PomeTheorem 19.8.5.1DFDCCEF325D If $\deg _\gamma L<0$, then $H^0(X,L^ m)=0$ for every $m\geq 1$. If $\deg _\gamma L=0$ and $H^0(X,L^ m)\neq 0$, then $L^ m$ is holomorphically trivial and $h^0(X,L^ m)=1$. At degree zero, each nonzero weighted subspace $\mathcal H_ m(u)$ is also one-dimensional.
19.8.4 Gauduchon degree and vanishingE5AAFD4B6553
The zero divisor of a holomorphic section is effective, and the Poincaré–Lelong formula expresses its integration current as the curvature form of the line bundle plus $\mathrm{dd}^{\mathrm{c}}$ of a global function. Pairing with a Gauduchon metric removes the $\mathrm{dd}^{\mathrm{c}}$ term, so the size of the divisor is bounded by a degree of the line bundle. Recall that a Hermitian form $\gamma $ is Gauduchon if $\mathrm{dd}^{\mathrm{c}}(\gamma ^{n-1})=0$ (definition 11.4.1); every Hermitian form on a curve is Gauduchon. Write $h^0(X,L^ m)=\dim H^0(X,L^ m)$.
For a Gauduchon metric $\gamma $ on $X$, define the degree
\[ \deg _\gamma L=\int _ X\beta \wedge \gamma ^{n-1}. \]
A change of the smooth metric on $L$ changes $\beta $ by $\mathrm{dd}^{\mathrm{c}}a$ for a smooth real function $a$, and the integral of $\mathrm{dd}^{\mathrm{c}}a$ against $\gamma ^{n-1}$ vanishes by Stokes' theorem. Hence the degree does not depend on the smooth metric on $L$.
Proof. Let $s$ be a nonzero section of $L^ m$ and let $D$ be its zero divisor. The function $\log |s|_{h^ m}^2$ is locally integrable, and the Poincaré–Lelong formula gives
\[ \mathrm{dd}^{\mathrm{c}}\log |s|_{h^ m}^2=[D]-m\beta . \]
Since $\gamma $ is Gauduchon, integration by parts against $\gamma ^{n-1}$ annihilates the left-hand side, and therefore
\[ m\deg _\gamma L=\int _ D\gamma ^{n-1}\geq 0, \]
where the components of $D$ are counted with multiplicities. If $D\neq 0$, the right-hand side is strictly positive, because $\gamma $ is positive on the regular locus of each component. Hence negative degree is impossible. At degree zero we obtain $D=0$, so $s$ has no zeros and trivializes $L^ m$. Every other section of $L^ m$ is then a holomorphic function times $s$, and holomorphic functions on the connected compact manifold $X$ are constant. Thus $h^0(X,L^ m)=1$, and every nonzero subspace $\mathcal H_ m(u)\subseteq H^0(X,L^ m)$ is one-dimensional. $\square$
The degree constrains positive curvature currents in the same way. If $T=\beta +\mathrm{dd}^{\mathrm{c}}u\geq 0$, the trace identity of proposition 11.5.1 gives
\[ \deg _\gamma L=\int _ X T\wedge \gamma ^{n-1}\geq 0. \]
If the degree is zero, the positivity of $T$ and the strict positivity of $\gamma $ imply $T=0$.
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