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I heard something about the triple intersection number $\text{mod}(2)$ (but maybe also $\text{mod}(n)$) of a surface in an orientable three-manifold but I couldn't find a precise definition. My guess is the following. Consider the homology with $\mathbb{Z}_n$ coefficients, and the Bockstein of the sequence $$ \mathbb{Z}_n\rightarrow \mathbb{Z}_{n^2}\rightarrow \mathbb{Z}_n $$ Given $\Sigma \in H_2(M_3,\mathbb{Z}_n)$, take its Poincarè dual $PD(\Sigma)\in H^1(M_3,\mathbb{Z}_n)$, and then I think this triple intersection number should be something like $$ \exp{\left(\frac{2\pi i}{n}\int _{M_3}PD(\Sigma)\cup \beta(PD(\Sigma))\right)}\in U(1) $$ which is an $n$'th root of unit. Is this definition somehow correct? Why is it called triple intersection number? Could somebody provide some discussion or maybe a reference. I also read that this is related with the Euler character of $\Sigma$ but I really don't see why....

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  • $\begingroup$ You could also just evaluate $\operatorname{PD}(\Sigma)^3$ on the fundamental class, giving an integer mod $n$. Is there a reason for preferring your formula? $\endgroup$
    – Mark Grant
    May 24, 2023 at 19:20
  • $\begingroup$ Since $PD(\Sigma)$ is a 1-cochain, unless $n=2$, $PD(\Sigma)^2=0$. $\endgroup$ May 24, 2023 at 19:52
  • $\begingroup$ When $n = 2$, your integrand is the same as what Mark Grant suggested. Also note that by graded commutativity $\operatorname{PD}(\Sigma)^3 = -\operatorname{PD}(\Sigma)^3$ so $2\operatorname{PD}(\Sigma)^3 = 0$, i.e. $\operatorname{PD}(\Sigma)^3$ is two-torsion. If $n$ is odd, then we must have $\operatorname{PD}(\Sigma)^3 = 0$, but if $n$ is even, it could be non-zero (there is a unique non-zero element of order two in $\mathbb{Z}_n$ when $n$ is even). $\endgroup$ May 24, 2023 at 20:09
  • $\begingroup$ I see. This would explain why "triple intersection". Is the formula with the Bockstein more general or not? $\endgroup$ May 24, 2023 at 21:33
  • $\begingroup$ If $\Sigma$ is an oriented surface in an oriented $3$-manifold, then it gives an element $PD(\Sigma)$ in $H_1(M)$ (with $\mathbb Z$ coefficients). If you pass to cohomology with $\mathbb Z_n$ coefficients, this new $PD(\Sigma)$ has zero Bockstein. $\endgroup$ May 24, 2023 at 23:46

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