The answer to this question should be obvious, but I can't seem to figure it out. Suppose we have a surface $F$, and a representation $\rho : \pi_1(F)\to SU(n)$. We can define the homology with local coefficients $H_*(F,\rho)$ straightforwardly as the homology of the twisted complex $$C_*(F,\rho):=C_*(\widetilde{F};\mathbf{Z})\otimes_{\mathbf{Z}[\pi_1(F)]} \mathbf{C}^n$$ where $\widetilde{F}$ is the universal cover, and $\mathbf{Z}[\pi_1(F)]$ acts on each side in the obvious way.
Now, this complex is actually very easy to compute explicitly: just lift a nice basis of cells in $F$ to $\widetilde{F}$, and write down the boundary maps explicitly. For example, if $F$ is a torus and we take $n=2$, say, we can choose a natural meridian-longitude basis $(x,y)$ for $H_1(F)$, and the twisted boundary map $\partial_1:C_1(F,\rho)=\mathbf{C}^4\to C_2(F,\rho)=\mathbf{C}^2$ is $$ \left( \begin{array}{ccc}
\rho(x)-Id \newline\rho(y)-Id\end{array} \right)$$
So, here's my question. Since $\rho$ is a unitary representation, we should get a twisted intersection form on $H_1(F)$, simply by combining the untwisted intersection form with the standard hermitian product on $\mathbf{C}^2$, right? And I would imagine this is also really easy to compute, in a similar basis, say? I can't seem to figure out how it would go. Could anyone help me, even show me how it works for the same torus example?
Or, if I've said anything wrong, tell me where?
Best Answer
For me it is easier to work with cohomology (just for psychological reasons). Also, I will distinguish the representation $\rho$ from the local system $V$ with fibres ${\mathbb C}^2$ that it gives rise to. So where you would write $H^1(F,\rho)$ I will write $H^1(F,V)$. I will let $\overline{V}$ denote the complex conjugate local system to $V$. (So it is the same underlying local system of abelian groups, but we give it the conjugate action of $\mathbb C$.)
The Hermitian pairing on the fibres of $V$ and $\overline{V}$ gives a pairing of local systems $V \times \overline{V} \to \mathbb R$, where $\mathbb R$ is the constant local system with fibre the real numbers. If you like we can think of this as an $\mathbb R$-linear map $V\otimes_{\mathbb C}\overline{V} \to \mathbb R.$ This pairing will induce a map on cohomology $H^2(F,V\otimes_{\mathbb C}\overline{V}) \to H^2(F,\mathbb R)$.
There will also be a cup product $H^1(F,V) \times H^1(F,\overline{V}) \to H^2(F, V\otimes_{\mathbb C} \overline{V})$. Composing this with the previous map on $H^2$ gives your twisted cup product $H^1(F,V)\times H^1(F,\overline{V}) \to H^2(F,\mathbb R)$.
This gives one perspective on your construction. To compute it, write down the twisted cochains $C^{\bullet}(\tilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2$, then write down the cup-product $$(C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2 ) \times (C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]}\mathbb C^2) \to C^{\bullet}(\widetilde{F})\otimes_{\mathbb Z[\pi_1(F)]} \mathbb R^2 = C^{\bullet}(F,\mathbb R).$$ The cup product will just be given by the usual formula, and then you will also pair the $\mathbb C^2$ parts of the cochains using the hermitian pairing.
Hopefully you can follow your nose and do this explicitly for the torus. Then you can just dualize everything to get to the homology version.