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Jul 18, 2021 at 20:45 comment added Will Sawin @GDR It is not true that $h^0(X, mD)$ is necessarily a polynomial function. One can, at least sometimes, prove it is polynomial for $m$ sufficiently large, but this is harder than this argument, and not sufficient anyways.
Jul 18, 2021 at 20:25 comment added Puzzled Thank you very much to both of you for the answer. Could one argue also by saying that $f(m) = h^0(X,mD)$ is a polynomial function? So if $f(m) = c $ for $m\gg 0$ then $f(m) = c$ for all $m\geq 0$, and hence $f(m) = 1$ for all $m\geq 0$. Is this correct?
Jul 18, 2021 at 19:31 history answered Will Sawin CC BY-SA 4.0