ScalingStacks

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

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 $A\geq 6C_{0}$. ∎