Relation between $H^0(\mathbb{P}^n,\mathcal{O}(-1))|_X$ and $H^0(X,O(-1)|_X))$.

Question

This question comes from a exercise 2.4.3 on D.Huybrechts ‘s book “Complex Geometry”,there we need compute the Kodaira dimension of some smooth hypersurface $X\subset \mathbb{P}^n$.

Actually we can compute the canonical bundle $K_X\cong \mathcal{O}(d-n-1)|_X$ (for $X\subset \mathbb{P}^n $ is a hypersurface of degree d) by adjunction formula.

And then,I was troubled by the relations between $H^0(\mathbb{P}^n, \mathcal{O}(k))|_X$ and $H^0(X, \mathcal{O}(k)|_X)$. First,we know that for a holomorphic map $f:X\to Y$ and $E$ a holomorphic vector bundle over Y,we have a natural homomorphism $H^0(Y,E)\to H^0(X,f^*E)$.

Here it just is the restriction $H^0(\mathbb{P}^n,O(k))\to H^0(X,O(k)|_X)$, so the question equals to ask the map is surjective or not. Of course,for $k=0$, i.e $O(k)$ is trivial bundle,it’s true,both side equal to $\mathbb{C}$.

So my question is

(1)for any holomorphic map $f:X\to Y$, is the natural homomorphism $H^0(Y,E)\to H^0(X,f^*E)$ always surjective?

(2) In my case $f:X\to Y$ is a submanifold, the homomorphism above is surjective or not?

(3) For $X\subset \mathbb{P}^n$, $E=\mathcal{O}_{\mathbb{P}^n}(k)$.

I believe (3) is true,but I can’t give a proof.

Any suggestion will be appreciated. Feel sorry for my poor English,and hope I have post my question clearly.

Answer 1

Consider the restriction sequence for $X$, twisted by $\mathcal O(k)$: $$0\to I_X\otimes\mathcal{O}_{\mathbb P^n}(k)\to \mathcal{O}_{\mathbb P^n}(k)\to \mathcal{O}_{X}(k)\to 0$$ and note that $I_X=\mathcal O_{\mathbb P^n}(-d)$ if $X$ is a hypersurface of degree $d$. Thus, we get an exact cohomology sequence $$H^0(\mathcal{O}_{\mathbb P^n}(k))\to H^0(\mathcal{O}_{X}(k))\to H^1(\mathcal O_{\mathbb P^n}(k-d)).$$ Therefore, it is sufficient if $H^1(\mathcal O_{\mathbb P^n}(k-d))=0$. This is always the case if $n\not=1$.

If $n=1$, however, then $X$ is zero-dimensional; thus, the restriction of the line bundle is necessarily trivial; hence, $h^0(\mathcal O_X(k))=h^0(\mathcal O_X)=d$ for all $k$, whereas $\mathcal O_{\mathbb P^1}(k)$ has no global sections if $k<0$. In particular, there is no chance for $H^0(\mathcal O_{\mathbb P^1}(k))$ to map surjectively onto $H^0(\mathcal O_X(k))$ if $k$ is negative.