Hello, let L be a holomorphic line bundle over a compact complex manifold of dimension 2. Suppose $\int_{X}c_{1}(L)^{2} > 0$ ($c_{1}$ means first Chern class). I would like to show $L^{\otimes m}$ or $L^{\otimes -m}$ admits a non vanishing holomorphic section for $m$ large enough.

I tried the following argument which failled. Using Hirzebruch Riemann Roch theorem, I get $h^{0}(X, L^{\otimes m}) = \frac{m^{2}}{2}\int_{X}c_{1}(L)^{2} + \frac{m}{2}\int_{X}c_{1}(X) c_{1}(L) + \chi(X, \mathcal{O}_{X}) + h^{1}(X, L^{\otimes m}) - h^{2}(X, L^{\otimes m})$. If it was $0$ for all $m$, then, for $m$ large enough I would have $h^{1}(X, L^{\otimes m}) < h^{2}(X, L^{\otimes m}) = h^{0}(X, K_{X} \otimes L^{\otimes -m})$ by Serre duality. The only thing I can deduce from this (applies to $-m$) is a non trivial holomorphic section of $K_{X} \otimes K_{X}$.

If $L$ was positive, Nakano Serre vanishing theorem would imply $h^{1}(X, L^{\otimes m}) - h^{2}(X, L^{\otimes m}) = 0$ and then the results would be obvious.

Any ideas?

I wish you a good day.

PS : Here $h^{k}(X, L)$ denotes the complex dimension of $H^{k}(X, L)$.

  • 1
    $\begingroup$ Does the hypothesis $\int_{X} c_{1}(X)^{2} > 0$ implies the bundle $L$ is positive? $\endgroup$
    – Analyse300
    Oct 29, 2021 at 15:47
  • 2
    $\begingroup$ Your RR shows, ignoring the first cohomology, for large $m$, $h^0+h^2>0$ and thus at least one of them is positive. Can you analyze this? Also, $c_1^2>0$ does not imply positivity. $\endgroup$
    – Mohan
    Oct 30, 2021 at 20:29
  • $\begingroup$ Thanks for your anwer. If it's $h^{0}$ I conclude. Else, it's $h^{2}$ and by Serre duality I have $0 \neq h^{2}(X, L^{\otimes m} ) = h^{0}(X, K_{X} \otimes L^{\otimes -m})$. So I obtain a non trivial holomorphic section of $K_{X} \otimes L^{\otimes -m}$. I don't see how to deduce from this a section of $L^{\otimes -m}$. $\endgroup$
    – Analyse300
    Oct 31, 2021 at 16:46
  • $\begingroup$ If $K_{X}^{*}$ would admit a non trivial section I could easily conclude but I don't know if it's true. $\endgroup$
    – Analyse300
    Oct 31, 2021 at 16:53
  • 1
    $\begingroup$ You are not using the full strength of your hypothesis. Try using that if $h^0=0$, then $h^2$ grows at least like $cm^2$, for a positive constant $c$. $\endgroup$
    – Mohan
    Nov 9, 2021 at 21:30

1 Answer 1


With your help we search the problem with a friend. The book I will mention is Huybrecht complex analysis an introduction. Here is what HE done :

The Hirzebruch-Riemann-Roch formula gives $h^{0}(X, L^{\otimes m}) =\frac{1}{2}\int_{X}c_{1}(L^{\otimes m})^{2} + \frac{1}{2}\int_{X}c_{1}(X) c_{1}(L^{\otimes m}) + \chi(X, \mathcal{O}_{X}) + h^{1}(X, L^{\otimes m}) - h^{2}(X, L^{\otimes m}) = \frac{m^{2}}{2}\int_{X}c_{1}(L)^{2} + \frac{m}{2}\int_{X}c_{1}(X) c_{1}(L) + \chi(X, \mathcal{O}_{X}) + h^{1}(X, L^{\otimes m}) - h^{2}(X, L^{\otimes m})$.

If $h^{0}(X, L^{\otimes m}) = 0$ for all $m$ different of $0$ then Serre duality implies $h^{0}(X, K_{X} \otimes L^{\otimes -m}) = \frac{m^{2}}{2}\int_{X}c_{1}(L)^{2} + \frac{m}{2}\int_{X}c_{1}(X) c_{1}(L) + \chi(X, \mathcal{O}_{X}) + h^{1}(X, L^{\otimes m})$. By remarks $2.3.171)$ page $82$, the twoo divisors $Z(m)$ and $Z(m')$ associated to not trivial holomorphic section of the twoo respectives bundles $K_{X} \otimes L^{\otimes m}$ and $K_{X} \otimes L^{\otimes m'}$ are effectivs and correspond via proposition $2.3.18 i)$ page $83$ at those bundles (via the map $Div(X) \mapsto Pic(X)$. Let $H$ be the line bundle corresponding to the divisor $Z := Z(m') - Z(m)$. We have $K_{X} \otimes L^{\otimes m} \otimes H$ isomorphic to $K_{X} \otimes L^{\otimes m'}$ so that $H$ is isomorphic to $L^{\otimes (m'-m)}$. For $k$ large enough $kZ$ is effective. So, $L^{\otimes k(m'-m)}$ admits a non trivial holomorphic section by Proposition $2.3.18$ $ii)$ page $83$. If $Z$ the result is clear. If not, $Z$ has a negative part $\sum_{j} n_{j} [Y_{j}]$ with $n_{j} \leq 0$. For $k$ large enough, $kZ + Z(m)$ has also a negative part. The line bundle corresponding to this divisor is $K_{X} \otimes L^{\otimes m} \otimes L^{\otimes k(m'-m)} = K_{X} \otimes L^{\otimes [m + k(m-m')]}$ which admits a non trivial section. But $Z( m + k(m-m'))$ and $kZ + Z(m)$ defines the same bundle so their differnces is a principal divisor (by Corollary $2.3.19$ page $83$) so define by a meromorphic section $f$. So, $Z(m) + kZ + (f) = Z(m + k(m-m'))$. Morover by the same corollary, $Z(m + k(m-m')) = Z(m) + Z(k(m-m')) + (g)$ for $g$ meromorphe. In particular $kZ + (f-g) = Z(k(m-m'))$ is effective so that $kZ + (f-g)$ has no negative part. So as $kZ + (f-g)$ corresponds to $L^{\otimes k(m'-m)} \otimes \mathcal{O} = L^{\otimes k(m'-m)}$, this bundle admits a non trivial holomorphic section (by Lemma $2.3.14$ page $81$ and by Proposition $2.3.18$ $ii)$ page $83$).

One more time this is not my argument, it's the argument of my friend.

Edit : $m$ and $m'$ are different.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .