ScalingStacks

Verified tagged author-source HTML · 0905.4718v1 · cited publication edition alignment unverified.

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​σ−λ​(log⁡trωX​ω~t−At​(φt−φt¯)),K=e^{-B\sigma^{-\lambda}}\left(\log\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}-\frac{A}{t}(\varphi_{t}-\underline{\varphi_{t}})\right),

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

(3.10) ΔωX​φt=trωX​ω~t−trωX​ω0−n​t≤trωX​ω~t≤C​eC​eB​σ−λ,\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 C2C^{2} estimates [Y1]

Δω~t​log⁡trωX​ω~t≥−C​trω~t​ωX−C,\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 CC. On the other hand

Δω~t​φt≤n−t⋅trω~t​ωX,\Delta_{\tilde{\omega}_{t}}\varphi_{t}\leq n-t\cdot\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X},

and so if AA is large enough we get

Δω~t​(log⁡trωX​ω~t−At​φt)≥trω~t​ωX−Ct.\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 ff is locally a submersion on X\SX\backslash S, the fiber integration formula

∂∂¯​φt¯=f∗​(∂∂¯​φt∧ωXn−m)\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) Δω~t​φt¯=trω~t​f∗​(−1​∂∂¯​φt∧ωXn−m)=trω~t​f∗​((ω~t−ωt)∧ωXn−m)≥−trω~t​f∗​(ωt∧ωXn−m)=−trω~t​f∗​(f∗​ωY∧ωXn−m)−t​trω~t​f∗​(ωXn−m+1)=−trω~t​ω0−t​trω~t​f∗​(ωXn−m+1).\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\f⁡(S)Y\backslash f(S) the Kähler form f∗​(ωXn−m+1)f_{*}(\omega_{X}^{n-m+1}) can be estimated by

(3.12) f∗​(ωXn−m+1)≤ωYm−1∧f∗​(ωXn−m+1)ωYm​ωY=f∗​(ω0m−1∧ωXn−m+1)ωYm​ωY≤C​f∗​(ωXn)ωYm​ωY=C​f∗​(H−1​ω0m∧ωXn−m)ωYm​ωY≤C​σ−λ​f∗​(ω0m∧ωXn−m)ωYm​ωY=C​σ−λ​ωY.\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

Δω~t​φt¯≥−C−t​C​σ−λ.\Delta_{\tilde{\omega}_{t}}\underline{\varphi_{t}}\geq-C-tC\sigma^{-\lambda}.

It follows that

(3.13) Δω~t​(log⁡trωX​ω~t−At​(φt−φt¯))≥trω~t​ωX−Ct−C​σ−λ.\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) |Δω~t​σ|≤C​trω~t​ω0≤C,|\Delta_{\tilde{\omega}_{t}}\sigma|\leq C\textrm{tr}_{\tilde{\omega}_{t}}\omega_{0}\leq C,
(3.15) |∇σ|ω~t2≤C​trω~t​ω0≤C,|\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) Δω~t​K≥e−B​σ−λ​(trω~t​ωX−Ct−C​σ−λ)+(log⁡trωX​ω~t−At​(φt−φt¯))​Δω~t​(e−B​σ−λ)+2​eB​σ−λ​Re​⟨∇K,∇e−B​σ−λ⟩ω~t−2​(log⁡trωX​ω~t−At​(φt−φt¯))​eB​σ−λ​|∇e−B​σ−λ|ω~t2.\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) Δω~t​(e−B​σ−λ)=B​λ​e−B​σ−λσλ+1​Δω~t​σ+B2​λ2​e−B​σ−λσ2​λ+2​|∇σ|ω~t2−B​λ​(λ+1)​e−B​σ−λσλ+2​|∇σ|ω~t2≥−C​e−B​σ−λσλ+1−C​e−B​σ−λσλ+2≥−C​e−B​σ−λσλ+2.\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 KK we may assume that K≥0K\geq 0, otherwise we have nothing to prove. Hence we can use (3.9) to estimate

(3.18) (log⁡trωX​ω~t−At​(φt−φt¯))​Δω~t​(e−B​σ−λ)≥−C​e−B​σ−λσλ+2​log⁡trωX​ω~t−Cσλ+2.\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) |∇e−B​σ−λ|ω~t2=B2​λ2​e−2​B​σ−λσ2​λ+2​|∇σ|ω~t2≤C​e−2​B​σ−λσ2​λ+2,\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) −2​(log⁡trωX​ω~t−At​(φt−φt¯))​eB​σ−λ​|∇e−B​σ−λ|ω~t2≥−C​e−B​σ−λσ2​λ+2​log⁡trωX​ω~t−Cσ2​λ+2.\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 KK we get

0≥trω~t​ωX−Ct−Cσλ−Cσ2​λ+2​log⁡trωX​ω~t−C​eB​σ−λσ2​λ+2.0\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 ω,ω~\omega,\tilde{\omega} we have

(3.21) trω​ω~≤(trω~​ω)n−1​ω~nωn,\textrm{tr}_{\omega}\tilde{\omega}\leq(\textrm{tr}_{\tilde{\omega}}\omega)^{n-1}\frac{\tilde{\omega}^{n}}{\omega^{n}},

we see that

trωX​ω~t≤C​tn−m​(trω~t​ωX)n−1≤C​(trω~t​ωX)n−1,\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 2​a​b≤ε​a2+b2/ε2ab\leq\varepsilon a^{2}+b^{2}/\varepsilon and (log⁡x)2≤x+C(\log x)^{2}\leq x+C we get

trω~t​ωX≤Ct+Cσλ+Cσ4​λ+4+C​eB​σ−λσ2​λ+2+12​trω~t​ωX,\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

trω~t​ωX≤Ct+C​eC​σ−λ.\textrm{tr}_{\tilde{\omega}_{t}}\omega_{X}\leq\frac{C}{t}+Ce^{C\sigma^{-\lambda}}.

At the same point we then get

trω~t​ωt=trω~t​(ω0+t​ωX)≤C+t​C​eC​σ−λ.\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) trωt​ω~t≤(trω~t​ωt)n−1​ω~tnωtn≤(C+t​C​eC​σ−λ)n−1​ω~tnωtn.\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) ω~tnωtn≤C​tn−m​ωXnω0m∧(t​ωX)n−m=CH≤Cσλ.\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

trωt​ω~t≤C​eC​σ−λ,\textrm{tr}_{\omega_{t}}\tilde{\omega}_{t}\leq Ce^{C\sigma^{-\lambda}},

for some uniform constant CC. But we also have ωt=ω0+t​ωX≤C​ωX\omega_{t}=\omega_{0}+t\omega_{X}\leq C\omega_{X} and so we get

trωX​ω~t≤C​eC​σ−λ.\textrm{tr}_{\omega_{X}}\tilde{\omega}_{t}\leq Ce^{C\sigma^{-\lambda}}.

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

K≤C+e−B​σ−λ​log⁡(C​eC​σ−λ)≤C.K\leq C+e^{-B\sigma^{-\lambda}}\log(Ce^{C\sigma^{-\lambda}})\leq C.

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