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Let $X$ be a normal irreducible closed subvariety of $\mathbb{P}^n$ and let $\Lambda$ be a linear system on $X$, without fixed component. Then, there exists a linear system $\Lambda'$ on $\mathbb{P}^n$ such that the restriction of this linear system to $X$ yields $\Lambda$, plus some fixed components.

Is it true that the degree of $\Lambda'$ is bounded if the degree of $\Lambda$ is bounded? (the degree is here the degree in $\mathbb{P}^n$).

Comment: Denoting by $D\in \mathrm{Pic}(X)$ the divisor of $\Lambda$ and by $H\in \mathrm{Pic}(X)$ the very ample divisor given by the restriction of an hyperplane, we are looking for an integer $d$ such that $dH-D$ is effective. The question corresponds to ask if $d$ is bounded if the degree of $D$ is.

EDIT: The bound asked here can depend on $X$.

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  • $\begingroup$ Can the bound depend on $X$? $\endgroup$ Oct 23, 2013 at 15:25
  • $\begingroup$ yes, I added a comment. $\endgroup$ Oct 23, 2013 at 15:57

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I think that with some mild(?) additional hypothesis this holds. I also think that with some work one can figure out whether or how much of that additional condition is necessary.

So, let $r=\dim X$ and let's assume that

  1. There is a $D\in \Lambda$ that is normal. (By the way, you probably meant to use this $D$ instead of $\Lambda$ in $dH-\Lambda$).
  2. $K_D\cdot H^{r-1}$ is bounded by the degree of $D$. (Alternatively you may say that the bound on $d$ depends on the degree of $D$ and this value.)

Now, a theorem of Kollár and Matsusaka (in Riemann-Roch type inequalities) prove that for every $m$ there exists an absolute polynomial of degree at most $m$ with coefficients that depend on the degree of the variety and the number that appears in 2) above such that the dimension of the global sections of a semi-ample Cartier divisor on a normal projective variety differ from the leading term of the Hilbert polynomial by that polynomial.

EDIT: Apparently Matsusaka in a later paper (On numerically effective divisors with positive self-intersection numbers. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 38 (1991), no. 1, 165–183) improved this: semi-ample is not necessary, it is enough if the divisor is nef and big.

If you apply this for $dH$ on $X$ and $dH|_D$ on $D$ you get that $d$ has to be large enough to make the difference of the two estimates positive. These estimates only depend on the parameters listed so far, hence so will this $d$. And of course what you are trying to make effective is the kernel of the restriction map, so this is exactly the value you are looking for.


Thoughts on removing the extra conditions

  1. The assumption that $D$ be normal is probably not necessary for you. It is kind of needed for the above description of that number, but it is really just the second coefficient of the Hilbert polynomial. That is probably enough for you. See the lemma immediately following the main theorem in the Kollár-Matsusaka paper. If you want to remove that assumption you might have to go through their proof and see that it does not really use normality.
  2. The other thing is that you want to bound the second coefficient of the Hilbert polynomial with the first one. According to Eisenbud in Commutative Algebra with a view... this is actually done by Kleiman (1971). (I don't have access to the latter at the moment). I suspect that this dependence is not polynomial, because then that lemma in Kollár-Matsusaka's paper probably wouldn't have that comment next to it. Anyway, if you find Kleiman's result you can possibly replace that with the Kollár-Matsusaka result in this argument, although the fact that they bound the difference by a polynomial of smaller degree means that we know that the larger guy will win eventually, so I am not entirely sure that that works without seeing what exactly Kleiman says.
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  • $\begingroup$ Thanks for the nice answer. I need to see, as you said, that the second coeff is also bounded. I will try to find Kleiman's text. By the way, is the condition semi-ample important? Is mobile sufficient? And also, the result of Kollár-Matsusaka is in characteristic 0, what happens in char. p? $\endgroup$ Oct 25, 2013 at 8:25
  • $\begingroup$ For semi-ampleness, perhaps you can blow up the base locus and take the mobile part. (What do you mean by "mobile"? I would expect basepoint-free, but that is better than semi-ample.) I don't know if this holds in characteristic $p$. You can look at this paper on MathSciNet and look at those papers that cited it. Any related char $p$ result will likely have cited this paper. $\endgroup$ Oct 25, 2013 at 9:32

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