Non trivial holomorphic section

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 $$H^{k}(X, L)$$.

• Does the hypothesis $\int_{X} c_{1}(X)^{2} > 0$ implies the bundle $L$ is positive? Oct 29, 2021 at 15:47
• 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. Oct 30, 2021 at 20:29
• 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 $L^{\otimes -m}$. Oct 31, 2021 at 16:46
• If $K_{X}^{*}$ would admit a non trivial section I could easily conclude but I don't know if it's true. Oct 31, 2021 at 16:53
• 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 $c$. Nov 9, 2021 at 21:30

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 $$83$$).
Edit : $$m$$ and $$m'$$ are different.