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?