ScalingStacks

Proof of Theorem 2.2. First we will show the right-hand side inequality in (2.9). We will apply the maximum principle to the quantity

$$K=e^{-B\sigma^{-\lambda}}\left(\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right),$$

where $A$ is a suitably chosen uniform large constant. The maximum of $K$ on $X\backslash S$ is obviously achieved, and we will show that $K\leq C$ for a uniform constant $C$. This together with (3.9) will show that on $X\backslash S$ we have

(3.10)

$$\Delta_{\omega_{X}}\varphi_{t}=\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\textrm{tr}_{\omega_{X}}\omega_{0}-nt\leq\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\leq Ce^{Ce^{B\sigma^{-\lambda}}},$$

which is half of (2.9) To do this, we first compute as in Yau’s $C^{2}$ estimates [Y1]

$$\Delta_{\tilde{\omega}_{t}}\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\geq-C\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-C,$$

for a uniform constant $C$. On the other hand

$$\Delta_{\tilde{\omega}_{t}}\varphi_{t}\leq n-t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X},$$

and so if $A$ is large enough we get

$$\Delta_{\tilde{\omega}_{t}}\left(\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{A}{t}\varphi_{t}\right)\geq\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}.$$

Since $f$ is locally a submersion on $X\backslash S$, the fiber integration formula

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

holds. So we can compute that

(3.11)

$$\begin{split}\Delta_{\tilde{\omega}_{t}}\underline{\varphi_{t}}&=\textrm{tr}_{\tilde{\omega}_{t}}f_{*}(\sqrt{-1}\partial\overline{\partial}\varphi_{t}\wedge\omega_{X}^{n-m})\\ &=\textrm{tr}_{\tilde{\omega}_{t}}f_{*}((\tilde{\omega}_{t}-\omega_{t})\wedge\omega_{X}^{n-m})\\ &\geq-\textrm{tr}_{\tilde{\omega}_{t}}f_{*}(\omega_{t}\wedge\omega_{X}^{n-m})\\ &=-\textrm{tr}_{\tilde{\omega}_{t}}f_{*}(f^{*}\omega_{Y}\wedge\omega_{X}^{n-m})-t\textrm{tr}_{\tilde{\omega}_{t}}f_{*}(\omega_{X}^{n-m+1})\\ &=-\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0}-t\textrm{tr}_{\tilde{\omega}_{t}}f_{*}(\omega_{X}^{n-m+1}).\end{split}$$

On $Y\backslash f(S)$ the Kähler form $f_{*}(\omega_{X}^{n-m+1})$ can be estimated by

(3.12)

$$\begin{split}f_{*}(\omega_{X}^{n-m+1})&\leq\frac{\omega_{Y}^{m-1}\wedge f_{*}(\omega_{X}^{n-m+1})}{\omega_{Y}^{m}}\omega_{Y}=\frac{f_{*}(\omega_{0}^{m-1}\wedge\omega_{X}^{n-m+1})}{\omega_{Y}^{m}}\omega_{Y}\\ &\leq C\frac{f_{*}(\omega_{X}^{n})}{\omega_{Y}^{m}}\omega_{Y}=C\frac{f_{*}(H^{-1}\omega_{0}^{m}\wedge\omega_{X}^{n-m})}{\omega_{Y}^{m}}\omega_{Y}\\ &\leq C\sigma^{-\lambda}\frac{f_{*}(\omega_{0}^{m}\wedge\omega_{X}^{n-m})}{\omega_{Y}^{m}}\omega_{Y}=C\sigma^{-\lambda}\omega_{Y}.\end{split}$$

and so using (3.1) we get

$$\Delta_{\tilde{\omega}_{t}}\underline{\varphi_{t}}\geq-C-tC\sigma^{-\lambda}.$$

It follows that

(3.13)

$$\Delta_{\tilde{\omega}_{t}}\left(\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)\geq\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-C\sigma^{-\lambda}.$$

Using (2.3) and (3.1) we have that

(3.14)

$$|\Delta_{\tilde{\omega}_{t}}\sigma|\leq C\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0}\leq C,$$

(3.15)

$$|\nabla\sigma|^{2}_{\tilde{\omega}_{t}}\leq C\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0}\leq C,$$

Using (3.13) we then compute

(3.16)

$$\begin{split}\Delta_{\tilde{\omega}_{t}}K&\geq e^{-B\sigma^{-\lambda}}\left(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-C\sigma^{-\lambda}\right)\\ &+\left(\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)\Delta_{\tilde{\omega}_{t}}\left(e^{-B\sigma^{-\lambda}}\right)\\ &+2e^{B\sigma^{-\lambda}}\mathrm{Re}\langle\nabla K,\nabla e^{-B\sigma^{-\lambda}}\rangle_{\tilde{\omega}_{t}}\\ &-2\left(\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)e^{B\sigma^{-\lambda}}|\nabla e^{-B\sigma^{-\lambda}}|^{2}_{\tilde{\omega}_{t}}.\end{split}$$

Using (2.3), (3.14) and (3.15), the second term in (3.16) can be estimated as follows

(3.17)

$$\begin{split}\Delta_{\tilde{\omega}_{t}}\left(e^{-B\sigma^{-\lambda}}\right)&=\frac{B\lambda e^{-B\sigma^{-\lambda}}}{\sigma^{\lambda+1}}\Delta_{\tilde{\omega}_{t}}\sigma+\frac{B^{2}\lambda^{2}e^{-B\sigma^{-\lambda}}}{\sigma^{2\lambda+2}}|\nabla\sigma|^{2}_{\tilde{\omega}_{t}}\\ &-\frac{B\lambda(\lambda+1)e^{-B\sigma^{-\lambda}}}{\sigma^{\lambda+2}}|\nabla\sigma|^{2}_{\tilde{\omega}_{t}}\\ &\geq-C\frac{e^{-B\sigma^{-\lambda}}}{\sigma^{\lambda+1}}-C\frac{e^{-B\sigma^{-\lambda}}}{\sigma^{\lambda+2}}\\ &\geq-C\frac{e^{-B\sigma^{-\lambda}}}{\sigma^{\lambda+2}}.\end{split}$$

At the maximum of $K$ we may assume that $K\geq 0$, otherwise we have nothing to prove. Hence we can use (3.9) to estimate

(3.18)

$$\begin{split}\left(\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)\Delta_{\tilde{\omega}_{t}}\left(e^{-B\sigma^{-\lambda}}\right)\\ \geq-C\frac{e^{-B\sigma^{-\lambda}}}{\sigma^{\lambda+2}}\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{C}{\sigma^{\lambda+2}}.\end{split}$$

The fourth term in (3.16) can be estimated using (3.15)

(3.19)

$$\begin{split}|\nabla e^{-B\sigma^{-\lambda}}|^{2}_{\tilde{\omega}_{t}}&=\frac{B^{2}\lambda^{2}e^{-2B\sigma^{-\lambda}}}{\sigma^{2\lambda+2}}|\nabla\sigma|^{2}_{\tilde{\omega}_{t}}\leq\frac{Ce^{-2B\sigma^{-\lambda}}}{\sigma^{2\lambda+2}},\end{split}$$

(3.20)

$$\begin{split}-2\left(\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)e^{B\sigma^{-\lambda}}|\nabla e^{-B\sigma^{-\lambda}}|^{2}_{\tilde{\omega}_{t}}\\ \geq-C\frac{e^{-B\sigma^{-\lambda}}}{\sigma^{2\lambda+2}}\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{C}{\sigma^{2\lambda+2}}.\end{split}$$

Plugging (3.18) and (3.20) in (3.16), at the maximum point of $K$ we get

$$0\geq\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-\frac{C}{\sigma^{\lambda}}-\frac{C}{\sigma^{2\lambda+2}}\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-C\frac{e^{B\sigma^{-\lambda}}}{\sigma^{2\lambda+2}}.$$

Since for any two Kähler metrics $\omega,\tilde{\omega}$ we have

(3.21)

$$\textrm{tr}_{\omega}\tilde{\omega}\leq(\textrm{tr}_{\tilde{\omega}}\omega)^{n-1}\frac{\tilde{\omega}^{n}}{\omega^{n}},$$

we see that

$$\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\leq Ct^{n-m}(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})^{n-1}\leq C(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})^{n-1},$$

and using this and the inequalities $2ab\leq\varepsilon a^{2}+b^{2}/\varepsilon$ and $(\log x)^{2}\leq x+C$ we get

$$\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq\frac{C}{t}+\frac{C}{\sigma^{\lambda}}+\frac{C}{\sigma^{4\lambda+4}}+C\frac{e^{B\sigma^{-\lambda}}}{\sigma^{2\lambda+2}}+\frac{1}{2}\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X},$$

whence

$$\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq\frac{C}{t}+Ce^{C\sigma^{-\lambda}}.$$

At the same point we then get

$$\textrm{tr}_{\tilde{\omega}_{t}}\omega_{t}=\textrm{tr}_{\tilde{\omega}_{t}}(\omega_{0}+t\omega_{X})\leq C+tCe^{C\sigma^{-\lambda}}.$$

and using (3.21) we get

(3.22)

$$\textrm{tr}_{\omega_{t}}\tilde{\omega}_{t}\leq(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{t})^{n-1}\frac{\tilde{\omega}^{n}_{t}}{\omega^{n}_{t}}\leq\left(C+tCe^{C\sigma^{-\lambda}}\right)^{n-1}\frac{\tilde{\omega}^{n}_{t}}{\omega^{n}_{t}}.$$

We now use (2.1), (2.7) and (2.4) to get

(3.23)

$$\frac{\tilde{\omega}^{n}_{t}}{\omega^{n}_{t}}\leq\frac{Ct^{n-m}\omega_{X}^{n}}{\omega_{0}^{m}\wedge(t\omega_{X})^{n-m}}=\frac{C}{H}\leq\frac{C}{\sigma^{\lambda}}.$$

Combining (3.22) and (3.23) we get

$$\textrm{tr}_{\omega_{t}}\tilde{\omega}_{t}\leq Ce^{C\sigma^{-\lambda}},$$

for some uniform constant $C$. But we also have $\omega_{t}=\omega_{0}+t\omega_{X}\leq C\omega_{X}$ and so we get

$$\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\leq Ce^{C\sigma^{-\lambda}}.$$

Using (3.9) again, this implies that at the maximum of $K$ we have

$$K\leq C+e^{-B\sigma^{-\lambda}}\log(Ce^{C\sigma^{-\lambda}})\leq C.$$

We now show the left-hand side inequality in (2.9). To this extent we apply the maximum principle to the quantity

$$K_{1}=e^{-B\sigma^{-\lambda}_{h}}\left(\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right),$$

where $A$ is a suitably chosen uniform large constant. The maximum of $K_{1}$ on $X\backslash S$ is obviously achieved, and we will show that $K_{1}\leq C$ for a uniform constant $C$. This together with (3.9) will show that on $X\backslash S$ we have

(3.24)

$$\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq\frac{C}{t}e^{Ce^{B\sigma^{-\lambda}}},$$

which is the other half of (2.9). To prove that $K_{1}\leq C$ we use the maximum principle and, as in (3.16), we compute

(3.25)

$$\begin{split}\Delta_{\tilde{\omega}_{t}}K_{1}&\geq e^{-B\sigma^{-\lambda}}\left(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-C\sigma^{-\lambda}\right)\\ &+\left(\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)\Delta_{\tilde{\omega}_{t}}\left(e^{-B\sigma^{-\lambda}}\right)\\ &+2e^{B\sigma^{-\lambda}}\mathrm{Re}\langle\nabla K_{1},\nabla e^{-B\sigma^{-\lambda}}\rangle_{\tilde{\omega}_{t}}\\ &-2\left(\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right)e^{B\sigma^{-\lambda}}|\nabla e^{-B\sigma^{-\lambda}}|^{2}_{\tilde{\omega}_{t}}.\end{split}$$

We estimate this in the same way as before and get

(3.26)

$$\begin{split}\Delta_{\tilde{\omega}_{t}}K_{1}&\geq e^{-B\sigma^{-\lambda}}\left(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-C\sigma^{-\lambda}\right)\\ &-C\frac{e^{-B\sigma^{-\lambda}}}{\sigma^{2\lambda+2}}\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-\frac{C}{\sigma^{2\lambda+2}}\\ &+2e^{B\sigma^{-\lambda}}\mathrm{Re}\langle\nabla K_{1},\nabla e^{-B\sigma^{-\lambda}}\rangle_{\tilde{\omega}_{t}}.\end{split}$$

At the maximum of $K_{1}$ we get

$$0\geq\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}-\frac{C}{t}-\frac{C}{\sigma^{\lambda}}-\frac{C}{\sigma^{2\lambda+2}}\log(t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})-Ce^{C\sigma^{-\lambda}},$$

and using the inequalities $2ab\leq\varepsilon a^{2}+b^{2}/\varepsilon$ and $(\log x)^{2}\leq x+C$ we get

$$\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq\frac{C}{t}+\frac{C}{\sigma^{\lambda}}+\frac{C}{\sigma^{4\lambda+4}}+Ce^{C\sigma^{-\lambda}}+\frac{1}{2}\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X},$$

whence

$$t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq C+tCe^{C\sigma^{-\lambda}}\leq Ce^{C\sigma^{-\lambda}},$$

and so at that point

$$K_{1}\leq C+e^{-B\sigma^{-\lambda}}\log(Ce^{C\sigma^{-\lambda}})\leq C,$$

and we are done. ∎