Proof. The proof follows verbatim [Reference Porta and SalaPS, Proposition 5.1.2] after replacing the relative cotangent complex $\mathbb L_{{\mathcal{X}}/{\mathcal{S}}}$ with the relative dualizing sheaf $\omega _{{\mathcal{X}}/{\mathcal{S}}}$ .

Proof. Proof of $\underline{\;\,\mathcal{O}}$ -compactness : Let us begin with recalling the notation for the following maps from Section 1.1.

(3.9)

From (3.7), we have

(3.10) \begin{align} Rq_*{\mathcal{O}}_{{\mathcal{X}}_{Dol/\mathcal{S}}}\cong {\mathcal{O}}_{{\mathcal{X}}}\oplus \omega _{{\mathcal{X}}/{\mathcal{S}}}[-1], \end{align}

which is clearly a perfect complex. Therefore, ${\mathcal{O}}_{{\mathcal{X}}_{Dol/\mathcal{S}}}$ is $\mathcal{O}$ -compact relative to $\mathcal{S}$ .

Proof of $\underline{\;\;\mathcal{O}}$ -orie n tation : Using (3.10), we have

\begin{align*} ( Rp_*{\mathcal{O}}_{{\mathcal{X}}_{Dol/\mathcal{S}}})^{\vee }[-2] & \cong ( Rr_*{\mathcal{O}}_{{\mathcal{X}}}\oplus Rr_*\omega _{{\mathcal{X}}/{\mathcal{S}}}[-1])^{\vee }[-2] \\ & \cong ( Rr_*{\mathcal{O}}_{{\mathcal{X}}})^{\vee }[-2]\oplus ( Rr_*\omega _{{\mathcal{X}}/{\mathcal{S}}}[-1])^{\vee }[-2]. \end{align*}

By Serre duality, we have

\begin{align*} Rr_*(\omega _{{\mathcal{X}}/{\mathcal{S}}}[-1])\cong ( Rr_*((\omega _{{\mathcal{X}}/{\mathcal{S}}}[-1])^{\vee }\otimes \omega _{{\mathcal{X}}/{\mathcal{S}}}[1]))^{\vee }\cong ( Rr_*({\mathcal{O}}_{{\mathcal{X}}}[2]))^{\vee }\cong (Rr_*{\mathcal{O}}_{{\mathcal{X}}})^{\vee }[-2]\\ \implies ( Rr_*(\omega _{{\mathcal{X}}/{\mathcal{S}}}[-1]))^{\vee }[-2]\cong Rr_*({\mathcal{O}}_{{\mathcal{X}}}[2])[-2]\cong Rr_*{\mathcal{O}}_{{\mathcal{X}}} \hspace {6em} \end{align*}

Note that $H^0(Rr_*{\mathcal{O}}_{{\mathcal{X}}})\cong {\mathcal{O}}_{{\mathcal{S}}}$ . We choose an isomorphism once and for all and denote it by $\eta$ . This defines an element

(3.11) \begin{align} \eta \in ( Rp_*{\mathcal{O}}_{{\mathcal{X}}_{Dol/\mathcal{S}}})^{\vee }[-2]=R\mathcal{H}om_{\mathcal{S}}(Rp_*{\mathcal{O}}_{{\mathcal{X}}_{Dol/\mathcal{S}}}, {\mathcal{O}}_{{\mathcal{S}}}[-2]). \end{align}

We want to show that it is an ${\mathcal{O}}$ -orientation.

Let $\mathcal{E}$ be a perfect complex on ${\mathcal{X}}_{Dol/\mathcal{S}}\times _{{\mathcal{S}}} \textbf {Spec}(A)$ , for an $A\in \textbf {cdga}^{\leq 0}_{{{\mathcal{S}}}}$ . We have to show that the following map

(3.12) \begin{align} (-\cap \eta _A): R{p_A}_*{\mathcal{E}}\longrightarrow (R{p_A}_*({\mathcal{E}}^{\vee }))^{\vee }[-2]=(R{p_A}_*({\mathcal{E}}^{\vee })[2])^{\vee } \end{align}

is a quasi-isomorphism of $A$ -dg-modules, where $\eta _A: Rp_{A*}{\mathcal{O}}_{{\mathcal{X}}_{Dol, A}}\longrightarrow A[-2]$ is the derived pullback of $\eta : Rp_*{\mathcal{O}}_{{\mathcal{X}}_{Dol/\mathcal{S}}}\longrightarrow {\mathcal{O}}_{{\mathcal{S}}}[-2]$ under the map $\textrm {Spec}\, (A)\longrightarrow {\mathcal{S}}$ and $(R{p_A}_*({\mathcal{E}}^{\vee }))^{\vee }$ is the derived $A$ -dual of $R{p_A}_*({\mathcal{E}}^{\vee })$ . Here,

\begin{align*} p_A: {\mathcal{X}}_{Dol/\mathcal{S}}\times _{{\mathcal{S}}} \textbf {Spec}(A) \longrightarrow \textbf {Spec}(A) \end{align*}

is the projection. The left-hand side is isomorphic to $ R{r_A}_*\mathsf{Dol}_{(E_A, \phi _A)}$ , where

\begin{align*} \mathsf{Dol}_{(E_A, \phi _A)}:=[E_A\xrightarrow {\phi _A} E_A\otimes \omega _{{\mathcal{X}}_A, A}] \\ 0 \hspace {4cm} 1 \end{align*}

corresponding to $R{q_A}_*{\mathcal{E}}_A$ (here, $E_A$ is in degree $0$ ). The right-hand side is isomorphic to $( R{r_A}_*(\mathsf{Dol}^{\mathsf{D}}_{(E_A, \phi _A)}))^{\vee }$ , where

(3.13) \begin{align} \mathsf{Dol}^{{\bf D}}_{(E_A, \phi _A)}:=[E^{\vee }_A\xrightarrow {-\phi ^{\vee }_A} E^{\vee }_A\otimes \omega _{{\mathcal{X}}_A, A}] \end{align}

(3.14) \begin{align} -2 \hspace {3.5cm} -1 \end{align}

where $\phi ^{\vee }_A$ is the image of $\phi _A$ under the natural isomorphism ${{\mathcal{E}}\mathsf{nd}} (E_A)\cong {{\mathcal{E}}\mathsf{nd}} (E^{\vee }_A)$ . Now since the Grothendieck–Serre dual of the complex $\mathsf{Dol}_{(E_A, \phi _A)}$ is the same as $\mathsf{Dol}^{{\bf D}}_{(E_A, \phi _A)}$ , the morphism (3.12) is quasi-isomorphism. (We have used the notation $\mathsf{Dol}^{{\bf D}}_{(E_A, \phi _A)}$ in (3.13) to disambiguate between the ordinary dual and the Serre dual of the complex $\mathsf{Dol}_{(E_A, \phi _A)}$ .)

Therefore, we have shown that ${\mathcal{X}}_{Dol/\mathcal{S}}$ is $\mathcal{O}$ -compact and $\mathcal{O}$ -oriented over ${\mathcal{S}}$ and from [Reference Pantev, Toën, Vaquié and VezzosiPTV⁺13, Theorem 2.5], it follows that $\mathsf{Map}_{{\mathcal{S}}}({\mathcal{X}}_{Dol/\mathcal{S}}, BGL_{n}(\mathcal{O}_{\mathcal{S}}))$ has a zero-shifted relative symplectic structure over ${\mathcal{S}}$ .

Proof. Let us denote $\mathsf{Map}_{_{\mathfrak M}}(\mathfrak{X}_{Dol/\mathfrak{M}}, BGL_n(\mathcal{O}_{\mathfrak{M}}))$ by $M$ , for simplicity of notation. As before, the natural map $p : \mathfrak{X}_{Dol,\mathfrak{M}} \to \mathfrak{M}$ factors as

(3.15)

The relative tangent complex of the morphism $M\longrightarrow \mathfrak M$ is given by

(3.16) \begin{align} \mathbb T_{M/\mathfrak M}\cong Rp_*{{\mathcal{E}}\mathsf{nd}} \,{\mathcal{E}}[1]\cong Rr_*\mathcal C(E, \phi )[1], \end{align}

where $\mathcal C(E, \phi )$ denotes the following complex

(3.17) \begin{align} \Big[ {{\mathcal{E}}\mathsf{nd}} (E)\xrightarrow {[-, \phi ]} {{\mathcal{E}}\mathsf{nd}} (E)\otimes \mathsf{pr}_{\mathfrak{X}}^{*}\omega _{\mathfrak{X}/\mathfrak{M}} \Big],\end{align}

with ${{\mathcal{E}}\mathsf{nd}} (E)$ sitting in the degree $0$ . Here $\mathcal{E}$ denotes the universal sheaf on $\mathfrak{X}_{Dol/\mathfrak{M}}\times _{\mathfrak{M}} M$ and $(E, \phi )$ denotes the corresponding universal relative Higgs complex on $\mathfrak{X}\times _{\mathfrak{M}} M$ . From [Reference ToënToë12, Theorem 0.3], it follows that $\mathbb L_{M/\mathfrak M}$ is a perfect complex because the map $r: \mathfrak{X} \longrightarrow \mathfrak{M}$ is a proper representable local complete intersection morphism. Also, using Grothendieck–Serre duality we can see that the complex $Rr_*\mathcal C(E, \phi )[1]$ has amplitude in $[-1, 1]$ . Therefore, we conclude that the morphism

\begin{align*} \mathsf{Map}_{_{\mathfrak M}}(\mathfrak{X}_{Dol/\mathfrak{M}}, BGL_n(\mathcal{O}_{\mathfrak M}))\longrightarrow \mathfrak M \end{align*}

is a quasi-smooth morphism of derived Artin stacks.

Proof. Recall that $\mathfrak{L}^{0}_{S,\mathsf{lf}}$ denotes the classifying stack of log morphisms to $S$ from schemes equipped with locally free log structures. The log structure on $\mathfrak M$ , which we discussed in Proposition 2.15, then gives a map $\mathfrak M\to \mathfrak{L}^{0}_{S,\mathsf{lf}}$ .

Consider the composite morphism:

\begin{align*} M:=\mathsf{Map}_{_{\mathfrak M}}(\mathfrak{X}_{Dol/\mathfrak{M}}, BGL_n(\mathcal{O}_{\mathfrak M})) \longrightarrow \mathfrak M\longrightarrow \mathfrak{L}^{0}_{S,\mathsf{lf}}. \end{align*}

The composite morphism induces a log structure on $M$ , which is the same as the pullback of the log structure from $\mathfrak M$ . Note that the map $M \to \mathfrak{M}$ is quasi-smooth (Proposition 3.9) and the map $\mathfrak{M} \to \mathfrak{L}^{0}_{S,\mathsf{lf}}$ is smooth with $\mathbb L_{\mathfrak M/\mathfrak{L}^{0}_{S,\mathsf{lf}}}\cong 0$ (Proposition 2.15). Therefore, it follows that

(3.18) \begin{align} \mathbb L_{M/\mathfrak{L}^{0}_{S,\mathsf{lf}}}\cong \mathbb L_{M/\mathfrak M}\cong Rr_*\mathcal C(E, \phi )[1]. \end{align}

Thus, the relative log-cotangent complex of the map $M\longrightarrow S$ is isomorphic to the relative cotangent complex of the map $M\longrightarrow \mathfrak M$ . From Theorem 3.7, it follows that the stack $M$ has a zero-shifted symplectic structure relative to the stack $\mathfrak M$ , which is, by Definition2.23, a zero-shifted log-symplectic structure on $M$ relative to $S$ .

The log-symplectic pairing can be described as follows. The stack $BGL_n$ has a $2$ -shifted symplectic form given by the trace pairing:

(3.19) \begin{align} \mathbb{T}_{BGL_n}\wedge \mathbb{T}_{BGL_n}=\mathfrak gl_n[1]\wedge \mathfrak gl_n[1]\xrightarrow {(\mathsf x,\mathsf y)\mapsto Trace(\mathsf x\cdot \mathsf y)} \unicode{x1D55C}[2]. \end{align}

Now let $A$ be an object of ${\rm cdga}^{\leq 0}$ and let $\textrm {Spec}(A) \to M$ be an $A$ -valued point of $M$ . Write $\mathfrak{X}_{A}$ for the pullback $\mathfrak{X}\times _{\mathfrak{M}} \textrm {Spec}(A)$ of the universal curve to $\textrm {Spec}(A) \to M \to \mathfrak{M}$ . We also write $\mathfrak{X}_{Dol, A}$ for the pullback $\mathfrak{X}_{Dol/\mathfrak{M}}\times _{\mathfrak{M}} \textrm {Spec}(A)$ for the pullback of the relative Dolbeault stack. Now, by the definition of $M$ , the map $\textrm {Spec}(A) \to M$ gives us a map

\begin{align*}f_A: \mathfrak{X}_{Dol,A}\longrightarrow BGL_n\times \textrm {Spec} (A),\end{align*}

which corresponds to a vector bundle $\mathcal{E}_A$ on $\mathfrak X_{Dol, A}$ . The vector bundle ${\mathcal{E}}_A$ on $\mathfrak X_{Dol, A}$ corresponds to a Higgs field

\begin{align*} (E_{A} \xrightarrow {\phi _A} E_{A}\otimes \omega _{\mathfrak{X}_A , A}). \end{align*}

on $\mathfrak X_A$ .

We have the following induced pairing by pulling back (3.19).

(3.20) \begin{align} f^*_A \mathbb{T}_{BGL_n}\otimes f^*_A \mathbb{T}_{BGL_n}= {{\mathcal{E}}\mathsf{nd}} \,\mathcal{E}_A[1]\otimes {{\mathcal{E}}\mathsf{nd}} \,\mathcal{E}_A [1]\xrightarrow {Tr:=Trace} {\mathcal{O}}_{\mathfrak{X}_{Dol, A}}[2] .\end{align}

Now by pushing it forward by the map $q_A$ , we get

(3.21)

The complex ${q_A}_*({{\mathcal{E}}\mathsf{nd}} \,\mathcal{E}_A [1])=[{{\mathcal{E}}\mathsf{nd}} \,E_A \xrightarrow {[-,\phi _A]} {{\mathcal{E}}\mathsf{nd}} \,E_A \otimes \omega _{\mathfrak{X}_A, A}]$ can be used to describe the pullback of the pairing in (3.21) by the natural map

\begin{align*} {q_A}_*({{\mathcal{E}}\mathsf{nd}} \,\mathcal{E}_A [1]) \otimes {q_A}_*({{\mathcal{E}}\mathsf{nd}} \,\mathcal{E}_A [1]) \to {q_A}_*({{\mathcal{E}}\mathsf{nd}} \,\mathcal{E}_A[1] \otimes {{\mathcal{E}}\mathsf{nd}} \,\mathcal{E}_A [1]) \end{align*}

as the following map of complexes.

(3.22)

Here the map $d_1:=([-, \phi _A], [-, \phi _A])$ and $d_2:=[-, \phi _A]\otimes {\unicode{x1D7D9}}+ {\unicode{x1D7D9}}\otimes [-, \phi _A]$ . Finally, by pushing forward (3.21) by the map $r_A$ and then composing with the orientation $\eta _A$ (see (3.12)), we get our log-symplectic pairing. From the above description it is evident that this log-symplectic form coincides with the log-symplectic form discussed in [Reference DasDas22].

Proof. Let $ T = \textrm {Spec} A$ be the spectrum of a discrete valuation ring over $ \unicode{x1D55C}$ , with fraction field $ K$ . Denote the generic point by $ T^o = \textrm {Spec} K$ . Suppose that we are given the following commutative diagram.

(4.3)

We must show that this diagram extends to a morphism $ T \to M^{cl}_{Gie}$ . Let $ \mathsf{TFH}({\mathcal{X}}/S)$ be the ordinary Artin stack of families of torsion-free $ \omega _{{\mathcal{X}}/S}$ -valued Higgs pairs on $ {\mathcal{X}}/S$ (see Definition A.1.2). There is a natural morphism

\begin{align*}\theta \colon M^{cl}_{Gie} \longrightarrow \mathsf{TFH}({\mathcal{X}}/S), \qquad (\pi \colon \mathfrak{X} \to {\mathcal{X}},\ \mathcal{E},\ \phi ) \mapsto ((\pi _* \mathcal{E}),\ (\pi _* \phi )).\end{align*}

This is induced by pushforward along the modification $ \mathfrak{X} \to {\mathcal{X}}$ . Applying $ \theta$ to the original diagram gives the following commutative triangle.

(4.4)

By [Reference LangtonLan75, Proposition 6] and [Reference NitsureNit91, Lemma 6.5], there exists a morphism $ T \to \mathsf{TFH}({\mathcal{X}}/S)$ making the triangle commute. Now consider the surface $ {\mathcal{X}}_T := {\mathcal{X}} \times _S T$ . There are two cases.

Case 1. The map $ T \to S$ is not faithfully flat. Then $ {\mathcal{X}}_T \cong X_0 \times T$ . The normalization of $ {\mathcal{X}}_T$ is $ \widetilde {X}_0 \times T$ . Let $ f \colon \widetilde {X}_0 \to X_0$ be the normalization map. By [Reference DasDas19, Proposition 3.2], the pullback of the generic Higgs bundle extends to a good generalized parabolic Hitchin pair on $ \widetilde {X}_0 \times T$ , which in turn gives a torsion-free Hitchin pair on $ {\mathcal{X}}_T$ .

Case 2. The map $ T \to S$ is faithfully flat. Then $ {\mathcal{X}}_T$ is a normal surface with an isolated singularity of type $ \unicode{x1D55C}[[x,y,t]]/(xy - t^d)$ . Let $ i \colon {\mathcal{X}}_{T^o} \hookrightarrow {\mathcal{X}}_T$ be the inclusion of the generic fibre. Suppose we have a Higgs bundle $ (\mathcal{E}^o, \phi ^o)$ on $ {\mathcal{X}}_{T^o}$ . Any vector bundle on the generic fibre extends to a torsion-free sheaf $ \mathcal{F}$ on $ {\mathcal{X}}_T$ . We wish to extend the Higgs field to $ \mathcal{F} \to \mathcal{F} \otimes \omega _{{\mathcal{X}}_T/T}$ . Consider the following diagram.

(4.5)

By [Reference NitsureNit91, Lemma 6.5], the Higgs field extends over $ {\mathcal{X}}_T$ except possibly at the node. The composite map $ \mathcal{F} \to (i_*\mathcal{E}^o / \mathcal{F}) \otimes \omega _{{\mathcal{X}}_T/T}$ vanishes generically on the closed fibre. Thus, the obstruction lies in the torsion part of $ i_*\mathcal{E}^o / \mathcal{F}$ (viewed as a sheaf on its support $ X_0$ ). Restricting the top sequence to the closed fibre gives

(4.6) \begin{align} 0\longrightarrow \mathsf{Tor}^1_{_{{\mathcal{X}}_T}}(i_*{\mathcal{E}}^{o}/{\mathcal F}, {\mathcal{O}}_{X_0})\longrightarrow \mathcal F|_{X_0}\longrightarrow (i_*{\mathcal{E}}^{o})|_{X_0}\longrightarrow i_*{\mathcal{E}}^{o}/{\mathcal F}\longrightarrow 0. \end{align}

Since $ \mathsf{Tor}^1_{_{{\mathcal{X}}_T}}(i_*{\mathcal{E}}^{o}/{\mathcal F}, {\mathcal{O}}_{X_0})= (i_*\mathcal{E}^o / \mathcal{F}) \otimes \mathcal{O}_{{\mathcal{X}}_T}(-X_0)$ and $ \mathcal{O}_{{\mathcal{X}}_T}(-X_0)$ is locally free, the sheaf $ i_*\mathcal{E}^o / \mathcal{F}$ is pure (torsion-free on its support) if and only if the Tor term is pure. But the Tor term is a subsheaf of $ \mathcal{F}|_{X_0}$ , hence torsion-free on $ X_0$ . This implies the obstruction vanishes, and the Higgs field extends everywhere.

In both cases we obtain the following commutative diagram.

(4.7)

Choose a connected component $ R^{\Lambda , m_T}_S$ of an atlas for $ \mathsf{TFH}({\mathcal{X}}/S)$ such that the image of $ T \to \mathsf{TFH}({\mathcal{X}}/S)$ lies in the image of $ R^{\Lambda , m_T}_S \to \mathsf{TFH}({\mathcal{X}}/S)$ . Let $ \mathcal{Y}^H_S$ be the corresponding Quot scheme of Gieseker–Hitchin pairs over $ R^{\Lambda , m_T}_S$ (see Appendix A.3), with projection $ \tilde {\theta } \colon \mathcal{Y}^H_S \to R^{\Lambda , m_T}_S$ . Both maps $ R^{\Lambda , m_T}_S \to \mathsf{TFH}({\mathcal{X}}/S)$ and $ \mathcal{Y}^H_S \to M^{cl}_{Gie}$ are principal $ GL_N$ -bundles for sufficiently large $ N$ , and $ \tilde {\theta }$ is $ GL_N$ -equivariant (see Appendix A.9).

(4.8)

Since $T$ is a spectrum of a discrete valuation ring, any $GL_N$ bundle on $T$ is trivial. Therefore, by choosing a trivialization of the $GL_N$ -bundle $R^{\Lambda , m_T}_S\times _{\mathsf{TFH}({\mathcal{X}}/S)} T\to T$ we get the following lift of the diagram (4.7).

(4.9)

From [Reference Balaji, Barik and NagarajBBN16, Proposition 5.11], it follows that the map $\tilde {\theta }: \mathcal Y^H_S\longrightarrow R^{\Lambda , m_T}_S$ is proper (we discuss the proof below in Lemma 4.5 and Corollary 4.6 for the reader’s convenience). Thus, there exists a lift $T\longrightarrow \mathcal Y^H_S$ . The composite map $T\longrightarrow \mathcal Y^H_S\longrightarrow M^{cl}_{Gie}$ is our desired extension. This completes the proof.

Proof. Take the closure of $Y$ inside some projective bundle (over $R$ ) to get a projective morphism $\overline {Y}\longrightarrow R$ . Take any element $y\in \overline {Y}\setminus Y$ . Then there exists a map $T\longrightarrow \overline {Y}$ (here, $T$ is a spectrum of a discrete valuation ring) passing through $y$ and flat over $S$ . Therefore, we get a map $T\longrightarrow R$ . Then, by our lifting assumption, this map can be lifted to a map $T\longrightarrow Y$ . Let $y^{\prime}\in Y$ be the image of the closed point of $T$ under this map. But, by the separateness of $\overline {Y}$ , we must have $y=y^{\prime}$ and therefore $\overline {Y}=Y$ , and the map $Y\longrightarrow R$ is proper.

Proof. Since the map $\tilde {\theta }$ is quasi-projective, we choose a closure $\overline {\mathcal Y^H_S}\longrightarrow R^{\Lambda , m_T}_S$ inside some projective bundle over $R^{\Lambda , m_T}_S$ . Let $y\in \overline {\mathcal Y^H_S}\setminus {\mathcal Y^H_S}$ . Then there exists a map $T\longrightarrow \overline {\mathcal Y^H_S}$ (here, $T$ is a spectrum of a discrete valuation ring) passing through $y$ and flat over $S$ . Consider the composite map $T\to \overline {\mathcal Y^H_S}\to R^{\Lambda , m_T}_S$ . Let $(\mathcal F_T, \phi _T)$ be the flat family of torsion-free Higgs pair on ${\mathcal{X}}_T:={\mathcal{X}}\times _S T$ corresponding to the map $T\to R^{\Lambda , m_T}_S$ .

Claim: The map $T\to R^{\Lambda , m_T}_S$ can be lifted to a map $T\to \mathcal Y^H_S$ .

Proof of the claim: Note that the surface ${\mathcal{X}}_T$ may have a singularity of type $ {\unicode{x1D55C}[|x,y,t|]}/{(xy-t^n)}$ because the map $T\to S$ is a faithfully flat map of spectrum of discrete valuation rings. Let $r_{_T}: {\mathcal{X}}^{{res}}_T\longrightarrow {\mathcal{X}}_T$ be the minimal resolution of singularities. Then from [Reference LipmanLip69, Proposition 6.5] and [Reference Balaji, Barik and NagarajBBN16, Section 4] it follows that the vector bundle ${\mathcal{E}}_T:= {r_{_T}^*\mathcal F_T}/{Torsion}$ has the property that $(r_{_T})_*{\mathcal{E}}_T\cong \mathcal F_T$ . Note also that by construction and by the property that $(r_{_T})_*{\mathcal{E}}_T\cong \mathcal F_T$ , it follows that the natural map $r_{_T}^*(r_{_T})_*{\mathcal{E}}_T\longrightarrow {\mathcal{E}}_T$ is a surjective map of sheaves, which means $({\mathcal{E}}_T)|_R$ is globally generated. Here $R$ denotes the chain of rational curves in ${\mathcal{X}}^{{res}}_T$ . Hence, by definition (DefinitionA.1 and Remark A.2), the vector bundle ${\mathcal{E}}_T$ is a Gieseker vector bundle. The pullback of the Higgs field $\phi _{_T}$ defines a Higgs field on ${\mathcal{E}}_T$ . The pair $({\mathcal{E}}_T, \phi _T)$ is the desired lift.

Therefore, from Lemma 4.5 it follows that the map $\tilde {\theta }: \mathcal Y^H_S\longrightarrow R^{\Lambda , m_T}_S$ is proper.

Proof. For every $i\in [1,k+1]$ , we have morphisms $\phi ^i: {\mathcal{E}}\longrightarrow {\mathcal{E}}\otimes \omega ^{\otimes i}_{X_r}$ and $\psi ^i: \mathcal F\longrightarrow \mathcal F\otimes \omega ^{\otimes i}_{X_0}$ . Moreover, $\ker (\phi ^i)={\mathcal{E}}^i$ and $\ker (\psi ^i)=\mathcal F^i$ . Let $\sigma$ be a local section of $(\pi _r)_*{\mathcal{E}}^i$ on a neighbourhood $U$ of the node of $X_0$ . It is an element of ${\mathcal{E}}^i((\pi _r)^{-1}(U))$ . Therefore, the section $\phi ^i(\sigma )=0$ on the open set $(\pi _r)^{-1}(U)\cap \tilde {X_0}=U^o$ , where $U^o$ denotes the complement of the node of $X_0$ . Therefore, $\psi ^i(\sigma )=0$ on $U^o$ , because $\psi =\phi$ on $U^o$ . Since $U^o$ is dense in $U$ , therefore $\psi ^i(\sigma )=0$ on $U$ and $\sigma \in \mathcal F^i$ . Therefore, $(\pi _r)_*{\mathcal{E}}^i\subseteq \mathcal F^i$ for all $i\in [1,k+1]$ .

For the converse, let $\sigma \in \mathcal F^i(U)$ . Since $\mathcal F^i\subseteq \mathcal F=(\pi _r)_*{\mathcal{E}}$ , we have $\sigma \in {\mathcal{E}}((\pi _r)^{-1}(U))$ . Since $\psi ^i(\sigma )=0$ , therefore $\phi ^i(\sigma )=0$ on $(\pi _r)^{-1}(U^o)$ and hence, by continuity, on $(\pi _r)^{-1}(U)\cap \widetilde {X_0}$ . Now note that $\phi ^i(\sigma )$ is a section of ${\mathcal{E}}\otimes \omega _{X_r}$ . The bundle ${\mathcal{E}}\otimes \omega _{X_r}$ is a Gieseker-vector bundle because $\omega _{X_r}|_R\cong {\mathcal{O}}_R$ , where $R$ the chain of $\mathbb P^1$ in $X_r$ . Therefore, it must vanish everywhere since the section vanishes at the two points $\widetilde {X_0}\cap R$ . This implies that $\sigma \in {\mathcal{E}}^i((\pi _r)^{-1}(U))=((\pi _r)_*{\mathcal{E}}^i)(U)$ . Therefore, $\pi _{r*}({\mathcal{E}}^i)=\mathcal F^i$ for all $i\in [1,k+1]$ .

Proof. Consider any connected component $C$ of ${\mathcal{N}ilp}^{sm, gen}_0$ to see this. Then we have the following exact sequence for any element $c:\textrm {Spec}\, \unicode{x1D55C}\longrightarrow C$ :

\begin{align*} 0\longrightarrow \textsf K_c^i\longrightarrow c^*{\mathcal{E}}\xrightarrow {(c^*\phi )^i} c^*{\mathcal{E}}\otimes \omega ^{\otimes i}_{c^*\mathfrak X}\longrightarrow \textsf {CK}^i_c\longrightarrow 0. \end{align*}

Here $\textsf K_c^i$ and $\textsf {CK}^i_c$ denote the kernel and cokernel of the map $(c^*\phi )^i$ , respectively. We also have the following exact sequence:

\begin{align*} 0\longrightarrow \underline{\textsf {K}}^i_c\longrightarrow (\pi _r)_*(c^*{\mathcal{E}})\xrightarrow {(\pi _r)_*((c^*\phi )^i)=((\pi _r)_*(c^*\phi ))^i} (\pi _r)_*(c^*{\mathcal{E}}\otimes \omega ^{\otimes i}_{c^*\mathfrak X})\longrightarrow \underline{\textsf {CK}}^i_c\longrightarrow 0. \end{align*}

Here $\underline{\textsf {K}}^i_c$ and $\underline{\textsf {CK}}^i_c$ are kernel and cokernel, respectively. Note from Lemma 5.5 that it follows that $\underline{\textsf {K}}^i_c=(\pi _s)_*(\textsf {K}^i_c)$ . Since the $\textsf {Type}\,\,(c^*{\mathfrak X}, c^*{\mathcal{E}}, c^*\phi )$ is constant over $C$ , therefore the Hilbert polynomial of $\underline{\textsf {K}}^i_c$ does not depend on $c\in C$ , which, in turn, implies that the Hilbert polynomial of $\underline{\textsf {CK}}^i_c$ does not depend on $c\in C$ . Consider next the sheaf cokernel $\underline{\textsf {CK}}^i:=Coker((\pi _r)_*((\phi )^i)$ . Since $\underline{\textsf {CK}}^i$ is a cokernel, we have that $c^*\underline{\textsf {CK}}^i=\underline{\textsf {CK}}^i_c$ . Therefore, the sheaf $\underline{\textsf {CK}}^i$ is flat over $C$ . Now consider the kernel sheaf $\underline{\textsf {K}}^i:=Ker((\pi _r)_*((\phi )^i))$ . Since in the four terms sequence

\begin{align*} 0\longrightarrow \underline{\textsf {K}}^i \longrightarrow (\pi _r)_*({\mathcal{E}})\xrightarrow {(\pi _r)_*((\phi )^i)=((\pi _r)_*(\phi ))^i} (\pi _r)_*({\mathcal{E}}\otimes \omega ^{\otimes i}_{\mathfrak X})\longrightarrow \underline{\textsf {CK}}^i\longrightarrow 0 \end{align*}

the last three terms are flat over $C$ , we conclude that $\underline{\textsf {K}}^i$ will also be flat over $C$ , and therefore $\textsf {K}^i$ is flat over $C$ .

Proof. We choose a trivialization $X_r=\bigcup\nolimits_{i\in I} V_i$ of the vector bundle $\mathcal{E}$ and the line bundle $\omega _{X_r}$ . Then a first-order infinitesimal deformation (as a Higgs bundle) of $( E, \phi )$ can be described as a pair $(s_{ij}, t_i)$ , where $s_{ij}\in \Gamma (V_{ij}, {{\mathcal{E}}\mathsf{nd}} E)$ and $t_i\in \Gamma (V_i, {{\mathcal{E}}\mathsf{nd}} E\otimes \omega _{X_r})$ satisfying $s_{ij}+s_{jk}=s_{ik}$ and $t_i-t_j=[s_{ij}, \phi ]$ (see [Reference Biswas and RamananBR94, Theorem 2.3] and [Reference BottacinBot95, Proposition 3.1.2]).

Let $\textrm {Spec} (\unicode{x1D55C}[\epsilon ]) \longrightarrow {\mathcal{N}ilp}^{sm, gen}_0$ be a map such that the image of the closed point is given by the Higgs bundle $(X_r, E, \phi )$ . Let us denote by the corresponding first-order infinitesimal deformation of the nilpotent Higgs bundle $(E, \phi )$ by $(E[\epsilon ], \phi [\epsilon ])$ . We assume that $(E[\epsilon ], \phi [\epsilon ])$ is a nilpotent Higgs bundle. We define $ E^i[\epsilon ]:=\ker \,(\phi [\epsilon ])^i$ for $i\in [1, k+1]$ . Since the induced flag $E^{\bullet }$ is flat over ${\mathcal{N}ilp}^{sm, gen}_0$ (see Lemma 5.7), therefore $ E^{\bullet }[\epsilon ]$ is flat over $\textrm {Spec}(\unicode{x1D55C}[\epsilon ])$ . Since $E^{i}[\epsilon ]$ is flat over $\textrm {Spec}(\unicode{x1D55C}[\epsilon ])$ , therefore it is an extension of $E^i$ by $E^i$ .

(5.7)

Now it is straightforward to check that for $(s_{ij}, t_i)$ to be an infinitesimal deformation of $(E, \phi )$ as a nilpotent Higgs field, it has to satisfy the extra condition that $s_{ij}(E^{\bullet })\subseteq E^{\bullet -1}$ and $t_{i}(E^{\bullet })\subseteq E^{\bullet -1}\otimes \omega _{X_r}$ , where $ E^{\bullet }$ is the flag (5.2). This means that $s_{ij}\in \Gamma (V_{ij}, SC(E, \phi ))$ for all $i,j$ and, therefore, the proposition follows.

Proof. Proof of part (1). Let $a_{\bullet }:=(a_1,\ldots , a_n)\in B$ be any point. This point $a_{\bullet }$ can either lie over the closed fibre of the map $B\to S$ or outside the closed fibre. If it lies outside the closed fibre, then it lies in the image of the Hitchin map. This fact follows from the spectral correspondence [Reference SimpsonSim95, Theorem 6.11]. If $a_{\bullet }$ lies over the closed fibre, we again use a version of spectral correspondence for the Hitchin map for nodal curves [Reference Balaji, Barik and NagarajBBN16, Lemma 2.4]. Let us consider the function $s(a_{\bullet }):=t^n+a_1t^{n-1}+\cdots +a_{n-1}t+a_n$ on the total space $\textrm {Tot}(\omega _{X_0})$ . Here $t$ denotes the canonical section of $f^*\omega _{X_0}$ , where $f: \textrm {Tot}(\omega _{X_0})\longrightarrow X_0$ is the projection map. Then the vanishing locus $V(s(a_{\bullet }))$ defines a closed subscheme of $\textrm {Tot}(\omega _{X_0})$ . Note that $\textrm {Tot}(\omega _{X_0})$ is only quasi-projective and it is an open subscheme of $Z:=\mathbb P(\omega _{X_0}^*\oplus {\mathcal{O}}_{X_0})$ . But since $s(a_{\bullet })$ is a monic polynomial, therefore the closure of $V(s(a_{\bullet }))$ in $Z$ is $V(s(a_{\bullet }))$ itself. In particular, $V(s(a_{\bullet }))$ is a closed subscheme in $Z$ such that $V(s(a_{\bullet }))\cap D_{\infty }=\emptyset$ , where $D_{\infty }:=Z\setminus \textrm {Tot}(\omega _{X_0})$ . Therefore, by spectral correspondence, it follows that any rank- $1$ locally free sheaf on $V(s(a_{\bullet }))$ corresponds to a Higgs bundle $(E, \phi )$ on the nodal curve $X_0$ whose characteristic polynomial is given by $a_{\bullet }:=(a_1,\ldots , a_n)\in B$ . Therefore, the Hitchin map is surjective.

Proof of part (2). We want to show that ${\mathcal{N}ilp}^{sm,gen}$ is relatively isotropic over $S$ . First, we note that the smooth locus of the generic fibre of ${\mathcal{N}ilp}^{red}\longrightarrow S$ is isotropic, and its dimension is equal to $n^2(g-1)$ , which is the same as the dimension of the stack of rank- $n$ vector bundles on a smooth projective curve of genus $g$ . Therefore, we have $ \dim {\mathcal{N}ilp}^{red}_0\geq n^2(g-1)$ . We want to show that the ${\mathcal{N}ilp}^{red,sm}_0$ is an isotropic substack in $M^{cl}_{Gie,0}$ . The tangent complex at a point $(X_r, E, \phi )$ of ${\mathcal{N}ilp}^{red, sm}_0$ is given by the derived global sections of (5.6). The tangent complex at the corresponding point of $M^{cl}_{Gie}$ is given by the derived global sections of the following complex (see (3.17) and Theorem 3.18):

(5.8) \begin{align} \mathcal{C}{(E, \phi )}:=[{{\mathcal{E}}\mathsf{nd}} E\xrightarrow {[-, \phi ]} {{\mathcal{E}}\mathsf{nd}} E\otimes \omega _{X_r}]. \end{align}

We want to show that the following composite morphism is homotopic to $0$ .

(5.9)

Note that the morphism $\omega ^{\flat }: R\Gamma (\mathcal{C}(E, \phi ))[1]\to (R\Gamma (\mathcal{C}(E, \phi ))[1])^{\vee }$ is given by the logarithmic-symplectic pairing (Theorem 3.11). If we show that the composite morphism (5.9) is homotopic to $0$ , this will imply that ${\mathcal{N}ilp}^{red, sm}_0$ is an isotropic substack. To show this, we choose a cover $X_r=\bigcup\nolimits_{i\in I} V_i$ for which the vector bundle $E$ and the line bundle $\omega _{X_r}$ trivialize. With respect to this cover, the complex $R\Gamma (\mathcal{SC}(E, \phi ))[1]$ is quasi-isomorphic to the following Čech complex

(5.10) \begin{align} \prod SC( E,\phi )(V_i)&\longrightarrow \prod SC( E,\phi )(V_{ij})\oplus \prod (SC(E,\phi )\otimes \omega _{X_r} )({V_i})\nonumber \\ &\longrightarrow \prod (SC( E,\phi )\otimes \omega _{X_r} )(V_{ij})\oplus \prod SC(E,\phi )(V_{ijk}), \end{align}

where the term $\prod SC(E,\phi )({V_i})$ is in degree $-1$ .

Note that the composite map can only be non-zero on three cohomologies: in the degree $-1$ , $0$ and $1$ .

$\underline{\textrm{Case 1: the induced map}\;H^0(\omega ^{\flat })}$ . An element of

\begin{align*}\prod SC(E,\phi )(V_{ij})\oplus \prod (SC(E,\phi )\otimes \omega _{X_r} )({V_i})\end{align*}

is given by $\{(s_{ij}, t_i)\}_{i,j\in I}$ , where $(s_{ij}, t_i)$ , where $s_{ij}\in \Gamma (V_{ij}, {{\mathcal{E}}\mathsf{nd}} E)$ and $t_i\in \Gamma (V_i, {{\mathcal{E}}\mathsf{nd}} E\otimes \omega _{X_r})$ such that $s_{ij}(E^{\bullet })\subseteq E^{\bullet -1}$ , and $t_{i}(E^{\bullet })\subseteq E^{\bullet -1}\otimes \omega _{X_r}$ , where $E^{\bullet }$ is the flag (5.2). Then from (5.9) it follows that the pairing of two such elements $\{(s_{ij}, t_i)\}$ and $\{(s^{\prime}_{ij}, t^{\prime}_i)\}$ is given by

\begin{align*}\{T_{ij}:=Trace(s_{ij}\circ t^{\prime}_j-t_i\circ s^{\prime}_{ij})\in \omega _{X_r}(V_{ij})\}_{i,j\in I}.\end{align*}

Claim. The section $T_{ij}$ vanishes for all $i,j\in I$ .

Proof of the claim. We first observe that, we can choose the trivialization $X_r=\bigcup\nolimits_{i\in I} V_i$ of $E$ and $\omega _{X_r}$ in such a way that:

1. (1) $V_i$ contains at most one node for each $i\in I$ ;

2. (2) $V_i$ are connected.

We note that $V_{ij}$ also contain at most one node. Suppose, $V_{ij}$ contains a node $p$ . Then $V_{ij}=V^1_{ij}\coprod V^2_{ij}$ , the union of two smooth irreducible components of $V_{ij}$ . We use the notation $V^o_{ij}:=V_{ij}\setminus p$ , $V^{1,o}_{ij}:=V^1_{ij}\setminus p$ and $V^{2,o}_{ij}:=V^2_{ij}\setminus p$ . We consider the restriction of the section $T_{ij}$ to the two open subsets $V^{1,o}_{ij}$ and $V^{2,o}_{ij}$ . Notice that the restrictions of the flag (5.2) to these two open subsets are all sub-bundles. We note that $SC(E|_{V^{1,o}_{ij}},\phi )$ is the nilpotent part of the parabolic sub-algebra of ${{\mathcal{E}}\mathsf{nd}} (E|_{V^{1,o}_{ij}})$ given by the natural flag of sub-bundles induced by $\phi$ . Therefore, it is clear that the trace pairing is $0$ , i.e. $(T_{ij})|_{V^{1,o}_{ij}}=0$ . Similarly, $(T_{ij})|_{V^{2,o}_{ij}}=0$ . Since $V^o_{ij}$ is a dense open subset of $V_{ij}$ , therefore, by continuity, $T_{ij}=0$ .

If a $V_{ij}$ does not contain any node, then the proof is similar.

$\underline{\textrm{Case 2: the induced map}\;H^{-1}(\omega ^{\flat }).}$ In this case, we look at the pairing of $\{t_i\in SC(E, \phi )\otimes \omega _{X_r}\}$ and $\{s_{ij}\in SC(E, \phi )\}$ . The pairing is given by $\{{\textrm {tr}}(t_i\circ s_{ij})\}$ . By similar arguments as before, we can show that ${\textrm {tr}}(t_i\circ s_{ij})=0\,\,\forall i, j\in I$ .

$\underline{\textrm{Case 3: the induced map}\;H^1(\omega ^{\flat }).}$ In this case, we look at the pairing of $\{t_{ij}\in (SC(E, \phi )\otimes \omega _{X_r})(V_{ij})\}$ and $\{s_j\in SC(E, \phi )(V_j)\}$ . The pairing is given by $\{{\textrm {tr}}(t_{ij}\circ s_j)\}$ . By similar arguments as before, we can show that ${\textrm {tr}}(t_{ij}\circ s_j)=0\,\,\forall i,j\in I$ .

Therefore, we conclude that the smooth locus of this particular component is isotropic.

Proof of part (3). From part (2), it follows that $ \dim {\mathcal{N}ilp}^{red}_0= n^2(g-1)$ . Now from Lemma A.12, it follows that the dimension of $M^{cl}_{Gie}$ is equal to $2n^2(g-1)+2$ . Once we know the dimension, it is easy to see that $M^{cl}_{Gie}$ is a local complete intersection (Corollary A.13). Since $M^{cl}_{Gie}$ is a local complete intersection, and all the fibres of the Hitchin map $h: M^{cl}_{Gie}\longrightarrow B$ are of dimension equal to $n^2(g-1)$ , therefore by the miraculous flatness criterion [Reference HartshorneHar77, Exercise 10.9, p. 276], it follows that the Hitchin map is flat.