# Relation between rank and dimension of global sections of a globally generated vector bundle

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

• 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. Nov 5, 2019 at 14:00

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.

• 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 Nov 5, 2019 at 13:34
• ,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? Nov 5, 2019 at 13:45
• I'm sorry, I completely missed the question about $h^1$! I've edited the answer. Nov 5, 2019 at 14:11
• @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? Nov 5, 2019 at 15:49
• 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. Nov 5, 2019 at 18:23