Proof. Parts (1) and (2) follow from the well-known fact (see the proof of [Reference Maulik and Poonen32, Prop. 1.13]) that the jumping locus

\begin{align*} S(k)_{\mathrm{jump}}=\{s\in S(k)\: :\: \operatorname{End}(J_{C_{\overline \eta}})\hookrightarrow \operatorname{End}(J_{C_s}) \text{is an isomorphism}\} \end{align*}

is the set of k-points of a union of countably many proper substacks of S.

Part (3) follows from the non-trivial result (proved in [Reference Maulik and Poonen32], [Reference Christensen12], [Reference Ambrosi1]) that $S(k)_{\mathrm{jump}}\neq S(k)$ if $k\neq \overline{\mathbb{F}_p}$.

Fact 2.6.

Let G be a connected and smooth linear algebraic group and let $\pi:C\to S$ be a family of curves.

1. (1) The morphism $\Phi_{G}(C/S)$ is locally of finite presentation, smooth, with affine and finitely presented relative diagonal.

2. (2) There is a functorial decomposition into a disjoint union

   (2.9)\begin{equation} \Phi_{G}(C/S): \coprod_{\delta\in \pi_1(G)} \operatorname{\mathrm{Bun}}_{G}^{\delta}(C/S)\stackrel{\Phi_{G}^{\delta}(C/S)}{\longrightarrow} S \end{equation}

   such that the morphism $\Phi_G^{\delta}(C/S): \operatorname{\mathrm{Bun}}_{G}^{\delta}(C/S)\to S$ has geometric integral fibres for every $\delta\in \pi_1(G)$.

3. (3) If G is reductive, then for any $\delta\in\pi_1(G)$, the morphism $\Phi_G^{\delta}(C/S):\operatorname{\mathrm{Bun}}_G^{\delta}(C/S)\to S$ is Stein, i.e. it is fpqc and cohomologically flat in degree zero (which means that the natural morphism $(\Phi_G^{\delta}(C/S)^{\sharp}:{\mathcal O}_S \to \Phi_G^{\delta}(C/S)_*({\mathcal O}_{\operatorname{\mathrm{Bun}}_G^{\delta}(C/S)})$ is a universal isomorphism), and of relative dimension equal to $(g-1)\dim G$.

Proof. Part (1): see [Reference Fringuelli and Viviani21, Thm. 3.1] and the references therein. Part (2): see [Reference Fringuelli and Viviani21, Thm. 3.1.1, Cor. 3.1.2] and the references therein. Part (3): see [Reference Fringuelli and Viviani21, Thm. 3.1.3] and the proof of [Reference Fringuelli and Viviani21, Prop. 3.3.2].

Fact 3.2.

Let G be a reductive group and fix $\delta\in\pi_1(G)$.

1. (1) There exists an exact sequence

   \begin{align*} 0\to \operatorname{Pic}(S) \to \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S)\to \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C_{\overline \eta}), \end{align*}

   where the first map is the pull-back map and the second one is the restriction to the fibre of C → S over a geometric generic point $\overline \eta$ of S.

2. (2) For any lift $d\in\pi_1(B_G)=\pi_1(T_G)=\Lambda(T_G)$ of δ (which is moreover generic if g = 0), the pull-back maps along the inclusions $\iota:T_G\hookrightarrow B_G \hookrightarrow G$

   \begin{align*}\iota_\sharp^*:\operatorname{Pic} \operatorname{\mathrm{Bun}}^{\delta}_G(C/S)\hookrightarrow \operatorname{Pic} \operatorname{\mathrm{Bun}}^d_{B_G}(C/S)\xrightarrow{\cong} \operatorname{Pic} \operatorname{\mathrm{Bun}}^d_{T_G}(C/S)\end{align*}

   are such that the first map is injective and the second one is an isomorphism.

3. (3) There exists a unique homomorphism (called transgression map)

   (3.1)\begin{equation} \tau_G^{\delta}=\tau^{\delta}_G(C/S):(\operatorname{Sym}^2 \Lambda^*(T_G))^{\mathscr W_G}\hookrightarrow \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S) \end{equation}

   that is functorial in S and G and fits in the following commutative diagram for any lift $d\in \pi_1(T_G)$ (which is generic if g = 0) of $\delta\in \pi_1(G)$:

   

   where $\langle -,-\rangle_{\pi}$ is the Deligne pairing of the family $\widehat \pi:C\times_S \operatorname{\mathrm{Bun}}_{T_G}^d(C/S)\times\operatorname{\mathrm{Bun}}_{T_G}^d(C/S) $ and ${\mathcal L}_{\chi}$ is the pull-back of the universal $\operatorname{\mathbb{G}}_{m}$-bundle (i.e. line bundle) over $C\times_S \operatorname{\mathrm{Bun}}_{\operatorname{\mathbb{G}}_{m}}(C/S)$ via the morphism $\operatorname{id} \times \chi_\#:C\times_S \operatorname{\mathrm{Bun}}_G^{\delta}(C/S)\to C\times_S \operatorname{\mathrm{Bun}}^{(\delta,\chi)}_{\operatorname{\mathbb{G}}_{m}}(C/S)$.

Proof. Part (1) follows from [Reference Fringuelli and Viviani21, Prop. 5.1.1]. Part (2) follows from [Reference Fringuelli and Viviani21, Cor. 5.1.2, Thm. 6.0.1]. Part (3) follows from [Reference Fringuelli and Viviani21, Thm. 5.0.1, Rmk. 5.2.2] (whose proof works also for g = 0).

Proof. This is proved as in [Reference Fringuelli and Viviani21, Prop. 4.1.2(1)].

Proof. Part (2): since $g(C/S)=0$, the rigidification $\operatorname{\mathrm{Bun}}_T^d(C/S)\textit{[\!]} T$ is isomorphic to S. Hence, the Leray spectral sequence for $\operatorname{\mathbb{G}}_{m}$ associated to the T-gerbe $\operatorname{\mathrm{Bun}}_T^d(C/S)\to \operatorname{\mathrm{Bun}}_T^d(C/S)\textit{[\!]} T$ gives the desired exact sequence

\begin{align*} 0\to \operatorname{Pic}(S)=\operatorname{Pic}(\operatorname{\mathrm{Bun}}_T^d(C/S)\textit{[\!]} T) \to \operatorname{Pic} \operatorname{\mathrm{Bun}}_T^d(C/S)\xrightarrow{w_T^d} \operatorname{Pic}(BT)=\Lambda^*(T). \end{align*}

Note that $\delta(C/S)=1$ or 2 according to whether $C/S$ is Zariski-locally trivial or not (see Remark 2.2(4)). Hence, using the above exact sequence, in order to prove parts (1) and (3), it is enough to prove the two equalities

(3.8)\begin{equation} w_T^d(\operatorname{Pic}^{\operatorname{taut}}\operatorname{\mathrm{Bun}}_T^d(C/S))=\text{Im}(w_T^d)= \begin{cases} \Lambda^*(T) & \text{if } \delta(C/S)=1,\\ \{\chi \in \Lambda^*(T)\: : \chi(d) \: \text{is even}\} & \text{if } \delta(C/S)=2. \end{cases} \end{equation}

We will distinguish two cases.

Case 1: $\delta(C/S)=1$.

In this case, the relative degree map (2.1) is an isomorphism. Hence, given any $\chi\in \Lambda^*(T)$, we can choose a line bundle $M_{\chi}\in \operatorname{Pic}(C)$ having relative degree equal to $g-\chi(d)$, and then formula (3.6) implies that

\begin{align*} w_T^d(d_{\widehat \pi}({\mathcal L}_{\chi}(M_{\chi})))=\chi. \end{align*}

This shows that

\begin{align*} w_T^d(\operatorname{Pic}^{\operatorname{taut}}\operatorname{\mathrm{Bun}}_T^d(C/S))=\Lambda^*(T), \end{align*}

which gives the desired two equalities in Equation (3.8).

Case 2: $\delta(C/S)=2$.

In this case, the image of the relative degree map (2.1) is the subgroup $2{\mathbb Z}\subset {\mathbb Z}$. We will divide the proof of the equalities in Equation (3.8) into three steps.

$\bullet$ $w_T^d(\operatorname{Pic}^{\operatorname{taut}}\operatorname{\mathrm{Bun}}_T^d(C/S))\subseteq \{\chi \in \Lambda^*(T)\: : \chi(d) \: \text{is even}\}$.

To prove this, since $\operatorname{Pic}^{\operatorname{taut}} \operatorname{\mathrm{Bun}}_T^d(C/S)$ is generated the tautological line bundles $d_{\widehat \pi}({\mathcal L}_{\chi}(M))$, it is enough to show, using formula 3.6, that

\begin{align*} d_{\widehat \pi}({\mathcal L}_{\chi}(M))(d)=[\chi(d)+\deg_{\pi}(M)+1]\chi(d)\: \text{is even} \end{align*}

for any $\chi\in \Lambda^*(T)$ and any $M\in \operatorname{Pic}(C)$. This is obvious if $\chi(d)$ is even. Otherwise, if $\chi(d)$ is odd, then it follows from the fact that $\chi(d)+\deg_{\pi}(M)+1$ is even because $\deg_{\pi}(M)$ is even.

$\bullet$ $w_T^d(\operatorname{Pic}^{\operatorname{taut}}\operatorname{\mathrm{Bun}}_T^d(C/S))\supseteq \{\chi \in \Lambda^*(T)\: : \chi(d) \: \text{is even}\}$.

Let $\chi\in \Lambda^*(T)$ such that $\chi(d)$ is even. Then, we can choice a line bundle $M_{\chi}\in \operatorname{Pic}(C)$ such that $\deg_{\pi}(M_{\chi})=-\chi(d)$. We then have, using formula (3.6), that

\begin{align*} d_{\widehat \pi}({\mathcal L}_{\chi}(M_{\chi}))=[\chi(d)+\deg_{\pi}(M_{\chi})+1]\chi=\chi, \end{align*}

which shows that $\chi \in w_T^d(\operatorname{Pic}^{\operatorname{taut}}\operatorname{\mathrm{Bun}}_T^d(C/S))$.

$\bullet$ $\text{Im}(w_T^d)\subseteq \{\chi \in \Lambda^*(T)\: : \chi(d) \: \text{is even}\}$.

Let L be a line bundle on $\operatorname{\mathrm{Bun}}_T^d(C/S)$ and set $\chi:=w_T^d(L)\in \Lambda^*(T)$. We want to show that $\chi(d)$ is even. Consider the following Cartesian diagram

The pull-back $\pi_T^*(L)$ on $\operatorname{\mathrm{Bun}}_T^d(C\times_S C/C)$ verifies $w_T^d(\pi_T^*(L))=\chi$, since the weight function is functorial with respect to base change. Consider now the line bundle $\langle {\mathcal L}_{\chi}, {\mathcal L}_0({\mathcal O}(\Delta))\rangle_{\widehat{\pi_2}}=\langle{\mathcal L}_{\chi}, {\mathcal O}(\widetilde \Delta)\rangle_{\widehat{\pi_2}}$ on $\operatorname{\mathrm{Bun}}_T^d(C\times_S C/C)$. By Equation (3.7) and the fact that $ {\mathcal O}(\Delta)$ has π ₂-relative degree equal to 1, we have that

\begin{align*} w_T^d(\langle {\mathcal L}_{\chi}, {\mathcal O}(\widetilde \Delta)\rangle_{\widehat{\pi_2}})=\chi. \end{align*}

Therefore, since $\pi_T^*(L)$ and $\langle {\mathcal L}_{\chi}, {\mathcal O}(\widetilde \Delta)\rangle_{\widehat{\pi_2}}$ have the same weight, part (2) implies that

\begin{align*} \pi_T^*(L)=\langle {\mathcal L}_{\chi}, {\mathcal O}(\widetilde \Delta)\rangle_{\widehat{\pi_2}}\otimes \Phi_C^*(M) \end{align*}

for some $M\in \operatorname{Pic}(C)$. Taking the relative degree with respect to π_T, we obtain that

\begin{align*} 0&=\deg_{\pi_T}(\pi_T^*(L))=\deg_{\pi_T}(\langle {\mathcal L}_{\chi}, {\mathcal O}(\widetilde \Delta)\rangle_{\widehat{\pi_2}})+\deg_{\pi_T}(\Phi_C^*(M)) =\\ &=\deg_{\pi_T}\widetilde \Delta^*({\mathcal L}_{\chi})+\deg_{\pi}(M)=\chi(d)+\deg_{\pi}(M), \end{align*}

where we have used the isomorphism $\langle {\mathcal L}_{\chi}, {\mathcal O}(\widetilde \Delta)\rangle_{\widehat{\pi_2}}\cong \widetilde \Delta^*({\mathcal L}_{\chi})$ has π_T-relative degree equal to $\chi(d)$. This implies that $\chi(d)$ is even since $\deg_{\pi}(M)$ is even because of the hypothesis that $\delta(C/S)=2$.

Proof. This is proved as in [Reference Fringuelli and Viviani21, Prop. 4.1.2(2)] (see also [Reference Fringuelli and Viviani22, Def./Lemma 3.5(2)]).

Proof. Let us first prove part (1). Observe that the diagram is commutative: the commutativity of the two left squares and he upper right square are obvious; the commutativity of the lower right square follows from the fact that $(\chi\otimes \chi)(x,y)=\chi(x)\chi(y)$.

We now divide the rest of the proof of part (1) into three steps.

Step 1: The middle row is exact.

First of all, observe that $i_T^d$ is well-defined since the Deligne pairing $\langle-,-\rangle_{\widehat \pi}$ is bilinear and it is trivial on the pull-back of line bundle on $\operatorname{\mathrm{Bun}}_T^d(C/S)$, and hence in particular on those coming from S. Consider now the following diagram

where the vertical maps are restriction maps to the geometric generic fibre $C_{\overline \eta}$ of $C/S$ and the map $i_T^d(C_{\overline \eta})$ is the restriction of $i_T^d$ to the the geometric generic fibre. Since the restriction maps are injective by Fact 3.2(1) and [Reference Fringuelli and Viviani21, Prop. 2.3.2] and the map $i_T^d(C_{\overline \eta})$ is injective by [Reference Biswas and Hoffmann7, § 3], we deduce that $i_T^d$ is also injective.

Lemma 3.7 implies that $\text{Im}(i_T^d)\subseteq \ker(\gamma_T^d)$ and that $\gamma_T^d$ is surjective, using that $(\Lambda^*(T)\otimes \Lambda^*(T))^s$ is generated by the elements $\{\chi\otimes \chi\: : \chi\in \Lambda^*(T)\}$ (see [Reference Fringuelli and Viviani21, § 2.2]).

It remains to prove that $\ker(\gamma_T^d)\subseteq \text{Im}(i_T^d)$. To this aim, we fix a basis $\{\chi_1,\ldots,\chi_r\}$ of $\Lambda^*(T)$. Then, $\operatorname{RPic}^{\operatorname{taut}}(\operatorname{\mathrm{Bun}}_T^d(C/S))$ is generated by the line bundles

\begin{align*} \{d_{\widehat \pi}({\mathcal L}_{\chi_i}), \langle {\mathcal L}_{\chi_j},{\mathcal L}_{\chi_k}\rangle_{\widehat \pi}, \langle {\mathcal L}_{\chi_h}, M\rangle_{\widehat \pi}\}, \end{align*}

with $1\leq i, h\leq r$, $1\leq j\leq k\leq r$ and $M\in \operatorname{RPic}(C/S)$. The line bundles $ \langle {\mathcal L}_{\chi_h}, M\rangle_{\widehat \pi}$ belong to $\text{Im}(i_T^d)\subseteq \ker(\gamma_T^d)$ as proved above. Suppose that we have a line bundle on $\operatorname{\mathrm{Bun}}_T^d(C/S)$

\begin{align*} K:=\bigotimes_i d_{\widehat \pi}({\mathcal L}_{\chi_i})^{a_i}\bigotimes_{j\leq k} \langle {\mathcal L}_{\chi_j},{\mathcal L}_{\chi_k}\rangle_{\widehat \pi}^{b_{jk}} \: \text{for some } a_i, b_{jk}\in {\mathbb Z} \end{align*}

whose class in the relative tautological Picard group belongs to $\ker(\gamma_T^d)$. Then, Lemma 3.7 implies that

(3.11)\begin{align} 0=\gamma_T^d\left(\bigotimes_i d_{\widehat \pi}({\mathcal L}_{\chi_i})^{a_i}\bigotimes_{j\leq k} \langle {\mathcal L}_{\chi_j},{\mathcal L}_{\chi_k}\rangle_{\widehat \pi}^{b_{jk}}\right)=\sum_i a_i \chi_i\otimes\chi_i+\sum_{j\leq k} b_{jk}(\chi_j\otimes \chi_k+\chi_k\otimes \chi_j).\nonumber\\ \end{align}

Since a basis of $\operatorname{Bil}^s(\Lambda(T))=(\Lambda^*(T)\otimes \Lambda^*(T))^s$ is given by $\{\{\chi_i\otimes \chi_i\}_i\cup \{\chi_j\otimes \chi_k+\chi_k\otimes \chi_j\}_{j \lt k}\}$, the relation Equation (3.11) is equivalent to

\begin{align*} \begin{cases} b_{jk}=0 & \text{for any }j \lt k,\\ a_i+2b_{ii}=0 & \text{for any }i. \end{cases} \end{align*}

Hence, the line bundle K is equal to

\begin{align*} K=\bigotimes_i [d_{\widehat \pi}({\mathcal L}_{\chi_i})^{-2}\bigotimes \langle {\mathcal L}_{\chi_i}, {\mathcal L}_{\chi_i}\rangle_{\widehat \pi}]^{b_{ii}}= \bigotimes_i [\langle {\mathcal L}_{\chi_i}, \omega_{\widehat \pi}\rangle_{\widehat \pi}\otimes d_{\widehat \pi}({\mathcal O})^{-2}]^{b_{ii}}, \end{align*}

where the last equality follows from [Reference Fringuelli and Viviani21, Rmk. 3.5.1]. This implies that the class of K in the relative Picard group belongs to $\text{Im}(i_T^d)$, and we are done.

Step 2: The first row is exact.

Indeed, the map $j_T^d$ is injective since it is the restriction of the injective map $i_T^d$. Using the exactness of the middle row, the kernel of $w_T^d\oplus \gamma_T^d$ is equal to

\begin{align*} \ker(w_T^d\oplus \gamma_T^d)=\left\{i_T^d(\chi\otimes M)\: : \: \operatorname{wt}_T^d(i_T^d(\chi\otimes M))=0\right\}. \end{align*}

By the definition of $i_T^d$ and Equation (3.7), we compute

\begin{align*} \operatorname{wt}_T^d(i_T^d(\chi\otimes M))=w_T^d(\langle {\mathcal L}_{\chi}, M\rangle_{\widehat \pi})=\deg_{\pi}(M)\chi. \end{align*}

Hence, the kernel of $w_T^d\oplus \gamma_T^d$ coincides with the image via $i_T^d$ (or equivalently via $j_T^d$) of $\Lambda^*(T)\otimes \operatorname{RPic}^0(C/S)$, and we are done.

Step 3: The last row is exact.

Indeed, the composition $\gamma_T^d\circ \tau_T^d$ is given by

\begin{align*} (\gamma_T^d\circ \tau_T^d)(\gamma\cdot \gamma^\prime)=\gamma_T^d(\langle {\mathcal L}_{\gamma}, {\mathcal L}_{\gamma^\prime}\rangle_{\widehat \pi})=\gamma\otimes \gamma^\prime+\gamma^\prime\otimes \gamma . \end{align*}

Hence, $\gamma_T^d\circ \tau_T^d$ is injective, and its image is the subgroup $\operatorname{Bil}^{s,\operatorname{ev}}(\Lambda(T))\subset \operatorname{Bil}^s(\Lambda(T))$ of even symmetric bilinear forms (see [Reference Fringuelli and Viviani21, § 2.2]). We now deduce the exactness of the last row from the exactness of the second row and the fact that $\operatorname{Bil}^{s,\operatorname{ev}}(\Lambda(T))$ is the kernel of the surjective map ϵ.

Let us prove now part (2). We first observe that the image of $w_T^d\oplus \gamma_T^d$ is contained in

\begin{align*} I:=\left\{(\chi, b)\in \Lambda^*(T)\oplus \operatorname{Bil}^s(\Lambda(T))\right.\:& : \: \delta(C/S)\vert \chi(x)-b(d,x)+(g-1)b(x,x) \\ &\left.\text{for every } x\in \Lambda(T)\right\}. \end{align*}

Indeed, for the generators of $\operatorname{RPic}^{\operatorname{taut}}\operatorname{\mathrm{Bun}}_T^d(C/S)$, we compute (using Equation (3.6) and Lemma 3.7) that

\begin{align*} (w_T^d,\gamma_T^d)(d_{\widehat \pi}({\mathcal L}_{\chi}(M)))=(\chi(d)+\deg_{\pi}(M)+1-g)\chi, \chi\otimes \chi). \end{align*}

Hence, for any $x\in \Lambda(T)$, we get

\begin{align*} [\chi(d)+\deg_{\pi}(M)+1-g] \chi(x) -\chi(d)\chi(x)+(g-1)\chi(x)^2&=[\deg_{\pi}(M)+1-g] \chi(x)\\&\quad +(g-1)\chi(x)^2=\\ =\deg_{\pi}(M) \chi(x) +(g-1)[\chi(x)^2-\chi(x)]& \equiv \deg_{\pi}(M) \chi(x) \mod 2g-2 \\ & \equiv 0 \mod \delta(C/S), \end{align*}

where in the last congruence relation we used the fact that $\deg_{\pi}(M) $ is a multiple of $\delta(C/S)$ (by definition of $\delta(C/S)$) and that $\delta(C/S)$ is a multiple of $2g-2$ (by Remark 2.2(1)).

Next, in order to show that the inclusion $\text{Im}(w_T^d\oplus \gamma_T^d)\subseteq I$ is an equality, we consider the following commutative diagram with exact vertical columns:

We now conclude using that

$\bullet$ $p_2(\text{Im}(w_T^d\oplus \gamma_T^d))=\operatorname{Bil}^s(\Lambda(T))$ as it follows from the right upper square of the diagram (3.10);

$\bullet$ the inclusion

\begin{align*}\text{Im}(w_T^d\oplus \gamma_T^d)\cap (\Lambda^*(T)\times \{0\}) \subseteq I \cap (\Lambda^*(T)\times \{0\})=\delta(C/S)\cdot \Lambda^*(T)\end{align*}

is an equality since it coincides with the boundary homomorphism coming from the snake lemma applied to the first two horizontal lines of the diagram (3.10).

Proof. We will distinguish three cases.

Case I: d = 0 and $\pi: C\to S$ admits a section σ.

Since d = 0, we have the canonical identifications

\begin{align*} \begin{cases} & \operatorname{\mathrm{Bun}}_T^0(C/S)= {\mathcal J}^0(C/S)\otimes \Lambda(T), \\ & \operatorname{\mathrm{Bun}}_T^0(C/S)\textit{[\!]} T=J^0(C/S)\otimes \Lambda(T),\\ \end{cases} \end{align*}

where ${\mathcal J}^0(C/S)$ is the Jacobian stack of $C/S$ and $J^0(C/S)={\mathcal J}^0(C/S)\textit{[\!]} T$ is the Jacobian scheme of $C/S$. The theorem is implied in this case by the following commutative diagram with exact rows:

where

• • the first line is exact by Theorem 3.8 using that $\delta(C/S)=1$ since $\pi: C\to S$ has a section by assumption;

• • the second exact line comes from the fact that the rigidification $\operatorname{\mathrm{Bun}}_T^0(C/S)\textit{[\!]} T=J^0(C/S)\otimes \Lambda(T)\to S$ is an abelian scheme (see e.g. the proof of [Reference Fringuelli and Pirisi19, Prop. 3.6]), where $(J^0(C/S)\otimes \Lambda(T))^{\vee}=J^0(C/S)^{\vee} \otimes \Lambda^*(T)$ is the dual abelian scheme and the last map is the restriction homomorphism onto the geometric generic fibre;

• • the left vertical isomorphism follows from the fact that $\operatorname{RPic}^0(C/S)=J^0(C/S)^{\vee}(S)$;

• • the right vertical isomorphism follows from the fact that (see [Reference Biswas and Hoffmann7, § 3.2])

  \begin{align*} \operatorname{NS}(\operatorname{\mathrm{Bun}}_T^0(C_{\overline \eta}))=\Lambda^*(T)\oplus \operatorname{Hom}^s(\Lambda(T)\otimes \Lambda(T), \operatorname{End}(J_{C_{\overline \eta}})), \end{align*}

  where $\operatorname{Hom}^s(\Lambda(T)\otimes \Lambda(T), \operatorname{End}(J_{C_{\overline \eta}}))$ is the set of bilinear forms $\operatorname{Hom}(\Lambda(T)\otimes \Lambda(T), \operatorname{End}(J_{C_{\overline \eta}}))$ that are symmetric with respect to the Rosati involution, together with the assumption that $\operatorname{End}(J_{C_{\overline \eta}}) ={\mathbb Z}$.

• • the commutativity of the left square follows from the proof of [Reference Fringuelli and Viviani21, Lemma 4.2.1]Footnote ¹ together with formula (3.4);

• • the commutativity of the right square follows from [Reference Fringuelli and Viviani21, Prop. 4.1.2]Footnote ².

Case II: $\pi: C\to S$ admits a section σ.

Fix an isomorphism $T\cong \operatorname{\mathbb{G}}_{m}^r$, which induces a canonical isomorphism $\Lambda(T)\cong {\mathbb Z}^r$ under which d corresponds to $(d_1,\ldots, d_r)$. Then, we have an isomorphism

\begin{align*}\operatorname{\mathrm{Bun}}_T^d(C/S)\cong \operatorname{\mathrm{Bun}}_{\operatorname{\mathbb{G}}_{m}}^{d_1}(C/S)\times_S \ldots \times_S \operatorname{\mathrm{Bun}}_{\operatorname{\mathbb{G}}_{m}}^{d_r}(C/S).\end{align*}

Using the section σ, we define an isomorphism

\begin{align*} \begin{aligned} \mathfrak t: \operatorname{\mathrm{Bun}}_{\operatorname{\mathbb{G}}_{m}}^{d_1}(C/S)\times_S \ldots \times_S \operatorname{\mathrm{Bun}}_{\operatorname{\mathbb{G}}_{m}}^{d_r}(C/S) & \xrightarrow{\cong} \operatorname{\mathrm{Bun}}_{\operatorname{\mathbb{G}}_{m}}^{0}(C/S)\times_S \ldots \times_S \operatorname{\mathrm{Bun}}_{\operatorname{\mathbb{G}}_{m}}^{0}(C/S)\\ &\cong \operatorname{\mathrm{Bun}}^0_T(C/S)\\ (L_1,\ldots, L_r) & \mapsto (L_1(-d_1\sigma), \ldots, L_r(-d_r\sigma)). \end{aligned} \end{align*}

By construction, we have that $(\mathfrak t\times \operatorname{id})^*({\mathcal L}_{e_i})={\mathcal L}_{e_i}(-d_i\sigma)$, where $\{e_i\}$ is the canonical basis of ${\mathbb Z}^r\cong \Lambda^*(T)$. Hence, using the functoriality of the determinant of cohomology, we deduce that the pull-back isomorphism

\begin{align*}\mathfrak t^*:\operatorname{Pic} \operatorname{\mathrm{Bun}}^0_T(C/S)\xrightarrow{\cong} \operatorname{Pic}\operatorname{\mathrm{Bun}}_T^d(C/S)\end{align*}

preserves the tautological subgroups. Hence, the conclusion is implied by Case I.

Case III: General case.

Let $\pi: C\to S$ be an arbitrary family of curves (i.e. possibly without a section). Fix a smooth cover $S^\prime\to S$ such that the family $C^\prime:=C\times_SS^\prime\to S^\prime$ admits a section (e.g. C → S). By Case II, we have the following morphism of exact sequences:

where $(C^{\prime\prime}\to S^{\prime\prime}):=(C^{\prime}\times_SS^{\prime}\to S^{\prime}\times_SS^{\prime})$ and $p_1,p_2:C^{\prime\prime}\to C^{\prime}$ denotes the natural projections.

Consider the following commutative diagram of abelian groups:

where all the rows are exact: the first one by Theorem 3.8, the second one by definition and the third one is obtained by taking the kernels of the vertical arrows in Diagram (3.13).

By applying the snake lemma at the first and the third rows, and using the assumption $\operatorname{RPic}(C/S)=\operatorname{Pic}_{C/S}(S)$, we deduce that the central vertical arrows in the middle must be equalities,and thus, we have the assertion.

Proof. The composition

\begin{align*}\phi:T_G\stackrel{\iota}{\hookrightarrow} G \stackrel{\mathrm{ab}}{\twoheadrightarrow} G^{\mathrm{ab}}\end{align*}

is a surjective homomorphism of tori, hence it admits a section. This implies that, for any lift $d\in \pi_1(T_G)$ of $\delta\in \pi_1(G)$, the homomorphism

\begin{align*} \phi_{\sharp}^*:\operatorname{Pic} \operatorname{\mathrm{Bun}}_{G_{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)\xrightarrow{\mathrm{ab}_\sharp^*} \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S) \xrightarrow{\iota_\sharp^*} \operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S) \end{align*}

is injective. Hence, also the homomorphism $\mathrm{ab}_{\sharp}^*$ must be injective.

Proof. This is easy and left to the reader.

Proof. Part (1) is proved in [Reference Fringuelli and Viviani22, Prop. 2.4]. Part (2) follows from the fact that we have finite index inclusions of lattices

\begin{align*} \Lambda(T_{G^{\mathrm{sc}}})\hookrightarrow \Lambda(T_{\mathscr D(G)})\hookrightarrow \Lambda(T_{G^{\mathrm{ss}}}) \end{align*}

together with [Reference Fringuelli and Viviani21, Lemma 2.2.1]. Part (3) follows from [Reference Biswas and Hoffmann7, Lemma 4.3.4].

Proof. The second assertion follows from the first one using Lemmas 3.12 and 3.13, and the assumption that $\mathscr D(G)$ is simply connected. Hence, it is enough to prove the first assertion.

Let $\overline \eta$ be a geometric generic point of S. The family $\pi:C\to S$ sits in the following Cartesian diagram of families of curves:

The above diagram gives rise to the following commutative diagram with exact rows:

where

• • the first line is exact by [Reference Fringuelli and Viviani22, Cor. 3.4] for $g\geq 1$ and by Lemma 3.15 for g = 0, where $\operatorname{\mathrm{Bun}}^{\delta}_{G,g}:=\operatorname{\mathrm{Bun}}_G^{\delta}({\mathcal C}_g/{\mathcal M}_g)$ and similarly for $G^{\mathrm{ab}}$;

• • ${\mathcal Q}$ is the cokernel of $\mathrm{ab}_\sharp^*(C/S)$, so that the second line is exact by definition;

• • the third line is exact as it follows by applying [Reference Biswas and Hoffmann7, Thm. 5.3.1] to the morphism $\mathrm{ab}$ and using [Reference Biswas and Hoffmann7, Prop. 5.2.11] together with our hypothesis that $\mathscr D(G)$ is simple connected;

• • the vertical morphisms in the left and central columns are injective by Fact 3.2(1);

• • the composition $\sigma\circ \rho$ is the identity, which implies that ρ is injective and σ is surjective.

It remains to show that σ is injective (or, equivalently, that ρ is surjective), and that the exact sequence Equation (3.17) admits a functorial splitting.

To this aim, consider the surjective morphisms of tori $T_G\hookrightarrow G\xrightarrow{\mathrm{ab}} G^{\mathrm{ab}}$ and pick a section $\widetilde \iota: G^{\mathrm{ab}}\hookrightarrow T_G$. The composition $\iota: G^{\mathrm{ab}}\stackrel{\widetilde \iota}{\hookrightarrow} T_G\hookrightarrow G$ is a section of $\mathrm{ab}$. The induced morphism

\begin{align*}\iota_{\sharp}(C/S):\operatorname{\mathrm{Bun}}_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)\hookrightarrow \operatorname{\mathrm{Bun}}_{G}^{\delta}(C/S)\end{align*}

is a section of the morphism $\mathrm{ab}_{\sharp}(C/S)$. By taking the pull-back at the level of the Picard groups, we deduce that the inclusion

\begin{align*} \mathrm{ab}_{\sharp}^*(C/S): \operatorname{Pic} \operatorname{\mathrm{Bun}}_{G_{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)\hookrightarrow \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S) \end{align*}

admits a splitting, which is functorial in S, namely

\begin{align*}\iota_{\sharp}(C/S)^*: \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S)\twoheadrightarrow \operatorname{Pic} \operatorname{\mathrm{Bun}}_{G_{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S).\end{align*}

This shows that the exact rows in Equation (3.15) admit compatible splittings, which implies that σ is injective, and we are done.

Proof. We will use the notation and results from [Reference Fringuelli and Viviani21, § 5]. Since the map $\pi_1(\mathrm{ab}):\pi_1(G)\xrightarrow{\cong} \pi_1(G^{\mathrm{ab}})=\Lambda(G^{\mathrm{ab}})$ is an isomorphism by the hypothesis that $\mathscr D(G)$ is simply connected (see Equation (2.6)), we can pick a lift $d\in \Lambda(T_G)$ of $\delta\in \pi_1(G)$, which is generic (see Definition/Lemma 3.1) and such that the following condition holds:

(*)\begin{equation} \delta^{\mathrm{ab}} \text{is 2-divisible in }\Lambda(G^{\mathrm{ab}}) \Longrightarrow d \text{is 2-divisible in }\Lambda(T_G). \end{equation}

By [Reference Fringuelli and Viviani21, Thm. 5.2], there exists a commutative diagram

where $\iota:T_G\hookrightarrow G$.

By the definition of $\Omega_d(T_G)$ and the fact that d is generic (see Definition/Lemma 3.1), the contraction homomorphism defines an isomorphism

(3.21)\begin{equation} \begin{aligned} (d^{\mathrm{ss}},-): \left(\operatorname{Sym}^2 \Lambda^*(T_{G^{\mathrm{sc}}})\right)^{\mathscr W_G} & \stackrel{\cong}{\longrightarrow} \frac{\Omega_d(T_G)}{\Lambda^*(G^{\mathrm{ab}})} \subseteq \frac{\Lambda^*(T_G)}{\Lambda^*(G^{\mathrm{ab}})}=\Lambda^*(T_{G^{\mathrm{sc}}}), \\ b &\mapsto b(d^{\mathrm{ss}}, -). \end{aligned} \end{equation}

Now, if $n\geq 1$, then the maps $\omega_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}$ and $\omega_G^{\delta}$ are isomorphisms by [Reference Fringuelli and Viviani21, Thm. 5.2(2)], and we conclude by putting together Equations (3.20) and (3.21). On the other hand, if n = 0, then [Reference Fringuelli and Viviani21, Thm. 5.2(2)] gives that

(3.22)\begin{equation} \begin{cases} & \text{Im}(\omega_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}})=\{\chi \in \Lambda^*(G^{\mathrm{ab}})\: : \: \chi(\delta^{\mathrm{ab}})\in 2{\mathbb Z}\}, \\ & \text{Im}(\omega_{G}^{\delta})=\{\chi \in \Omega^*_d(T_G)\: : \: \chi(d)\in 2{\mathbb Z}\}. \\ \end{cases} \end{equation}

This implies that the commutative diagram of abelian groups is a pull-back diagram

since if $\chi \in \Lambda^*(G^{\mathrm{ab}})$, then $\chi(d)=\chi(\delta^{\mathrm{ab}})$. Hence, we get an injective homomorphism

(3.23)\begin{equation} \alpha: \frac{\Lambda^*(G^{\mathrm{ab}})}{\text{Im}(\omega_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}})}\hookrightarrow \frac{\Omega^*_d(T_G)}{\text{Im}(\omega_{G}^{\delta})}. \end{equation}

Now if $\delta^{\mathrm{ab}}$ is not 2-divisible, then

\begin{align*}\frac{\Lambda^*(G^{\mathrm{ab}})}{\text{Im}(\omega_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}})}\cong {\mathbb Z}/2{\mathbb Z}\cong \frac{\Omega^*_d(T_G)}{\text{Im}(\omega_{G}^{\delta})}.\end{align*}

If, instead, $\delta^{\mathrm{ab}}$ is 2-divisible, then condition (*) implies that

\begin{align*} \frac{\Lambda^*(G^{\mathrm{ab}})}{\text{Im}(\omega_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}})}=0= \frac{\Omega^*_d(T_G)}{\text{Im}(\omega_{G}^{\delta})}. \end{align*}

In any case, the morphism α of Equation (3.23) is an isomorphism, which implies, by the snake lemma applied to Equation (3.22), that

\begin{align*} \frac{\text{Im}(\omega_{G}^{\delta})}{\text{Im}(\omega_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}})}\xrightarrow{\cong} \frac{\Omega^*_d(T_G)}{\Lambda^*(G^{\mathrm{ab}})}. \end{align*}

Hence, we get the conclusion using again Equations (3.20) and (3.21).

Proof. The proof follows the same lines of the proof of [Reference Fringuelli and Viviani21, Thm. C(2)], as we now indicate.

First of all, because of the assumptions, the Picard group of $\operatorname{\mathrm{Bun}}_{T_G}^d(C/S)$ (where $d\in \pi_1(T_G)$ is a fixed lift of $\delta\in \pi_1(G)$) is generated by tautological classes by Theorem 3.9. Hence, using Theorem 3.8, we can define an action of the Weyl group $\mathscr W_G$ on $\operatorname{Pic}^d(\operatorname{\mathrm{Bun}}_{T_G}^d(C/S))$ by setting

\begin{align*} w\cdot d_{\widehat \pi}({\mathcal L}_{\chi}(M)):=d_{\widehat \pi}({\mathcal L}_{w\cdot \chi}(M)) \quad \text{for any }\chi\in \Lambda^*(T) \: \text{and any } M\in \operatorname{Pic}(C/S), \end{align*}

in such a way that the diagram (3.10) for $\operatorname{RPic}^d(\operatorname{\mathrm{Bun}}_{T_G}^d(C/S))$ is $\mathscr W_G$-equivariant.

Now, using the bottom exact sequence of Equation (3.10) and arguing as in the proof of [Reference Fringuelli and Viviani21, Lemma 5.3.3] for the case m = 2, we get that the subgroup of invariants of the above action of $\mathscr W_G$ sit in the following push-out diagram:

Therefore, it remains to show that the natural inclusion

(3.25)\begin{equation} \psi:\left(\operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)\right)^{\mathscr W_G} \hookrightarrow \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S) \end{equation}

induced by the push-out diagram (3.24) and the commutative diagram (3.14) is an isomorphism. This will be achieved in two steps:

Step 1: $\operatorname{coker} \psi$ is torsion.

Indeed, consider the two inclusions

\begin{align*} \mathrm{ab}_{\sharp}^*: \operatorname{Pic} \operatorname{\mathrm{Bun}}_{G_{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)\stackrel{(\mathrm{ab}\circ \iota)_{\sharp}^*}{\hookrightarrow} \left(\operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)\right)^{\mathscr W_G}\stackrel{\psi}{\hookrightarrow} \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S). \end{align*}

We get an induced exact sequence

\begin{align*} 0\to \frac{\left(\operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)\right)^{\mathscr W_G}}{\operatorname{Pic} \operatorname{\mathrm{Bun}}_{G_{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)}\to \frac{\operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S)}{\operatorname{Pic} \operatorname{\mathrm{Bun}}_{G_{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)}\to \operatorname{coker} \psi\to 0. \end{align*}

By the push-out diagram (3.24) and using Lemmas 3.12 and 3.13, we get that

\begin{align*} \frac{\left(\operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)\right)^{\mathscr W_G}}{\operatorname{Pic} \operatorname{\mathrm{Bun}}_{G_{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)}\cong {\mathbb Z}^s, \end{align*}

where s is the number of quasi-simple factors of $G^{\mathrm{sc}}$. Moreover, arguing as in the proof of [Reference Fringuelli and Viviani21, Lemma 5.3.4], we deduce that

\begin{align*} \operatorname{rk} \frac{\operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S)}{\operatorname{Pic} \operatorname{\mathrm{Bun}}_{G_{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)}\leq s. \end{align*}

By putting everything together, we deduce that $\operatorname{coker} \psi$ is a finite group.

Step 2: $\operatorname{coker} \psi$ is torsion-free.

Indeed, the natural inclusion of the subgroup of $\mathscr W_G$-invariants of $\operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)$ factors is

\begin{align*} \left(\operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)\right)^{\mathscr W_G} \stackrel{\psi}{\hookrightarrow} \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S) \stackrel{\iota_\sharp^*}{\hookrightarrow} \operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S), \end{align*}

where $\iota_{\sharp}^*$ is injective by Fact 3.2(2). Therefore, $\operatorname{coker} \psi$ injects into the quotient

\begin{align*} Q:=\frac{\operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)}{\left(\operatorname{Pic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)\right)^{\mathscr W_G}}=\frac{\operatorname{RPic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)}{\left(\operatorname{RPic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)\right)^{\mathscr W_G}} \end{align*}

Because of our three assumptions on the family $C/S$, Theorems 3.9 and 3.8 imply that $\operatorname{RPic} \operatorname{\mathrm{Bun}}_{T_G}^{d}(C/S)$ is torsion-free. Hence, the quotient Q is torsion-free by [Reference Fringuelli and Viviani21, Lemma 5.3.6], which implies that $\operatorname{coker} \psi$ is torsion-free.

Proof. Observe that our assumptions imply that Equation (3.14) is a push-out diagram (by Theorems 3.14 and 3.16), and that $\operatorname{Pic} \operatorname{\mathrm{Bun}}_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)= \operatorname{Pic}^{\operatorname{taut}} \operatorname{\mathrm{Bun}}_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)$ by Theorem 3.9.

Let us first describe the maps appearing in the above diagram (3.27):

$\bullet$ the map $\iota_G^{\delta}$ is defined by

\begin{align*} i_G^{\delta}(\chi\otimes M):=\langle {\mathcal L}_{\chi}, M \rangle_{\widehat \pi} \end{align*}

and $j_G^{\delta}$ is the restriction of $i_G^{\delta}$.

$\bullet$ the morphism $\omega_G^{\delta}$ is induced by the weight homomorphism

\begin{align*} \operatorname{wt}_G^{\delta}:\operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S)\to \Lambda^*(\mathscr Z(G))=\frac{\Lambda^*(T_G)}{\Lambda^*(T_{G^{\mathrm{ad}}})} \end{align*}

coming from the exact sequence Equation (4.1).

$\bullet$ the morphism $\gamma_G^{\delta}$ is uniquely determined, using that the diagram (3.14) is a push-out, by

1. (1) The composition $\gamma_G^{\delta}\circ \tau_G^{\delta}$ is the inclusion

   \begin{align*} \alpha: \left(\operatorname{Sym}^2\Lambda^*(T_G)\right)^{\mathscr W_G}\cong \operatorname{Bil}^{s,\operatorname{ev}}(\Lambda(T_G))^{\mathscr W_G} \hookrightarrow \operatorname{Bil}^{s, \mathscr D-\operatorname{ev}}(\Lambda(T_G))^{\mathscr W_G}. \end{align*}

2. (2) The composition $\gamma_G^{\delta}\circ \mathrm{ab}_\#^*$ is equal to the following composition:

   \begin{align*} \operatorname{RPic}^{\operatorname{taut}} \operatorname{\mathrm{Bun}}_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S) &\xrightarrow{\gamma_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}} (\Lambda^*(G^{\mathrm{ab}})\otimes \Lambda^*(G^{\mathrm{ab}}))^s \cong \operatorname{Bil}^{s}(\Lambda(G^{\mathrm{ab}}))\\ & \xrightarrow{B_{\mathrm{ab}}^*} \operatorname{Bil}^{s, \mathscr D-\operatorname{ev}}(\Lambda(T_G))^{\mathscr W_G}, \end{align*}

   where $\gamma_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}$ is the homomorphism of Lemma 3.7 and $B_{\mathrm{ab}}^*$ is obtained by pull-back.

Now the proof is deduced from Theorems 3.8, 3.9 and 3.14, arguing as in the proofs of [Reference Fringuelli and Viviani22, Thm. 3.6, Thm. 3.12].

Proof. Note that $C/S$ is Zariski locally trivial if and only if $C/S$ admits a section. Therefore, for a choice of a geometric generic point $\overline \eta$ of S, we get the following diagram:

with Cartesian squares, where

\begin{align*}n= \begin{cases} 1 & \text{if }C/S \text{is Zariski locally trivial,} \\ 0 & \text{if }C/S \text{is not Zariski locally trivial.} \end{cases} \end{align*}

The above diagram induces the following homomorphisms, functorial in S and in G:

(3.30)\begin{equation} \operatorname{RPic}\operatorname{\mathrm{Bun}}_{G,0,n}^{\delta} \hookrightarrow \operatorname{RPic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S) \hookrightarrow \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}({\mathbb P}^1_{\overline \eta})\xrightarrow[\cong]{c_G^{\delta}({\mathbb P}^1_{\overline \eta})} \operatorname{NS}\operatorname{\mathrm{Bun}}_G^{\delta}({\mathbb P}^1), \end{equation}

where the first two pullback homomorphisms are injective by Fact 3.2(1) (note that $\overline \eta$ is also a geometric generic point of ${\mathcal M}_{0,n}$) and the last isomorphism is the one of [Reference Biswas and Hoffmann7, Thm. 5.3.1(iii)]. Therefore, we get our desired injective homomorphism, functorial in S and G

\begin{align*} c_G^{\delta}(C/S): \operatorname{RPic} \operatorname{\mathrm{Bun}}_G^{\delta}(C/S) \hookrightarrow \operatorname{Pic} \operatorname{\mathrm{Bun}}_G^{\delta}({\mathbb P}^1_{\overline \eta})\xrightarrow[\cong]{c_G^{\delta}({\mathbb P}^1_{\overline \eta})} \operatorname{NS}\operatorname{\mathrm{Bun}}_G^{\delta}({\mathbb P}^1), \end{align*}

which coincides with the weight homomorphism if G is a torus, because this is the case for the isomorphism $c_G^{\delta}({\mathbb P}^1_{\overline \eta})$ (see the proof of [Reference Biswas and Hoffmann7, Thm. 5.3.1(iii)]).

It remains to prove part (2) by distinguishing the two cases $C/S$ Zariski locally trivial or not.

If $C/S$ is Zariski locally trivial (and hence n = 1 in the above diagram), then we get that $c_G^{\delta}(C/S)$ is surjective since the composite homomorphism in Equation (3.30)

\begin{align*} c_G^{\delta}({\mathcal C}_{0,1}/{\mathcal M}_{0,1}): \operatorname{RPic}\operatorname{\mathrm{Bun}}_{G,0,1}^{\delta} \to \operatorname{NS}\operatorname{\mathrm{Bun}}_G^{\delta}({\mathbb P}^1) \end{align*}

is surjective by [Reference Fringuelli and Viviani21, Thm. 5.0.2].

If $C/S$ is not Zariski locally trivial (and hence n = 0 in the above diagram), then we get that

\begin{equation*} \{(l,b)\in \operatorname{NS} \operatorname{\mathrm{Bun}}_G^{\delta}({\mathbb P}^1) : l(\delta^{\mathrm{ab}})+b(d^{\mathrm{ss}}, d^{\mathrm{ss}}) \text{is even}\}\subseteq \text{Im}(c_G^{\delta}(C/S)), \end{equation*}

since the same is true for the image of $c_G^{\delta}({\mathcal C}_0/{\mathcal M}_0)$ by [Reference Fringuelli and Viviani21, Thm. 5.0.2]. In order to prove the other inclusion, consider the following commutative diagram (for a chosen lift $d\in \Lambda(T_G)$ of δ):

where the bottom map is defined by (see [Reference Biswas and Hoffmann7, Def. 5.2.5])

\begin{align*} \begin{aligned} \iota_{\sharp}^{\operatorname{NS},*}: \operatorname{NS}\operatorname{\mathrm{Bun}}_G^{\delta}({\mathbb P}^1) & \to \operatorname{NS}\operatorname{\mathrm{Bun}}_{T_G}^{d}({\mathbb P}^1)=\Lambda^*(T_G)\\ (l,b) &\mapsto l+b(d^{\mathrm{ss}},-). \end{aligned} \end{align*}

The above commutative diagram, together with Theorem 3.6(2), implies that

\begin{align*} \text{Im}(c_G^{\delta}(C/S))&\subseteq (\iota_{\sharp}^{\operatorname{NS},*})^{-1}(\text{Im}(c_{T_G}^d(C/S)))\\ &=\{(l,b)\in \operatorname{NS} \operatorname{\mathrm{Bun}}_G^{\delta}({\mathbb P}^1) : l(\delta^{\mathrm{ab}})+b(d^{\mathrm{ss}}, d^{\mathrm{ss}}) \text{is even}\}, \end{align*}

and we are done.

Proof. Theorem 3.20 implies that the following diagram

is Cartesian, where the map $\mathrm{ab}_{\sharp}^{\operatorname{NS},*}$ sends χ into $(\chi, 0)$. Therefore, we get that the cokernel of $\mathrm{ab}_{\sharp}^*$ injects into the cokernel of $\mathrm{ab}_{\sharp}^{\operatorname{NS},*}$, which is equal (by [Reference Biswas and Hoffmann7, Prop. 5.2.11]) to

\begin{align*} \operatorname{coker}(\mathrm{ab}_{\sharp}^{\operatorname{NS},*})=\left\{b\in \left(\operatorname{Sym}^2 \Lambda^*(T_{G^{\mathrm{sc}}})\right)^{\mathscr W_G}: b^{{\mathbb Q}}(d^{\mathrm{ss}},-)_{|\Lambda(T_{\mathscr D(G)})}\text{ is integral }\right\}. \end{align*}

This proves the existence of the exact sequence Equation (3.31), as well as of the following exact sequence:

\begin{align*} 0\to \operatorname{coker} \operatorname{wt}_{G^{\mathrm{ab}}}^{\delta^{\mathrm{ab}}}(C/S)\to \operatorname{coker} c_{G}^{\delta}(C/S)\to \operatorname{coker} p_G^{\delta} \to 0. \end{align*}

Since $\operatorname{coker} c_{G}^{\delta}(C/S)$ is a finite group of order at most two by Theorem 3.20 and it is trivial if $C/S$ is Zariski locally trivial, we deduce that the same is true for $\operatorname{coker} p_G^{\delta}$. Moreover, $\operatorname{coker} p_G^{\delta}$ is trivial if $\mathscr D(G)$ is simply connected by Theorem 3.14.