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What is the exact geometric meaning of the Simpson's correspondence between Higgs bundles and local systems ? I know that it should have a rich geometric content but don't know an explicit geometric interpretation which reveals the significance f this correspondence.

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The nonabelian Hodge theorem, i.e. Simpson's correspondence, for smooth projective varieties refines a number of earlier results by several authors (Narasimhan-Seshadri, Donaldson...) where things can be understood more explicitly. For example, a unitary local system $L$ gives rise to polystable vector bundle $E= L\otimes \mathcal{O}_X$ with zero Higgs field. In general, the correspondence is highly transcendental, and I think it would be fair say that the geometric meaning is a very deep mystery. We can see this even in rank one. On the local system side, the moduli space is the character variety $$M_{Betti}=Hom(\pi_1(X), \mathbb{C}^*);$$ on the Higgs bundle side the moduli space is the cotangent bundle $$M_{Dol}=T^*Pic(X) = Pic(X)\times H^0(X,\Omega_X^1).$$ As algebraic varieties or even as complex manifolds these are very different, $M_{Betti}$ is Stein and the other is not. Yet they correspond as sets or topological spaces because they can both be identified using polar coordinates/Hodge theory with $$Hom(\pi_1(X),U(1))\times Hom(\pi_1(X),\mathbb{R})$$ I could go on, but perhaps I've made my point that the algebro-geometric meaning of the correspondence is by no means clear.

But I should add the significance should not be underestimated. For example, Simpson's work shows that among all representations of the fundamental group of a variety, the ones having Hodge theoretic (e.g. geometric) origin hold special status in this framework. In particular, he showed that any representation can be deformed to such a representation, which I think was totally unexpected.

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    $\begingroup$ However, a nice analytical consequence is that $M_{Dol}$ is hyperKaehler (by using both Betti and Dolbeault complex structures). $\endgroup$ Nov 1, 2011 at 21:08
  • $\begingroup$ Yes, sure. I agree there is much more to the story. $\endgroup$ Nov 1, 2011 at 21:14

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