[proofplan]
The proof is the dual form of Kodaira vanishing. Serre duality identifies the dual of $H^q(X,M)$ with the cohomology group $H^{n-q}(X,K_X \otimes M^*)$. Since $M$ is negative, its dual $M^*$ is positive, and since $q<n$, the degree $n-q$ is positive. Kodaira vanishing therefore annihilates the Serre-dual group, and finite-dimensional duality forces the original cohomology group to vanish.
[/proofplan]
[step:Pass from $H^q(X,M)$ to its Serre-dual cohomology group]
Fix an integer $q$ with $0 \leq q < n$. Let $K_X := \Lambda^n T_X^*$ denote the canonical holomorphic line bundle of $X$, and let $M^* \to X$ denote the dual holomorphic line bundle of $M$.
Since $X$ is compact and complex, the cohomology group $H^q(X,M)$ is finite-dimensional over $\mathbb{C}$. By Serre duality (citing a result not yet in the wiki: Serre duality for compact complex manifolds), applied to the holomorphic line bundle $M \to X$, there is a natural isomorphism of complex vector spaces
\begin{align*}
H^q(X,M)^* \cong H^{n-q}(X,K_X \otimes M^*).
\end{align*}
[guided]
Fix an integer $q$ satisfying $0 \leq q < n$. The goal is to prove that the complex [vector space](/page/Vector%20Space) $H^q(X,M)$ is zero. The standard way to use positivity here is to pass to the dual bundle, because Kodaira vanishing applies to line bundles of the form $K_X \otimes L$ with $L$ positive.
Let $K_X := \Lambda^n T_X^*$ be the canonical holomorphic line bundle of $X$, and let $M^* \to X$ be the holomorphic dual line bundle of $M$. Since $X$ is compact, the sheaf cohomology group $H^q(X,M)$ is finite-dimensional over $\mathbb{C}$. Serre duality applies to compact complex manifolds and holomorphic vector bundles; in the present line bundle case, applied to $M \to X$, it gives a natural vector-space isomorphism
\begin{align*}
H^q(X,M)^* \cong H^{n-q}(X,K_X \otimes M^*).
\end{align*}
This replaces the desired vanishing of $H^q(X,M)$ by a vanishing statement for a cohomology group involving the positive bundle $M^*$.
[/guided]
[/step]
[step:Apply Kodaira vanishing to $K_X \otimes M^*$]
Because $M \to X$ is negative, its dual line bundle $M^* \to X$ is positive. Since $q<n$, the integer $n-q$ satisfies $n-q>0$. Therefore Kodaira vanishing (citing a result not yet in the wiki: Kodaira vanishing theorem) applies to the positive holomorphic line bundle $M^* \to X$ on the compact Kähler manifold $X$, giving
\begin{align*}
H^{n-q}(X,K_X \otimes M^*) = 0.
\end{align*}
[guided]
The hypothesis that $M$ is negative is used exactly here. By definition of negativity for holomorphic line bundles, $M$ is negative if and only if its dual bundle $M^*$ is positive. Thus $M^* \to X$ is a positive holomorphic line bundle.
Kodaira vanishing applies on compact Kähler manifolds to bundles of the form $K_X \otimes L$, where $L \to X$ is a positive holomorphic line bundle, and it gives
\begin{align*}
H^r(X,K_X \otimes L)=0
\end{align*}
for every integer $r>0$. In the present situation, take $L := M^*$ and $r := n-q$. The positivity hypothesis is satisfied because $M^*$ is positive, and the degree hypothesis is satisfied because $q<n$, hence $n-q>0$. Therefore Kodaira vanishing gives
\begin{align*}
H^{n-q}(X,K_X \otimes M^*) = 0.
\end{align*}
[/guided]
[/step]
[step:Conclude that the original cohomology group vanishes]
Combining Serre duality with the Kodaira vanishing result gives
\begin{align*}
H^q(X,M)^* \cong 0.
\end{align*}
Hence $H^q(X,M)^*=0$. Since $H^q(X,M)$ is a finite-dimensional complex vector space, its dual is zero only when the space itself is zero. Therefore
\begin{align*}
H^q(X,M)=0.
\end{align*}
Because $q$ was arbitrary with $0 \leq q<n$, the theorem follows.
[/step]
