ScalingStacks

Proof. We first prove that $\tilde{\omega}_{t}$ converges to $\omega$ in the weak topology of currents. Since the cohomology class of $\tilde{\omega}_{t}$ is bounded, weak compactness of currents implies that from any sequence $t_{i}\to 0$ we can extract a subsequence so that $\tilde{\omega}_{t_{i}}$ converges weakly to a limit closed positive $(1,1)$-current $\hat{\omega}$, which a priori depends on the sequence. If we write $\hat{\omega}=\omega_{0}+\sqrt{-1}\partial\overline{\partial}\hat{\varphi}$, it follows that $\varphi_{t_{i}}\to\hat{\varphi}$ in $L^{1}$, and from the bound (2.8) we infer that $\hat{\varphi}$ is in $L^{\infty}$. Moreover restricting $\hat{\omega}$ to any smooth fiber $X_{y}$ we see that

$$\sqrt{-1}\partial\overline{\partial}\hat{\varphi}|_{X_{y}}\geq 0,$$

and the maximum principle implies that $\hat{\varphi}$ is constant on each fiber, and so descends to a bounded function $\hat{\varphi}$ on $Y\backslash f(S)$. We will show that $\hat{\varphi}$ satisfies the same equation (4.3) as $\psi$, and so by uniqueness $\hat{\varphi}=\psi$. To this end we fix an arbitrary compact set $K\subset Y\backslash f(S)$, and we wish to show that $\hat{\varphi}$ satisfies (4.3) on $K$.

We then fix $\eta$ a smooth function with support contained in $K$, and we will also denote by $\eta$ its pullback to $X$ via $f$. Recall that we have called $\tilde{\omega}_{1}$ the Ricci-flat metric in the class $[\omega_{1}]$, and $\Omega=\tilde{\omega}_{1}^{n}$. Then from the Monge-Ampère equation (2.5) we have

(4.6)

$$\int_{X}\eta\Omega=\frac{1}{a_{t}}\int_{X}\eta(\omega_{0}+t\omega_{X}+\sqrt{-1}\partial\overline{\partial}\varphi_{t})^{n},$$

where the constants $a_{t}$ are equal to

$$\frac{\int_{X}\omega_{t}^{n}}{\int_{X}\omega_{1}^{n}},$$

and behave like (2.6). We can also write

(4.7)

$$\int_{X}\eta\Omega=\int_{X}\eta F\omega_{SF}^{n-m}\wedge\omega_{0}^{m}.$$

We are now going to estimate $\frac{1}{a_{t}}\int_{X}\eta(\omega_{0}+t\omega_{X}+\sqrt{-1}\partial\overline{\partial}\varphi_{t})^{n}.$ We have

$$\begin{split}&\frac{1}{a_{t}}\int_{X}\eta(\omega_{0}+t\omega_{X}+\sqrt{-1}\partial\overline{\partial}\varphi_{t})^{n}\\ &=\frac{1}{a_{t}}\int_{X}\eta\left((\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})+(t\omega_{X}+\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}})\right)^{n}\\ &=\frac{1}{a_{t}}\int_{X}\eta\sum_{k=0}^{n}\binom{n}{k}(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k}\wedge(t\omega_{X}+\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}}))^{n-k}\end{split}$$

First of all observe that the form $\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}}$ is the pullback of a form on $Y$, and it can be wedged with itself at most $m$ times, so all terms in the sum with $k>m$ are zero. Next, we claim that all the terms with $k<m$ go to zero as $t\to 0$. To see this, start by observing that on the compact set $K$ the estimate (3.29) gives a constant $C$ (that depends on $K$) such that

(4.8)

$$-C\omega_{X}\leq\sqrt{-1}\partial\overline{\partial}\varphi_{t}\leq C\omega_{X}.$$

Moreover from the equation

$$\partial\overline{\partial}\underline{\varphi_{t}}=f_{*}(\partial\overline{\partial}\varphi_{t}\wedge\omega_{X}^{n-m})$$

together with (4.8), (3.12), we see that on $f(K)$ we have

(4.9)

$$-C\omega_{Y}\leq\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}}\leq C\omega_{Y}.$$

We also need to use (3.9) which on $K$ gives

(4.10)

$$\sup_{K}|\varphi_{t}-\underline{\varphi_{t}}|\leq Ct.$$

Then any term with $k<m$ is equal to

$$\frac{\binom{n}{k}}{a_{t}}\int_{X}\eta(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k}\wedge(t\omega_{X}+\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}}))^{n-k},$$

and it can be expanded into

$$\frac{\binom{n}{k}}{a_{t}}\sum_{i=0}^{n-k}\binom{n-k}{i}\int_{X}\eta(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k}\wedge(t\omega_{X})^{n-k-i}\wedge(\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}}))^{i}.$$

On $K$ the $(1,1)$-form $\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}}$ is bounded by (4.9). Since $a_{t}=O(t^{n-m})$ from (2.6), we see that the term in this sum with $i=0$ goes to zero. Any term with $i>0$ is comparable to

(4.11)

$$\frac{1}{t^{n-m}}\int_{X}(\varphi_{t}-\underline{\varphi_{t}})\sqrt{-1}\partial\overline{\partial}\eta\wedge(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k}\wedge(t\omega_{X})^{n-k-i}\wedge(\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}}))^{i-1}.$$

Notice that all the $(1,1)$-forms appearing inside the integral are bounded by (4.8), (4.9), and that the function $\varphi_{t}-\underline{\varphi_{t}}$ is $O(t)$ by (4.10). On $K$ the estimate (2.10) gives

(4.12)

$$-Ct\omega_{y}\leq(\sqrt{-1}\partial\overline{\partial}\varphi_{t})|_{X_{y}}=(\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}}))|_{X_{y}}\leq Ct\omega_{y}.$$

The form $\sqrt{-1}\partial\overline{\partial}\eta\wedge(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k}$ is the pullback of a form from $Y$, and so we can use (4.12) to estimate

$$\left|\frac{\sqrt{-1}\partial\overline{\partial}\eta\wedge(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k}\wedge(t\omega_{X})^{n-k-i}\wedge(\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}}))^{i-1}}{\omega_{X}^{n}}\right|\leq Ct^{n-m},$$

and so the term (4.11) goes to zero. This proves our claim.

We are then left with only the term with $k=m$, which is

$$\frac{1}{a_{t}}\int_{X}\eta\binom{n}{m}(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{m}\wedge(t\omega_{X}+\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}}))^{n-m},$$

and if we expand the term $(t\omega_{X}+\sqrt{-1}\partial\overline{\partial}(\varphi_{t}-\underline{\varphi_{t}}))^{n-m}$, we get

$$\begin{split}\frac{1}{a_{t}}&\int_{X}\eta\binom{n}{m}(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{m}\wedge(t\omega_{X})^{n-m}\\ &+\frac{1}{a_{t}}\int_{X}\sqrt{-1}\partial\overline{\partial}\eta\wedge(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{m}\wedge\dots,\end{split}$$

and the second term is zero because $\partial\overline{\partial}\eta$ is the pullback of a form from the base. We are then left with the term

(4.13)

$$\frac{1}{a_{t}}\int_{X}\eta\binom{n}{m}(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{m}\wedge(t\omega_{X})^{n-m},$$

which we need to further estimate. Using (4.8) we see that, up to taking a further subsequence, the functions $\varphi_{t_{i}}$ converge to $\hat{\varphi}$ in the $C^{1,\beta}(K)$ topology, and (4.10) implies that the functions $\underline{\varphi_{t_{i}}}$ also converge to $\hat{\varphi}$ uniformly. We can then rewrite (4.13) as

$$\frac{t^{n-m}\binom{n}{m}}{a_{t}}\int_{X}\eta(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{m}\wedge\omega_{X}^{n-m}.$$

Using (2.6) we see that as $t$ goes to zero the coefficient $\frac{t^{n-m}\binom{n}{m}}{a_{t}}$ converges to

$$\frac{\int_{X}\omega_{1}^{n}}{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}.$$

On the other hand we have

$$\int_{X}\eta(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{m}\wedge\omega_{X}^{n-m}=\sum_{k=0}^{m}\binom{m}{k}\int_{X}\eta\omega_{0}^{m-k}\wedge(\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k}\wedge\omega_{X}^{n-m}.$$

The term with $k=0$ is independent of $t$, while any term with $k>0$ can be written as

(4.14)

$$\int_{X}\underline{\varphi_{t}}\sqrt{-1}\partial\overline{\partial}\eta\wedge\omega_{0}^{m-k}\wedge(\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k-1}\wedge\omega_{X}^{n-m}.$$

The $(n,n)$-form $\sqrt{-1}\partial\overline{\partial}\eta\wedge\omega_{0}^{m-k}\wedge(\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k-1}\wedge\omega_{X}^{n-m}$ is supported in $K$ and is uniformly bounded by (4.9), and the functions $\underline{\varphi_{t_{i}}}$ converge uniformly to $\hat{\varphi}$, and so along the sequence $t_{i}$ the term (4.14) has the same limit as

$$\int_{X}\hat{\varphi}\sqrt{-1}\partial\overline{\partial}\eta\wedge\omega_{0}^{m-k}\wedge(\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k-1}\wedge\omega_{X}^{n-m}.$$

But this is equal to

$$\int_{X}\underline{\varphi_{t}}\sqrt{-1}\partial\overline{\partial}\eta\wedge\omega_{0}^{m-k}\wedge(\sqrt{-1}\partial\overline{\partial}\underline{\varphi_{t}})^{k-2}\wedge\sqrt{-1}\partial\overline{\partial}\hat{\varphi}\wedge\omega_{X}^{n-m},$$

and repeating the same argument $k-1$ times we see that along the sequence $t_{i}$ the term (4.14) converges to

$$\int_{X}\eta\omega_{0}^{m-k}\wedge(\sqrt{-1}\partial\overline{\partial}\hat{\varphi})^{k}\wedge\omega_{X}^{n-m}.$$

It follows that along the sequence $t_{i}$ the term (4.13) converges to

$$\frac{\int_{X}\omega_{1}^{n}}{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}\int_{X}\eta(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\hat{\varphi})^{m}\wedge\omega_{X}^{n-m},$$

and using (4.6), (4.7) we get

$$\int_{X}\eta F\omega_{SF}^{n-m}\wedge\omega_{0}^{m}=\frac{\int_{X}\omega_{1}^{n}}{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}\int_{X}\eta(\omega_{0}+\sqrt{-1}\partial\overline{\partial}\hat{\varphi})^{m}\wedge\omega_{X}^{n-m}.$$

We then integrate first along the fibers and get

$$\begin{split}\int_{Y}&\eta F\omega_{Y}^{m}\left(\int_{X_{y}}\omega_{SF,y}^{n-m}\right)\\ &=\frac{\int_{X}\omega_{1}^{n}}{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}\int_{Y}\eta(\omega_{Y}+\sqrt{-1}\partial\overline{\partial}\hat{\varphi})^{m}\left(\int_{X_{y}}\omega_{y}^{n-m}\right),\end{split}$$

and since $\omega_{y}$ is cohomologous to $\omega_{SF,y}$, we get

$$\int_{Y}\eta F\omega_{Y}^{m}=\frac{\int_{X}\omega_{1}^{n}}{\int_{X}\omega_{0}^{m}\wedge\omega_{X}^{n-m}}\int_{Y}\eta(\omega_{Y}+\sqrt{-1}\partial\overline{\partial}\hat{\varphi})^{m},$$

which is just the weak form of (4.3). This shows that any weak limit $\hat{\omega}$ of $\tilde{\omega}_{t}$ as $t\to 0$ satisfies (4.3) weakly, and we have already remarked that we can write $\hat{\omega}=\omega_{Y}+\sqrt{-1}\partial\overline{\partial}\hat{\varphi}$ with $\hat{\varphi}$ in $L^{\infty}$. By Kołodziej’s uniqueness of $L^{\infty}$ weak solutions of (4.3) (see [ST2, Theorem 3.2] and [EGZ1, Z]), we must have $\hat{\varphi}=\psi$, and so the whole sequence $\tilde{\omega}_{t}$ converges weakly to $\omega$ as $t\to 0$. Then the bound (2.9) implies that $\varphi_{t}$ actually converges to $\psi$ in the $C^{1,\beta}_{loc}$ topology on $X\backslash S$. ∎