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Let $C$ be smooth projective curve, $\mathcal{M}$ be the moduli space of semistable Higgs bundles, $H:\mathcal{M}\rightarrow W= \bigoplus H^0(C,K_{C}^{\otimes i})$ be the Hitchin map, and $\pi :C_s\rightarrow C$ be the smooth spectral curve associated to $s=(s_1,...,s_n)\in W$.

I'm trying to show the well-known fact,

There is a natural bijective correspondence isomorphism classes of line bundles on $C_s$ and isomorphism classed of pairs $(E,\phi)$ where $E$ is a locally free sheaf of rank $n$ on $C$ and $\phi:E\rightarrow K_C\otimes E$ a homomorphism with characteristic coefficients $s_i$

Given $L$ line bundle on $C_s$, then $\pi_*L$ is a vector bunlde of rank $n$ on $C$ with a $\pi_*\mathcal{O}_{C_s}$ module structure.
Conversely, given a Higgs pair $(E,\phi)$, I want to construct the line bundle $L_E$ on the spectral curve $C_s$ such that $\pi_*L_E\cong E$ But how is this possible ?

$C_s$ can be interpreted as $\operatorname{Spec}(\operatorname{Sym}(K_C^{-1})/\mathcal{I}_s)\subset \operatorname{Spec}(\operatorname{Sym}(K_C^{-1}))=T^*C$, and $\phi: \operatorname{Sym}(K_C^{-1})/\mathcal{I}_s\rightarrow \operatorname{End}(E)$. Then, $E$ can be seen as a module over $\pi_*\mathcal{O}_{C_s}=\operatorname{Sym}(K_C^{-1})/\mathcal{I}_s$.
I'm stuck here.

Thanks in advance.

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  • $\begingroup$ Note that in general (for an affine morphism $\pi$) the functor $\pi_\ast$ induces an equivalence of categories between quasi-coherent sheaves on $C_s$ andquasi-coherent $\pi_\ast(\mathcal O_{C_s})$-modules on $C$. Thus $E$ corresponds to a quasi-coherent sheaf $L_E$ on $C_s$ with $\pi_\ast (L_E) \cong E$. It remains to show that $L_E$ is an invertible sheaf. You can do this by first showing that $L_E$ must be torsion-free of rank 1. $\endgroup$
    – Sam Gunningham
    Commented Oct 14, 2020 at 14:27
  • $\begingroup$ There are details in Section 3 of this paper math.unice.fr/~beauvill/pubs/bnr.pdf $\endgroup$
    – Sam Gunningham
    Commented Oct 14, 2020 at 14:27
  • $\begingroup$ Thanks, your comment is very helpful. I read the paper of BNR, but I don't understand the reason $L_E$ is torsion-free of rank $1$. $\endgroup$
    – Aoki
    Commented Oct 14, 2020 at 14:48
  • $\begingroup$ The condition that $C_a$ is irreducible means that we have a degree $n$ extension of function fields $K(C) \hookrightarrow K(C_a)$. Then $L_E$ restricted to the generic point must be a rank $1$ $K(C_a)$-vector space, as we know that it is rank $n$ over $K(C)$. Thus $L_E$ is generically rank 1. $\endgroup$
    – Sam Gunningham
    Commented Oct 15, 2020 at 12:13
  • $\begingroup$ Moreover, for any open affine $U\subseteq C$, we have that $\Gamma(U;E) = \Gamma(\pi^{-1}(U);L_E)$ injects in to the generic stalk $E_{\eta_C} = (L_E)_{\eta_{C_a}}$ (as $E$ is torsion free). Thus $L_E$ must be torsion free. I hope this makes sense! $\endgroup$
    – Sam Gunningham
    Commented Oct 15, 2020 at 12:15

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