# Do finitely many plurigenera determine the Kodaira dimension?

Let $X$ be a smooth projective variety over a field of characteristic $0$ and let $K_X$ be the canonical bundle. Recall that the Kodaira dimension $\kappa(X)$ is defined as the number $\kappa$ such that $$\alpha m^{\kappa}\le h^0(X,mK_X) \le \beta m^{\kappa}$$ for $\alpha,\beta>0$ and $m$ sufficiently large and divisible (or $\kappa=-\infty$ if $h^0(X,mK_X)=0$ for all $m$). It is well-known that $\kappa(X)$ is a birational invariant. A natural question is how large $m$ we need to take to determine $\kappa$. More precisely:

Is there an integer $M>0$, depending only $\dim X$, such that the values $h^0(X,K_X),h^0(X,2K_X),\ldots,h^0(X,MK_X)$ determine $\kappa(X)$?

In particular, is there an $M>0$, depending on $\dim X$, such that $h^0(K_X)=\ldots=h^0(MK_X)=0$ implies that $\kappa=-\infty$?

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In the case of surfaces one has $h^0(12K_X)=0$ if and only if $X$ is ruled. I think this was proven first by Enriques. – rita Mar 2 '13 at 11:24
$h^0(N K_X)$ can be arbitrarily large with Kodaira dimension $1$. If $A$ is an abelian variety and $C$ is a curve, then the canonical ring of $A \times C$ is the canonical ring of $C$, so as the genus goes to $\infty$, the plurigenera do as well. Thus a more subtle argument is is probably required. – Will Sawin Mar 4 '13 at 0:19

I don't know the answer, but the survey Boundedness results in birational geometry by Hacon and McKernan cites some results and conjectures that might be useful.

I think your second question would have an affirmative answer if Conjecture 3.6 there is true:

There exists an integer $N = N(n,\kappa)$ such that if $X$ is a smooth variety of dimension $n$ and Kodaira dimension $\kappa \geq 0$, then $\phi_{|kNK_X|}$ is birational to the Iitaka fibration for any $k \geq 1$.

In particular we must have $h^0(NK_X) > 0$, so if you have $h^0(cK_X) = 0$ for some $c$ divisible by $N(n,0)$,...$N(n,n)$, then it must be that $\kappa(X) = -\infty$. Asking for $h^0(NK_X) > 0$ is of course weaker than asking that $\phi_{|NK_X|}$ be birational, so presumably the bounds for your question should be smaller.

The conjecture apparently known in dimension 2 and 3. For $n = 3$ and $\kappa = 0$, the bound cited is $P_{2^5 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19}(X) > 0$.

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I guess in general the answer is NO.

For instance if you take $X$ to be an $n$ dimensional variety which is $Y\times E$, where $E$ is an elliptic curve and $Y$ is an $n-1$ dimensional variety of general type with fast growing pluricanonical general, say $Y$ is a hypersurface of large degree $d$, then for any fixed $M$, if you choose $d$ sufficiently large, $h^0(X,iK_X)$ can beat any sequence of numbers $a_i$ $(0\le i\le M)$ which you wrote for a fixed $n$-dimensional general type variety.

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Nice construction. Any ideas for the $\kappa=−\infty$ case? – J.C. Ottem Mar 9 '13 at 14:47
As the above post said, the negative &-\kappa& case is implied by the effective Iitaka fibration conjecture， which I think is expected to be true but still widely open in general. – CYXU Mar 9 '13 at 14:56