5

Here is an exercise V.1.11.12.1 in J. Kollar's book Rational curves on algebraic varieties:

Let $X$ be a smooth projective variety of dimension $n$ over an algebraically closed field $k$. Let $\mathscr{L}$ be a line bundle on $X$ such that $\mathscr{L}^{n+1}\cong \omega_X^{-1}$ and $h^0(X,\mathscr{L})\geq n+1$. Then $X\cong\mathbb{P}^n_k$ and $\mathscr{L}\cong\mathscr{O}(1)$. (Hint: induction on $n$)

Actually before this exercise he proved a classical result: if $X$ is a smooth Fano variety with index $n+1$, then $X\cong\mathbb{P}^n$. So this exercise is some kind of generalization of it. It seems right but I don't know why or why not? If it is not right, are there some counter-examples?

Here is what I thought (has been edited follows Cranium Clamp's remark): We can prove it under three assumptions.

We know this is right for $n=1$. We assume that this is right for $n-1$. Now consider $n\geq 2$. Pick $D\in|\mathscr{L}|$ (I don't know why this is smooth. But I will assume ($\star$) it is), then by adjunction formula we have $$\omega_D\cong\omega_X|_D\otimes\mathscr{L}|_D\cong \mathscr{L}^n|_D.$$ Moreover, by $0\to\mathscr{O}_X\to\mathscr{L}\to\mathscr{L}|_D\to0$, we have $h^0(D,\mathscr{L}|_D)\geq n$. Hence by induction $(D,\mathscr{L}|_D)\cong(\mathbb{P}^{n-1},\mathscr{O}(1))$. Hence we have $\int_Xc_1(\mathscr{L})^n=1$. As $0\to\mathscr{O}_X\to\mathscr{L}\to\mathscr{L}|_D\to0$, we find that $\mathscr{L}$ is base-point free if $H^1(X,\mathscr{O}_X)=0$ ($\star$$\star$). Hence this induce $\phi:X\to\mathbb{P}^n$. Hence $1=\int_Xc_1(\mathscr{L})^n=\deg\phi\deg\mathrm{Im}\phi$. Hence $\phi$ is of degree $1$ which is also surjective. Hence if we assume ($\star$$\star$$\star$) $\mathscr{L}$ is ample, then $f$ is also finite. Hence $f$ is an isomorphism. Hence $X\cong\mathbb{P}^n_k$ and $\mathscr{L}\cong\mathscr{O}(1)$.

So I solved this problem with some assumptions:

Assumption ($\star$): $\mathscr{L}$ has smooth divisor.

It seems a hard condition... We can not use Bertini's theorem here. Is there some Bertini-type theorem can slove this?

Assumption ($\star$$\star$): $H^1(X,\mathscr{O}_X)=0$.

If $\mathrm{char}k=0$ and $\mathscr{L}$ is ample, then this follows from Kodaira vanishing theorem.

Assumption ($\star$$\star$$\star$): $\mathscr{L}$ is ample in this case.

As here we already proved (using the first assumption) that $\mathscr{L}$ is base-point free, then by Nakai-Moishizon theorem, to show $\mathscr{L}$ is ample we just need to show $\mathscr{L}\cdot C>0$ for any integral curves $C\subset X$. But how?

So if we let $\mathscr{L}$ is very ample at beginning, these two assumptions will hold automatically. But unfortunately we have no this beautiful assumption.

Thank you for your help!!!

WakeUp-X.Liu
  • 693
  • 4
  • 10
  • 1
    You do not need vanishing of $ H^1(X, \mathcal{O}_X) $. Since $ H^0(X,L)/H^0(X, \mathcal{O}_X) $ injects into $ H^0(D, L|_D) $, it follows that this vector space has dimension at least $ n $. In addition, if we know that $ L $ is ample, then a general section of $ L $ gives a smooth divisor by Bertini's theorem. Therefore, I believe that the hard part is to prove $ L $ ample. – Cranium Clamp Nov 25 '23 at 08:47
  • @CraniumClamp Yes you are right, I will edit this. Note that I did not actually use this condition $H^0(X,\mathscr{L})\geq n+1$. But I don't know how to use it. – WakeUp-X.Liu Nov 25 '23 at 08:56
  • We used it in deducing that $ h^0(D, L|_D) \ge n $ but that was a very easy application of that condition. The condition should be telling us, at least morally, that $ L $ has no base points - this is because $ n+1 $ general divisors have empty intersection on $ X $ as it is of dimension $ n $. – Cranium Clamp Nov 25 '23 at 09:44
  • @CraniumClamp Oh! So in this case we can not use $0\to\mathscr{O}_X\to\mathscr{L}\to\mathscr{L}|_D\to0$ and $(D,\mathscr{L}|_D)\cong(\mathbb{P}^{n-1},\mathscr{O}(1))$ to deduce that $\mathscr{L}$ is base-point free since $H^1(X,\mathscr{O}_X)$ may not vanish. This is my fault. But we can not use $h^0(X,\mathscr{L})\geq n+1$ to deduce that $\mathscr{L}$ is base-point free. – WakeUp-X.Liu Nov 25 '23 at 13:03
  • By the way, I'm not sure your application of Nakai-Moishezon is correct here. The condition in the criterion isn't just for curves, but you need to check for any positive dimensional subvariety $Y$, that $L^{dim(Y)}\cdot Y>0$. – Shrugs Nov 25 '23 at 14:43
  • Actually I'm not sure whether this exercise is right or not. There is an exercise in this book is not right (see Exercise V.3.7.10). And the condition in this exercise is much harder in the main theorem before the exercise... – WakeUp-X.Liu Nov 25 '23 at 16:11
  • @K02 Yes I know. But in that step we have showed $\mathscr{L}$ is base-point free (under some assumptions), in this case we just need to show this for curves (see Corollary 1.2.15 in Lazarsfeld's Positivity I). – WakeUp-X.Liu Nov 25 '23 at 16:20
  • @DiamondVillager Thanks, I was unaware of that corollary. The more I look at this particular exercise, the more I am uncertain of its truth. It has very weak hypotheses in comparison to the results of the section. Could it be possible that Koll'ar intended to assume Fano + characteristic 0? (Actually, can you prove it under these additional assumptions? I suspect this just uses V.3.1.11 and induction). – Shrugs Nov 25 '23 at 17:19
  • @K02 You are welcome! Yes. If we assume $X$ is Fano over characteristic $0$ field $k$, this is right directly from Theorem V.1.11(.2) in this book. This is because by bend-and-break lemma, $X$ is covered by rational curves such that $-C\cdot K_X\leq n+1$ where $n=\dim X$. As $\mathscr{L}^{n+1}\cong \omega_X^{-1}$ which is ample, then $\mathrm{index}(X)=n+1$. Hence $X\cong\mathbb{P}^n$. So we even do not need $h^0(X,\mathscr{L})\geq n+1$ now since this can be deduced from Kodaira's vanishing theorem (see the proof of this theorem). – WakeUp-X.Liu Nov 26 '23 at 03:37
  • 2
    @DiamondVillager Well then I’m quite pleased and wonder if Koll\’ar has a trick in mind. If you do not mind, I will put a bounty on this question as I would love to know how one circumvents these assumptions. – Shrugs Nov 26 '23 at 17:47

0 Answers0