# Holomorphic vector bundle obtained from exceptional divisor has one section.

I've been reading Huybrecht's Complex Geometry, and ran across this question in section 2.5:

If $X$ is a compact complex manifold, and $Y$ a hypersurface, show that $\mathcal{O}(E)$ ( where $E = \sigma^{-1}(Y)$ is the exceptional divisor) has exactly one global section up to scaling.

I suspect this problem is quite simple, yet I frustratingly cannot make much progress on it. There is a hint given that one could consider the case of the blow-up of a point on a Riemann surface, but in this case, since the point is a smooth divisor, $\text{Bl}_Y(X)= \hat X \cong X$. Since in addition $\mathcal{O}(E)|_E \cong \mathcal{O}(-1)$, the obvious candidate for our supposed unique section is the defining function for the divisor (as we must vanish along $E$ as $\mathcal{O}(-1)$ has no sections). Does this line of reasoning hold for higher dimensions (even if $Y$ is not smooth)? I didn't need compactness so I imagine that this is wrong. If so, can someone point me in the right direction?

• Small remark : when you say $\rm{Bl}_Y(X) \cong X$, it's not true. Blowing-up up a surface will change the surface, even if it is at a smooth point. For example, the blow-up of $\Bbb P^2$ at a point is the projectivized bundle $\Bbb P_{\Bbb P^1} (\mathcal O(1) \oplus \mathcal O)$. – Nicolas Hemelsoet Oct 6 '17 at 4:48
• I should have clarified. I meant one complex dimension, not two. (Edited) – DKS Oct 6 '17 at 4:57
• You have at least one global section cutting out $E$. If there was a second independent section then you should be able to show that its zero locus would intersect $E$ generically transversally, violating negativity of the normal bundle of $E$. Can you make this precise? Alternatively, you could try playing around with basic exact sequences... – Tabes Bridges Oct 6 '17 at 17:29
• I figured it out. Thanks Tabes for the bit above (even if I am unable to make it precise currently). I'll sketch it here for posterity. $E$ is a hypersurface, and as such is an effective divisor. We know that we can then construct a global section of $\mathcal{O}(E)$ from the cocycles. Using Tabes's comment, this section is unique up to multiplication by a global holomorphic function. Since $X$ is compact, these global holomorphic functions are just scalars. Hence the claim. – DKS Oct 7 '17 at 16:25