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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?

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2 Answers 2

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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.

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Ah, I think I see what you're saying... I can just use the cellular (simplicial) formulation of cup product, in the universal cover. Ok this should work! Thanks. – Sam Lewallen Feb 23 '10 at 3:07

What you say is right, and makes sense on any even dimensional manifold. Computing it can be tricky: a useful approach is to use a regular cell complex and the dual complex, then on the chain level the intersection form is given by the identity matrix (see the first couple pages of Milnor's "a duality theorem for Reidemeister torsion").

One suggestion for calculation is to assume $\rho$ is irreducible, since if $C^n$ splits invariantly under the $\pi_1F$ action so does the cohomology. In your torus example, since $\pi_1=Z\oplus Z$ is abelian, the only irreducible reps are 1-dimensional. For Euler characteristic reasons (and Poincare duality) in this case it turns out either the rep is trivial in which case you know the answer, or else the rep is non-trivial in which case the homology vanishes and the intersection form is trivial. For higher genus surfaces you will get something non-zero, but in this dimension you get a skew-hermitian form, which is determined up to iso by its rank, if I'm thinking clearly. For dimensions divisible by 4, you can get interesting (i.e. non-zero) signature, but for a closed manifold it will just equal n times the ordinary signature by the twisted form of the Hirzebruch signature theorem.

But the twisted intersection form is interesting when your manifold has non-empty boundary, since it gives invariants of the boundary. Hundreds of papers are based on this observation. Even when your surface has non-empty boundary you get something interesting.

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I'll just add that if you use the $su(n)$ (Lie algebra) coefficients instead of $C^n$ coefficients, then the pairing you describe (on cohomology as Emerton suggests) gives you the symplectic structure on the moduli space of flat $SU(n)$ connections on your surface. There hundreds of papers on this topic too! – Paul Feb 23 '10 at 2:32
Thanks a lot! Actually, what I really want to do is use a surface $F$ with boundary, and instead of the (skew-hermitian) intersection form on the surface, use the symmetric bilinear linking form induced by thinking of $F$ as a Seifert surface for a link. I also want my $\rho$ to be induced by a representation of the link group (by pushing the surface either upward or downward into the link complement, we get a map on fundamental groups, and then compose with $\rho$. I'm not sure if pushing up or down will give equivalent signatures). Is there anything in the literature about these signatures? – Sam Lewallen Feb 23 '10 at 2:59
I don't quite understand. Your form is $H_1(F;C^n)\times H_1(F;C^n)\to C$ induced by "tensoring" the linking pairing (ie $lk(a,b^+)$ ) with the twisted coefficients? I'm not sure how this would work since $\rho$ doesnt extend to $S^3$. – Paul Feb 23 '10 at 3:45
Also, did you work it out in the fibered case, eg the trefoil? If so, check out the last section of Milnor's "infinite cyclic covers" that shows how to "pretend" your generic knot is fibered, and to recover knot signatures (see eg see the last section of for the twisted version). There has been a lot of work recently on twisting everything in classical knot theory, but I dont recall seeing this construction. – Paul Feb 23 '10 at 3:50
I agree that the construction is a little strange... let me think about it some more, then I may ask you another question (if that's alright). Indeed, it was that paper of Kirk and Livingston that I was thinking about... the numbers I'm trying to compute might just be twisted signatures for dihedral representations of the knot group, but I'm not sure. By the way, do you know of some interpretation of twisted signatures (from the Kirk+Livingston paper, say), in terms of the linking form on $H_2(M)$, where $M$ is a 4-manifold which bounds the branched double cover of $S^3-K$? – Sam Lewallen Feb 23 '10 at 5:56

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