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This question is about how varieties of general type over an algebraically closed field of characteristic zero $k$ behave under generization in families.

Definition. An integral projective scheme X over k is of general type if some desingularization of X is of general type. A reduced projective scheme X over k is of general type if every (reduced) irreducible component of $X$ is of general type. Finally, a projective scheme X is of general type if X_{red} is of general type.

My question is as follows.

Let S be an integral normal noetherian scheme of characteristic zero with function field $K=K(S)$. Let $X\to S$ be a projective flat morphism. Suppose that there is a closed point $s$ in $S$ with residue field $k(s) = k$ such that $X_s$ has an irreducible component which is of general type over $k$. Is the generic fibre $X_K$ of $X\to S$ of general type over $K$?

Please note that I do not make any assumptions on the singularities of $X_s$, nor do I assume $X_s$ to be irreducible.

To answer this question, we may assume that $S$ is the spectrum of $\mathbb{C}[[t]]$.

A partial (positive) answer follows from deep theorems of Siu, Kawamata, and Nakayama on the constancy of plurigenera. But, as far as I know, these theorems require some conditions on the singularities of $X_s$ (e.g., canonical singularities). Can one reduce to this situation using semi-stable reduction, maybe?

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  • $\begingroup$ The general fiber is non-uniruled. The abundance conjecture then predicts that the Kodaira dimension is nonnegative. $\endgroup$ Jan 21, 2019 at 12:33
  • $\begingroup$ I agree that the generic fibre is not uniruled. Assume it is uniruled. Take a general point on the special fiber $X_s$. Let $x\in X$ be a point specializing to this point. First, since $X_K$ is uniruled, there is a rational curve going through $x_K$. Second, this rational curve specializes to a rational curve through $x_s$. $\endgroup$ Jan 21, 2019 at 13:25
  • $\begingroup$ One method to solve similar questions is using specialization map for the Grothendieck ring of varieties, as in arxiv.org/abs/1708.02790 and arxiv.org/abs/1708.05699v1. $\endgroup$ Jan 23, 2019 at 19:19
  • $\begingroup$ By the way, what about more general statement: if X has Kodaira dimension m, then every component of a degeneration of X has Kodaira dimension bounded by m? Is this false? $\endgroup$ Jan 23, 2019 at 19:24
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    $\begingroup$ It would be interesting to know if there's an unconditional proof of the statement about Kodaira dimension. @EvgenyShinder do you have an idea how one can use your motivic specialisation maps to pass information about plurigenera instead of (stable) birational type? $\endgroup$
    – Frank
    Jan 29, 2019 at 12:30

1 Answer 1

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The answer is yes to the original question and is a theorem of Noboru Nakayama in his book "Zariski decomposition and abundance" Theorem VI.4.3, which I state here for convenience:

Theorem (Nakayama): Let $\mathcal{X}\to S$ a projective surjective morphism with connected fibres from a normal complex analytic variety onto a smooth curve and $0\in S$. Let $\mathcal{X}_0=\cup_{i\in I}\Gamma_i$ the decomposition into irreducible components. If there is at least one irreducible component $\Gamma_j$ which is of general type (i.e. its desingularisation is such), then for any $m>0$ and for $s\in S$ general we have $$p_m(\mathcal{X}_s) \geq \sum_{i\in I} p_m(\Gamma_i)$$ where $p_m(Y):=h^0(\tilde{Y}, mK_{\tilde{Y}})$ are the plurigenera of a resolution.

To address the question (and one should be able to reduce to the following in general), let's assume that the generic fibre is smooth and irreducible (or say with canonical singularities so that the plurigenera of the resolution don't change) and that at least one of the general type components $\Gamma_j$ of the special fibre has the same dimension as the generic fibre. Then the sections of multiples of the canonical divisor $K_{\mathcal{X}_s}$ grow so that $\mathcal{X}_s$ is also of general type.

To motivate why something like this is true in a smooth family, note that even though dimension of cohomology is upper semicontinuous (which works against us in this question), the volume of the canonical divisor (even of a resolution more generally) is lower semicontinuous (see Kollár's new book Theorem 5.10 and Nakayama, 1985), implying that if $\rm{vol}(K_{\mathcal{X}_0})>0$ then $\rm{vol}(K_{\mathcal{X}_s})>0$ giving that $\mathcal{X}_s$ is of general type - think that if $D$ is relatively ample on $\mathcal{X}$, then $\rm{vol}(D_s)=D^n_s$ is constant in the family as it's an euler characteristic.

As for the question in the comments of whether Kodaira dimension in general is lower semicontinuous, this is conjectured (see Nakayama's book again just before Conjecture VI.1.2) and is proven assuming the abundance conjecture: (see Nakayama VI.4.1) in the notation of the theorem, assuming abundance for the generic fibre $\mathcal{X}_s$, then $\kappa(\mathcal{X}_s)\geq \max\kappa(\Gamma_i)$.

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    $\begingroup$ That is fantastic! $\endgroup$ Jan 24, 2019 at 15:28
  • $\begingroup$ Thank you Frank! $\endgroup$ Jan 29, 2019 at 18:03

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