# Non Abelian Hodge theory: underlying structure holomorphic vector bundles

Let $$X$$ be a compact Riemann surface. We fix a complex vector bundle $$E$$ of rank $$n$$ and degree $$d$$ (unique up to diffeomorphism). From results coming originally (I think at least) by Simpson,Donaldson and Hitchin one gets an equivalence between:

• Polystable Higgs field $$(\bar{\partial_E},\phi )$$ where $$\bar{\partial_E}$$ is a $$(0,1)$$ connection giving an holomorphic structure to $$E$$

-Projectively flat connections $$\nabla$$ over $$E$$ with semisimple monodromy.

The correspondence more or less is based on the following fact. We fix an Hermitian metric $$h$$ on $$E$$ and we decompose $$\nabla=\nabla^h+ \Phi$$ where $$\nabla^h$$ is an Hermitian connection and $$\Phi$$ an adjoint operator. Then we take the pair $$((\nabla^h)^{(0,1)},\Phi^{(1,0)})$$.In the semisimple monodromy situation this is a (poly)stable Higgs field.

However, I've not understood a fact: starting from a projectively flat $$\nabla$$ we can look at its $$(0,1)$$ part $$\nabla^{(0,1)}$$: this defines an holomorphic structure over $$E$$ as $$\nabla^{(0,1)}\nabla^{(0,1)}=0$$. Is this holomorphic structure the same (or better said isomorphic) to the structure induced by the couple $$((\nabla^h)^{(0,1)},\Phi^{(1,0)})$$. I tried to prove this by computing everything by hand, but I'm not able to provide an answer, neither a possible counterexample.

In general, the complex structures associated to $$\nabla$$ and $$\nabla^h$$ are different. One case where things can be made explicit is when the Higgs bundle arise from a (complex) variation of Hodge structure, c.f. Simpson, Higgs bundles and local systems. Then $$(E,\nabla^{0,1})$$ is a flat bundle associated to the local system underlying the VHS. While $$(E,(\nabla^h)^{0,1})$$ is isomorphic to the associated graded of $$E$$ with respect to the Hodge filtration, and this won't be flat except in special cases. One case worth noting is when flat connection $$\nabla$$ (or local system $$\ker \nabla$$) is unitary. Then it preserves a Hermitian metric $$h$$. So in this case $$\nabla=\nabla^h$$, and the Higgs field $$\Phi=0$$. When one specializes the nonabelian Hodge correspondence to this case, one recovers the theorem of Narasimhan-Seshadri.
• Can we at least say that for example the connection $\nabla$ is holomorphic with respect to the structure induced by $\nabla^h$ or this is not at all true? May 10, 2021 at 19:57
• Can we say something in some more specific cases? For example what happens what $\nabla^{(0,1)}$ induces the trivial holomorphic structure? May 12, 2021 at 6:17