| 估计一个general type的复投影代数簇的体积的下界是一个棘手的open problem，这个问题初看起来有点像spectral gap，或者估计第一个特征值的大小类似的问题，但是其实棘手的多，已有的结果表明 $1 / n^{n \log n}=1 / e^{n(\log n)^{2}}$ 是okey的，但是Tao的这篇最新文章表明，实际上下界可以比这小得多，而且从他们的手法上来看似乎还有相当大的优化空间。因为他们是限制在一族特定的投影代数簇里面，并且结果依赖于奇点消解。 |
The volume $\operatorname{vol}(X)$ of an $n$ -dimensional complex projective variety $X$ measures the asymptotic growth of the plurigenera, and more precisely is defined by the formula
$$\operatorname{vol}(X):=\limsup {m \rightarrow \infty} h^{0}\left(X, m K{X}\right) /\left(m^{n} / n !\right)$$
where $K_{X}$ is the canonical bundle. If $K_{X}$ is ample (or, more generally, nef), then $\operatorname{vol}(X)$ is also equal to the intersection number $K_{X}^{n}$, by the asymptotic Riemann-Roch theorem. However, in general the volume need not be a natural number. A variety is said to be of general type if its volume is positive.
It is a challenge to construct smooth projective varieties of general type with small volume, as they exhibit extreme behavior among all algebraic varieties. The known examples in high dimensions are:
- Ballico-Pignatelli-Tasin gave examples of $n$ -folds of general type with volume roughly $1 / n^{n}=1 / e^{n \log n}$
- Totaro and Wang obtained an improved family of examples of volume roughly $1 / n^{n \log n}=1 / e^{n(\log n)^{2}}$
In this paper, we go quite a bit further, giving examples of $n$ -folds of volume roughly $1 / e^{n^{3 / 2}}$ (Theorem 0.1). The examples will be resolutions of singularities of general hypersurfaces of degree $d$ with canonical singularities in a weighted projective space $P\left(a_{0}, \ldots, a_{n+1}\right)$. The volume can be calculated explicitly in terms of these numbers; the problem then becomes one of optimizing this volume over all admissible parameters of a certain form. In the class of examples we consider, we optimize the constant factor in the exponent by reducing to a purely analytic problem about equidistribution on the
unit circle
We now use Theorem $2.2$ to prove Theorem $0.1$ as stated. We build on the ideas of section 1 , where we proved a weaker version of parts (i) and (ii). Part (iii) of Theorem $0.1$ was already proved there.
Part (ii) of Theorem $0.1$ follows from part (i). To prove part (i), we have to produce varieties of general type with small volume. Let $b$ be an integer at least 2. Let $p_{0}, \ldots, p_{b-1}$ be the first $b$ prime numbers, so $p_{0}=2$. For each $0 \leq i \leq b-1$, let $k_{i}$ be the largest power of $p_{i}$ that is at most $p_{b-1}$. For most aspects of our example, the numbers $k_{0}, \ldots, k_{b-1}$ behave much like the sequence $p_{0}, \ldots, p_{b-1}$.
In particular, $k_{0}, \ldots, k_{b-1}$ are pairwise relatively prime, the sum $\sum_{j=0}^{b-1} k_{j}$ is asymptotic to $\left(b^{2} / 2\right) \log b$, and $\log \left(k_{0} \cdots k_{b-1}\right)$ is asymptotic to $b \log b$. These are the same asymptotics as for $p_{0}, \ldots, p_{b-1}$ because $k_{j}$ is equal to $p_{j}$ for most $0 \leq j \leq b-1$. Indeed, only primes less than $p_{b-1}^{1 / 2}$ are affected (less than $b^{1 / 2}$ primes for large $b$ ) and these are multiplied by less than $p_{b-1} \sim b \log b$. The sum $\sum_{j=0}^{b-1} k_{j}$ is therefore increased by less than $b^{3 / 2} \log b$, and the quantity $\log \left(k_{0} \cdots k_{b-1}\right)$ by less than $b^{1 / 2} \log (b \log b)$.
Let $a=\left\lceil k_{b-1} / 2\right\rceil .$ In section 1, for $0 \leq j \leq b-1$, we defined $c_{j} \in\left[k_{j} / 2,3 k_{j} / 2\right)$ by the congruence $c_{j} k_{0} \cdots \widehat{k}{j} \cdots k{b-1} \equiv-1-a\left(\bmod k_{j}\right) .$ One would expect $\sum_{j} c_{j}$ to be
roughly $\sum_{j} k_{j} \sim\left(b^{2} / 2\right) \log b$, but a priori it could be as big as $3 / 2$ times that, which would make our example worse. We get around that problem as follows.
Let $\lambda$ be any positive constant. Let us modify our example to make $\sum_{j} c_{j}$ at most $(1+\lambda)\left(b^{2} / 2\right) \log b$, in every sufficiently large dimension $n$. Let $\mu$ be the probability measure on the real line with mass $1 / b$ on each of the $b$ numbers $c_{j} / k_{j} .$ By Theorem 2.2, there is a positive integer $s$ at most $2^{1 / \lambda}$ such that the expected value $\mathbb{E}_{\mu}(g(s x))$ is at most $\lambda$, where $g$ is the sawtooth function $g(x)=x-\lfloor x-1 / 2\rfloor$.
Therefore, if we multiply each $c_{j}$ by $s\left(\right.$ modulo $\left.k_{j}\right)$, keeping it in the interval $\left[k_{j} / 2,3 k_{j} / 2\right)$, then $\sum_{j} c_{j}$ becomes at most $(1+\lambda) \sum_{j} k_{j}$.
We remark that this method fails for the case $a=0$ used to prove part (iii) of Theorem 0.1. In this case, each $c_{j}$ belongs to $\left[k_{j}, 2 k_{j}\right)$, so we might look at the expected value of the “shifted sawtooth” $g(x-1 / 2)$. However, it does not enjoy a property analogous to Theorem $2.2$ since it does not approach 0 near $x=0$. Multiplying by some $s$, therefore, will not necessarily yield the same improvement on the bound for $\sum_{j} c_{j}$.
We can make that change to the $c_{j}$ by dividing each number $k_{j}$ (for $0 \leq j \leq b-1$ ) by the appropriate power of $p_{j}$ so that the whole product $k_{0} \cdots k_{b-1}$ is divided by $s$. (This is possible because we are considering a sufficiently large dimension $n$, so that $b$ is also as large as we like.) Let $i$ be the largest natural number such that $p_{i}$ divides $s$. (Note that $i$ is bounded above by a constant depending on $\lambda$, whereas $b$ is arbitrarily large; so $b$ is much bigger than $i$.)
For $0 \leq j \leq b-1$, define new integers $c_{j}$ in $\left[k_{j} / 2,3 k_{j} / 2\right)$ such that $c_{j} k_{0} \cdots \widehat{k_{j}} \cdots k_{b-1} \equiv$ $-1-a\left(\bmod k_{j}\right)$, for the new values of $k_{j}$. This has a simple effect for all $i0$ was arbitrary here, it is equivalent to say that in every sufficiently large dimension $n$, we can choose integers $b$ and $k_{0}, \ldots, k_{b-1}$ as above, determining $a=\left\lceil k_{b-1} / 2\right\rceil$ and $c_{0}, \ldots, c_{b-1}$ as above, such that $-2+a+\sum c_{j} \leq n, \sum c_{j} \leq(1+o(1))\left(b^{2} / 2\right) \log b$,
and $n$ is close to $\sum c_{j}$ in the sense that $n \leq(1+o(1))\left(b^{2} / 2\right) \log b$.
Therefore, $b^{2} \geq(1+o(1)) 4 n / \log n$, and hence
$$
b \geq(1+o(1)) 2 n^{1 / 2}(\log n)^{-1 / 2}
$$
So
$$
b \log b \geq(1+o(1)) n^{1 / 2}(\log n)^{1 / 2}
$$
As in the proof of Theorem $0.1$, the hypersurface associated to the weights above has $-\log \operatorname{vol}(X)$ asymptotic to $n b \log b$. So
$$
-\log \operatorname{vol}(X) \geq(1+o(1)) n^{3 / 2}(\log n)^{1 / 2}
$$
In particular, for sufficiently large $n, X$ has volume less than $1 / e^{0.99 n^{3 / 2}(\log n)^{1 / 2}}$. (Here $0.99$ could be replaced by any number less than 1.) The variety $X$ has canonical singularities and ample canonical class, and so a resolution of singularities $W$ of $X$ is a smooth projective $n$ -fold of general type with the same volume. Theorem $0.1$ is proved.
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