# On a Conjecture of Drinfeld

SARBESWAR PAL<EMAIL_ADDRESS><EMAIL_ADDRESS>IISER
Thiruvananthapuram, Maruthamala P. O., Kerala 695551

###### Abstract.

Let $C$ be smooth irreducible projective curve of genus $g\geq 2$. Let
$\mathcal{M}_{C}(n,\delta)$ be the moduli space of stable vector bundles on
$C$ of rank $n$ and fixed determinant $\delta$ of degree $d$. Then the locus
of wobbly bundles is known to be closed in $\mathcal{M}_{C}(n,\delta)$.
Drinfeld has conjectured that the wobbly locus is pure of co-dimension one,
i.e., they form a divisor in $\mathcal{M}_{C}(n,\delta)$. In this article, we
will give a proof of the conjecture.

###### Key words and phrases:

rational curve, wobbly bundles, Fano manifold

###### 2010 Mathematics Subject Classification:

14J60

##  1\. Introduction

Let $C$ be a smooth projective complex curve of genus $g\geq 2$ and let
$K_{C}$ be its canonical line bundle. Fixing a line bundle $\delta$ of degree
$d$ we consider the coarse moduli space $\mathcal{M}_{C}(n,\delta)$
parameterizing semi-stable rank-$n$ vector bundles of fixed determinant of
degree $d$ over $C$. We recall that a vector bundle $E$ is called very stable
if $E$ has no nonzero nilpotent Higgs field $\phi\in
H^{0}(C,\text{End}(E)\otimes K_{C})$. A stable vector bundle is called wobbly
if it is not very stable. G. Laumon [3, Proposition 3.5] proved, assuming
$g\geq 2$, that a very stable vector bundle is stable and that the locus of
very stable bundles is a non-empty open subset of $\mathcal{M}_{C}(n,\delta)$.
Hence the locus of wobbly bundles is a closed subset
$\mathcal{W}_{C}\subset\mathcal{M}_{C}(n,\delta).$

######  Conjecture 1.1 (Drinfeld).

[3, Remarque 3.6 (ii)]
$\mathcal{W}_{C}$ is of pure codimension $1$.

The aim of this article is to give a proof of the above conjecture.
The term “wobbly" was introduced in the paper [2]. The conjecture has been
proved completely for rank $2$ case by current author of this article and
jointly with C. Pauly in [4] and [5].
Our approach in this article is completely different from [4] and [5].
Note that the stable locus $\mathcal{M}^{s}_{C}(n,\delta)$ is smooth. It is
also known that $\mathcal{M}^{s}_{C}(n,\delta)$ is a Fano manifold of Picard
number one. Hence, it can be covered by rational curves. It is known that a
general rational curve in $\mathcal{M}^{s}_{C}(n,\delta)$ is free (for
definition of free rational curves, see Section 2). Our main idea to prove the
conjecture is to study the nonfree rational curves in
$\mathcal{M}_{C}(n,\delta)$. In fact we show that a vector bundles on $C$ is
wobbly if and only if there is a nonfree rational curve through it and then we
show that the locus of nonfree rational curves forms a divisor.
Organization of the paper:
In section 2, we recall some basic facts that we need in the subsequent
sections.
In section 3, we give a necessary and sufficient criterion of a stable vector
bundle being wobbly. More precisely, we prove the following theorem:

######  Theorem 1.1.

A stable vector bundle $E\in\mathcal{M}_{C}(n,\delta)$ is wobbly if and only
if it passes through a nonfree minimal rational curve.

.

In section 4, we will study the locus of nonfree rational curves in
$\mathcal{M}_{C}(n,\delta)$. In fact, we will show that this locus forms a
divisor. Finally, using theorem 1.1 we prove Conjecture 1.1.

##  2\. Preliminaries

In this section we recall some basic definitions and facts about the Fano
manifold which can be found in [1]. Let $X$ be a Fano manifold of dimension
$n$. A parametrized rational curve in $X$ is a morphism $\mathbb{P}^{1}\to X$
which is birational over its image. We will not distinguish parametrized
rational curves from its image $f(\mathbb{P}^{1})$ and we call $f$, a rational
curve in $X$. Recall that any vector bundle on $\mathbb{P}^{1}$ is a direct
sum of line bundles. Thus, given a rational curve $f:\mathbb{P}^{1}\to X$, the
pull back of the tangent bundle $TX$ splits as

(2.1)
$f^{*}TX=\mathcal{O}(a_{1})\oplus\mathcal{O}(a_{2})\oplus...\oplus\mathcal{O}(a_{n}),a_{1}\geq
a_{2}\geq...\geq a_{n}.$

The rational curve $f$ is said to be free if all the integers
$a_{1},a_{2},...,a_{n}$ in the above equation are non-negative. We say a
rational curve is nonfree if it is not free.
From now on, we will assume that $X$ is a Fano manifold of dimension $n$ with
Picard number $1$.
It is known that through every point of $X$, there is a rational curve in $X$.
The degree of $f^{*}{K_{X}}^{-1}$ will be called the anti-canonical degree of
the curve $f:\mathbb{P}^{1}\to X$.
Following the conventions of [1], Let $\text{Hom}(\mathbb{P}^{1},X)$ be the
quasi projective scheme parametrizing the morphisms $\mathbb{P}^{1}\to X$. An
irreducible component $\mathcal{K}$ of $\text{Hom}(\mathbb{P}^{1},X)$, is said
to be dominating component if the evaluation map
$\mathbb{P}^{1}\times\mathcal{K}\to X$ is dominant.

Let $X$ be a smooth quasi-projective variety defined over a field $k$. Let
$f:\mathbb{P}^{1}\to X$ be a rational curve and let

(2.2) $\phi(p,f):H^{0}(\mathbb{P}^{1},f^{*}TX)\to T_{p}X$

be the natural evaluation map.
Let

$F:\mathbb{P}^{1}\times\text{Hom}(\mathbb{P}^{1},X)\to X$

be the evaluation map . Then we have

######  Proposition 2.1.

$dF(p,f)=df(p)+\phi(p,f)$

###### Proof.

See [1, Ch. II, Proposition 3.4]. ∎

##  3\. Wobbly locus on moduli of vector bundles on curves

Let $C$ be a smooth projective curve of genus $g\geq 2$ over complex numbers.
Let $\mathcal{M}_{C}(n,\delta)$ be the moduli space of semi-stable vector
bundles on $C$ of rank $r$ with fixed determinant $\delta$ of degree $d$. Let
$h$ be the gcd of $(r,d)$.

######  Definition 3.1.

A vector bundle $E$ is called very stable if $E$ does not admit any nonzero
Higgs field. A vector bundle is called wobbly bundle if it is not very stable.
In other words, a vector bundle is wobbly if and only if it admits a nonzero
nilpotent Higgs field.

######  Lemma 3.2.

If a vector bundle $E$ of rank $n$ with determinant $\delta$ admits a trace
free nonzero nilpotent Higgs field, then $E$ can be obtained as successive
extensions of the form

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{1}=E^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{m-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E=E^{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{m+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

such that $H^{0}(C,{E_{m+1}}^{*}\otimes E_{m}\otimes K_{C})\neq 0$ and
$H^{0}(C,{E_{m+1}}^{*}\otimes E^{i}\otimes K_{C})=0,i=0,1,\cdots m-1$.

###### Proof.

Let us assume that $E$ admits a nonzero trace free nilpotent Higgs field
$\varphi$. Since $\varphi$ is nonzero nilpotent, $\varphi^{m+1}=0$ for some
integer $m\geq 1$. Thus, $\text{ker}\varphi$ and $\text{im}\varphi$ are
nonzero. Let $\text{ker}{\varphi}^{i}=E^{i-1}$. Note that
$\text{ker}{\varphi}^{i}\subset\text{ker}{\varphi}^{i+1}$. Thus, we have the
following diagram:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{1}:=E^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{m-2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{m-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{m-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{m}=E\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\varphi}^{m+1}}$$\textstyle{E_{m+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{m}\otimes
K_{C}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{m+1}\otimes
K_{C}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(E_{m+1}/E_{m})\otimes
K_{C}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

Note that $E_{m+1}\otimes K_{C}\subset E\otimes K_{C}^{m+1}$. Since
$\varphi^{m+1}=0,$ the morphism $E_{m+1}\to(E_{m+1}/E_{m})\otimes K_{C}$
factors through the map $E_{m+1}\to E_{m}\otimes K_{C}$ and hence
$H^{0}(X,K_{C}\otimes E_{m}\otimes{E_{m+1}}^{*})\neq 0$. The last part of the
lemma follows from the minimality of $m$.

∎

######  Theorem 3.3.

A stable vector bundle $E\in\mathcal{M}_{C}(n,\delta)$ is wobbly if and only
if there is a nonfree rational curve $l$ in $\mathcal{M}_{C}(n,\delta)$
containing $E$.

###### Proof.

Let $l$ be a nonfree rational curve in $\mathcal{M}_{C}(n,\delta)$ containing
$E$. Let $h:T^{*}\mathcal{M}_{C}(n,\delta)\to\mathcal{W}$ be the restriction
of the Hitchin map to the co-tangent bundle of $\mathcal{M}_{C}(n,\delta)$,
where $\mathcal{W}$ is the Hitchin space which is known to be affine space of
dimension $N$, where $N$ is the dimension of $\mathcal{M}_{C}(n,\delta).$ Let
$h_{\mid_{l}}$ be the restriction of $h$ to ${T^{*}M}_{\mid_{l}}$. Since $l$
is nonfree, ${T^{*}M}_{\mid_{l}}$ contains a line subbundle
$\mathcal{O}_{\mathbb{P}^{1}}(m)$ as a direct summand with $m\geq 1$. Let $s$
be a nonzero section of ${T^{*}M}_{\mid_{l}}$ corresponding to the direct
summand $\mathcal{O}_{\mathbb{P}^{1}}(m)$. Since $l$ is complete, $h(s(l))$ is
constant. On the other hand, as there is a point $p\in l$ such that
$s(p)=0,h(s(l))=0$. Since $\mathcal{O}_{\mathbb{P}^{1}}(m)$ is globally
generated, for each point $p\in l$, there is a section $s$ such that $s(E)\neq
0$, but $h(s(E))=0$. Thus, the restriction of the Hitchin map to the cotangent
space $T_{E}^{*}\mathcal{M}_{C}(n,\delta)\to\mathcal{W}$ has positive
dimensional fiber over zero for every $p\in l.$ Thus, $p$ is not very stable
bundle [7, Corollary 1.2]. In particular, $E$ is wobbly.

Conversely, let $E$ be a wobbly bundle. We will show that there is a nonfree
rational curve in $\mathcal{M}_{C}(n,\delta)$ through $E$. Note that if
$f:\mathbb{P}^{1}\to\mathcal{M}_{C}(n,\delta)$ is a free rational curve of
anti-canonical degree $p$, then
$h^{0}(\mathbb{P}^{1},f^{*}T\mathcal{M}_{C}(n,\delta))=N+p$ and,
$h^{0}(\mathbb{P}^{1},f^{*}T\mathcal{M}_{C}(n,\delta)\otimes\mathcal{O}(-1))=p$
where $N$ is the dimension of $\mathcal{M}_{C}(n,\delta)$. Our aim is to
produce a rational curve $f:\mathbb{P}^{1}\to\mathcal{M}_{C}(n,\delta)$
through $E$ such that
$h^{0}(\mathbb{P}^{1},f^{*}T\mathcal{M}_{C}(n,\delta)\otimes\mathcal{O}(-1))>p$.
By Lemma 3.2, $E$ can be obtained by successive extensions of the form

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{1}=E^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{\cdots}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E^{m-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E=E^{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{m+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

such that $H^{0}(C,{E_{m+1}}^{*}\otimes E_{m}\otimes K_{C})\neq 0$ and
$H^{0}(C,{E_{m+1}}^{*}\otimes E^{i}\otimes K_{C})=0$ for all $i$. Let
$p_{1},p_{2}$ be the projections of $C\times\mathbb{P}^{1}$ to the first and
second factors respectively. Consider a rational curve through $E$ which is
constructed by successive extensions as follows:

(3.1) $0\to p_{1}^{*}E_{1}\otimes
p_{2}^{*}{\mathcal{O}(1)}\to\mathcal{E}_{1}\to p_{1}^{*}E_{2}\to 0,$
$0\to\mathcal{E}_{1}\otimes p_{2}^{*}\mathcal{O}(1)\to\mathcal{E}_{2}\to
p_{1}^{*}E_{3}\to 0$ $\cdots\cdots\cdots$ $0\to\mathcal{E}_{m-1}\otimes
p_{2}^{*}\mathcal{O}(1)\to\mathcal{E}\to p_{1}^{*}E_{m+1}\to 0$

Then $\mathcal{E}$ defines a rational curve
$f:\mathbb{P}^{1}\to\mathcal{M}_{C}(n,\delta)$ through $E$ in
$\mathcal{M}(n,\delta)$. Let $r_{i},d_{i}$ be the rank and degree of $E_{i}$.
Then the anticanonical degree of the rational curve is
$p:=\sum_{i<j}(r_{i}d_{j}-r_{j}d_{i})(j-i)$ [9, Lemma 3.6]. Note that
$f^{*}T\mathcal{M}_{C}(n,\delta)$ can be identified with
$R^{1}{p_{2}}_{*}(\text{Ad}(\mathcal{E})$, where $\text{Ad}(\mathcal{E})$
denotes the bunde of trace free endomorphisms of $\mathcal{E}$. We will show
that
$h^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{Ad}(\mathcal{E})\otimes\mathcal{O}(-1))=p+\sum_{i=1}^{m}h^{1}(C,{E_{i}}^{*}\otimes
E_{i+1})$.
Let $\text{
Ad}^{{}^{\prime}}(\mathcal{E}_{1})\subset\text{Ad}(\mathcal{E}_{1})$ be the
subbundle of the bundle of endomorphisms of trace zero which preserves the
exact sequence 3.1. Then we have the following exact sequences:

(3.2) $0\longrightarrow\text{
Ad}^{\prime}(\mathcal{E}_{1})\longrightarrow\text{Ad}(\mathcal{E}_{1})\longrightarrow{(p_{1}^{*}E_{1})}^{*}\otimes
p_{1}^{*}E_{2}\otimes p_{2}^{*}\mathcal{O}(-1)\longrightarrow 0$

and

(3.3) $0\longrightarrow p_{1}^{*}E_{1}\otimes{(p_{1}^{*}E_{2})}^{*}\otimes
p_{2}^{*}\mathcal{O}(1)\longrightarrow\text{
Ad}^{{}^{\prime}}(\mathcal{E})\longrightarrow\text{End}(p_{1}^{*}E_{2})\oplus\text{Ad}(p_{1}^{*}E_{1})\longrightarrow
0.$

Similarly we have,

(3.4) $0\longrightarrow\text{
Ad}^{\prime}(\mathcal{E}_{2})\longrightarrow\text{Ad}(\mathcal{E}_{2})\longrightarrow{(\mathcal{E}_{1})}^{*}\otimes
p_{1}^{*}E_{3}\otimes p_{2}^{*}\mathcal{O}(-1)\longrightarrow 0$

and

(3.5) $0\longrightarrow\mathcal{E}_{1}\otimes{(p_{1}^{*}E_{3})}^{*}\otimes
p_{2}^{*}\mathcal{O}(1)\longrightarrow\text{
Ad}^{{}^{\prime}}(\mathcal{E}_{2})\longrightarrow\text{End}(p_{1}^{*}E_{3})\oplus\text{Ad}(\mathcal{E}_{1})\longrightarrow
0$

and so on. We will prove the theorem by induction. Let $m=1$.// Note that
$H^{1}(\text{ Ad}(\mathcal{E}))=H^{0}(R^{1}p_{*}\text{ Ad}(\mathcal{E}))$ (by
Leray Spectral Sequence). Thus tensoring 3.2 by ${\mathcal{O}}(-1)$ and
considering the exact sequence of cohomologies we have the following exact
sequence,

(3.6) $\displaystyle 0\to H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{
Ad}^{{}^{\prime}}(\mathcal{E}_{1})\otimes{\mathcal{O}}(-1))\to
H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{
Ad}(\mathcal{E}_{1})\otimes{\mathcal{O}}(-1))\to H^{0}(C,E_{1}^{*}\otimes
E_{2})\otimes H^{1}(\mathbb{P}^{1},{\mathcal{O}}(-2))$ $\longrightarrow
H^{2}(\text{ Ad}^{{}^{\prime}}(\mathcal{E}_{1})\otimes
p_{2}^{*}{\mathcal{O}}(-1))\\\ \longrightarrow H^{2}(\text{
Ad}(\mathcal{E}_{1})\otimes p_{2}^{*}{\mathcal{O}}(-1))\longrightarrow
H^{1}(C,E_{1}^{*}\otimes E_{2})\otimes
H^{1}(\mathbb{P}^{1},{\mathcal{O}}(-2))\longrightarrow 0$

Tensoring (3.3) by $p_{2}^{*}{\mathcal{O}}(-1)$ and considering the long exact
sequence of cohomology we have,

(3.7) $H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}(\text{
Ad}^{{}^{\prime}}(\mathcal{E}_{1})\otimes p_{2}^{*}{\mathcal{O}}(-1)))\simeq
H^{1}(C,E_{1}\otimes E_{2}^{*}),H^{2}(\text{
Ad}^{{}^{\prime}}(\mathcal{E}_{1})\otimes p_{2}^{*}{\mathcal{O}}(-1))=0,$

Therefore, from the exact sequence 3.6 and 3.7 it follows that

$H^{0}(R^{1}{p_{2}}_{*}(\text{ Ad}(\mathcal{E})\otimes
p_{2}^{*}{\mathcal{O}}(-1))=H^{1}(C,E_{1}\otimes
E_{2}^{*})+H^{0}(C,E_{1}^{*}\otimes E_{2})\otimes
H^{1}(\mathbb{P}^{1},{\mathcal{O}}(-2)).$

Since $E$ is stable, $h^{0}(C,E_{1}\otimes E_{2}^{*})=0$. Thus, by Riemann-
Roch we have, $h^{1}(C,E_{1}\otimes
E_{2}^{*})=r_{1}r_{2}(g-1)+(r_{1}d_{2}-r_{2}d_{1})$ and
$h^{0}(C,E_{1}^{*}\otimes
E_{2})=(r_{1}d_{2}-r_{2}d_{1})-r_{1}r_{2}(g-1)+h^{0}(C,E_{1}\otimes
E_{2}^{*}\otimes K_{C})$.
Therefore,

(3.8)
$h^{0}(\mathbb{P}^{1},R^{1}(p_{2})_{*}(\text{Ad}(\mathcal{E}_{1})\otimes\mathcal{O}(-1)))=2(r_{1}d_{2}-r_{2}d_{1})+h^{0}(C,E_{1}\otimes
E_{2}^{*}\otimes K).$

Let $m=2$:
Similarly, tensoring 3.4 by ${\mathcal{O}}(-1)$ and considering the exact
sequence of cohomologies we have the following exact sequence,

(3.9) $\displaystyle 0\to H^{0}(\mathcal{E}_{1}^{*}\otimes
p_{2}^{*}\mathcal{O}(-2)\otimes p_{1}^{*}E_{3})\to
H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{
Ad}^{{}^{\prime}}(\mathcal{E}_{2})\otimes{\mathcal{O}}(-1))\to
H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{
Ad}(\mathcal{E}_{2})\otimes{\mathcal{O}}(-1))$ $\to
H^{1}(\mathcal{E}_{1}^{*}\otimes p_{2}^{*}\mathcal{O}(-2)\otimes
p_{1}^{*}E_{3})\to H^{2}(\text{ Ad}^{{}^{\prime}}(\mathcal{E}_{2})\otimes
p_{2}^{*}{\mathcal{O}}(-1))\\\ \to H^{2}(\text{ Ad}(\mathcal{E}_{2})\otimes
p_{2}^{*}{\mathcal{O}}(-1))$

Taking the dual exact sequence of 3.2 and tensoring by $p_{1}^{*}E_{3}\otimes
p_{2}^{*}\mathcal{O}(-2)$ and considering the cohomology sequence we get

(3.10) $H^{0}(\mathcal{E}_{1}^{*}\otimes p_{2}^{*}\mathcal{O}(-2)\otimes
p_{1}^{*}E_{3})=0.$

Now tensoring (3.5) by $p_{2}^{*}{\mathcal{O}}(-1)$ and considering the long
exact sequence of cohomology we have,

(3.11) $\displaystyle 0\to H^{1}(\mathcal{E}_{1}\otimes p_{1}^{*}E_{3}^{*})\to
H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}(\text{
Ad}^{{}^{\prime}}(\mathcal{E}_{2})\otimes p_{2}^{*}{\mathcal{O}}(-1)))\to
H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{Ad}(\mathcal{E}_{1}\otimes
p_{2}^{*}\mathcal{O}(-1)$ $\to H^{2}(\mathcal{E}_{1}\otimes
p_{1}^{*}E_{3}^{*})\to H^{2}(\text{ Ad}^{{}^{\prime}}(\mathcal{E}_{2}\otimes
p_{2}^{*}{\mathcal{O}}(-1))\to H^{2}(\text{Ad}(\mathcal{E}_{1}\otimes
p_{2}^{*}\mathcal{O}(-1))\to 0.$

Tensoring 3.2 by $p_{1}^{*}E_{3}^{*}$ and taking the cohology,we have
$H^{2}(\mathcal{E}_{1}\otimes p_{1}^{*}E_{3}^{*})=0$. Thus

(3.12) $H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}(\text{
Ad}^{{}^{\prime}}(\mathcal{E}_{2})\otimes
p_{2}^{*}{\mathcal{O}}(-1)))=H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{Ad}(\mathcal{E}_{1}\otimes
p_{2}^{*}\mathcal{O}(-1)\oplus H^{1}(\mathcal{E}_{1}\otimes p_{1}^{*}E_{3}).$

and

(3.13) $H^{2}(\text{Ad}^{{}^{\prime}}(\mathcal{E}_{2})\otimes
p_{2}^{*}\mathcal{O}(-1))=H^{2}(\text{Ad}(\mathcal{E}_{1})\otimes
p_{2}^{*}\mathcal{O}(-1)).$

Since $\mathcal{E}_{1}\otimes
p_{2}^{*}\mathcal{O}(1)\subset\mathcal{E},\text{Ad}(\mathcal{E}_{1})\subset\text{Ad}(\mathcal{E}_{2})$.
Now considering its cokernel and using the cohomology sequence one can show
that $H^{2}(\text{Ad}(\mathcal{E}_{1})\otimes p_{2}^{*}\mathcal{O}(-1))\to
H^{2}(\text{Ad}(\mathcal{E}_{2})\otimes p_{2}^{*}\mathcal{O}(-1))$ is
injective. Therefore, from 3.9, we have

(3.14) $H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}(\text{
Ad}(\mathcal{E}_{2})\otimes
p_{2}^{*}{\mathcal{O}}(-1)))=H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{Ad}^{{}^{\prime}}(\mathcal{E}_{2})\otimes
p_{2}^{*}\mathcal{O}(-1)\oplus H^{1}(\mathcal{E}_{1}^{*}\otimes
p_{1}^{*}E_{3}\otimes p_{2}^{*}\mathcal{O}(-2)).$

Hence by 3.12 and 3.14, we have

(3.15) $H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}(\text{
Ad}(\mathcal{E}_{2})\otimes
p_{2}^{*}{\mathcal{O}}(-1)))=H^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}\text{Ad}(\mathcal{E}_{1}\otimes
p_{2}^{*}\mathcal{O}(-1)\oplus H^{1}(\mathcal{E}_{1}\otimes
p_{1}^{*}E_{3}^{*})\oplus H^{1}(\mathcal{E}_{1}^{*}\otimes
p_{1}^{*}E_{3}\otimes p_{2}^{*}\mathcal{O}(-2)).$

Tensoring 3.2, by $p_{1}^{*}E_{3}^{*}$ and considering the cohomology, we have

(3.16) $H^{1}(\mathcal{E}_{1}\otimes
p_{1}^{*}E_{3}^{*})=H^{0}(\mathbb{P}^{1},\mathcal{O}(1))\otimes
H^{1}(C,E_{1}\otimes E_{3}^{*})\oplus H^{1}(C,E_{2}\otimes E_{3}^{*})$

and tensoring by $p_{1}^{*}E_{3}\otimes p_{2}^{*}\mathcal{O}(-2)$ to the dual
of 3.2 and considering the cohomologies, we have

(3.17) $\displaystyle 0\to H^{0}(C,E_{3}\otimes E_{2}^{*})\to
H^{1}(\mathcal{E}_{1}^{*}\otimes p_{1}^{*}E_{3}\otimes\mathcal{O}(-2))\to
H^{0}(C,E_{1}^{*}\otimes E_{3})\otimes H^{1}(\mathbb{P}^{1},\mathcal{O}(-3))$
$\to H^{1}(C,E_{3}\otimes E_{2}^{*})\to H^{2}(\mathcal{E}_{1}^{*}\otimes
p_{1}^{*}E_{3}\otimes p_{2}^{*}\mathcal{O}(-2))\to H^{1}(E_{1}^{*}\otimes
E_{3})\otimes H^{1}(\mathbb{P}^{1},\mathcal{O}(-3))\to 0.$

Since $H^{0}(E_{3}^{*}\otimes E_{1}\otimes K_{c})=0$, the map

$H^{1}(C,E_{3}\otimes E_{2}^{*})\to H^{2}(\mathcal{E}_{1}^{*}\otimes
p_{1}^{*}E_{3}\otimes p_{2}^{*}\mathcal{O}(-2))$

in 3.17, is surjective. The kernel of this map is the co-kernel of the dual
map

$H^{0}(\mathcal{E}_{1}\otimes p_{1}^{*}E_{3}^{*}\otimes p_{1}^{*}K_{C})\to
H^{0}(E_{3}^{*}\otimes E_{2}\otimes K_{C}).$

Tensoring 3.1 by $p_{1}^{*}E_{3}^{*}\otimes p_{1}^{*}K_{C}$ and sonsidering
the cohomology, we have

$H^{0}(\mathcal{E}_{1}\otimes p_{1}^{*}E_{3}^{*}\otimes p_{1}^{*}K_{C})\simeq
H^{0}(E_{3}^{*}\otimes E_{2}\otimes K_{C}).$

(3.18) $H^{1}(\mathcal{E}_{1}^{*}\otimes
p_{1}^{*}E_{3}\otimes\mathcal{O}(-2))=H^{0}(C,E_{3}\otimes E_{2}^{*})\oplus
H^{0}(C,E_{1}^{*}\otimes E_{3})\otimes H^{1}(\mathbb{P}^{1},\mathcal{O}(-3)).$

By 3.8, 3.15, 3.16 and 3.18, we have

(3.19) $\displaystyle h^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}(\text{
Ad}(\mathcal{E})\otimes
p_{2}^{*}{\mathcal{O}}(-1)))=2(r_{1}d_{2}-r_{2}d_{1})+h^{0}(C,E_{1}\otimes
E_{2}^{*}\otimes K_{C})+$ $2h^{1}(C,E_{1}\otimes E_{3}^{*})\oplus
h^{1}(C,E_{2}\otimes E_{3}^{*})+h^{0}(C,E_{3}\otimes E_{2}^{*})\oplus
2h^{0}(C,E_{1}^{*}\otimes E_{3}).$

By stability of $E,h^{0}(E_{3}^{*}\otimes E_{1})=h^{0}(E_{2}\otimes
E_{3}^{*})=0$, and by typothesis, $h^{0}(E_{3}^{*}\otimes E_{1}\otimes
K_{C})=0$. Thus we have,

$h^{1}(E_{1}\otimes E_{3}^{*})=r_{1}r_{3}(g-1)+(r_{1}d_{3}-r_{3}d_{1}),$
$h^{1}(E_{2}\otimes E_{3}^{*})=r_{2}r_{3}(g-1)+(r_{2}d_{3}-r_{3}d_{2}),$
$h^{0}(E_{3}\otimes E_{1}^{*})=(r_{1}d_{3}-r_{3}d_{1})-r_{1}r_{3}(g-1)$
$h^{0}(E_{3}\otimes
E_{2}^{*})=(r_{2}d_{3}-d_{2}r_{3})-r_{2}r_{3}(g-1)+h^{0}(E_{3}^{*}\otimes
E_{2}\otimes K_{C}).$

Substituting all these in 3.19, we get,

(3.20) $\displaystyle h^{0}(\mathbb{P}^{1},R^{1}{p_{2}}_{*}(\text{
Ad}(\mathcal{E})\otimes
p_{2}^{*}{\mathcal{O}}(-1)))=2(r_{1}d_{2}-r_{2}d_{1})+4(r_{1}d_{3}-d_{1}r_{3})+2(r_{2}d_{3}-d_{2}r_{3})$
$+h^{0}(E_{3}^{*}\otimes E_{2}\otimes K_{C})+h^{0}(E_{2}^{*}\otimes
E_{1}\otimes K_{C})=p+h^{0}(E_{3}^{*}\otimes E_{2}\otimes
K_{C})+h^{0}(E_{2}^{*}\otimes E_{1}\otimes K_{C}).$

For arbitrary $m$, using the following lemma and by induction one can conclude
the theorem. Since, $h^{0}(C,E_{2},\otimes E_{3}^{*}\otimes K_{C})\geq
1,h^{0}(\mathbb{P}^{1},R^{1}(p_{2})_{*}(\text{Ad}(\mathcal{E})\otimes\mathcal{O}(-1)))>p$.
Thus, by the discussion in the first paragraph of the converse part of the
theorem, the rational curve $f:\mathbb{P}^{1}\to\mathcal{M}_{C}(n,\delta)$
through $E$ is not-free.

######  Lemma 3.4.

Consider the sequence of exact sequences of vector bundles on
$C\times\mathbb{P}^{1}$, obtained by successive extensions as in the above
theorem:

(3.21) $0\to p_{1}^{*}E_{1}\otimes
p_{2}^{*}{\mathcal{O}(1)}\to\mathcal{E}_{1}\to p_{1}^{*}E_{2}\to 0,$ (3.22)
$0\to\mathcal{E}_{1}\otimes p_{2}^{*}\mathcal{O}(1)\to\mathcal{E}_{2}\to
p_{1}^{*}E_{3}\to 0$ $\cdots\cdots\cdots$ (3.23) $0\to\mathcal{E}_{m-1}\otimes
p_{2}^{*}\mathcal{O}(1)\to\mathcal{E}\to p_{1}^{*}E_{m+1}\to 0$

Then for all $i+1<j$ we have
(1). $H^{0}(\mathcal{E}_{i}^{*}\otimes p_{2}^{*}\mathcal{O}(-2)\otimes
p_{1}^{*}E_{j})=H^{2}(\mathcal{E}_{i}\otimes
p_{1}^{*}E_{j}^{*})=H^{2}(\mathcal{E}_{i}\otimes
p_{2}^{*}\mathcal{O}(1)\otimes p_{1}^{*}E_{j}^{*})=0$ .
(2). $H^{1}(\mathcal{E}_{i}\otimes p_{2}^{*}\mathcal{O}(k)\otimes
p_{1}^{*}E_{j}^{*})=H^{1}(C,E_{i+1}\otimes E_{j}^{*})\otimes
H^{0}(\mathbb{P}^{1},\mathcal{O}(k))\oplus H^{1}(\mathcal{E}_{i-1}\otimes
p_{2}^{*}\mathcal{O}(k+1)\otimes p_{1}^{*}E_{j}^{*})$.
Hence $H^{1}(\mathcal{E}_{i}\otimes
p_{1}^{*}E_{j}^{*})=\oplus_{k=0}^{i}H^{1}(C,E_{j}^{*}\otimes E_{i+1-k})\otimes
H^{0}(\mathbb{P}^{1},\mathcal{O}(k))$.
(3). $H^{1}(\mathcal{E}_{i}^{*}\otimes p_{2}^{*}\mathcal{O}(-2)\otimes
p_{1}^{*}E_{j})=\oplus_{k=0}^{i}H^{0}(C,E_{j}\otimes E_{i+1-k})\otimes
H^{1}(\mathbb{P}^{1},\mathcal{O}(-2-k))$.

###### Proof.

We will prove the lemma by induction on $i$. For $i=1$, the lemma follows from
the proof of the above theorem.
(1). Consider the dual exact sequence of $(i-1)$-th exact sequence in the
sequence in the lemma

$0\to E_{i+1}^{*}\to\mathcal{E}_{i}^{*}\to\mathcal{E}_{i-1}^{*}\otimes
p_{2}^{*}\mathcal{O}(-1)\to 0.$

Tensoring by $p_{1}^{*}E_{j}\otimes p_{2}^{*}\mathcal{O}(-2)$ we have,

$0\to p_{1}^{*}E_{i+1}^{*}\otimes p_{1}^{*}E_{j}\otimes
p_{2}^{*}\mathcal{O}(-2)\to\mathcal{E}_{i}^{*}\otimes p_{1}^{*}E_{j}\otimes
p_{2}^{*}\mathcal{O}(-2)\to\mathcal{E}_{i-1}^{*}\otimes
p_{2}^{*}\mathcal{O}(-3)\otimes p_{1}^{*}E_{j}\to 0.$

By induction hypothesys, $H^{0}(\mathcal{E}_{i-1}^{*}\otimes
p_{2}^{*}\mathcal{O}(-3)\otimes p_{1}^{*}E_{j})=0$. Therefore, we have
$H^{0}(\mathcal{E}_{i}^{*}\otimes p_{2}^{*}\mathcal{O}(-2)\otimes
p_{1}^{*}E_{j})=0$.
Again tensoring the exact sequence

$0\to\mathcal{E}_{i-1}\otimes p_{2}^{*}\mathcal{O}(1)\to\mathcal{E}_{i}\to
p_{1}^{*}E_{i+1}\to 0.$

by $p_{1}^{*}E_{j}^{*}$, we have

$0\to\mathcal{E}_{i-1}\otimes p_{2}^{*}\mathcal{O}(1)\otimes
p_{1}^{*}E_{j}^{*}\to\mathcal{E}_{i}\otimes p_{1}^{*}E_{j}^{*}\to
p_{1}^{*}E_{i+1}\otimes p_{1}^{*}E_{j}^{*}\to 0.$

Note that $H^{2}(p_{1}^{*}E_{i+1}\otimes p_{1}^{*}E_{j}^{*})=0$. Thus, we have
a surjection from

$H^{2}(\mathcal{E}_{i-1}\otimes p_{1}^{*}E_{j}^{*}\otimes
p_{2}^{*}\mathcal{O}(1))\to H^{2}(\mathcal{E}_{i}\otimes p_{1}^{*}E_{j}^{*}).$

Thus inductively we have a surjection from

$H^{2}(p_{1}^{*}E_{1}\otimes p_{1}^{*}E_{j}\otimes p_{2}^{*}\mathcal{O}(i))\to
H^{2}(\mathcal{E}_{i}\otimes p_{1}^{*}E_{j}^{*}).$

But $H^{2}(p_{1}^{*}E_{1}\otimes p_{1}^{*}E_{j}\otimes
p_{2}^{*}\mathcal{O}(i))=0$. Hence, $H^{2}(\mathcal{E}_{i}\otimes
p_{1}^{*}E_{j}^{*})=0$, which also proves the last part of (1).
(2). Consider the exact sequence

$0\to\mathcal{E}_{i-1}\otimes p_{2}^{*}\mathcal{O}(1)\to\mathcal{E}_{i}\to
p_{1}^{*}E_{i+1}\to 0.$

Tensoring by $p_{1}^{*}E_{j}\otimes p_{2}^{*}\mathcal{O}(k)$, we have,

$0\to\mathcal{E}_{i-1}\otimes p_{2}^{*}\mathcal{O}(k+1)\otimes
p_{1}^{*}E_{j}\to\mathcal{E}_{i}\otimes p_{1}^{*}E_{j}\otimes
p_{2}^{*}\mathcal{O}(k)\to p_{1}^{*}E_{i+1}\otimes p_{1}^{*}E_{j}\otimes
p_{2}^{*}\mathcal{O}(k)\to 0.$

Note that $H^{0}(C,E_{i+1}\otimes E_{j}^{*})=0$ and
$H^{2}(\mathcal{E}_{i-1}\otimes p_{2}^{*}\mathcal{O}(k+1)\otimes
p_{1}^{*}E_{j})=0$ by (1),

$H^{1}(\mathcal{E}_{i}\otimes p_{1}^{*}E_{j}\otimes
p_{2}^{*}\mathcal{O}(k))=H^{1}(\mathcal{E}_{i-1}\otimes
p_{2}^{*}\mathcal{O}(k+1)\otimes p_{1}^{*}E_{j})\oplus H^{1}(C,E_{i+1}\otimes
E_{j}^{*})\otimes H^{0}(\mathbb{P}^{1},\mathcal{O}(k)).$

(3). Similar as (2).

∎

∎

######  Remark 3.5.

Note that if $h^{0}(C,E_{m}\otimes E_{m+1}^{*}\otimes K_{C})=1$ and
$h^{0}(C,E_{i}\otimes E_{i+1}^{*}\otimes K_{C})=0$ for all $i<m$, then
$h^{0}(\mathbb{P}^{1},R^{1}(p_{2})_{*}(\text{Ad}(\mathcal{E})\otimes\mathcal{O}(-1)))=p+1$.
Hence, the restriction of the tangent bundle $T\mathcal{M}_{C}(n,\delta)$ to
the rational curve through $E$ in the above theorem has exactly one direct
summand of negative degree.

##  4\. Proof of the Conjecture 1.1

Let $\mathcal{W}$ be the set of wobbly bundles. Then it is known that
$\mathcal{W}$ is a closed subset of $\mathcal{M}_{C}(n,\delta)$.

Let $\mathcal{W}_{1}$ be the set of all wobbly bundles $E$ such that
$r_{m}d_{m+1}-r_{m+1}d_{m}<r_{m}r_{m+1}(g-1)$, where
$r_{m},r_{m+1},d_{m},d_{m+1}$ are respective rank and degree of the vector
bundles $E_{m}$ and $E_{m+1}$ appearing in lemma 3.2. Let
$\mathcal{W}_{2}=\mathcal{W}\setminus\mathcal{W}_{1}$.

######  Proposition 4.1.

$\mathcal{W}_{1}\subset\overline{\mathcal{W}_{2}}$, where
$\overline{\mathcal{W}_{2}}$ denotes the closure of $\mathcal{W}_{2}$.

###### Proof.

Let $E\in\mathcal{W}_{1}$. Then $E$ fits in the following exact sequence:

$0\to E^{m-1}\to E\to E_{m+1}\to 0,$

and $E^{m-1}$ fits in

$0\to E^{m-2}\to E^{m-1}\to E_{m}\to 0$

such that $h^{0}(C,E_{m}\otimes K_{C}\otimes E_{m+1}^{*})\geq 1$ and
$s(E):=r_{m}d_{m+1}-r_{m+1}d_{m}<r_{m}r_{m+1}(g-1)$. Then as in [8,
Proposition 1.9], one can deform $E$ to a stable vector bundle $\tilde{E}$
such that $\tilde{E}$ fits in an exact sequence of the form

$0\to\tilde{E}^{m-1}\to\tilde{E}\to\tilde{E}_{m+1}\to 0$

and $\tilde{E}^{m-1}$ fits in an exact sequence of the form

$0\to E^{m-2}\to\tilde{E}^{m-1}\to\tilde{E}_{m}\to 0,$

where rank and degree of $\tilde{E}$ are same as that of $E$, degrees of
$\tilde{E}^{m-1}$ and $\tilde{E}_{m}$ are one less than that of $E^{m-1}$ and
$E_{m}$ and the degree of $\tilde{E}_{m+1}$ is one more than that of $E_{m+1}$
with $h^{0}(C,\tilde{E}_{m}\otimes K_{C}\otimes\tilde{E_{m+1}}^{*})\neq 0$.
Note that $s_{\tilde{E}}=s_{E}+r_{m}+r_{m+1}$. If
$s_{E}+r_{m}+r_{m+1}<r_{m}r_{m+1}(g-1)$, then we continue the process to
deform $E$ to a stable bundle $E^{{}^{\prime}}$ such that $E^{{}^{\prime}}$
can be obtained as extensions

$0\to{E^{m-1}}^{{}^{\prime}}\to E^{{}^{\prime}}\to E^{{}^{\prime}}_{m+1}\to
0,$

$0\to E^{m-2}\to{E^{m-1}}^{{}^{\prime}}\to E^{{}^{\prime}}_{m}\to 0$

with $H^{0}(C,{E^{{}^{\prime}}_{m+1}}^{*}\otimes E^{{}^{\prime}}_{m}\otimes
K_{C})\neq 0$ and $s(E^{{}^{\prime}})\geq r_{m}r_{m+1}(g-1)$. Thus,
$\mathcal{W}_{1}\subset\overline{\mathcal{W}_{2}}$. ∎

######  Lemma 4.2.

Let $E\in\mathcal{W}_{2}$ be a general point. Then there is a rational curve
in $f:\mathbb{P}^{1}\to\mathcal{M}_{C}(n,\delta)$ through $E$ such that
$f^{*}T\mathcal{M}_{C}(n,\delta)$ has exactly one direct summand of negative
degree.

###### Proof.

Let $E\in\mathcal{W}_{2}$, then $E$ can be obtained by successive extensions
as in lemma 3.2, with $H^{0}(C,{E_{m+1}}^{*}\otimes E_{m}\otimes K_{C})=\neq
0$ $s_{E}\geq r_{m}r_{m+1}(g-1)$. Let $E_{m}$ be a generic vector bundle of
rank $r_{m}$ and degree $d_{m}$ such that $r_{m}d_{m+1}-r_{m+1}d_{m}\geq
r_{m}r_{m+1}(g-1)$. Let $W^{0}_{r_{m+1},d_{m+1}}(E_{m}\otimes K_{C})$ be the
twisted Brill-Noether loci of stable bundles of rank $r_{m+1}$ and degree
$d_{m+1}$ such that $h^{0}(C,E_{m+1}^{*}\otimes E_{m}\otimes K_{C})\geq 1$.
Then the expected dimension $\rho^{0}$ of the loci is given by

(4.1)
$\begin{split}\rho^{0}&=r_{m+1}(r_{m+1}-r_{m})(g-1)+r_{m+1}(r_{m}(2g-2)+d_{m})-r_{m}d_{m+1}\\\
&={r_{m+1}}^{2}(g-1)+r_{m}r_{m+1}(g-1)-(r_{m}d_{m+1}-r_{m+1}d_{m})\end{split}$

By [8, Theorem 0.3], if
$(r_{m}d_{m+1}-r_{m+1}d_{m})-r_{m}r_{m+1}(g-1)<{r_{m+1}}^{2}(g-1)$, then the
dimension of $W^{0}_{r_{m+1},d_{m+1}}(E_{m}\otimes K_{C})$ is the expected
dimension $\rho^{0}$. Note that for a general element $E_{m+1}\in
W^{0}_{r_{m+1},d_{m+1}}(E_{m}\otimes K_{C})$ one has
$h^{0}(C,{E_{m+1}}^{*}\otimes E_{m}\otimes K_{C})=1$ and for general
$E_{m-i},E_{m-i+1},h^{0}(E_{{m-i+1}}^{*}\otimes E_{m-i}\otimes K_{C})=0$ for
$i\geq 1$. Thus, if $E\in\mathcal{W}_{2}$ is general,
$h^{0}(C,{E_{m+1}}^{*}\otimes E_{m}\otimes K_{C})=1$. Therefore, by remark
3.5, the rational curve through $E$ constructed in previous section is non-
free and the restriction of the tangent bundle to it has exactly one direct
summand of negative degree. ∎

######  Proposition 4.3.

Let $E\in\mathcal{W}_{2}$ be general. Then there is an irreducible component
of $\mathcal{W}$ containing $E$ of co-dimension one in
$\mathcal{M}_{C}(n,\delta)$.

###### Proof.

Let $E\in\mathcal{W}_{2}$ be a general element. Then by Lemma 4.2, there is a
rational curve $f:\mathbb{P}^{1}\to\mathcal{M}_{C}(n,\delta)$ through $E$ such
that $f^{*}T\mathcal{M}_{C}(n,\delta)$ has exactly one direct summand of
negative degree. Let
$\text{Hom}^{\text{nonfree}}(\mathbb{P}^{1},\mathcal{M}_{C}(n,\delta))$ be the
closed subset of $\text{Hom}(\mathbb{P}^{1},\mathcal{M}_{C}(n,\delta))$
consisting of nonfree rational curves in $\mathcal{M}_{C}(n,\delta)$. Let $V$
be an irreducible component of
$\text{Hom}^{\text{nonfree}}(\mathbb{P}^{1},\mathcal{M}_{C}(n,\delta))$
containing $f$. Then for a general element
$g:\mathbb{P}^{1}\to\mathcal{M}_{C}(n,\delta)$ in
$V,g^{*}T\mathcal{M}_{C}(n,\delta)$ has exactly one direct summand of negative
degree. Thus, for a general $g\in V$, the map $\phi(p,g)$ is of co-rank one
and hence, by proposition 2.1, $dF(p,g)$ has co-rank one. Therefore,
$\overline{F(\mathbb{P}^{1}\times V)}$ has co-dimension one in
$\mathcal{M}_{C}(n,\delta)$. Since $f\in
V,E\in\overline{F(\mathbb{P}^{1}\times V)}$. Note that every element in $V$ is
nonfree. Thus, $\overline{F(\mathbb{P}^{1}\times V)}\subset\mathcal{W}$, which
proves the proposition.

∎

###### Proof.

(of 1.1) The conjecture 1.1 now follows immediately from Propositions 4.1 and
4.3. ∎

Acknowledgement: I would like to thank to Dr. Suratno Basu for many useful
discussions, comments and suggestions. I also would like to thank Prof.
Montserrat Teixidor I Bigas for many important discussions.

## References

  * [1] Kollár, J.: Rational curves on algebraic varieties, Erg. d. Math. 3 Folge 32, Springer Verlag 1996.
  * [2] R. Donagi, T. Pantev, Geometric Langlands and non-abelian Hodge theory, Surveys in differential geometry, Vol. XIII. Geometry, analysis, and algebraic geometry: forty years of the Journal of Differential Geometry, 85–116, International Press, Somerville, MA, 2009.
  * [3] G. Laumon, Un analogue global du cône nilpotent, Duke Math. Jour. 57 (1988), 647-671.
  * [4] S. Pal, Locus of non-very stable bundles and its geometry, Bulletin des Sciences Mathématiques 141 (2017), 747-765.
  * [5] Pal, Sarbeswar; Pauly, Christian The wobbly divisors of the moduli space of rank-2 vector bundles. Adv. Geom. 21 (2021), no. 4, 473–482.
  * [6] Pal, Sarbeswar Moduli of rank 2 stable bundles and Hecke curves. [Paging previously given as 1–13]. Canad. Math. Bull. 59 (2016), no. 4, 865–877.
  * [7] Pauly, Christian; Peón-Nieto, Ana Very stable bundles and properness of the Hitchin map. Geom. Dedicata 198 (2019), 143–148.
  * [8] B.Russo, M.Teixidor, On a conjecture of Lange, J.Alg.Geom.8(1999), p. 483-496.
  * [9] Yusuf Mustopa, Montserrat Teixidor I Bigas, Rational Curves on Moduli Spaces of Vector Bundles, Pre-print, https://arxiv.org/pdf/2007.10511.pdf