Let $X$ be a projective scheme and $\mathcal{L}$ be a very ample line bundle on $X$ with respect to some projective embedding $X \hookrightarrow \mathbb{P}^n_{\mathbb{C}}$ (for some $n$). Given any closed point $x \in X$, denote by $\mathcal{L}(x):=\mathcal{L}_x \otimes_{\mathcal{O}_{X,x}} k_x$, where $k_x$ is the residue field of the local ring $\mathcal{O}_{X,x}$.

Fix an integer $m \ge 0$. Denote by $t:=h^0(\mathcal{L}^m)$. Assume $t>0$. Then,

1) Is it true that given any closed point $x \in X$, the natural morphism from $H^0(\mathcal{L}^m)$ to $H^0(\mathcal{L}^m(x))$ is surjective?

2) If $(1)$ is true, can we say anything about what number of closed points (not necessarily distinct), say $x_1,...,x_{t_o}$ for which the natural morphism from $H^0(\mathcal{L}^m)$ to $\oplus_{i=1}^{t_0} H^0(\mathcal{L}^m(x_i))$ is surjective?

  • $\begingroup$ I think you can do $t_0=m$ as long as the points are distinct. $\endgroup$
    – Will Sawin
    Mar 3 '15 at 18:25
  • $\begingroup$ @Sawin: Thanks. Could you please give an idea. $\endgroup$
    – user43198
    Mar 3 '15 at 19:38

It seems to me that the $m$ is a red herring. Since you already assume that $\mathcal L$ is very ample, why do you need a power? I'll ignore $m$ here.

For the questions: In case the base field is algebraically closed (1) is a direct consequence of $\mathcal L$ being very ample. That surjectivity means that you can choose a global section of $\mathcal L$ that is zero and another one that is not zero at a given point.

For (2), I don't understand what you mean by "not necessarily distinct". If the same point appears more than once, then this will certainly not be surjective. Any section will either be zero or not zero at any given point.

If the points are distinct, then this seems like a strong criterium for large $t_0$.

Let's say that $X$ is a smooth curve. Then it seems to me that what you are asking for is essentially the same as asking that if you embed $X$ via the global sections of $\mathcal L$, then the images of that set of points is in general position. In other words the set of points that might satisfy this condition cannot even be chosen to be simply general, but they have to be consecutively chosen to be general with respect to the points already chosen.

  • $\begingroup$ @Kovacs: Thank you very much for the answer. For the second question, I understand that for large $t_0$ this is not possible. But for small $t_0$, like less than $t$, is it possible? For your comment on distinct points I do not get it yet. As you say it is possible that a section is zero at point, but then there is another section that is non-zero at that point. Won't this suffice for surjectivity? $\endgroup$
    – user43198
    Mar 3 '15 at 17:57
  • 1
    $\begingroup$ To get a surjective map to a direct sum you need to get a single element that maps to the various choices. If the same point is repeated, then on the right hand side you can choose an element which has a component that is zero at that point and another component that is non-zero. Then to get surjectivity you would have to find a single section on the left hand side that maps to this element, but it could only map onto one of those components. $\endgroup$ Mar 3 '15 at 18:11
  • $\begingroup$ @Kovacs: Sorry, I think I understand now. $\endgroup$
    – user43198
    Mar 3 '15 at 19:37
  • $\begingroup$ Karl, you are right. I was only thinking about 0 or not zero. $\endgroup$ Mar 3 '15 at 21:06
  • $\begingroup$ In the spirit of Karl and Sandor, I will also stick to algebraically closed fields. I'll go further and stick to curves because RR is so exact for a curve !. Certainly the number $t_0$ is bounded by $h^0(\mathcal{L})$. But you are asking for how many points are always guaranteed to be linearly independent. If $\mathcal{L} = K_X$ is very ample, this is the gonality of the curve X and it can be any integer between 3 and appx $\frac{g-1}{2} $ ( i forget the exact number the Riemann count gives). So even in this very well understood case, there isn't a great answer. $\endgroup$
    – meh
    Mar 3 '15 at 22:53

Maybe it's worth saying something more about question (2) (like Sándor, I'll stay in the algebraically closed case). This gets very close to the realm of Seshadri constants, see Positivity in Algebraic Geometry I, Section 5.

In particular, suppose that we do allow some of the points $x_i$ to be the same. Let $I_{x_i}$ be the ideal defining $x_i$. Then let's define $\mathcal{L}(x_1, \ldots, x_n) = \mathcal{L} \Big/ \Big(\prod_i I_{x_i} \Big)$. In the case that $x_1 = x_2$, a surjectivity $H^0(\mathcal{L}) \to H^0(\mathcal{L}(x_1, \ldots, x_n)) = \mathcal{L}(x_1, \ldots, x_n)$ would in particular mean that the global sections of $\mathcal{L}$ generate $\mathcal{O}_{X,x_1}\big/I_{x_1}^2$ as a vector space (ie, in particular this implies that the global sections separate tangent vectors). If more $x_i$ are equal, then this means that higher order jets (measures of tangency) are separated.

Sehsadri constants

The notion of Seshadri constants explores the above asymptotically. Indeed, let $Z = \{x_1, \ldots, x_n \}$ denote a collection of closed points (possibly with repeats as above). Define $$ s(\mathcal{L}^n, Z) $$ to be the largest integer $k$ such that $$ H^0(X, \mathcal{L}^n) \to H^0\Big(X, \mathcal{L^n}\big/ \big(\prod_i I_{x_i}^k\big) \Big) $$ surjects (maybe if we define $kZ$ to be $k$ of the $Z$'s, then this could be compactly written as $H^0(X, \mathcal{L}^n) \to H^0(X, \mathcal{L^n}(kZ))$ using the above notation).

Then we define $$ \varepsilon(X, \mathcal{L}; Z) = \lim_{k \to \infty} \frac{s(\mathcal{L}^k, Z)}{k}. $$

The point is, the more positive (ample) $\mathcal{L}$ is at the various points, the bigger the Seshadri constant will be. The question is what this limit


Frequently, this is used to actually do the opposite of what you are doing (ie, prove certain ample divisors are in fact very ample, or globally generated). Some common statements include ones like if $X$ is smooth and projective and $\varepsilon(X, \mathcal{L}, x) > \dim X$ for all points $x$ then $\mathcal{L} \otimes \mathcal{O}_X(K_X)$ is globally generated and if $\varepsilon(X, \mathcal{L}, x) > 2 \dim X$ for all $x \in X$ then $\mathcal{L} \otimes \mathcal{O}_X(K_X)$ is very ample.

There are other applications too, like restatements/generalizations of Nagata's conjecture. It's certainly important to also try to understand lower bounds for these constants for ample line bundles: there aren't any in general, but there are for sufficiently general points. Anyways, the aforementioned section of Lazarsfeld's book has a nice survey.

  • $\begingroup$ @Schwede: Thank you very very much for the detailed answer. This is very helpful. $\endgroup$
    – user43198
    Mar 3 '15 at 21:52

I propose that this is what you "really" mean to say by 2): "what is the maximum length $t(L)$ such that for any $0$-dimensional scheme $Z$ of length $t(L)$, the evaluation morphism $H^0(L) \otimes \mathcal{O}_X \to L_{|{_Z}}$ is surjective". This, I think, is what you mean by letting the points come together. For an general very ample line bundle $L$, the answer is of course $t(L)=2$ (very ampleness is equivalent to $t(L) \geq 2$, and it is not hard to find examples where $t(L)=2$).

Here is the exact terminology I think you are looking for: a line bundle is called $k$ very ample if $t(L) \geq k+1$. This is a notion due to Sommese and there is a huge amount of literature about it. A good primer on it is Ch. 5 of Göttsche's paper "A conjectural generating function for numbers of curves on surfaces". This is closely related to the notion of Seshadri constants as in Karl Schwede's answer, but it is slightly different in general.

For example a globally generated line bundle is $0$ very ample and a very ample line bundle is $1$ very ample.

Seeing as you were taking powers of $L$, perhaps the result you are looking for is the following: suppose $L$ is very ample, then $L^m$ is $m$ -very ample. In particular, if $L$ is very ample, then the answer to 2) in general, with my interpretation of the question, is $t_0=m+1$.

EDIT: I originally stated that $L$ $k$-very ample implies $L^m$ is $km$ very ample. This does seem to be true but may be less well-known than I recalled: it follows from the paper of Hinohara, Takahashi, Terakawa "On tensor products of $k$-very ample line bundles". For line bundles on K3 surfaces (where I usually work) or Enriques surfaces, the result is more well-known; it follows from Knutsen's paper "On kth order embeddings.." plus Lemma 0.5.3 of Beltrametti and Sommese, "On k-spannedness for projective surfaces". The weaker statement $L$ very ample implies $L^m$ $m$-very ample, which is all we need, follows from Beltrametti and Sommese, "On k-jet ampleness".

It is worth having a quick look at these papers above. It turn out there are three related notions related to how one lets points come together: $k$ very ample, $k$ spanned and $k$ jet ample, which can sometimes coincide but not always. In general, $k$ jet ample is the strongest, $k$ spanned is the weakest and $k$ very ample sits in the middle. For $k=1$ they all coincide. For some surfaces like K3s, $k$ spanned and $k$ very ample are the same.


Elaborating on my comment:

Take $t_0=m+1$ for distinct points.

By very ampleness, for any point $x_i$ we can find a section of $\mathcal L$ that vanishes on $x_i$ but not on any of the other points - just take a hyperplane through that point that doesn't pass through any of the other points.

By taking the product of $m$ such sections, we obtain a section of $\mathcal L^m$ that vanishes at all but one of the $x_i$, and with the nonzero $x_i$ of our choice. By scaling the section, it is equal to $1$ at $x_i$ and $0$ for all other $x_j$. By taking linear combinations of these sections, we obtain surjectivity.

This is generalized to the non-distinct case by mkemeny.


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