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Let $X$ be a smooth projective variety over an algebraically closed field $K$ of characteristic zero and fix a point $x\in X(K)$.

We can associate to $X$ two Tannakian categories: the category of Higgs bundles on $X$ and the category of vector bundles with flat connection on $X$. The fiber functor in both cases is given by taking the fiber at $x$. Denote by $\pi_1^\text{Higgs}(X,x)$ and $\pi_1^\text{loc. sys.}(X,x)$ the respective pro-algebraic groups associated to these Tannakian categories.

If $K=\mathbb{C}$, then these two Tannakian categories are both equivalent to the category of representations of the topological fundamental group $\pi_1(X,x)$ by the nonabelian Hodge correspondence and the Riemann-Hilbert correspondence, respectively. Therefore, we have isomorphisms $$\pi_1^\text{Higgs}(X,x) \cong \pi_1^\text{loc. sys.}(X,x) \cong \widehat{\pi_1}(X,x)$$ where $\widehat{\pi_1}(X,x)$ denotes the pro-algebraic completion of $\pi_1(X,x)$ (and I think that these isomorphisms are canonical but I'm not totally sure). Hence we have an isomorphism between two pro-algebraic groups with a purely algebraic definition but whose construction a priori crucially uses the complex analytic topology.

If $K$ is not taken to be $\mathbb{C}$, how do the two candidates for a pro-algebraic fundamental group $\pi_1^\text{Higgs}(X,x)$ and $\pi_1^\text{loc. sys.}(X,x)$ compare? Are they still always (canonically) isomorphic?

The same question can also be asked if we replace the two Tannakian categories with their full subcategories of nilpotent objects, which corresponds to replacing the Tannakian groups with their pro-unipotent quotients. These pro-unipotent $\pi_1$'s are somewhat nicer than the pro-algebrac $\pi_1$'s because they are are compatible with base change (at least I know that this is true for the local systems $\pi_1$, I'm not totally about the Higgs $\pi_1$). I am also interested in an answer to the question above for the pro-unipotent $\pi_1$'s.

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  • $\begingroup$ For the pro-unipotent quotients, you can deduce this from Morgan's "The algebraic topology of smooth algebraic varieties"; it follows from formality of the de Rham algebra over $K$. For the Tannakian groups, you can't expect an isomorphism: just consider the case where $X$ is an elliptic curve and look at the respective spaces of $1$-dimensional representations; you know that $\check{X} \times \mathbb{A}^1$ isn't isomorphic as a scheme to $\mathbb{G}_m^2$. $\endgroup$ Commented Nov 8 at 17:33
  • $\begingroup$ Can you say a bit more on how formality of the de Rham algebra yields an isomorphism of the pro-unipotent quotients? Regarding your second point, I'm not sure why that forbids an isomorphism of the pro-algebraic groups; in general the set of 1d representations of a pro-algebraic group doesn't come with a natural moduli space structure. $\endgroup$ Commented Nov 8 at 19:55
  • $\begingroup$ I'm also a bit confused about where the $\mathbb{G}_m^2$ comes from. The moduli space of $1$-dimensional vector bundles with flat connection on an elliptic curve $X$ is an affine space bundle over $\operatorname{Jac}(X)=X$, not $\mathbb{G}_m^2$. $\endgroup$ Commented Nov 8 at 19:58
  • $\begingroup$ Such an isomorphism would have to be pretty exotic. As you note, an isomorphism of pro-algebraic groups arising from Tannakian categories whose objects have moduli spaces need not give an isomorphism of the moduli spaces. But it gives a bijection between the $K$-points of one moduli space and the $K$-points of the other. For $K \subset \mathbb C$ there is a natural complex-analytic bijection between the $K$-points of one and the $K$-points of the other, but it is transcendental and hence usually won't send $K$-points to $K$-points. $\endgroup$
    – Will Sawin
    Commented Nov 8 at 20:05
  • $\begingroup$ One can prove the existence of a bijection (even a bijection that is a restriction of a continuous map on $\mathbb C$-points) but this will just be an arbitrary choice, there surely won't be a canonical bijection. $\endgroup$
    – Will Sawin
    Commented Nov 8 at 20:06

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