[proofplan] The proof computes the Euler characteristic by Hirzebruch-Riemann-Roch and then extracts the top-degree part of the resulting characteristic-class expression. The Chern character of $L^m$ is $\exp(m c_1(L))$, so the only contribution of degree $n$ in $m$ comes from the term $c_1(L)^n/n!$. All other top-degree cohomology contributions involve lower powers of $m$. Finally, Serre vanishing for the positive line bundle $L$ kills the higher cohomology of $L^m$ for large $m$, so the Euler characteristic becomes $h^0(X,L^m)$. [/proofplan] [step:Apply Hirzebruch-Riemann-Roch to $L^m$] Let $\alpha := c_1(L)\in H^2(X,\mathbb R)$ denote the first Chern class of $L$. For each integer $m\ge 0$, the bundle $L^m\to X$ is a holomorphic line bundle, and its first Chern class is \begin{align*} c_1(L^m)=m\alpha. \end{align*} By the Hirzebruch-Riemann-Roch theorem for compact complex manifolds applied to the holomorphic line bundle $L^m$, we have \begin{align*} \chi(X,L^m)=\left\langle \left(\operatorname{ch}(L^m)\operatorname{td}(X)\right)_{2n},[X]\right\rangle. \end{align*} Here $\operatorname{ch}(L^m)$ is the Chern character of $L^m$, $\operatorname{td}(X)$ is the Todd class of the holomorphic tangent bundle $T_X$, $(\cdot)_{2n}$ denotes the degree-$2n$ component, and $[X]$ is the fundamental class of $X$. [guided] We want to turn the cohomological invariant $\chi(X,L^m)$ into an explicit expression in the integer $m$. The tool for this is Hirzebruch-Riemann-Roch. Its input is a compact complex manifold and a holomorphic vector bundle. These hypotheses are satisfied: $X$ is compact complex by assumption, and $L^m$ is a holomorphic line bundle because tensor powers of a holomorphic line bundle are holomorphic line bundles. Let \begin{align*} \alpha := c_1(L)\in H^2(X,\mathbb R) \end{align*} be the first Chern class of $L$. Since first Chern classes add under [tensor product](/page/Tensor%20Product) of line bundles, the $m$-fold tensor power satisfies \begin{align*} c_1(L^m)=m\alpha. \end{align*} Hirzebruch-Riemann-Roch for compact complex manifolds gives \begin{align*} \chi(X,L^m)=\left\langle \left(\operatorname{ch}(L^m)\operatorname{td}(X)\right)_{2n},[X]\right\rangle. \end{align*} The expression on the right is the standard characteristic-class pairing: take the degree-$2n$ part of the cohomology class $\operatorname{ch}(L^m)\operatorname{td}(X)$ and evaluate it on the fundamental class $[X]$ of the compact complex manifold $X$. This is exactly the form of Riemann-Roch needed because the dependence on $m$ appears inside $\operatorname{ch}(L^m)$. [/guided] [/step] [step:Extract the leading term from the Chern character expansion] Write the Todd class as \begin{align*} \operatorname{td}(X)=\sum_{j=0}^{n}\tau_j, \end{align*} where $\tau_j\in H^{2j}(X,\mathbb Q)$ is the degree-$2j$ component and $\tau_0=1$. Since $L^m$ is a line bundle, \begin{align*} \operatorname{ch}(L^m)=\exp(c_1(L^m))=\exp(m\alpha). \end{align*} Only terms of total cohomological degree $2n$ contribute to the integral. Therefore \begin{align*} \chi(X,L^m)=\sum_{k=0}^{n}\frac{m^k}{k!}\left\langle \alpha^k\tau_{n-k},[X]\right\rangle. \end{align*} The term with $k=n$ is \begin{align*} \frac{m^n}{n!}\left\langle \alpha^n,[X]\right\rangle. \end{align*} Every remaining term has power $m^k$ with $0\le k\le n-1$. Hence, with \begin{align*} C:=\sum_{k=0}^{n-1}\left|\frac{1}{k!}\left\langle \alpha^k\tau_{n-k},[X]\right\rangle\right|, \end{align*} we have for all integers $m\ge 1$ \begin{align*} \left|\chi(X,L^m)-\frac{m^n}{n!}\left\langle \alpha^n,[X]\right\rangle\right|\le C m^{n-1}. \end{align*} Since $\alpha=c_1(L)$, this proves \begin{align*} \chi(X,L^m)=\frac{1}{n!}\left\langle c_1(L)^n,[X]\right\rangle m^n+O(m^{n-1}). \end{align*} [guided] The purpose of this step is to identify exactly which part of the Riemann-Roch expression can produce the highest power of $m$. We decompose the Todd class into its homogeneous cohomology components: \begin{align*} \operatorname{td}(X)=\sum_{j=0}^{n}\tau_j, \end{align*} where $\tau_j\in H^{2j}(X,\mathbb Q)$ and $\tau_0=1$. Components above degree $2n$ cannot contribute when integrated over a complex $n$-fold, so this finite sum is enough. For a line bundle, the Chern character is the exponential of the first Chern class. Since $c_1(L^m)=m\alpha$, we get \begin{align*} \operatorname{ch}(L^m)=\exp(m\alpha). \end{align*} Expanding the exponential in cohomology and discarding terms of degree greater than $2n$ gives \begin{align*} \operatorname{ch}(L^m)=\sum_{k=0}^{n}\frac{m^k\alpha^k}{k!}. \end{align*} Now multiply by $\operatorname{td}(X)$. The class $\alpha^k$ has degree $2k$, and $\tau_{n-k}$ has degree $2(n-k)$. Their product has degree $2n$, so it is exactly the part that can be evaluated on $X$. Thus Hirzebruch-Riemann-Roch becomes \begin{align*} \chi(X,L^m)=\sum_{k=0}^{n}\frac{m^k}{k!}\left\langle \alpha^k\tau_{n-k},[X]\right\rangle. \end{align*} Which summand gives the leading asymptotic term? The largest exponent of $m$ occurs when $k=n$. In that case $\tau_0=1$, so the leading contribution is \begin{align*} \frac{m^n}{n!}\left\langle \alpha^n,[X]\right\rangle. \end{align*} All other summands have exponent at most $n-1$. Define \begin{align*} C:=\sum_{k=0}^{n-1}\left|\frac{1}{k!}\left\langle \alpha^k\tau_{n-k},[X]\right\rangle\right|. \end{align*} This is a finite constant depending only on $X$ and $L$. For every integer $m\ge 1$, each power $m^k$ with $0\le k\le n-1$ is bounded above by $m^{n-1}$, so \begin{align*} \left|\chi(X,L^m)-\frac{m^n}{n!}\left\langle \alpha^n,[X]\right\rangle\right|\le C m^{n-1}. \end{align*} Substituting back $\alpha=c_1(L)$ gives the asserted Euler characteristic asymptotic. [/guided] [/step] [step:Use Serre vanishing to identify the Euler characteristic with $h^0$ in high degree] Since $L$ is positive in the sense required by [[Serre Vanishing for Positive Line Bundles](/theorems/9109)][citetheorem:9109], that theorem applies to the coherent sheaf $\mathcal O_X$ and the positive line bundle $L$. Therefore there exists an integer $m_0\ge 1$ such that \begin{align*} H^q(X,L^m)=0 \end{align*} for every $q>0$ and every integer $m\ge m_0$. For such $m$, the Euler characteristic of $L^m$ is \begin{align*} \chi(X,L^m)=\sum_{q=0}^{n}(-1)^q h^q(X,L^m)=h^0(X,L^m), \end{align*} because $h^q(X,L^m)=\dim_{\mathbb C}H^q(X,L^m)=0$ for every $q>0$. Combining this equality with the Euler characteristic asymptotic already proved gives, for all sufficiently large integers $m$, \begin{align*} h^0(X,L^m)=\frac{1}{n!}\left\langle c_1(L)^n,[X]\right\rangle m^n+O(m^{n-1}). \end{align*} This is the desired high-degree asymptotic for the dimension of global sections. [/step]