes through $v$ obeying the right hand rule, which gives coefficient $-1$, hence $[f_a] \cup [f_b] = [-f_T]$ (I'm using square braces to denote cohomology classes of cocycles).

Regarding what may or may not look weird, I should add that different gluing patterns are going to give different matrices for the cup product. Your gluing pattern gives this matrix $$\begin{pmatrix} 0 & -1 & -1 & -1 \\ 1 & 0 & -1 & -1 \\ 1 & 1 & 0 & -1 \\ 1 & 1 & 1 & 0 \end{pmatrix} $$ whereas the diagram on the right of the post in your question will give this matrix (which perhaps looks less weird): $$\begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} $$

Lee Mosher
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