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For $G = GL_n$, it is known that the generic fibers of the Hitchin fibration are the Picard stacks of line bundles on the corresponding spectral curves and the duality of Hitchin fibrations in this case amounts to the self-duality of the Jacobian of an algebraic curve. Note however that these statements are valid on an open subset of the Hitchin base. My question is about this open subset, See S4.2, S4.4 and Cor 4.5 of Bezrukavnikov-Braverman for an overview (and all notation that is used in the next paragraph): https://arxiv.org/pdf/math/0602255.pdf

Prop 4.3 of that paper states that there is a non-empty open subset $Hitch_n^0$ of $Hitch_n$ over which $pr_1$ is smooth. How can we explicitly write down the open subset $Hitch_n^0$ in this context? The proof of Prop 4.3 is unclear to me at this stage, so any clarifications would be appreciated (in particular, I'm looking for an explicit description of this open subset). Note that $Hitch_n$ is explicitly defined in S4.1.

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The relevant open subset is simply the open subset where the spectral curve is smooth. Thus it is the nonvanishing locus of some discriminant polynomial.

The Hitchin fiber is the moduli space of bundles on the spectral curve, generically free of rank one, whose pushforward to the base curve is locally free of rank $n$.

For a smooth curve, the only way a generically free bundle can fail to be locally free is if it contains torsion, in which case its pushforward is never locally free, so the Hitchin fiber is simply the space of line bundles on the spectral curve generically free of rank one, i.e. the Jacobian, and thus is smooth.

However, for a non-smooth curves, one will need to compactify the Jacobian to get the Hitchin fiber, and this will introduce singularities.

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  • $\begingroup$ Thanks! What does the discriminant polynomial look like in low genus (i.e. for elliptic curves, and also in genus 2)? Is it easy to write down? $\endgroup$
    – Puraṭci Vinnani
    Commented Sep 19, 2022 at 0:09
  • $\begingroup$ @PuraṭciVinnani For elliptic curves, since the canonical bundle is trivial, the Hitchin base merely parameterizes polynomials, and the relevant discriminant is just the polynomial discriminant. For genus $2$ curves, one can view the Hitchin base as parameterizing monic polynomials in $x$ of degree $n$ with coefficients in $k[t, \sqrt{f(t)}]$ where $f$ is a fixed polynomial of degree $6$ and the coefficient of $x^{n-i}$ has degree at most $i$ in $t$. $\endgroup$
    – Will Sawin
    Commented Sep 19, 2022 at 0:18
  • $\begingroup$ @PuraṭciVinnani One can take the polynomial discriminant in $x$, then norm from $k[t, \sqrt{f(t)}]$ to $k[t]$, then take the polynomial discriminant in $t$, and then this will be a multiple of the discriminant I mean, so the true answer is obtained by taking some factor of this. $\endgroup$
    – Will Sawin
    Commented Sep 19, 2022 at 0:19
  • $\begingroup$ Thanks! I'll think more about this soon, it's intriguing.. $\endgroup$
    – Puraṭci Vinnani
    Commented Sep 20, 2022 at 13:15

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