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Let,$E$ be a nontrivial vector bundle of rank $r$ on a projective variety $X$ over the field of complex numbers.Also let's assume that $E$ is globally generated.

Then my question is the following :Is it true that $h^i(E) \geq r+1$ , for $i =0,1 $ ?

My attempt : Sine $E$ is globally generated, so we have the evaluation map given by $\mathcal O_S^{\oplus h^{0}(E)} \to E$ is surjective . Taking stalk at any point we have $ h^0(E) \geq r $. But I don't see how $r+1$ comes.I also tried to compute via long exact cohomology sequence but I don't think that gives the desired inequality.

At this point my next question is :Is it true atleast in the case when $X$ is a smooth irreducible curve and $E$ is a line bundle on it?

Any help from anyone is welcome

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    $\begingroup$ This is true for $i=0$ and false for $i=1$. There are many globally generated non-trivial vector bundles with $h^1=0$. For your argument for $h^0$, assume $h^0=r$ and then show (as the answer below) that $E$ must be trivial. $\endgroup$
    – Mohan
    Nov 5, 2019 at 14:00

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Suppose $h^0(E) = r$. Since $E$ is globally generated, there is a surjection $\mathcal O^{\oplus h^0(E)} = \mathcal O^r \to E$, and let $K$ be its kernel. Take $p\in X$ a closed point, and consider the restriction to $p$. Since $E$ is a vector bundle, it is flat, so there is a short exact sequence of vector spaces: $$ 0\to K|_p \to \mathcal O^r|_p \to E|_p \to 0. $$ Since $\mathcal O^r$ and $E$ are both vector bundles, their restrictions are both isomorphic to $\mathbb C^r$, so the above sequence shows that $K|_p=0$.

Therefore, $K|_p=0$ for any closed point $p$, which implies by Nakayama's lemma that $K=0$. So the surjection $\mathcal O^r\to E$ must be an isomorphism.

It is not necessarily true that $h^1(E) \ge r + 1$. For example, $\mathcal O(1)$ on $\mathbb P^1$ has vanishing first cohomology and it is globally generated.

More generally, let $X$ be a smooth curve of genus $g$ and $E$ be a line bundle of degree at least $2g$. Serre duality tells you that $h^1(L) = h^0(K_X - L)$, where $K_X$ is the canonical bundle of $X$. Since $\operatorname{deg} K_X = 2g-2$, $h^1(L) = 0$, whenever $\deg L \ge 2g-1$. In particular, if $E$ has degree $2g$ then its higher cohomology vanishes.

Let us show that $E$ is globally generated. Consider $p\in X$ and look at the short exact sequence $$ 0\to E(-p) \to E \to E|_p \to 0. $$ Its cohomology long exact sequence starts as follows: $$ 0\to H^0(X,E(-p)) \to H^0(X,E) \to \mathbb C \to H^1(X,E(-p)). $$ $E(-p)$ is a line bundle of degree at least $2g-1$, so by the discussion above its first cohomology vanishes. Therefore, the map $H^0(X,E(-p)) \to H^0(X,E)$ is not surjective. This means that for any $p\in X$, there is a global section of $E$ that doesn't vanish at $p$, i.e. $E$ is base-point free, which for a line bundle means it is globally generated.

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  • $\begingroup$ If we denote $f : \mathcal O_S^{\oplus h^{0}(E)} \to E$ as evaluation map and $K$ be its kernel ,then from the S.E.S $0 \to K \to \mathcal O_S^{\oplus h^{0}(E)} \to E \to 0$ taking stalk at a point $x \in X$ and then tensoring by the residue field we end up having a right exact sequence of $K(x)$ vector space from where we have from rank nullity theorem $h^0(E) = r +c$ where $c =$ the dimension of the image of the leftmost vector space, $c$ can't be $0$ ,because otherwise $h^0(E) =r$ and $E$ becomes isomorphic to $r$ copies of $\mathcal 0_X$, a contradiction as $E$ is nontrivial $\endgroup$
    – HARRY
    Nov 5, 2019 at 13:34
  • $\begingroup$ ,according to your answer since $E$ is nontrivial so by contrapositive argument there is no such surjection from $\mathcal O^r \to E$ ,but how does that imply the inequalities involving cohomology? $\endgroup$
    – HARRY
    Nov 5, 2019 at 13:45
  • $\begingroup$ I'm sorry, I completely missed the question about $h^1$! I've edited the answer. $\endgroup$
    – Moisés
    Nov 5, 2019 at 14:11
  • $\begingroup$ @Mohan,the motivation behind asking this question was the definition of clifford index of a curve.It's the infimum of clifford index of line bundles satisfying the above $2$ cohomology conditions.What happens if for a smooth irreducible curve there are no such line bundles.How do we define it's clifford index then? $\endgroup$
    – HARRY
    Nov 5, 2019 at 15:49
  • $\begingroup$ I don't know. I would guess it is infinite. It seems to me that Wikipedia has answers concerning the existence of such line bundles. $\endgroup$
    – Moisés
    Nov 5, 2019 at 18:23

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