Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let us work with projective algebraic varieties over $k= \mathbb{C}$. If necessary we can also assume smoothness of the varieties.

Of course it is not in general true that given two line bundles $L, M$ on a variety $X$, we have $$ \Gamma(X,L \otimes M) = \Gamma(X,L)\otimes \Gamma(X,M), $$ an easy example is $\mathcal{O}(1)$ and $\mathcal{O}(-1)$ on projective space.

I was wondering if there are conditions that one can place on the variety or the bundles, such that the above equality does hold. Maybe if the bundles admit global sections, or are even generated by global sections?

As a second question, when we know the dimension of $\Gamma(X,L \otimes M)$, can we translate this back into information on $\Gamma(X,L)$ or $\Gamma(X, M)$? (assuming for the moment anything that you wish to assume.)

I know this last question is vague, so as an answer basically any general observation, or anything in a direction of a technique for calculation would be great!

share|improve this question
2  
You mean the canonical map $\Gamma(X, L)\otimes\Gamma(X, M)\to \Gamma(X, L\otimes M)$ is an isomorphism. –  user18119 Feb 18 '13 at 20:28
    
It is almost never an isomorphism, even when $L,M$ are globally generated. In some sense, Serre's results about coherent sheaves imply that it is true "up to shift" with an ample sheaf. –  Martin Brandenburg Feb 18 '13 at 21:08
1  
There was a very similar question on math.SE: math.stackexchange.com/questions/18209 ... have a look at the answers there. –  Martin Brandenburg Feb 18 '13 at 22:45

2 Answers 2

up vote 7 down vote accepted

The paper Global sections and tensor products of line bundles over a curve by David C. Butler quotes and proves some interesting results in that direction. For example:

On a smooth projective curve of genus $g$ let $L_1$ be a line bundle of degree $\geq 2g$ and $L_2$ a line bundle of degree $>2g$. Then $\tau : \Gamma(L_1) \otimes \Gamma(L_2) \to \Gamma(L_1 \otimes L_2)$ is surjective (this is due to Mumford). This also holds when $L_1$ and $L_2$ are globally generated and $\mathrm{deg}(L_1) + \mathrm{deg}(L_2) \geq 4g+1$. The paper also contains refinements about the image of $\tau$ (Theorems 1 and Theorem 2 in loc. cit). The proofs use Riemann-Roch.

If you are also interested in vector bundles instead of just line bundles, see the paper On the tensor product of sections of vector bundles on an algebraic curve by M. Baiesi and E. Ballico, and the references there.

I don't know if anything is known beyond curves.

share|improve this answer

If $X=Spec(A)$ is any affine scheme and if $L,M$ are arbitrary line bundles, then the canonical map $$ \Gamma(X,L)\otimes_{\mathcal O_X(X)} \Gamma(X,M) \to \Gamma(X,L \otimes_{\mathcal O_X} M) \quad (\bigstar)$$ is always an isomorphism.

Indeed $L=\tilde P$ and $M=\tilde{Q}$ are associated to the projective $A$-modules of rank one $P=\Gamma(X,L)$ and $Q=\Gamma(X,M)$.
The morphism $(\bigstar)$ then becomes $$P\otimes_A Q\to \Gamma(X,\tilde P \otimes_{\mathcal O_X} \tilde X) \quad (\bigstar \bigstar)$$
To conclude, it suffices to show that $$\tilde P \otimes_{\mathcal O_X} \tilde X=\widetilde {P\otimes _A Q}\quad (KEY)$$ because then in $(\bigstar \bigstar)$ $$\Gamma(X,\tilde P \otimes_{\mathcal O_X} \tilde M)=\Gamma(X,\widetilde {P\otimes _A Q})=P\otimes_AQ$$ too.
And here is the good news: Dieudonné and Grothendieck proved $(KEY)$ for you half a century ago!
Just check EGA I, Corollaire (1.3.12 (i)), page 88.

A Conjecture
I am pretty confident that the corresponding result holds for two arbitrary holomorphic line bundles on a Stein manifold.

share|improve this answer
2  
Isn't this merely a comment than an answer? This has nothing to do with line bundles. It is well-known that for an affine scheme $X$ the functor of global sections $\mathsf{Qcoh}(X) \to \mathrm{Mod}(\Gamma(X,\mathcal{O}_X))$ is an equivalence of monoidal categories. In the question $X$ is projective. –  Martin Brandenburg Feb 18 '13 at 22:39
    
Oops, you're right: for some reason I overlooked the hypothesis that $X$ is projective. I know that the result is more general than for line bundles, but since Joachim asked about line bundles and not about quasi-coherent sheaves, I gave an answer for line bundles. Apologies to all for my sloppy reading of the question. –  Georges Elencwajg Feb 18 '13 at 23:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.