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The divisor is understood to be on $X$; the printed “on $D$” is a typo. Combining asymptotic Riemann–Roch with cohomology growth for nef twists givesSince a nef divisor has nonnegative top self-intersection number, this limit is positive exactly when $D^n>0$. This is the volume of a nef divisor criterion.

Each $M_n$ is a nef divisor. If $M_n^2>0$, the volume of a nef divisor formula gives quadratic growth of $h^0(X,q{M}_n)$. Multiplication by the section of $q{F}_n$ injects this space into $H^0(X,qnH)$, so the birational linear system criterion for bigness makes the Iitaka dimension of $H$ two.

Conversely, if $\kappa (H)=2$, some $|nH|$, and hence its movable part $|M_n|$, has a two-dimensional image. If $M_n^2=0$, the previous part would make $|M_n|$ basepoint-free with image a curve; this is impossible. Since $M_n$ is nef, $M_n^2\ge 0$, hence $M_n^2>0$.

If all $M_n^2=0$ and one $|M_{n^{\prime }}|$ has positive dimension, the preceding part supplies a curve image. The just-proved equivalence excludes dimension two, so $\kappa (H)=1$. Conversely $\kappa (H)=1$ excludes every positive $M_n^2$ and requires a positive-dimensional linear system. Thus