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Let $A:={\mathbb{C}}[x_0,\ldots,x_n]=\oplus_{d=0}^{\infty} {\mathbb{C}}[x_0,\ldots, x_n]_d$ be the graded complex algebra of polynomials in $n+1$ variables, graded by degree.

Suppose $L$ is a line bundle over a projective manifold $X$ such that the ring of plurisections of $L$, i.e. $\oplus_{d=0}^{\infty} H^0(X,L^d)$ is isomorphic to $A$ as graded complex algebras.

Does that imply that $X$ is necessarily complex projective space ${\mathbb{CP}}^n$ and $L$ is the tautological line bundle?

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  • $\begingroup$ thanks for this nice question. Both question and answer are useful $\endgroup$ – gauss Mar 18 '12 at 9:31
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The answer is no. Take any projective manifold $X$ mapping $\pi\colon X\rightarrow\mathbb P^n$ to such that $\pi_*\mathcal O_X=\mathcal O_{\mathbb P^n}$ and let $L$ be $\pi^*\mathcal O(1)$. Then $H^0(X,L^{\otimes m})=H^0(\mathbb P^n,\mathcal O(m))$ Examples are $X=Y\times\mathbb P^n$ or a blowing up of a closed smooth subvariety of $\mathbb P^n$.

I do not know if some power of $L$ is necessarily without base points (in which case $L$ itself is). In any case we can blow up $X$ such that there is a map to $\mathbb P^n$. The line bundle that gives that mapping may I guess have more sections.

Addendum: I was really rambling in the last paragraph. What I meant was that suppose we only know that $\bigoplus_nH^0(X,L^{\otimes n})$ is a polynomial ring. Does that mean that $L$ is base-point free? It may very well be true but I don't see it at the moment.

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    $\begingroup$ your answer is unexpected and surprising! (not to say the least, very helpful). Thanks! $\endgroup$ – user2529 Nov 6 '10 at 13:58
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    $\begingroup$ @Torsten: thanks for the clarification. I was sure it had to be something different. :) This is a good question. Maybe you should pose it as a question. $\endgroup$ – Sándor Kovács Nov 6 '10 at 18:54
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    $\begingroup$ @Colin: here is why this is not surprising: The section ring of $\mathcal L$, $R=\oplus_n H^0(X,\mathcal L^n)$ describes $Y={\rm Proj}R$ and not $X$. Your condition simply means that the rational map given by some high power of $\mathcal L$ maps onto a projective space. Exactly as in Torsten's example. $\endgroup$ – Sándor Kovács Nov 6 '10 at 23:47
  • $\begingroup$ @Torsten: an example (I think) with a not-basepoint-free system is below. $\endgroup$ – Sándor Kovács Nov 6 '10 at 23:48
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Here is an example for Torsten's Addendum: a line bundle with the required properties which is not semi-ample (no multiple is globally generated):

Take an $\pi:X\to \mathbb P^n$ as in Torsten's example, that is, with $\pi_*\mathcal O_X\simeq \mathcal O_{\mathbb P^n}$. Also assume that there is an exceptional divisor $E$, so let's just say it is a birational morphism contracting the divisor $E$ to a point. Now take $\mathcal L=\pi^*\mathcal O_{\mathbb P^n}(1)$ and consider the short exact sequence:

$$ 0\to \mathcal L^{\otimes n} \to (\mathcal L(E))^{\otimes n} \to \mathcal O_{nE}(nE) \to 0 $$ The sheaf on the right has no global sections, so the other two have the same $H^0$. Therefore $$ H^0(X,(\mathcal L(E))^{\otimes n}) = H^0(\mathbb P^n, \mathcal O_{\mathbb P^n}(n)) $$

Finally $(\mathcal L(E))^{\otimes n}$ cannot be globally generated, because $H^0(X,\mathcal L^{\otimes n})=H^0(X,(\mathcal L(E))^{\otimes n})$, so ${\rm supp}\,E$ is always in the base locus (or in other words, if it had a section whose zero section did not contain ${\rm supp}\,E$, then its restriction to $nE$ would give a nonzero section of $\mathcal O_{nE}(nE)$).


And here is what you can assert based on your condition:

Naive statement: Essentially the only reason your desired claim can fail is via Torsten's example.

More precise statement: Let $k$ be a field, $S:=k[x_0,…,x_n]$ and suppose $\mathcal L$ is a line bundle on a projective variety $X$ (over $k$) such that $R(X,{\mathcal L})=\oplus_{d=0}^\infty H^0(X,\mathcal L^d)\simeq S$ as graded $k$-algebras. Then there exists a diagram $$ X \overset\sigma\longleftarrow \widetilde X \overset\phi\longrightarrow \mathbb P^n $$ such that $\sigma$ is a projective birational morphism and $\phi$ is surjective. Furthermore, $\sigma^*\mathcal L\supseteq \phi^*\mathcal O_{\mathbb P^n}(1)$. Finally if $n=\dim X$, then $\phi$ is also birational and hence $X$ is a rational variety.

Proof: By the assumption ${\rm Proj}\, R(X,\mathcal L)\simeq \mathbb P^n$ and there exists a dominant rational map $\psi: X\dashrightarrow \mathbb P^n$. Let $$ X \overset\sigma\longleftarrow \widetilde X \overset\phi\longrightarrow \mathbb P^n $$ be the resolution of indeterminacies of $\psi$ via blow-ups on $X$. Then $\sigma$ is a projective birational morphism and hence $\widetilde X$ is also projective and $\sigma^*\mathcal L\supseteq \phi^*\mathcal O_{\mathbb P^n}(1)$. It follows that the induced morphism $\phi$ has to be surjective (since $\widetilde X$ is projective and the morphism is defined everywhere). Since this morphism is given by global sections of sufficiently high powers of a line bundle (i.e., $\phi^*\mathcal O_{\mathbb P^n}(m)$ for some $m>0$), it follows that it has connected fibers and $\phi_*\mathcal O_{\widetilde X}\simeq \mathcal O_{\mathbb P^n}$ and hence if $n=\dim X$, then $\phi$ is birational.

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