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Let $V$ be a fixed vector space. Let $X$ be a smooth curve. Consider the Quot scheme $Q$ of Quotients of $V\otimes \mathcal{O}_X$ of degree $d$ and rank $r$. Let $R$ be the open subscheme of $Q$ consisiting only vector bundles with following property: $H^1$ vanishes and $H^0\cong V$ via the natural map. Its clear that every such quotient $E$ in $R$ is globally generated vector bundle.

Now in Atiyah's Paper he shows that there is following exact sequence:

$0\rightarrow \mathcal{O}_X^{\oplus (r-1)}\rightarrow E\rightarrow det ~E\rightarrow 0$, for $E$ globally generated.

My Question is does it hold in the described case i.e, is there a exact sequence like this:

$0\rightarrow \mathcal{O}_{X\times R}^{\oplus (r-1)}\rightarrow \mathcal{E}\rightarrow det ~\mathcal{E}\rightarrow 0$, where $\mathcal{E} $ is the universal bundle on $R$?

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No, this is not true in general. Actually, the exact sequence on $X$ is not canonical, but is determined by an $(r-1)$-dimensional vector subspace in $V$. Moreover, to extend to such a sequence the first morphism should be a fiberwise monomorphism. For this, there is an obstruction given by $c_2(E)$. Eventually, since $X$ is a curve, the obstruction vanishes, and such a sequence exists.

However, over the product $X \times R$ there is no reason for $c_2(\mathcal{E})$ to be zero, so for a general morphism $\mathcal{O}_{X \times R}^{\oplus(r-1)} \to \mathcal{E}$ the cokernel is not a line bundle.

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  • $\begingroup$ Is there a surjective morphism from $\mathcal{E}\rightarrow det~\mathcal{E}$ over $X\times R$? $\endgroup$
    – user100841
    Commented Dec 4, 2017 at 9:28
  • $\begingroup$ @S.D. I don't see a way to construct a canonical morphism like that. Definitely, one can consider the space $H^0(X\times R,\mathcal{E}^\vee \otimes \det(\mathcal{E}))$ of all morphisms (this space is possibly zero), and ask whether some of them is surjective. Note that for the surjectivity there is also a cohomological obstruction, namely $c_r(\mathcal{E}^\vee \otimes \det(\mathcal{E}))$. $\endgroup$
    – Sasha
    Commented Dec 4, 2017 at 11:19
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I am trying to answer this in Rank 2 case (over $R$):

Fix a basis $\{v_1,v_2,....v_n\}$ of $V$. Then its clear that there is a morphism $\mathcal{E}\rightarrow det~\mathcal{E}$. Then consider the exact sequence $0\rightarrow Ker\rightarrow \mathcal{E}\rightarrow det~\mathcal{E}\rightarrow 0$. Then we have $det~\mathcal{E}=Ker \otimes det~\mathcal{E}$. So $Ker=\mathcal{E}$. So we have an exact sequence $0\rightarrow \mathcal{O}\rightarrow \mathcal{E}\rightarrow det~\mathcal{E}\rightarrow 0$.

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    $\begingroup$ What is not clear at all is that this "clear" morphism is surjective -- in fact, this is not true. $\endgroup$
    – abx
    Commented Dec 3, 2017 at 19:01

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