ScalingStacks

Proof of the diameter upper bound in Theorem 1.1. Let $\omega^{\prime}_{t}$ be the Kähler metric defined in Proposition 3.1, and let $\omega_{{\rm FS},t}=\frac{1}{|\log|t||}\omega_{\rm FS}|_{X_{t}},$ where $\omega_{\rm FS}$ is a suitable Fubini-Study metric on $\mathbb{P}^{N}$ scaled so that $\omega_{\rm FS}|_{X_{t}}\in c_{1}(\mathfrak{L})|_{X_{t}}$. Then the metrics $\omega^{\prime}_{t}$ and $\omega_{{\rm FS},t}$ are cohomologous, and on $B_{t}\subset X_{t}$ in any adapted coordinate chart we have

(4.1)

$$\omega^{\prime}_{t}+\omega_{{\rm FS},t}\geqslant C^{-1}\frac{i}{|\log|t||^{2}}\sum_{j=1}^{m}\frac{dz_{j}}{z_{j}}\wedge\frac{d\overline{z_{j}}}{\overline{z_{j}}}+C^{-1}\frac{i}{|\log|t||}\sum_{j=m+1}^{n}dz_{j}\wedge d\overline{z_{j}}.$$

Let us then fix a point $x\in E^{0}_{J}$, an adapted coordinate chart near $x$, and in these coordinates define a local Kähler metric $\tilde{\omega}_{t}$ on $X_{t}$ by the RHS of (4.1). Inside this coordinate chart intersected $X_{t}$, define also $\tilde{B}_{t}$ to be a Euclidean rectangle which is contained inside $B_{t}$ so that in the metric $\tilde{\omega}_{t}$, in the first $m$ complex directions $\tilde{B}_{t}$ has length $\sim 1$ in the radial directions and length $\sim|\log|t||^{-1}$ in the logarithmic angular directions, and in the other $n-m$ complex directions $\tilde{B}_{t}$ has length $\sim|\log|t||^{-\frac{1}{2}}$. Therefore, given any $x,y\in\tilde{B}_{t}$, if we denote by $\gamma_{x,y}$ the Euclidean straight line in $\tilde{B}_{t}$ joining them, parametrized linearly by $0\leqslant s\leqslant 1$, then $|\dot{\gamma}_{x,y}(s)|_{\tilde{\omega}_{t}}\leqslant C$ for a uniform constant $C$ independent of $t$ and $s$.

On $\tilde{B}_{t}$ we have

$$\frac{C^{-1}}{|\log|t||^{n}}\mu_{t}\leqslant\tilde{\omega}_{t}^{n}\leqslant\frac{C}{|\log|t||^{n}}\mu_{t},$$

and again by direct computation in polar coordinates (as in [3]), thanks to the definition of $\tilde{B}_{t}$ and to (3.1) we have that

(4.2)

$$C^{-1}\leqslant\int_{\tilde{B}_{t}}d\mu_{t}\leqslant 1,$$

and so

$$\frac{C^{-1}}{|\log|t||^{n}}\leqslant\int_{\tilde{B}_{t}}\tilde{\omega}_{t}^{n}\leqslant\frac{C}{|\log|t||^{n}}.$$

If we then define

$$\tilde{\mu}_{t}=\frac{\tilde{\omega}^{n}_{t}}{\int_{\tilde{B}_{t}}\tilde{\omega}_{t}^{n}},$$

then $\tilde{\mu}_{t}$ is uniformly comparable to $\mu_{t}$ on $\tilde{B}_{t}$ and

(4.3)

$$\mu_{t}(\tilde{B}_{t})\geqslant C^{-1}\mu_{t}(B_{t})\geqslant C^{-1}.$$

Thanks to (4.1) we have

(4.4)

$$\begin{split}\int_{\tilde{B}_{t}}\Tr_{\tilde{\omega}_{t}}\omega_{t}d\tilde{\mu}_{t}&=\frac{n\int_{\tilde{B}_{t}}\omega_{t}\wedge\tilde{\omega}_{t}^{n-1}}{\int_{X_{t}}\tilde{\omega}^{n}_{t}}\leqslant C|\log|t||^{n}\int_{X_{t}}\omega_{t}\wedge(\omega^{\prime}_{t}+\omega_{{\rm FS},t})^{n-1}\\ &\leqslant C\int_{X_{t}}c_{1}(L)\cdot c_{1}(\mathfrak{L})^{n-1}\leqslant C.\end{split}$$

We wish to use this to prove that

(4.5)

$$\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\text{dist}_{\omega_{t}}(x,y)d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)\leqslant C.$$

Indeed, since $|\dot{\gamma}_{x,y}(s)|_{\tilde{\omega}_{t}}\leqslant C$, we can estimate

$$\mathrm{dist}_{\omega_{t}}(x,y)\leqslant C\int_{0}^{1}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}ds,$$

and

$$\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\text{dist}_{\omega_{t}}(x,y)d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)\leqslant C\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\int_{0}^{1}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}ds\,d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y).$$

We can then argue as in [5, Lemma 1.3], using Fubini

$$\begin{split}&\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\int_{0}^{1}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}ds\,d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)\\ &=\int_{0}^{1}\int_{\tilde{B}_{t}\times\tilde{B}_{t}}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)ds\\ &=\int_{0}^{\frac{1}{2}}\int_{\tilde{B}_{t}\times\tilde{B}_{t}}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}d\tilde{\mu}_{t}(x)d\tilde{\mu}_{t}(y)ds\\ &+\int_{\frac{1}{2}}^{1}\int_{\tilde{B}_{t}\times\tilde{B}_{t}}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(sy+(1-s)x))^{\frac{1}{2}}d\tilde{\mu}_{t}(y)d\tilde{\mu}_{t}(x)ds\end{split}$$

and in the two innermost integrals we change variable from $x$ (resp. $y$) to $z=sy+(1-s)x$ with $0\leqslant s\leqslant\frac{1}{2}$ (resp. $\frac{1}{2}\leqslant s\leqslant 1$), noting that $\tilde{\mu}_{t}(x)\leqslant C\tilde{\mu}_{t}(z)$ (resp. $\tilde{\mu}_{t}(y)\leqslant C\tilde{\mu}_{t}(z)$). Thus both of these innermost integrals can be bounded above by

$$C\int_{\tilde{B}_{t}}(\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(z))^{\frac{1}{2}}d\tilde{\mu}_{t}(z)\leqslant C\left(\int_{\tilde{B}_{t}}\Tr_{\tilde{\omega}_{t}}{\omega_{t}}(z)d\tilde{\mu}_{t}(z)\right)^{\frac{1}{2}}\leqslant C,$$

by (4.4), and (4.5) follows.

But (4.5) is equivalent to

$$\int_{\tilde{B}_{t}\times\tilde{B}_{t}}\text{dist}_{\omega_{t}}(x,x^{\prime})d\mu_{t}(x)d\mu_{t}(x^{\prime})\leqslant C,$$

hence for some $x^{\prime}\in\tilde{B}_{t}$ we have

$$\int_{\tilde{B}_{t}}\text{dist}_{\omega_{t}}(x,x^{\prime})d\mu_{t}(x)\leqslant C,$$

which gives a weak $L^{1}$-estimate

$$\mu_{t}\{x\in\tilde{B}_{t}\ |\ \text{dist}_{\omega_{t}}(x,x^{\prime})\geqslant r\}\leqslant\frac{C}{r}.$$

Taking $r\gg 1$ independent of $t$, we can ensure that

$$\begin{split}\mu_{t}(B_{\omega_{t}}(x^{\prime},r))&\geqslant\mu_{t}(\tilde{B}_{t}\cap B_{\omega_{t}}(x^{\prime},r))\\ &=\mu_{t}(\tilde{B}_{t})-\mu_{t}\{x\in\tilde{B}_{t}\ |\ \text{dist}_{\omega_{t}}(x,x^{\prime})\geqslant r\}\\ &\geqslant C^{-1}-\frac{C}{r}\\ &\geqslant C^{-1}=C^{-1}\mu_{t}(X_{t}),\end{split}$$

using here (4.3). The Bishop-Gromov volume comparison theorem then gives us that $\mu_{t}(B_{\omega_{t}}(x^{\prime},1))\geqslant C^{-1}\mu_{t}(X_{t}),$ or equivalently

$$\frac{\int_{X_{t}}\omega_{t}^{n}}{\int_{B_{\omega_{t}}(x^{\prime},1)}\omega_{t}^{n}}\leqslant C.$$

This bound can then be inserted into a well-known result of Yau (see e.g. [22, Lemma 3.2]): given a complete $d$-dimensional Riemannian manifold $(M,g)$ with nonnegative Ricci curvature, for any $x\in M$ and $1<R<\mathrm{diam}(M,g)$ we have

$$\frac{R-1}{2d}\leqslant\frac{\mathrm{Vol}(B_{g}(x,2R+2))}{\mathrm{Vol}(B_{g}(x,1))}.$$

Indeed, it suffices to choose $R=\mathrm{diam}(X_{t},\omega_{t})-1$ to obtain the desired uniform diameter upper bound for $(X_{t},\omega_{t})$. ∎