Section 3. (00WC) · ScalingStacks

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3. A priori estimates

In this section we prove a priori $C^{2}$ estimates for the degenerating complex Monge-Ampère equations that we are considering, and we also prove $C^{3}$ estimates along the fibers of $f$ .

We start with a few lemmas.

00VX

Lemma 3.1. There is a uniform constant $C$ so that for all $0<t\leq 1$ we have

(3.1)

\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0}\leq C.

00VY

Proof. Recall that we are assuming that $\omega_{0}=f^{*}\omega_{Z}$ where $f:X\to Z$ is a holomorphic map. We can then use the Chern-Lu formula that appears in Yau’s Schwarz lemma computation [Y2, To1] and get

\Delta_{\tilde{\omega}_{t}}\log\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0}\geq-A\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0},

for a uniform constant $A$ . Noticing that

\Delta_{\tilde{\omega}_{t}}\varphi_{t}=n-\textrm{tr}_{\tilde{\omega}_{t}}\omega_{t}\leq n-\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0},

we see that

(3.2)

\Delta_{\tilde{\omega}_{t}}(\log\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0}-(A+1)\varphi_{t})\geq\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0}-n(A+1).

Then the maximum principle applied to (3.2), together with the estimate (2.8), gives (3.1). ∎

The next lemma, which gives a Sobolev constant bound, is due independently to Allard [A] and Michael-Simon [MS].

00VZ

Lemma 3.2. There is a uniform constant $C$ so that for any $0<t\leq 1$ , for any $y\in Y\backslash f(S)$ and for any $u\in C^{\infty}(X_{y})$ we have

(3.3)

\left(\int_{X_{y}}|u|^{\frac{2(n-m)}{n-m-1}}\omega_{y}^{n-m}\right)^{\frac{n-m-1}{n-m}}\leq C\int_{X_{y}}(|\nabla u|^{2}_{\omega_{y}}+|u|^{2})\omega_{y}^{n-m}.

00W0

Proof. For any $y\in Y\backslash f(S)$ the fiber $X_{y}$ is a smooth $(n-m)$ -dimensional complex submanifold of $X$ . Since $X$ is Kähler, it follows that $X_{y}$ is a minimal submanifold, and so it has vanishing mean curvature vector. We then use the Nash embedding theorem to isometrically embed $(X,\omega_{X})$ into Euclidean space, and so we have an isometric embedding $X\to\mathbb{R}^{N}$ . The length of the mean curvature vector of the composite isometric embedding $X_{y}\to X\to\mathbb{R}^{N}$ is then uniformly bounded independent of $y$ , since it depends only on the second fundamental form of $X\to\mathbb{R}^{N}$ . Then (3.3) follows from the uniform Sobolev inequality of [A, MS]. Notice that they prove an $L^{1}$ Sobolev inequality, but this implies the stated $L^{2}$ Sobolev inequality thanks to the Hölder inequality. ∎

One can easily avoid the Nash embedding theorem by using a partition of unity to reduce directly to the Euclidean case, but the above proof is perhaps cleaner.

We note here that the volume of $X_{y}$ with respect to $\omega_{y}$ , $\int_{X_{y}}\omega_{y}^{n-m}$ , is a homological constant independent of $y\in Y\backslash f(S)$ , and up to scaling $\omega_{X}$ we may assume that it is equal to $1$ . The next step is to prove a diameter bound for $\omega_{y}$ :

00W1

Lemma 3.3. There is a uniform constant $C$ so that for any $0<t\leq 1$ , for any $y\in Y\backslash f(S)$ we have

(3.4)

\mathrm{diam}(X_{y},\omega_{y})\leq C.

00W2

Proof. As above we embed $(X,\omega_{X})$ isometrically into $\mathbb{R}^{N}$ and we get that the length of the mean curvature vector of the composite isometric embedding $X_{y}\to X\to\mathbb{R}^{N}$ is then uniformly bounded independent of $y$ . We can then apply Theorem 1.1 of [Tp] and get the required diameter bound.

Alternatively, first one observes that (3.3) implies that there is a uniform constant $\kappa$ so that that geodesic balls in $X_{y}$ of radius $r<1$ have volume at least $\kappa r^{2(n-m)}$ (Lemma 3.2 in [H]). Since the total volume of $X_{y}$ is constant equal to $1$ , an elementary argument gives the required diameter bound. ∎

The next step is to prove a Poincaré inequality for the restricted metric $\omega_{y}$ . This time the constant will not be uniformly bounded, but it will blow up like a power of $\frac{1}{H}$ . To this end, we first estimate the Ricci curvature of $\omega_{y}$ . Fix a point $y\in Y\backslash f(S)$ and choose local coordinates $z^{1},\dots,z^{n-m}$ on the fiber $X_{y}$ , which extend locally to coordinates in a ball in $X$ . Then pick local coordinates $w^{n-m+1},\dots,w^{n}$ near $y\in Y\backslash f(S)$ , so that $z^{1},\dots,z^{n-m},z^{n-m+1}=f^{*}(w^{n-m+1}),\dots,z^{n}=f^{*}(w^{n})$ give local holomorphic coordinates on $X$ . We can also assume that at the point $y$ the metric $\omega_{Y}$ is the identity. At any fixed point of $X_{y}$ we then have

(3.5)

\begin{split}\mathrm{Ric}(\omega_{y})&=-\sqrt{-1}\partial\overline{\partial}\log\frac{\omega_{y}^{n-m}}{dz^{1}\wedge\dots\wedge d\overline{z}^{n-m}}\\ &=-\sqrt{-1}\partial\overline{\partial}\log\frac{\omega_{X}^{n-m}\wedge\omega_{0}^{m}}{dz^{1}\wedge\dots\wedge d\overline{z}^{n}}\\ &=-\sqrt{-1}\partial\overline{\partial}\log H-\sqrt{-1}\partial\overline{\partial}\log\frac{\omega_{X}^{n}}{dz^{1}\wedge\dots\wedge d\overline{z}^{n}}\\ &\geq-\frac{\sqrt{-1}\partial\overline{\partial}H}{H}+\mathrm{Ric}(\omega_{X})|_{X_{y}}\\ &\geq-\left(\frac{C}{H}+C\right)\omega_{y}\geq-\frac{C}{H}\omega_{y},\end{split}

where all derivatives are in fiber directions. Combining (3.5) and (2.4) we see that the Ricci curvature of $\omega_{y}$ is bounded below by $-C\sigma^{-\lambda}$ . Since the diameter of $\omega_{y}$ is bounded by Lemma 3.3, a theorem of Li-Yau [LY] then shows that the Poincaré constant of $\omega_{y}$ is bounded above by $Ce^{B\sigma^{-\lambda}}$ . This proves the following

00W3

Lemma 3.4. There are uniform constants $\lambda,B,C$ so that for any $0<t\leq 1$ , for any $y\in Y\backslash f(S)$ and for any $u\in C^{\infty}(X_{y})$ with $\int_{X_{y}}u\omega_{y}^{n-m}=0$ we have

(3.6)

\int_{X_{y}}|u|^{2}\omega_{y}^{n-m}\leq Ce^{B\sigma^{-\lambda}}\int_{X_{y}}|\nabla u|^{2}_{\omega_{y}}\omega_{y}^{n-m}.

We now let $\tilde{\omega}_{y}$ be the restriction $\tilde{\omega}_{t}|_{X_{y}}$ . We have the following estimate for the volume form of $\tilde{\omega}_{y}$ on $X_{y}$ :

(3.7)

\begin{split}\frac{\tilde{\omega}_{y}^{n-m}}{\omega_{y}^{n-m}}&=\frac{\tilde{\omega}_{t}^{n-m}\wedge\omega_{0}^{m}}{\omega_{X}^{n-m}\wedge\omega_{0}^{m}}=\frac{\tilde{\omega}_{t}^{n-m}\wedge\omega_{0}^{m}}{\tilde{\omega}_{t}^{n}}\cdot\frac{\tilde{\omega}_{t}^{n}}{H\omega_{X}^{n}}\\ &\leq\left(\frac{\tilde{\omega}_{t}^{n-1}\wedge\omega_{0}}{\tilde{\omega}_{t}^{n}}\right)^{m}\frac{c_{t}t^{n-m}e^{E}}{H}\\ &=(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0})^{m}\frac{c_{t}t^{n-m}e^{E}}{H}\leq\frac{Ct^{n-m}}{\sigma^{\lambda}}.\end{split}

Notice that when we restrict to $X_{y}$ we have

\tilde{\omega}_{y}=(\omega_{0}+t\omega_{X}+\sqrt{-1}\partial\overline{\partial}\varphi_{t})|_{X_{y}}=t\omega_{y}+(\sqrt{-1}\partial\overline{\partial}\varphi_{t})|_{X_{y}}.

It is convenient to define a function $\underline{\varphi_{t}}$ on $Y\backslash f(S)$ by

\underline{\varphi_{t}}(y)=\int_{X_{y}}\varphi_{t}\omega_{y}^{n-m}.

This is just the “integration along the fibers” of $\varphi_{t}$ , and we will also denote by $\underline{\varphi_{t}}$ its pullback to $X\backslash S$ via $f$ . We also define a function on $X\backslash S$ by

\psi=\frac{1}{t}\left(\varphi_{t}-\underline{\varphi_{t}}\right),

so that we have $\int_{X_{y}}\psi\omega_{y}^{n-m}=0$ and on $X_{y}$ we have

(3.8)

(\omega_{y}+\sqrt{-1}\partial\overline{\partial}\psi)^{n-m}=\frac{\tilde{\omega}_{y}^{n-m}}{t^{n-m}}\leq\frac{C}{\sigma^{\lambda}}\omega_{y}^{n-m}.

We can then apply Yau’s $L^{\infty}$ estimate for complex Monge-Ampère equations [Y1] to the inequality (3.8). Since the volume of $X_{y}$ is constant equal to $1$ , the Sobolev constant of $\omega_{y}$ is uniformly bounded (Lemma 3.2) and the Poincaré constant is controlled by Lemma 3.4, Yau’s $L^{\infty}$ estimate gives

(3.9)

\sup_{X_{y}}\left|\varphi_{t}-\underline{\varphi_{t}}\right|=t\sup_{X_{y}}|\psi|\leq tCe^{B\sigma(y)^{-\lambda}},

where we increased the constant $B$ to absorb the term $\sigma^{-\lambda}$ in (3.8). Recall that from (2.8) we have a uniform bound for the oscillation of $\varphi_{t}$ .

00W4

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. ∎

00W5

Proof of Theorem 2.3. We will first show (2.10), which is an easy consequence of (2.9). The left-hand side follows immediately from (2.9), which implies

(3.27)

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

Then (3.21) and (3.7) give

(3.28)

\textrm{tr}_{\omega_{y}}\tilde{\omega}_{y}\leq(\textrm{tr}_{\tilde{\omega}_{y}}\omega_{y})^{n-m-1}\frac{\tilde{\omega}_{y}^{n-m}}{\omega_{y}^{n-m}}\leq t\frac{Ce^{Ce^{B\sigma^{-\lambda}}}}{\sigma^{\lambda}}\leq tCe^{Ce^{B\sigma^{-\lambda}}},

which proves (2.10).

Next, we show (2.11). Recall from (3.10) and (3.24) that on $X\backslash S$ we have

(3.29)

\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\leq Ce^{C_{0}e^{B\sigma^{-\lambda}}}.

(3.30)

\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq\frac{C}{t}e^{C_{0}e^{B\sigma^{-\lambda}}},

for uniform constants $B,C,C_{0}$ . We apply the maximum principle to the quantity

K_{2}=e^{-Ae^{B\sigma^{-\lambda}}}\left(\mathcal{S}+C\frac{e^{3C_{0}e^{B\sigma^{-\lambda}}}}{t^{5/2}}\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\right),

for suitable constants $A,C$ , where the quantity $\mathcal{S}$ is the same quantity as in [Y1]:

\mathcal{S}=|\nabla\tilde{\omega}_{t}|^{2}_{\tilde{\omega}_{t}},

where $\nabla$ is the covariant derivative associated to the metric $\omega_{X}$ . Using $\varphi_{t}$ we can write

\mathcal{S}=\tilde{g}_{t}^{i\overline{p}}\tilde{g}_{t}^{q\overline{j}}\tilde{g}_{t}^{k\overline{r}}\varphi_{i\overline{j}k}\varphi_{\overline{p}q\overline{r}},

where again lower indices are covariant derivatives with respect to $\omega_{X}$ . We are going to show that $K_{2}\leq\frac{C}{t^{5/2}}$ , and using (3.29) this implies that

(3.31)

\mathcal{S}\leq\frac{Ce^{Ae^{B\sigma^{-\lambda}}}}{t^{5/2}}.

We now use (3.28), which says that on $X_{y}$ we have

(3.32)

\textrm{tr}_{\omega_{y}}\tilde{\omega}_{y}\leq tCe^{C_{0}e^{B\sigma^{-\lambda}}},

At any given point of $X_{y}$ we can assume that $\omega_{X}$ is the identity and $\tilde{\omega}_{t}$ is diagonal with positive entries $\lambda_{i}$ , $1\leq i\leq n$ , so that the first $n-m$ directions are tangent to the fiber $X_{y}$ . Then (3.32) gives that

(3.33)

\lambda_{i}\leq tCe^{C_{0}e^{B\sigma^{-\lambda}}},

for $1\leq i\leq n-m$ . Then using (3.31) we see that

\sum_{i,j,k=1}^{n-m}\frac{1}{\lambda_{i}\lambda_{j}\lambda_{k}}|\varphi_{i\overline{j}k}|^{2}\leq\sum_{i,j,k=1}^{n}\frac{1}{\lambda_{i}\lambda_{j}\lambda_{k}}|\varphi_{i\overline{j}k}|^{2}=\mathcal{S}\leq\frac{Ce^{Ae^{B\sigma^{-\lambda}}}}{t^{5/2}},

and using (3.33) we get

|\nabla\tilde{\omega}_{y}|^{2}_{\omega_{y}}=\sum_{i,j,k=1}^{n-m}|\varphi_{i\overline{j}k}|^{2}\leq t^{1/2}Ce^{(A+3C_{0})e^{B\sigma^{-\lambda}}},

and this is (2.11).

We now prove that $K_{2}\leq\frac{C}{t^{5/2}}$ . To simplify the computation, we will use the notation

\mathcal{F}(x)=e^{xe^{B\sigma^{-\lambda}}},

where $x$ is a real number, and we note here that $\mathcal{F}$ is increasing. The starting point is a precise formula for $\Delta_{\tilde{\omega}_{t}}\mathcal{S}$ . This is just Yau’s $C^{3}$ estimate [Y1], but without assuming that the metrics $\tilde{\omega}_{t}$ and $\omega_{X}$ are equivalent, and it is done in a more general setting in [TWY] (see also [PSS]). With the notation of [TWY] we can write

\mathcal{S}=\sum_{i,j,k}|a^{i}_{jk}|^{2}.

We then choose local unitary frames $\{\theta^{1},\dots,\theta^{n}\}$ for $\omega_{X}$ and $\{\tilde{\theta}^{1},\dots,\tilde{\theta}^{n}\}$ for $\tilde{\omega}_{t}$ , and write

\tilde{\theta}^{i}=\sum_{j}a^{i}_{j}\theta^{j},

\theta^{i}=\sum_{j}b^{i}_{j}\tilde{\theta}^{j},

for some local matrices of functions $a^{i}_{j},b^{i}_{j}$ . Notice that at any given point we can choose the frames and arrange that

(3.34)

a^{i}_{j}=\sqrt{\lambda_{i}}\delta^{i}_{j},

(3.35)

b^{i}_{j}=\frac{1}{\sqrt{\lambda_{i}}}\delta^{i}_{j}.

Then in our case [TWY, (4.3)] reads

(3.36)		$\displaystyle\Delta_{\tilde{\omega}_{t}}\mathcal{S}$	$\displaystyle\geq$	$\displaystyle 2\mathrm{Re}\biggl(\overline{a^{i}_{k\ell}}\biggl(b^{m}_{k}b^{q}_{\ell}\overline{b^{s}_{p}}R^{j}_{mq\overline{s}}a^{i}_{rp}a^{r}_{j}-a^{i}_{j}b^{q}_{\ell}\overline{b^{s}_{p}}R^{j}_{mq\overline{s}}a^{r}_{kp}b^{m}_{r}$
(3.36)				$\displaystyle\mbox{}-a^{i}_{j}b^{m}_{k}\overline{b^{s}_{p}}R^{j}_{mq\overline{s}}a^{r}_{\ell p}b^{q}_{r}+a^{i}_{j}b^{m}_{k}b^{q}_{\ell}\overline{b^{s}_{p}}b_{p}^{u}R^{j}_{mq\overline{s},u}\biggr)\biggr),$

where we are summing over all indices, $R^{j}_{mq\overline{s}}$ represents the curvature of $\omega_{X}$ and $R^{j}_{mq\overline{s},u}$ its covariant derivative (with respect to $\omega_{X}$ ). Since these are fixed tensors, we can use the Cauchy-Schwarz inequality and (3.34), (3.35) to estimate the first term on the right hand side of (3.36) by

\begin{split}&\left|2\mathrm{Re}\left(\overline{a^{i}_{k\ell}}b^{m}_{k}b^{q}_{\ell}\overline{b^{s}_{p}}R^{j}_{mq\overline{s}}a^{i}_{rp}a^{r}_{j}\right)\right|\leq C\sum_{i,k,\ell,r,p}|a^{i}_{k\ell}a^{i}_{rp}|\sqrt{\frac{\lambda_{r}}{\lambda_{k}\lambda_{\ell}\lambda_{p}}}\\ &\leq C\left(\sum_{j}\lambda_{j}\right)^{\frac{1}{2}}\left(\sum_{q}\frac{1}{\lambda_{q}}\right)^{\frac{3}{2}}\sum_{k,\ell,r,p}\left(\sum_{i}|a^{i}_{k\ell}|^{2}\right)^{\frac{1}{2}}\left(\sum_{i}|a^{i}_{rp}|^{2}\right)^{\frac{1}{2}}\\ &=C(\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t})^{\frac{1}{2}}(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})^{\frac{3}{2}}\left(\sum_{i,k,\ell}|a^{i}_{k\ell}|^{2}\right)^{\frac{1}{2}}\left(\sum_{i,r,p}|a^{i}_{rp}|^{2}\right)^{\frac{1}{2}}\\ &=C\mathcal{S}(\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t})^{\frac{1}{2}}(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})^{\frac{3}{2}}.\end{split}

The second and third term in (3.36) are estimated similarly, while the fourth term can be bounded by

\begin{split}&\left|2\mathrm{Re}\left(\overline{a^{i}_{k\ell}}a^{i}_{j}b^{m}_{k}b^{q}_{\ell}\overline{b^{s}_{p}}b_{p}^{u}R^{j}_{mq\overline{s},u}\right)\right|\leq C\sum_{i,k,\ell,p}|a^{i}_{k\ell}|\sqrt{\frac{\lambda_{i}}{\lambda_{k}\lambda_{\ell}\lambda_{p}^{2}}}\\ &\leq C\left(\sum_{j}\lambda_{j}\right)^{\frac{1}{2}}\left(\sum_{q}\frac{1}{\lambda_{q}}\right)^{2}\sum_{i,k,\ell}|a^{i}_{k\ell}|\\ &\leq C\sqrt{\mathcal{S}}(\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t})^{\frac{1}{2}}(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})^{2}.\end{split}

Overall we can estimate

(3.37)

\Delta_{\tilde{\omega}_{t}}\mathcal{S}\geq-C\mathcal{S}(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})^{3/2}(\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t})^{1/2}-C\sqrt{\mathcal{S}}(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})^{2}(\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t})^{1/2}.

On the other hand from [TWY, Lemma 3.3] we see that

(3.38)

\begin{split}\Delta_{\tilde{\omega}_{t}}\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}&=a^{i}_{k\ell}\overline{a^{i}_{p\ell}}a^{k}_{j}\overline{a^{p}_{j}}+\overline{a^{i}_{j}}a^{i}_{r}b^{q}_{\ell}\overline{b^{s}_{\ell}}R^{r}_{jq\overline{s}}\\ &\geq\sum_{i,j,\ell}|a^{i}_{j\ell}|^{2}\lambda_{j}-C\sum_{i,\ell}\frac{\lambda_{i}}{\lambda_{\ell}}\\ &\geq\left(\sum_{k}\frac{1}{\lambda_{k}}\right)^{-1}\sum_{i,j,\ell}|a^{i}_{j\ell}|^{2}-C\left(\sum_{p}\lambda_{p}\right)\left(\sum_{q}\frac{1}{\lambda_{q}}\right)\\ &=\frac{\mathcal{S}}{\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}}-C(\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X})(\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}).\end{split}

We now insert (3.29), (3.30) in (3.37), (3.38) and get

(3.39)

\Delta_{\tilde{\omega}_{t}}\mathcal{S}\geq-\frac{C\mathcal{F}(2C_{0})}{t^{3/2}}\mathcal{S}-\frac{C\mathcal{F}(5C_{0}/2)}{t^{2}}\sqrt{\mathcal{S}},

\Delta_{\tilde{\omega}_{t}}\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\geq\frac{t\mathcal{F}(-C_{0})}{C}\mathcal{S}-\frac{C\mathcal{F}(2C_{0})}{t}.

We then compute

(3.40)

\begin{split}\Delta_{\tilde{\omega}_{t}}\left(\frac{\mathcal{F}(3C_{0})}{t^{5/2}}\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\right)&\geq\frac{\mathcal{F}(2C_{0})}{Ct^{3/2}}\mathcal{S}-\frac{C\mathcal{F}(5C_{0})}{t^{7/2}}\\ &+\frac{2}{t^{5/2}}\mathrm{Re}\langle\nabla\mathcal{F}(3C_{0}),\nabla\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\rangle_{\tilde{\omega}_{t}}\\ &+\frac{1}{t^{5/2}}(\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t})\Delta_{\tilde{\omega}_{t}}\mathcal{F}(3C_{0}),\end{split}

and estimate

\mathrm{Re}\langle\nabla\mathcal{F}(3C_{0}),\nabla\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\rangle_{\tilde{\omega}_{t}}\geq-|\nabla\mathcal{F}(3C_{0})|_{\tilde{\omega}_{t}}|\nabla\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}|_{\tilde{\omega}_{t}}.

Using [TWY, (3.20)] we see that

|\nabla\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}|_{\tilde{\omega}_{t}}\leq\sqrt{\mathcal{S}}(\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}).

On the other hand a direct computation using (3.14) and (3.15) shows that there is a constant $C$ such that for any real number $x$ we have

|\nabla\mathcal{F}(x)|_{\tilde{\omega}_{t}}\leq C\mathcal{F}(x+1),

|\Delta_{\tilde{\omega}_{t}}\mathcal{F}(x)|\leq C\mathcal{F}(x+1),

and so we have

(3.41)

\begin{split}\Delta_{\tilde{\omega}_{t}}\left(\frac{\mathcal{F}(3C_{0})}{t^{5/2}}\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\right)&\geq\frac{\mathcal{F}(2C_{0})}{Ct^{3/2}}\mathcal{S}-\frac{C\mathcal{F}(5C_{0})}{t^{7/2}}-\frac{C\mathcal{F}(5C_{0})}{t^{5/2}}\sqrt{\mathcal{S}}\\ &-\frac{C\mathcal{F}(5C_{0})}{t^{5/2}}.\end{split}

This and (3.39) give

\begin{split}\Delta_{\tilde{\omega}_{t}}\left(\mathcal{S}+\frac{C\mathcal{F}(3C_{0})}{t^{5/2}}\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\right)&\geq\frac{\mathcal{F}(2C_{0})}{t^{3/2}}\mathcal{S}-\frac{C\mathcal{F}(5C_{0}/2)}{t^{2}}\sqrt{\mathcal{S}}\\ &-\frac{C\mathcal{F}(5C_{0})}{t^{7/2}}-\frac{C\mathcal{F}(5C_{0})}{t^{5/2}}\sqrt{\mathcal{S}}-\frac{C\mathcal{F}(5C_{0})}{t^{5/2}}\\ &\geq\frac{\mathcal{F}(2C_{0})}{t^{3/2}}\mathcal{S}-\frac{C\mathcal{F}(5C_{0})}{t^{7/2}}-\frac{C\mathcal{F}(5C_{0})}{t^{5/2}}\sqrt{\mathcal{S}},\end{split}

and

(3.42)

\begin{split}\Delta_{\tilde{\omega}_{t}}K_{2}&\geq\mathcal{F}(-A)\biggl(\frac{\mathcal{F}(2C_{0})}{t^{3/2}}\mathcal{S}-\frac{C\mathcal{F}(5C_{0})}{t^{7/2}}-\frac{C\mathcal{F}(5C_{0})}{t^{5/2}}\sqrt{\mathcal{S}}\\ &-C\mathcal{F}(1)\mathcal{S}-\frac{C\mathcal{F}(4C_{0}+1)}{t^{5/2}}\biggr)+2\mathcal{F}(A)\mathrm{Re}\langle\nabla K_{2},\nabla\mathcal{F}(-A)\rangle_{\tilde{\omega}_{t}}\\ &\geq\mathcal{F}(-A)\biggl(\frac{\mathcal{F}(2C_{0})}{Ct^{3/2}}\mathcal{S}-\frac{C\mathcal{F}(5C_{0})}{t^{7/2}}-\frac{C\mathcal{F}(5C_{0})}{t^{5/2}}\sqrt{\mathcal{S}}\biggr)\\ &+2\mathcal{F}(A)\mathrm{Re}\langle\nabla K_{2},\nabla\mathcal{F}(-A)\rangle_{\tilde{\omega}_{t}}.\end{split}

At the maximum of $K_{2}$ we then get

\mathcal{S}\leq\frac{C\mathcal{F}(3C_{0})}{t}\sqrt{\mathcal{S}}+\frac{C\mathcal{F}(3C_{0})}{t^{2}},

which implies that

\mathcal{S}\leq\frac{C\mathcal{F}(6C_{0})}{t^{2}},

and so

K_{2}=\mathcal{F}(-A)\left(\mathcal{S}+\frac{C\mathcal{F}(3C_{0})}{t^{5/2}}\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\right)\leq\mathcal{F}(-A)\frac{C\mathcal{F}(6C_{0})}{t^{5/2}}\leq\frac{C}{t^{5/2}},

if we choose $A\geq 6C_{0}$ . ∎

Remark. In the estimates proved in this section we have repeatedly used the fact that the metrics $\tilde{\omega}_{t}$ are Ricci-flat. If instead one is dealing with the general equation (2.7), the only estimate that does not generalize immediately is (3.1) (which requires that the Ricci curvature of $\tilde{\omega}_{t}$ be nonnegative). On the other hand, all of the estimates have parabolic analogues in the case of Kähler-Ricci flow as in [ST2].