用户: Junyu*Cao/杂项/草稿

1Gauduchon 泛函

(参考了 2205.12480 的计算)

对于紧复流形 $X$, 我们对它的任何一个 Hermite 度量定义如下的 Gauduchon 泛函:

定义 1.1. $$G(\omega )=V^{\frac{1-n}{n}}{\int }_X|\eta {|}^2\mathrm{d}{\mathrm{v}\mathrm{o}\mathrm{l}}_{\omega }=\frac{1}{2}{V}^{\frac{1-n}{n}}{\int }_X|\theta {|}^2\mathrm{d}{\mathrm{v}\mathrm{o}\mathrm{l}}_{\omega }$$这里 $\mathrm{d}{\mathrm{v}\mathrm{o}\mathrm{l}}_{\omega }=\frac{{\omega }^n}{n!}$, ${\theta }^{1,0}=-\eta$, $\partial ({\omega }^{n-1})=-\eta \wedge \omega$

这个泛函的极小点如下

命题 1.2. $G(\omega )$ 的 Euler-Lagrange 方程为$$\sqrt{-1}(\partial \overline{\eta }-\overline{\partial }\eta -\eta \wedge \overline{\eta })=a\omega$$这里的 $a=\frac{1}{nV}{\int }_X|\eta {|}^2\mathrm{d}{\mathrm{v}\mathrm{o}\mathrm{l}}_{\omega }$

证明. 取如下的变分轨迹: $${\omega }_t=\omega +tH$$其中 $H$ 满足 ${\mathrm{t}\mathrm{r}}_{\omega }H=0$.

还需注意$$\mathrm{d}({\omega }_t^{n-1}{)}^{\prime }={\theta }_t^{\prime }\wedge {\omega }_t^{n-1}+{\theta }_t\wedge ({\omega }_t^{n-1}{)}^{\prime }$$这里 ${}^{\prime }$ 表示对 $t$ 求导.

细节之后补.

$a$ 的值如此看出: $$\sqrt{-1}(\partial \overline{\eta }-\overline{\partial }\eta -\eta \wedge \overline{\eta })\wedge {\omega }^{n-1}=a{\omega }^n$$注意$$\partial \overline{\eta }\wedge {\omega }^{n-1}=\mathrm{d}(\overline{\eta }\wedge {\omega }^{n-1})+\overline{\eta }\wedge \partial ({\omega }^{n-1})=\mathrm{d}(\overline{\eta }\wedge {\omega }^{n-1})+\eta \wedge \overline{\eta }\wedge {\omega }^{n-1}$$$$-\overline{\partial }\eta \wedge {\omega }^{n-1}=-\mathrm{d}(\eta \wedge {\omega }^{n-1})-\eta \wedge \overline{\partial }{\omega }^{n-1}=-\mathrm{d}(\eta \wedge {\omega }^{n-1})+\eta \wedge \overline{\eta }\wedge {\omega }^{n-1}$$积分可得 (回忆 $n\sqrt{-1}\eta \wedge \overline{\eta }\wedge {\omega }^{n-1}=|\eta {|}^2{\omega }^n$): $$a=\frac{{\int }_X|\eta {|}^2{\omega }^n}{n{\int }_X{\omega }^n}$$

$\square$

更近一步, 上述极小点是均衡度量, 亦即 $\eta =0$, $\mathrm{d}({\omega }^{n-1})=0$.

命题 1.3. 满足上述 Euler-Lagrange 方程的 $\omega$ 是均衡度量.

证明. 只需证 $a=0$. 否则 $a>0$, 取$${\omega }_0=\sqrt{-1}(\partial \overline{\eta }-\overline{\partial }\eta )=a\omega +\sqrt{-1}\eta \wedge \overline{\eta }>0$$记 $\psi =\sqrt{-1}(\overline{\eta }-\eta ),\alpha =-\sqrt{-1}\partial \eta$, 则 $\alpha \wedge \overline{\alpha }$ 是弱正的 (weakly positive, 限制在子空间上看), 且$$\mathrm{d}\psi ={\omega }_0+\alpha +\overline{\alpha }$$$$(\mathrm{d}\psi {)}^n={\omega }_0^n+(\genfrac{}{}{0ex}{}{n}{2}){\omega }_0^{n-2}\wedge \alpha \wedge \overline{\alpha }+(\genfrac{}{}{0ex}{}{n}{4}\left)\right(\genfrac{}{}{0ex}{}{4}{2}){\omega }_0^{n-4}\wedge (\alpha \wedge \overline{\alpha }{)}^2+\cdots \ge {\omega }_0^n$$于是$${\int }_X(\mathrm{d}\psi {)}^n\ge {\int }_X{\omega }_0^n>0$$而 $(\mathrm{d}\psi {)}^n$ 是恰当形式, 这导出矛盾, 证毕.

$\square$