---
abstract: 'Let $X,S$ be two smooth projective varieties and $X\rightarrow S$ be a smooth family of projective curves of genus $\geq 2$ over an algebraically closed field $k$ and let $E$ be a vector bundle of rank $r\geq 3$ over $X$ and $\mathbb{P}(E)$ be its projectivization. Fix $d\geq 1$. Let $\mathcal{Q}(E,d)$ be the relative quot scheme of torsion quotients of $E$ of degree $d$. Then we show that the identity component of the group of automorphisms of $\mathcal{Q}(E,d)$ over $S$ is isomorphic to the identity component of the group of automorphisms of $\mathbb{P}(E)$ over $S$. As a corollary, the identity component of the automorphism group of flag scheme of filtrations of torsion quotients of $\mathcal{O}^{r}_{C}$, where $r\geq 3$ and $C$ a smooth projective curve of genus $\geq 2$ is computed.'
address: 'School of Mathematics, Tata Institute of Fundamental Research, Homi Bhabha Road, Mumbai 400005, India'
author:
- Chandranandan Gangopadhyay
title: Automorphisms of relative Quot schemes
---

Introduction
============

Let $p_{S}:X\rightarrow S$ be a smooth family of projective curves over an algebraically closed field $k$. Assume $X$ and $S$ are smooth projective varieties. Let $E$ be a vector bundle over $X$ of rank $r\geq 3$.\
Fix $d \geq 1$. Let $\pi_{S}:\mathcal{Q}(E,d)\rightarrow S$ be the relative Quot scheme \[1,Th.2.2.4\] whose closed points correspond to quotients $E|_{X_{s}} \rightarrow B_{d}$, $\forall s\in S$ where $X_{s}$ is the fibre of $p_{S}$ over $S$, and $B_{d}$ is a torsion sheaf over the smooth projective curve $X_{s}$ of degree $d$.\
\
**Notation.** If $f:Y\rightarrow Z$ is a smooth morphism between two projective varieties, let us denote the automorphism group of $Y$ over $Z$ by $\text{Aut}(Y/Z)$. It is known that $\text{Aut}(Y/Z)$ is a group scheme and let its identity component be denoted by $\text{Aut}^{o}(Y/Z)$. Let $\mathcal{T}_{Y/Z}$ be the relative tangent bundle of $f$. Then Lie($\text{Aut}^{o}(Y/Z))=H^{0}(Y,\mathcal{T}_{Y/Z})$ \[3,(1)\].\
\
Let us denote the projective bundle associated to $E$ by $\mathbb{P}(E)$. Note that $\mathcal{Q}(E,1)\cong \mathbb{P}(E)$.\
Then, we will be proving the following theorem:\
\
**Theorem 2.1** *(i)$ \text{Aut}^{o}(\mathcal{Q}(E,d))\cong \text{Aut}^{o}(\mathbb{P}(E)/S)$\
(ii) $H^{0}(X,\mathcal{T}_{\mathbb{P}(E)/S})\cong H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)/S})$*\
In the third section, we will use Theorem 2.1 to compute the automorphism group in certain specific cases.

Main Theorem
============

(i)$ \text{Aut}^{o}(\mathcal{Q}(E,d))\cong \text{Aut}^{o}(\mathbb{P}(E)/S)$\
(ii) $H^{0}(X,\mathcal{T}_{\mathbb{P}(E)/S})\cong H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)/S})$

Note that by \[3, Cor 2.2\], any automorphism $g\in \text{Aut}^{o}(\mathbb{P}(E)/S)$ descends to an automorphism $g'\in \text{Aut}^{o}(X/S)$. Therefore we have the following diagram:\

(E)((g’)\^[\*]{}E) & (E)\
X & X

\
Then, $E\cong (g')^{*}  E\bigotimes p^{*}\mathcal{L}$ for some line bundle $\mathcal{L}$ on $X$. Let us denote this isomorphism of bundles by $\Psi_{g}$.\
It is clear that $\Psi_{g}$ induces an automorphism of $\mathcal{Q}(E,d)$ by sending $[E|_{p^{-1}_{S}(s)} \rightarrow B_{d}\rightarrow 0]$ to $[E_{p^{-1}_{S}(s)} \xrightarrow{\Psi|_{p^{-1}_{S}(s)}} (g')^{*}(E|_{p^{-1}_{S}}(s)\rightarrow B_{d}\rightarrow 0)\bigotimes \mathcal{L}|_{p^{-1}_{S}(s)}]$.\
Hence, we have a homomorphism $\text{Aut}^{o}(\mathbb{P}(E)/S) \rightarrow \text{Aut}^{o}(\mathcal{Q}(E,d)/S)$ and clearly it is injective.\
Hence, we have morphism of lie algebras: $H^{0}(X,\mathcal{T}_{\mathbb{P}(E)/S})\hookrightarrow H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)/S})$.\
Now, by Theorem 2.2(iii), both are of same dimension as vector spaces. Therefore,\
$H^{0}(X,\mathcal{T}_{\mathbb{P}(E)/S})\cong H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)/S})$. Hence, $ \text{Aut}^{o}(\mathcal{Q}(E,d))\cong \text{Aut}^{o}(\mathbb{P}(E)/S)$.

Let $\mathcal{Z}$ be the fibered product of $d$ copies of $\mathbb{P}(E)$ over $S$. Then

(i)There exists an open subset $\mathcal{U}\subseteq \mathcal{Z}$ and a dominant map $\Phi :\mathcal{U}\rightarrow \mathcal{Q}(E,d)$ over $S$ such that codim$_{\mathcal{Z}}(\mathcal{Z}\setminus \mathcal{U})\geq 2$\
(ii)$H^{0}(\mathcal{U},\Phi^{*}\mathcal{T}_{\mathcal{Q}(E,d)/S})=\bigoplus\limits_{i=1}^{d}H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)/S})$\
(iii)The natural map $H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)/S})\rightarrow H^{0}(\mathcal{U},\Phi^{*}\mathcal{T}_{\mathcal{Q}(E,d)/S})=\bigoplus\limits_{i=1}^{d}H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)/S})$ is an injection and is invariant under permutation of the components of $\bigoplus\limits_{i=1}^{d}H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)/S})$ i.e. we have an injection $H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)/S})\hookrightarrow H^{0}(X,\mathcal{T}_{\mathbb{P}(E)/S})$.

\(i) Let $p:\mathbb{P}(E)\rightarrow X$ be the projection. Let $p_{i}$ be the $i$-th projection of $\mathcal{Z}$ to $\mathbb{P}(E)$. For, $i\neq j$, let $\Delta_{i,j}$ be the closed subscheme of $\mathcal{Z}$ given by the equation $p_{i}=p_{j}$, and for $i,j,k$ all distinct, let $\Delta_{i,j,k,X}$ be the subscheme given by the equation $p\circ p_{i}=p\circ p_{j}=p \circ p_{k}$. Let $\mathcal{U}=\mathcal{Z} \setminus (\bigcup \limits_{i,j} \Delta_{i,j} \bigcup\bigcup\limits_{i,j,k } \Delta_{i,j,k,X} )$.\
Now, let $\pi_{1}$ and $\pi_{2}$ are the first and secong projections from $X\times_{S} \mathcal{Z}$ to $X$ and $\mathcal{Z}$ respectively, and let $p_{i}\circ \pi_{2}$ be denoted by $\pi_{2,i}$. Let $\Delta_{i}$ be the closed subscheme of $X\times_{S} \mathcal{Z}$ given by the equation $\pi_{1}=p\circ \pi_{2,i}$.\
Then, consider the following morphism of sheaves over $X\times_{S}\mathcal{Z}$:\

\^[\*]{}\_[1]{}E & \_[i=1]{}\^[d]{} \^[\*]{}\_[2,i]{}(1)|\_[\_[i]{}]{}

\
Here, $\pi^{*}_{1}E \rightarrow \pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}}$ is given by the composition of morphisms $\pi^{*}_{1}E \rightarrow \pi^{*}_{1}E|_{\Delta_{i}}$ followed by $\pi^{*}_{1}E|_{\Delta_{i}} \cong \pi^{*}_{2,i}E|_{\Delta_{i}} \rightarrow \pi^{*}_{2,i} \mathcal{O}(1)|_{\Delta_{i}}$.\
Let $u \in \mathcal{U}$. Then, $(\bigoplus\limits_{i=1}^{d} \pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}})|_{X\times_{S}\{u\}}=\bigoplus\limits_{i=1}^{d} k_{p\circ p_{i}(u)}$, and the surjection $E|_{X\times_{S}\{u\}}\rightarrow k_{p\circ p_{i}(u)}$ is given by the element $p_{i}(u)$.\
By definition of $\mathcal{U}$, for any $1\leq i \leq d$, there can exist atmost one $j\neq i$ such that $p\circ p_{i}(u)=p\circ p_{j}(u)$, and for such a pair $(i,j)$, $p_{i}(u)\neq p_{j}(u)$, i.e. the homomorphism $E|_{X\times_{S}\{u\}}\rightarrow k_{p\circ p_{i}(u)}\bigoplus k_{p\circ p_{j}(u)}$ is a surjection.\
Therefore, $q|_{X\times_{S} \{u\}}$ is surjective $\forall u\in \mathcal{U}$, and clearly $(\bigoplus\limits_{i=1}^{d} \pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}})|_{X\times_{S}\{u\}}=\bigoplus\limits_{i=1}^{d} k_{p\circ p_{i}(u)}$ has length $d$. Hence, by universal property of $\mathcal{Q}(E,d)$, $q$ induces a map $\Phi:\mathcal{U} \rightarrow \mathcal{Q}(E,d)$, which is dominant \[6,Th.2.1\].\
(ii)Let us denote the projection $X\times_{S} \mathcal{Q}(E,d) \rightarrow X$ by $\pi_{X}$ and the projection $X\times_{S} \mathcal{Q}(E,d) \rightarrow \mathcal{Q}(E,d)$ by $p_{\mathcal{Q}}$ i.e. we have the following diagram:\

X\_[S]{} (E,d) & X\
(E,d) & S

\
Over $X\times_{S} \mathcal{Q}(E,d)$, we have the universal exact sequence:\

0 & (E,d) & \^[\*]{}\_[X]{} E & (E,d) & 0

\
\
Then, it is known that $\mathcal{A}(E,d)$ is a vector bundle of rank $r$ \[4, Lemma 2.2\] and\
$\mathcal{T}_{\mathcal{Q}(E,d)/S}=(p_{\mathcal{Q}})_{*}\mathcal{H}om(\mathcal{A}(E,d),\mathcal{B}(E,d))$. \[1, Prop.2.2.7\]\
Now, we have the following diagram:\

X\_[S]{} & X\_[S]{} (E,d)\
& (E,d)

\
By Grauert’s theorem \[7, Cor.12.9\], We get that\
$\Phi^{*}\mathcal{T}_{\mathcal{Q}(E,d)/S}=(\Phi)^{*}(p_{\mathcal{Q}})_{*}\mathcal{H}om(\mathcal{A}(E,d),\mathcal{B}(E,d)) \cong(\pi_{1})_{*}(id_{X}\times \Phi)^{*}\mathcal{H}om(\mathcal{A}(E,d),\mathcal{B}(E,d))$\
Now, since $\mathcal{A}(E,d)$ is a vector bundle, we have\
$(id_{X}\times_{S}\Phi)^{*}\mathcal{H}om(\mathcal{A}(E,d),\mathcal{B}(E,d))=\mathcal{H}om(\Phi^{*}\mathcal{A}(E,d),\Phi^{*}\mathcal{B}(E,d))$\
Let us denote the kernel of $q$ by $\mathcal{F}(E,d)$. Then, by the definition of the map $\Phi$, we have $\Phi^{*}\mathcal{B}(E,d)\cong (\bigoplus\limits_{i=1}^{d} \pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}})|_{X\times_{S}\mathcal{U}}$. Also, $\Phi^{*}\mathcal{A}(E,d)\cong \mathcal{F}(E,d)|_{X\times_{S} \mathcal{U}}$, since by \[4, Lemma 2.2\] $\mathcal{F}(E,d)|_{X\times_{S}\mathcal{U}}$ is again a vector bundle of rank $r$, and there exists a surjection $\Phi^{*}\mathcal{A}(E,d)\twoheadrightarrow \mathcal{F}(E,d)|_{X\times_{S} \mathcal{U}}$.\
Therefore, $\Phi^{*}\mathcal{T}_{\mathcal{Q}(E,d)/S}=\mathcal{H}om(\mathcal{F}(E,d),\bigoplus\limits_{i=1}^{d} \pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}} )|_{X\times_{S} \mathcal{U}}$.\
Now (ii) follows from Lemma 2.3.\
(iii)Note that since $\Phi:\mathcal{U}\rightarrow \mathcal{Q}(E,d)$ is dominant, then induced map\
$H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)/S})\rightarrow H^{0}(\mathcal{U},\Phi^{*}\mathcal{T}_{\mathcal{Q}(E,d)/S})=\bigoplus\limits_{i=1}^{d}H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)/S})$ is an injection.\
Now, since $\Phi : \mathcal{U}\rightarrow \mathcal{Q}(E,d)$ is invariant under any permutation of the various $\mathbb{P}(E)$ factors of $\mathcal{U}$, hence, the map $H^{0}(\mathcal{Q}(E,d), \mathcal{T}_{\mathcal{Q}(E,d)/S})\hookrightarrow \bigoplus\limits_{i=1}^{d }H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)/S})$ factors through the $H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)/S})  \hookrightarrow \bigoplus\limits_{i=1}^{d }H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)/S})$.

$H^{0}(X\times_{S} \mathcal{U}, \mathcal{H}om(\mathcal{F}(E,d), \pi^{*}_{2,j}\mathcal{O}(1)|_{\Delta_{j}} )|_{X\times_{S} \mathcal{U}})= H^{0}(\mathbb{P}(E), \mathcal{T}_{\mathbb{P}(E)/S})$ $\forall j$

Over $X\times_{S} \mathcal{U}$, we have the following commutative diagram:\

& & & 0 &\
& 0 & & \_[i=1,ij]{}\^[d]{} \^[\*]{}\_[2,i]{}(1)|\_[\_[i]{}]{} &\
0 & (E,d) & \^[\*]{}\_[1]{}E & \_[i=1]{}\^[d]{} \^[\*]{}\_[2,i]{} (1)|\_[\_[i]{}]{} & 0\
0 & (\_[1]{}\_[2,j]{})\^[\*]{}(E,1) & \^[\*]{}\_[1]{}E & \^[\*]{}\_[2,j]{}(1)|\_[\_[j]{}]{} & 0\
& & & 0 &\

\
Now, using snake lemma for the above diagram, we get the following exact sequence over $X\times_{S} \mathcal{U}$:\
$$\begin{tikzcd}
0 \arrow[r] & \mathcal{F}(E,d) \arrow[r] & (\pi_{1} \times \pi_{2,j})^{*}\mathcal{F}(E,1) \arrow[r] & \bigoplus\limits_{i=1,i\neq j}^{d} \pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}}  \arrow[r] & 0
\end{tikzcd}\\$$ Applying $\mathcal{H}om($ $ ,\pi^{*}_{2,j}\mathcal{O}(1)|_{\Delta_{j}} )$ and Lemma 2.4(i) to the exact sequence $(2.3)$, we get that over $X\times_{S} \mathcal{U}$, we have the following exact sequence:\

0 & om((\_[1]{}\_[2,j]{})\^[\*]{}(E,1),\_[2,j]{}\^[\*]{}(1)|\_[\_[j]{}]{}) & om((E,d),\_[2,j]{}\^[\*]{}(1)|\_[\_[j]{}]{})\
& 0 & xt\^[1]{}(\_[i=1,ij]{}\^[d]{}\^[\*]{}\_[2,i]{}(1)|\_[\_[i]{}]{},\_[2,j]{}\^[\*]{}(1)|\_[\_[j]{}]{})

\
Applying $H^{0}$ to the above exact sequence and using Lemma 2.4(i) and (ii), we get that\
$H^{0}(X\times_{S} \mathcal{U}, \mathcal{H}om(\mathcal{F}(E,d),\pi_{2,j}^{*}\mathcal{O}(1)|_{\Delta_{j}})=H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)/S})$.

\(i) $\mathcal{H}om(\bigoplus\limits_{i=1,i\neq j}^{d}\pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}}, \pi^{*}_{2,j}\mathcal{O}(1)|_{\Delta_{j}})=0$\
(ii) $H^{0}(X\times_{S}\mathcal{U}, \mathcal{E}xt^{1}(\bigoplus\limits_{i=1,i\neq j}^{d}\pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}}, \pi^{*}_{2,j}\mathcal{O}(1)|_{\Delta_{j}}))=0$\
(iii) $H^{0}(X\times_{S} \mathcal{U},\mathcal{H}om ((\pi_{1}\times \pi_{2,j})^{*}\mathcal{F}(E,1),\pi_{2,j}^{*}\mathcal{O}(1)|_{\Delta_{j}}))= H^{0}(\mathbb{P}(E), \mathcal{T}_{\mathbb{P}(E)/S})$

\(i) $\mathcal{H}om(\bigoplus\limits_{i=1,i\neq j}^{d}\pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}}, \pi^{*}_{2,j}\mathcal{O}(1)|_{\Delta_{j}})=\bigoplus\limits_{i=1,i\neq j}^{d} \mathcal{H}om(\pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}},\pi^{*}_{2,j} \mathcal{O}(1)|_{\Delta_{j}})= \\ \bigoplus\limits_{i=1,i\neq j}^{d} (\pi^{*}_{2,i}\mathcal{O}(-1) \bigotimes \pi^{*}_{2,j}\mathcal{O}(1)) \mathcal{H}om(\mathcal{O}_{\Delta_{i}}, \mathcal{O}_{\Delta_{j}})$.\
Now, by adjunction, $\mathcal{H}om_{\mathcal{O}_{X\times_{S} \mathcal{Z}}}(\mathcal{O}_{\Delta_{i}},\mathcal{O}_{\Delta_{j}})=\mathcal{H}om_{\mathcal{O}_{\Delta_{j}}}(\mathcal{O}_{\Delta_{i}\cap \Delta_{j}}, \mathcal{O}_{\Delta_{j}})$.\
Now, since $\Delta_{j}$ is an integral scheme, and $\Delta_{i}\cap \Delta_{j}$ is a proper subset of $\Delta_{j}$, so\
$\mathcal{H}om_{\mathcal{O}_{\Delta_{j}}}(\mathcal{O}_{\Delta_{i}\cap \Delta_{j}}, \mathcal{O}_{\Delta_{j}})=0$.\
(ii) $\mathcal{E}xt^{1}(\bigoplus\limits_{i=1,i\neq j}^{d}\pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}}, \pi^{*}_{2,j}\mathcal{O}(1)|_{\Delta_{j}})=\bigoplus\limits_{i=1,i\neq j}^{d} \mathcal{E}xt^{1}(\pi^{*}_{2,i}\mathcal{O}(1)|_{\Delta_{i}}, \mathcal{O}(1)|_{\Delta_{j}})= \\ \bigoplus\limits_{i=1,i\neq j}^{d} (\pi^{*}_{2,i}\mathcal{O}(-1) \bigotimes \pi^{*}_{2,j}\mathcal{O}(1)) \mathcal{E}xt^{1}(\mathcal{O}_{\Delta_{i}}, \mathcal{O}_{\Delta_{j}})$.\
Now, consider the exact sequence:\

0 & (-\_[i]{}) & \_[X\_[S]{}]{} & \_[\_[i]{}]{} & 0

\
\
Applying $\mathcal{H}om(\text{  }, \mathcal{O}_{\Delta_{j}})$ to the above exact sequence, we get:\

0 & \_[\_[j]{}]{} & (\_[i]{})|\_[\_[j]{}]{} & xt\^[1]{}(\_[\_[i]{}]{},\_[\_[j]{}]{}) & 0

\
\
i.e. $\mathcal{E}xt^{1}(\mathcal{O}_{\Delta_{i}},\mathcal{O}_{\Delta_{j}})\cong \pi^{*}_{1}\mathcal{T}_{X/S}|_{\Delta_{i}\cap \Delta_{j}} $.\
\
Therefore, $H^{0}(X\times_{S}\mathcal{U}, (\pi^{*}_{2,i}\mathcal{O}(-1) \bigotimes \pi^{*}_{2,j}\mathcal{O}(1)) \mathcal{E}xt^{1}(\mathcal{O}_{\Delta_{i}}, \mathcal{O}_{\Delta_{j}}))=\\ H^{0}(X\times_{S}\mathcal{U}, (\pi^{*}_{2,i}\mathcal{O}(-1) \bigotimes \pi^{*}_{2,j}\mathcal{O}(1))\bigotimes \pi^{*}_{1}\mathcal{T}_{X/S}|_{\Delta_{i}\cap \Delta_{j}})\\
= H^{0}((X\times_{S}\mathcal{U}) \cap \Delta_{i} \cap \Delta_{j}, (\pi^{*}_{2,i}\mathcal{O}(-1) \bigotimes \pi^{*}_{2,j}\mathcal{O}(1))\bigotimes \pi^{*}_{1}\mathcal{T}_{X/S}|_{\Delta_{i}\cap \Delta_{j}})$.\
\
Without loss of generality, assume $i=1, j=2$. Then, $\Delta_{1} \cong \mathcal{Z}$, where this isomorphism is given by $\Delta_{1}\hookrightarrow X\times_{S}Z \rightarrow \mathcal{Z}$. Under this ismorphism $(X\times_{S} \mathcal{U})\cap \Delta_{1}\cong \mathcal{U}$ and $\Delta_{1}\cap \Delta_{2} \cong (\mathbb{P}(E)\times_{X} \mathbb{P}(E))\times_{S}(\mathbb{P}(E))^{d-2}$, and the line bundle $(\pi^{*}_{2,i}\mathcal{O}(-1) \bigotimes \pi^{*}_{2,j}\mathcal{O}(1))\bigotimes \pi^{*}_{1}\mathcal{T}_{X/S}|_{\Delta_{1}} \cong p^{*}_{1}\mathcal{O}(-1)\bigotimes p^{*}_{2}\mathcal{O}(1)\times (p\circ p_{1})^{*}\mathcal{T}_{X/S}$.\
Let us denote the subscheme $(\mathbb{P}(E)\times_{X} \mathbb{P}(E))\times_{S}(\mathbb{P}(E))^{d-2}\hookrightarrow \mathcal{Z}$ by $Y$, and the line bundle $p^{*}_{1}\mathcal{O}(-1)\bigotimes p^{*}_{2}\mathcal{O}(1)\times (p\circ p_{1})^{*}\mathcal{T}_{X/S}|_{Y}$ by $\mathcal{L}$.\
\
We need to show that $H^{0}(Y\cap \mathcal{U},\mathcal{L})=0$.\
\
Now, $Y \setminus \mathcal{U}=\bigcup\limits_{i,j} (\Delta_{i,j}\cap Y) \bigcup\limits_{i,j,k} (\Delta_{i,j,k,X}\cap Y)$.\
**Claim.** (a) codim$_{Y}(\Delta_{i,j}\cap Y)\geq 2$\
(b)If $\{1,2\}\nsubseteq \{i,j,k\}$, then codim$_{Y}(\Delta_{i,j,k,X})\geq 2$.\
(c) $\Delta_{1,2,k,X}$ has codimension $1$ in $Y$.\
*Proof of Claim.*(a) If $\{i,j\}\neq \{1,2\}$, then $\Delta_{i,j}\cap Y\cong (\mathbb{P}(E)\times_{X}\mathbb{P}(E))\times_{S}(\mathbb{P}(E))^{d-3}$, and hence has codimension $\geq 2$ subset in $Y$.\
If $(i,j)=(1,2)$, then $\Delta_{1,2}\cong\mathbb{P}(E)\times_{S}(\mathbb{P}(E))^{d-2}$, and since $\text{rank }E\geq 3$, it has codimension $\geq 2$ in $Y$.\
(b) If $\{1,2\}\cap \{i,j,k\}=\phi$, then\
$\Delta_{i,j,k,X}\cap Y\cong (\mathbb{P}(E)\times_{X}\mathbb{P}(E))\times_{S}(\mathbb{P}(E) \times_{X}\mathbb{P}(E)\times_{X}\mathbb{P}(E))\times_{S}(\mathbb{P}(E))^{d-5}$\
If $i=1$ and $\{j,k\}\cap \{1,2\}=\phi$, then\
$\Delta_{i,j,k,X}\cap Y\cong(\mathbb{P}(E)\times_{X}\mathbb{P}(E)\times_{X}\mathbb{P}(E)\times_{X}\mathbb{P}(E))\times_{S}(\mathbb{P}(E))^{d-4}$. Hence, both in these two cases we have codimension of $\Delta_{i,j,k,X}\cap Y \geq 2$ in $Y$.\
(c) $\Delta_{1,2,k,X}\cong \mathbb{P}(E)\times_{X} \mathbb{P}(E) \times_{X} \mathbb{P}(E)\times_{S} (\mathbb{P}(E))^{d-3}$. Hence it has codimension $1$ in $Y$.

Let us denote the open set $Y\setminus ( \bigcup\limits_{k\geq 3} \Delta_{1,2,k,X})$ by $V$. Then, by the above claim\
$H^{0}(Y\cap \mathcal{U}, \mathcal{L})=H^{0}(V,\mathcal{L})$.\
Now, if $s \in H^{0}(V,\mathcal{L}|_{V})$, then for some $t\in H^{0}(Y,\mathcal{O}(\sum\limits_{k=3}^{d}\Delta_{1,2,k,X}))$, $st^{n}$ extends to a global section of $\mathcal{L}\bigotimes \mathcal{O}(\sum\limits_{k=3}^{d}n\Delta_{1,2,k,X})$, for some $n\in \mathbb{N}$. So, it is enough to show that $H^{0}(Y, \mathcal{L}\bigotimes \mathcal{O}(\sum\limits_{k=3}^{d}n\Delta_{1,2,k,X}))=0$ $\forall n\in \mathbb{N}$. Now, this follows from the next claim.\
**Claim:** $(p_{2}\times p_{3}\times ..\times p_{d})_{*}(\mathcal{L}\bigotimes \mathcal{O}(\sum\limits_{k=3}^{d}\Delta_{1,2,k,X}))=0$\
*Proof of claim.* Let us denote the $\mathbb{P}(E)\times_{S} (\mathbb{P}(E))^{d-2} \rightarrow \mathbb{P}(E)$ the $k$-th projection to $\mathbb{P}(E)$ by $p'_{k}$, where $3\leq k \leq d$.\
Now, consider the following diagram:\

((E) \_[X]{} (E))\_[S]{} ((E))\^[d-2]{}\
((E))\_[S]{} ((E))\^[d-2]{}\
X\_[S]{} X\^[d-2]{}

\
Let us denote the map $p_{2}\times p_{3}\times ..\times p_{d}$ by $f$, and $(p\circ p'_{2})\times (p\circ p'_{3})\times ... \times (p\circ p'_{d}))$ by $g$. Then, if we denote by $X_{k}\subseteq X\times_{S} X^{d-2}$ the closed subscheme whose points are of the form $(x_{1},x_{2},..,x_{d-1})$ with $x_{1}=x_{k}$, then $\mathcal{O}(\sum\limits_{k=3}^{d}n\Delta_{1,2,k,X}))=(g\circ f)^{*}\mathcal{O}(\sum\limits^{d}_{k=2}nX_{k})$.\
Hence, by projection formula we have\
$$\begin{split}
f_{*}(\mathcal{L}\bigotimes \mathcal{O}(\sum\limits_{k=3}^{d} n\Delta_{1,2,k,X}))=f_{*}((p_{1})_{*}\mathcal{O}(-1)\bigotimes(p_{2})^{*}\mathcal{O}(1)\bigotimes (p\circ p_{2})^{*}\mathcal{T}_{X/S}\bigotimes \mathcal{O}(\sum\limits_{k=3}^{d} n\Delta_{1,2,k,X})) \\
=(f_{*}((p_{1})^{*}\mathcal{O}(-1)))\bigotimes ((p'_{2})^{*}\mathcal{O}(1))\bigotimes (p\circ p'_{2})^{*}\mathcal{T}_{X/S}\bigotimes g^{*}\mathcal{O}(\sum\limits^{d}_{k=2}nX_{k}))
\end{split}$$\
Hence, it is enough to show that $f_{*}(p_{1})^{*}\mathcal{O}(-1)=0$.\
Now consider the following fibered diagram:\

((E)\_[X]{}(E))\_[S]{}((E))\^[d-2]{} & (E)\
(E)\_[S]{} ((E))\^[d-2]{} & X

\
Since $p\circ p'_{2}$ is flat, we have by \[7, Prop. 9.3\]\
$f_{*}p^{*}_{1}\mathcal{O}(1)\cong (p\circ p'_{2})^{*}p_{*}\mathcal{O}(-1)=0$.\
(iii) As before, we identify $\Delta_{j}$ with $\mathcal{Z}$. Then, we have,\
$H^{0}(X\times_{S} \mathcal{U}, \mathcal{H}om((\pi_{1}\times \pi_{2,j})^{*}\mathcal{F}(E,1),\pi^{*}_{2,j}\mathcal{O}(1)|_{\Delta_{j}}))\\
=H^{0}(\mathcal{U},\mathcal{H}om (p^{*}_{j}\mathcal{F}(E,1)|_{\Delta_{1}},p^{*}_{j}\mathcal{O}(1) ))=H^{0}(\mathcal{U},p^{*}_{j}(\mathcal{F}(E,1)^{\vee}|_{\Delta_{1}}\bigotimes \mathcal{O}(1)))$.\
Now, since $p^{*}_{j}(\mathcal{F}(E,1)^{\vee}|_{\Delta_{1}}\bigotimes \mathcal{O}(1))$ is vector bundle over $\mathcal{Z}$ and codimension of $\mathcal{Z} \setminus \mathcal{U}\geq 2$, hence,\
$H^{0}(\mathcal{U},p^{*}_{j}(\mathcal{F}(E,1)^{\vee}|_{\Delta_{1}}\bigotimes \mathcal{O}(1)))=H^{0}(\mathcal{Z},p^{*}_{j}(\mathcal{F}(E,1)^{\vee}|_{\Delta_{1}}\bigotimes \mathcal{O}(1)))$.\
Using projection formula for the morphism $p_{j}$, we get that\
$H^{0}(\mathcal{U},p^{*}_{j}(\mathcal{F}(E,1)^{\vee}|_{\Delta_{1}}\bigotimes \mathcal{O}(1)))=H^{0}(\mathbb{P}(E),(\mathcal{F}(E,1)^{\vee}|_{\Delta_{1}}\bigotimes \mathcal{O}(1)))$.\
Now, over $\mathbb{P}(E)$, $\mathcal{F}(E,d)^{\vee}|_{\Delta_{1}}\bigotimes \mathcal{O}(1)\cong \mathcal{T}_{\mathbb{P}(E)/S}$. Hence, we have the result.

Applications
============

We have the following left exact sequence of algebraic groups: $$\begin{tikzcd}
0 \arrow[r] & \text{GL}(E)/k^{*} \arrow[r] & \text{Aut}^{o}(\mathcal{Q}(E,d)/S) \arrow[r] & \text{Aut}^{o}(X/S)
\end{tikzcd}$$ The corresponding sequence of lie algebras is given by:\
$$\begin{tikzcd}
0 \arrow[r] & H^{0}(X,\text{ad }E) \arrow[r] & H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q})(E,d)/S}) \arrow[r] & H^{0}(X,\mathcal{T}_{X/S})
\end{tikzcd}$$

The left exactness of the above sequences follow from Theorem 2.1 and from the fact that $\text{Aut}^{o}(\mathbb{P}(E)/S)$ and its lie algebra fits into the above exact sequences.\

If genus of the fibres of $X\rightarrow S$ is $\geq 2$, then\
(i)Aut$^{o}(\mathcal{Q}(E,d)/S)=\text{GL}(E)/k^{*}$\
(ii)$H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)/S})=H^{0}(X,\text{ad }E)$\

If genus of each fibre is $\geq 2$, then $(p_{S})_{*}\mathcal{T}_{X/S}=0$. In particular, $H^{0}(X,\mathcal{T}_{X/S})=0$. Hence, $\text{Aut}^{o}(X/S)=0$. Now the corollary follows from Theorem 2.1

Let $C$ is a smooth projective curve of genus $\geq 2$ over an algebraically closed field $k$. Fix $\bar{d}=(d_{1},d_{2},..,d_{k})\in \mathbb{N}^{k}$ with $d_{1}>d_{2}>..>d_{k}$ and $r\geq 1$. Let $\mathcal{D}(r,\bar{d})$ be the flag scheme of filtration of quotients of $\mathcal{O}^{r}_{C}\rightarrow B_{1}\rightarrow B_{2}\rightarrow..\rightarrow B_{d}$, where $\mathcal{O}^{r}_{C}\rightarrow B_{i}$ is a torsion quotient of degree $d_{i}$\[1,2.A.1\]. Then,\
(i)$H^{0}(\mathcal{D}(r,\bar{d}), \mathcal{T}_{D(r,\bar{d})})=sl(r)$\
(ii)Aut$^{o}(\mathcal{D}(r,\bar{d}))=$ PGL$(r)$

Over $C\times \mathcal{D}(r,(d_{2},d_{3},..,d_{k}))$ we have the universal chain of filtrations:\

(r,d\_[2]{})(r,d\_[3]{})..(r,d\_[k]{})\^[r]{}\_[C(r,(d\_[1]{},d-[2]{},..,d\_[k]{}))]{}

Then $D(r,\bar{d})$ is the relative quot scheme of torsion quotients of degree $d_{1}-d_{2}$ of the vector bundle $\mathcal{A}(r,d_{1})$ for the map $C\times \mathcal{D}(r,(d_{2},d_{3},..,d_{k}))\rightarrow \mathcal{D}(r,(d_{2},d_{3},..,d_{k}))$. Then, by Corollary 3.1, we get that\
$H^{0}(\mathcal{D}(r,\bar{d}),\mathcal{T}_{\mathcal{D}(r,\bar{d})/\mathcal{D}(r,(d_{2},d_{3,..,d_{k}))}})=H^{0}(C\times \mathcal{D}(r,(d_{2},d_{3},..,d_{k})), \text{ad }\mathcal{A}(r,d_{2}))$. Now, we know that for $k\geq 1$, $\mathcal{A}(r,d_{2})$ is stable with respect to certain polarisations on\
$C\times \mathcal{D}(r,(d_{2},d_{3},..,d_{k}))$ \[4, Th.3.2.4, Th. 5.1\], so $H^{0}(C\times \mathcal{D}(r,(d_{2},d_{3},..,d_{k})), \text{ad }\mathcal{A}(r,d_{2}))=0$. Now, by induction on k, we get that $H^{0}(\mathcal{D}(r,\bar{d}),\mathcal{T}_{D}(r,\bar{d}))=H^{0}(C,\text{ad }\mathcal{O}^{r}_{C})=sl(r)$.(ii) follows immediately from this fact.

Let $C$ be a smooth projective curve over an algebraically closed field $k$. Let $E$ be a vector bundle of rank $\geq 3$ over $C$. Fix $d\geq 1$. Let $\mathcal{Q}(E,d)$ be the quot scheme of torsion quotients of $E$ of degree $d$. Then,\
(i)(a)If genus of $C=0$,i.e. $C\cong \mathbb{P}^{1}$, then\
Aut$^{o}(\mathcal{Q}(E,d))=$PGL$(2,k)\ltimes$GL$(E)/k^{*}$.\
(b) If genus of $C=1$ and if $E$ is semistable, then\
We have the following sequence of algebraic groups:\

0 & (E)/k\^[\*]{} & \^[o]{}((E,d)) & \^[o]{}(C) & 0

\
In both of these cases, we have the exact sequence of lie algebras:\

0 & H\^[0]{}(C, E) & H\^[0]{}((E,d),\_[(E,d)]{}) & H\^[0]{}(C,\_[C]{}) & 0

\
(ii) If $E$ is not semistable, then $\text{Aut}^{o}(\mathcal{Q}(E,d))=\text{GL}(E)/k^{*}$ and $H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)})=H^{0}(C,\text{ad }E)$

(i)(a) If $C\cong \mathbb{P}^{1}$, then any vector bundle $E$ admits a $\text{GL}(2)$ linearisation, in paricular we have a homomorphism GL$(2)\rightarrow \text{Aut}^{o}(\mathcal{Q}(E,d))$. This homomorphism factors through $\text{PGL}(2,k)$ and gives a section to the map $\text{Aut}^{o}(\mathcal{Q}(E,d))\rightarrow \text{PGL}(2,k)$. Therefore, the left exact sequence (3.1) is exact in this case and it splits.\
(b) We show that for any $g\in \text{Aut}^{o}(C)$ $g^{*}E\cong E$. This will show that the sequence (3.1) is also right exact in this case.\
We know that $E\cong \bigoplus E_{i}$, where $E_{i}$’s are indecomposable vector bundles. Since, $E$ is semistable, hence $\mu(E_{i})=\mu(E_{j})$ $\forall i,j$. Since the $\mathcal{Q}(E,d)\cong \mathcal{Q}(E\bigotimes \mathcal{L},d)$ canonically for any line bundle $\mathcal{L}$, after twisting $E$ by a line bundle of appropriate degree, we can assume $\mu(E)=\mu(E_{i})=0$ $\forall i$. By \[5,Th.5(i)\], $E_{i}\cong F_{r_{i}}\bigotimes \mathcal{L}_{i}$, where $F_{r_{i}}$ is the unique indecomposable vector bundle of rank $r_{i}$ with $H^{0}(C,E_{r_{i}})\neq 0$ and $\mathcal{L}$ is a line bundle of degree $0$.\
It follows that for any $g\in \text{Aut}^{o}(C)$, $g^{*}F_{r}\cong F_{r}$, since $g^{*}F_{r}$ is also indecomposable bundle of degree $0$ and rank $r$ with $H^{0}(C,g^{*}F_{r})\neq 0$.\
So, we need to show that $g^{*}\mathcal{L}\cong \mathcal{L}$ for any $\mathcal{L}\in \text{Pic}^{o}(C)$.\
Fix a base point $x_{0}\in C$. Then, under the group structure of $C$ with $x_{0}$ as the identity, $\text{Aut}^{o}(C)\cong C$ i.e. any element of $\text{Aut}^{o}(C)$ is given by $y \mapsto y+_{C}x$ for a fixed $x\in C$.(Here we denote the group addition of $C$ by $+_{C}$) Fix such an automorphism $x\in C\cong \text{Aut}^{0}(C)$\
Now, $(C,x_{0},+_{C})\cong \text{Pic}^{o}(C)$, with the morphism given by $z \mapsto \mathcal{O}(z-x_{0})$.\
Let us assume $\mathcal{L}=\mathcal{O}(z-x_{0})$ for $z\in C$.\
Then $x^{*}\mathcal{L}=\mathcal{O}((z-_{C}x)-(x_{0}-_{C}x))$.\
Since, $(C,x_{o},+_{C})\cong \text{Pic}^{0}(C)$ is a homomorphism, it follows that\
$x^{*}\mathcal{L}\cong \mathcal{O}((z-_{C}x)-(x_{0}-_{C}x))=\mathcal{O}(((z-_{C}x)-x_{0})-((x_{0}-_{C}x)-x_{0}))\\
=\mathcal{O}((z-x_{0})-(x-x_{0})-(x_{0}-x_{0})+(x-x_{0}))=\mathcal{O}(z-x_{0})=\mathcal{L}$.\
Hence, it follows that for any $g\in \text{Aut}^{o}(C)$, $g^{*}E\cong E$ for any $E$ semistable.\
(ii) By \[6, Proposition 6.13\], every semi-homogenous vector bundle\[6, Definition 5.2\] is semistable. In particular, if $E$ is not semistable, then the map $H^{0}(\mathbb{P}(E),\mathcal{T}_{\mathbb{P}(E)})\rightarrow H^{0}(C,\mathcal{T}_{C})$ is zero. Hence, using sequence (3.2), we get $H^{0}(C,\text{ad }E)\rightarrow H^{0}(\mathcal{Q}(E,d),\mathcal{T}_{\mathcal{Q}(E,d)})$ is an isomorphism. From this, it follows that $\text{Aut}^{o}(\mathcal{Q}(E,d))=\text{GL}(E)/k^{*}$.

[AAA]{}

D. Huybrechts and M. Lehn, The Geometry of Moduli Spaces of Sheaves, Cambridge University Press, Second Edition(2010).

I.Biswas, A.Dhillon and J.Hurtubise, Automorphisms of the Quot Schemes Associated to Compact Riemann Surfaces, International Mathematics Research Notices, Vol. 2015, No. 6, pp.1445-1460 M.Brion, On automorphism groups of fiber bundles, arXiv: https://arxiv.org/pdf/1012.4606 (2011) C.Gangopadhyay, Stability of sheaves over Quot Schemes, Bulletin des Sciences Mathématiques, Volume 149, (November 2018), pgs. 66-85 M. F. Atiyah, Vector bundles over an elliptic curve. Proc. London Math. Soc. (3), 414-452, 1957 S.Mukai, Semi-homogeneous vector bundles on an abelian variety, J.Math. Kyoto Univ. (JMKYAZ) 18-2(1978) 239-272 R.Hartshorne, Algebraic Geometry, Graduate Texts in Mathematics, vol. 52, Springer-Verlag, NewYork–Heidelberg, 1977.
