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We say that a set of varieties $S$ lives in a bounded family if there exists a projective morphism $\mathcal{X} \to T$ between varieties of finite type, such that for any $X \in S$, there exists a closed point $t \in T$, such that its fibre $\mathcal{X}_t$ is isomorphic to $X$.

It seems that the following two facts related to bounded families used quite often without proof in the literature of birational geometry:

(1) If $S$ is a bounded family of $\mathbb{Q}$-Gorenstein varieties (we can assume the elements in $S$ are normal, have log terminal singularities), then there exists a universal $m$, such that for any $X \in S$, $mK_X$ is a Cartier divisor, i.e. the Gorenstein index is bounded.

(2) Granted (1) is true, then the volume (suppose $n=\dim X$) $${\rm vol}(mK_X): = \lim_{k \to +\infty}\frac{n! h^0(X, kmK_X)}{k^n}$$ is bounded for any $X \in S$.

If I know the universal family $\mathcal{X}$ is $\mathbb Q$-Gorenstein, then the above two results follows easily. But I don't know if we can assume this because such universal families typically come from Hilbert schemes (I don't know if Hilbert schemes are $\mathbb Q$-Gorenstein or not). The closest result related (1) which I can find is Theorem B.1 in the paper Log canonical thresholds on varieties with bounded singularities. But it requires more than what I have: the fibres are required to be normal.

However, I do think one needs something extra which natural comes from the construction of bounded family in order to make (1) and (2) holds. Any suggestion?

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    $\begingroup$ Assuming that all S you consider is normal and with rational singularities, the you can use Theorem B.1 in the paper "Log canonical thresholds on varieties with bounded singularities". Because by normalizing the family $\mathcal{X}$, you can get a new family with all fibers normal. $\endgroup$
    – Chen Jiang
    Commented Nov 12, 2016 at 11:51
  • $\begingroup$ Sorry, but I don't understand. I can assume elements in $S$ is normal and have rational singularities. Since I can always resolve $\mathcal X$ and shrink $T$ such that the generic fibre is smooth. But don't have the property that for any $X \in S$, $X$ is isomorphic to a fibre of this family. So how to I apply Thm B.1? $\endgroup$
    – Li Yutong
    Commented Nov 13, 2016 at 3:07
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    $\begingroup$ Can we assume something about the singularities of $\mathcal X_t$? eg. canonical? log terminal? Can I at least assume that the $\mathcal X_t$ are normal? BTW there is no harm in assuming that $T$ is normal, so if we assume $\mathcal X_t$ is normal, then after shrinking $T$ (which is ok by Noetherian induction) we can assume that $\mathcal X$ is normal (EGA IV, cor. 5.12.7). If moreover $\mathcal X_t$ has rational sings. Then we apply Thm b.1 as Chen says. Are you really interested in the case that the fibers are not normal? $\endgroup$
    – Hacon
    Commented Nov 14, 2016 at 17:01
  • $\begingroup$ Yes, we can assume that over a dense set of the base $T$, $\mathcal{X}_t$ is normal and log terminal. But do you think under these assumptions, we can shrink $T$ further, such that $\mathcal{X}_T$ becomes normal? I only know that over a dense set of $T$, the fibre is normal. It seems that in order to apply EGA IV, cor. 5.12.7, one needs to know that for an open set, the fibre is normal... $\endgroup$
    – Li Yutong
    Commented Nov 15, 2016 at 2:18
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    $\begingroup$ I',m confused. It seems to me that EGA IV, cor. 5.12.7 says that if $(A,m)$ is local, $A/tA$ is integrally closed then so is $A$. Since $f$ is projective, doesn't it follow that if a fiber $\mathcal X_t$ is normal, then there is an open subset $T^0\subset T$ such that $\mathcal X\times _T T^0$ is normal? The proof is sort of obvious: since $\mathcal X_t$ is $R_1$ then so is $\mathcal X$ (near $\mathcal X _t$), since $\mathcal X _t$ is $S_2$, then so is $\mathcal X $ (near $\mathcal X _t$). By serre's Criterion, $\mathcal X $ is normal (near $\mathcal X _t$). What did I miss? $\endgroup$
    – Hacon
    Commented Nov 15, 2016 at 3:51

1 Answer 1

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By Noetherian induction, it suffices to show that indicies and volumes are bounded over an open subset of any irreducible component of $T$. We may assume that $T$ is smooth and there is a dense set $\{t_i\}\subset T$ such that the corresponding fibers $\mathcal X_{t_i}$ are normal. By EGA IV Theorem 12.2.4(iv), after shrinking $T$, we may assume that $\mathcal X$ is normal. Since $\mathcal X_{t_i}$ is log terminal, it is $\mathbb Q$-Gorenstein (by definition) and has rational singularities (see eg. ́Koll\'ar-Mori). By Theorem B.1 in the paper Log canonical thresholds on varieties with bounded singularities https://arxiv.org/pdf/1004.3336.pdf, $X$ is $\mathbb Q$-Gorenstein. Thus (1) holds. Let $\nu:\mathcal X'\to \mathcal X$ be a log resolution and write $K_{\mathcal X'}+B'=\nu ^* K_{\mathcal X}+E$ where $B'$ and $E$ are effective with no common components. Shrinking $T$ we may assume that $(\mathcal X',B')$ is log smooth over $T$ (so that every stratum of the support of $B'$ is smooth over $T$) and that $\mathcal X '_t\to \mathcal X _t$ is a log resolution for every $t\in T$. Note that since $\lfloor B'_{t_i}\rfloor =0$ then also $\lfloor B'\rfloor =0$. By Theorem 4.2 in https://arxiv.org/pdf/1412.1186.pdf, $h^0(mK_{\mathcal X _t})=h^0(m(K_{\mathcal X_{t}}+B_t))$ is independent of $t\in T$. In particular ${\rm vol}(K_{\mathcal X _t})$ is independent of $t\in T$. Thus (2) also holds.

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    $\begingroup$ Thank you for your terrific answer!! May be the "$X$" in the fifth and sixth lines should be $\mathcal X$? $\endgroup$
    – Li Yutong
    Commented Nov 16, 2016 at 4:23

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