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Let $S$ be a compact complex manifold and $L_1, L_2 \longrightarrow S$ be two holomorphic line bundles. Under what conditions (hopefully something that is easy to check) on $L_1$ and $L_2$ is the following fact true:

Let $f_1, f_2 \ldots, f_n$ be a basis for $H^0(L_1)$ and $g_1, g_2, \ldots g_m$ be a basis for $H^0(L_2)$. Then every element $s \in H^0(L_1 \otimes L_2)$ can be expressed as a $\mathbb{C}$-linear combination of $f_{i} \otimes g_j$, i.e. any $h \in H^0(L_1 \otimes L_2)$ is of the form $$ h = \Sigma c_{ij} f_{i} \otimes g_j $$ for some complex numbers $c_{ij}$.

Note that I am not asking for this expression to be unique (i.e. I am not asking for the collection $\{f_{i} \otimes g_j\} $ to be linearly independent in $H^0(L_1 \otimes L_2)$).

$\textbf{Remark:}$ 1) In general this is not true. Take $S:= \mathbb{CP}^N$, $L_1:= \mathcal{O}(2)$ and $L_2:= \mathcal{O}(-1)$.

I was wondering if it is it true if for instance when $L_1$ and $L_2$ are very ample?

2) If this question has been asked in too much generality, is anything known if $L_1= L_2 = L$? The statement does happen to be true if $L:= \mathcal{O}(1) \longrightarrow \mathbb{CP}^N$.

$\textbf{Remark:}$ I had meant to ask is the statement true when $L_1$ and $L_2$ are very ample (as opposed to ample). By this I mean the Kodiara map from $S$ to the projectivization of the dual of $H^0(L)$ is a well defined map and an embedding.

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This issue arises often and in general it is false. For a simple example, take $S$ to be an elliptic curve, $L$ a line bundle of degree two. Then, you can easily check that the map $H^0(L)\otimes H^0(L)\to H^0(L^2)$ is not surjective.

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    $\begingroup$ I would add to what Mohan, a man of inestimable reputation, said. For curves one can sometimes use the bpf pencil trick. That is if $L_1$ is bpf, then the vanishing of of $H^1(L_2 \otimes L_1^{-1} ) $ is a sufficient criteria. Of course S doesn't have to be a curve for this criteria to be true, but for curves that vanishing is usually easy to see. $\endgroup$ – aginensky Jun 10 '18 at 16:00
  • $\begingroup$ easier to see :) $\endgroup$ – aginensky Jun 10 '18 at 16:11
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    $\begingroup$ @aginensky: What is bpf? $\endgroup$ – Ritwik Jun 10 '18 at 18:21
  • $\begingroup$ base point free, I believe. $\endgroup$ – roy smith Jun 10 '18 at 19:33
  • $\begingroup$ @ Ritwik Sorry, bpf = base point free as Roy says. The first reference I found on the web for the lemma is math.lsa.umich.edu/~mmustata/hw9.pdf which gives an application of the sort you are looking for. $\endgroup$ – aginensky Jun 10 '18 at 21:19
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There are ample line bundles without any non-zero global sections. E.g., on any smooth projective non-rational non-hyperelliptic curve $X$ and pairwise different points $P,Q,R\in X$, the line bundle $\mathscr O_X(P+Q-R)$ is ample but has no global sections. Then let $\mathscr A$ be such a line bundle and $n\in \mathbb N$ the largest integer such that $H^0(X,\mathscr A^{2^n})=0$. Finally, let $\mathscr L=\mathscr A^{2^n}$. Then $H^0(X,\mathscr L)\otimes H^0(X,\mathscr L)=0$, while $H^0(X,\mathscr L^{2})\neq 0$.

In your example of $X=\mathbb P^n$, it works, because all line bundles of positive degree on $\mathbb P^n$ are very ample and give a projectively normal embedding.

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  • $\begingroup$ I actually meant to say Very Ample (i.e. the Kodiara map from S to P(H^0(L)^dual) is an embedding ). Are you saying that the statement is true when L is very ample (as opposed to just ample)? $\endgroup$ – Ritwik Jun 10 '18 at 18:14
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    $\begingroup$ No, definitely not. For a simple example, take a curve $C$ of genus $g\geq 4$ and a general line bundle $L$ on $C$ of degree $g+3$. Then it is standard to prove that the linear system $|L|$ embeds $C$ into $\mathbb{P}^3$. But $h^0(L^2)= g+7$ is greater than $\dim \mathsf{S}^2H^0(L)=10$. $\endgroup$ – abx Jun 10 '18 at 19:35
  • $\begingroup$ @Ritwik: I didn't say that very ample is enough. I said that very ample and giving a projectively normal embedding is. $\endgroup$ – Sándor Kovács Jun 11 '18 at 16:25
  • $\begingroup$ @Sandor: I understand now. $\endgroup$ – Ritwik Jun 12 '18 at 17:39

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