Lemma 10.7.1A0C109B78052

Proof. In linear coordinates at $x$ with $\mathrm dz_ j=\sqrt2\, \theta _ j$ we have $\sigma =\frac{\mathrm i}{2}\sum _ j\mathrm dz_ j\wedge \mathrm d\bar z_ j=\sum _ j\mathrm dx_ j\wedge \mathrm dy_ j$ at $x$, so the Riemannian metric of $\sigma $ at $x$ is Euclidean in the real coordinates, and $\langle \mathrm dx_ j+\mathrm i\, \mathrm dy_ j,\mathrm dx_ k\pm \mathrm i\, \mathrm dy_ k\rangle =(1\pm 1)\delta _{jk}$ shows that the covectors $\theta _ j,\bar\theta _ k$ form an orthonormal basis of the complexified cotangent space at $x$. Hence the products $\theta _ I\wedge \bar\theta _ J$ form an orthonormal basis of the complex covectors at $x$. Let $\iota _ j$ and $\bar\iota _ j$ be the adjoints of the exterior products with $\theta _ j$ and $\bar\theta _ j$; thus $\iota _ j$ maps $\theta _ j\wedge \theta _ K\wedge \bar\theta _ L$ to $\theta _ K\wedge \bar\theta _ L$ for $j\notin K$ and annihilates the basis covectors that do not contain $\theta _ j$, and similarly for $\bar\iota _ j$. Since $\sigma \wedge {}$ is $\mathrm i\sum _ j(\theta _ j\wedge )(\bar\theta _ j\wedge )$, its adjoint is $\Lambda _\sigma =-\mathrm i\sum _ j\bar\iota _ j\iota _ j$.

(1) The covector $\gamma \wedge v$ has bidegree $(n+1,2)$ and vanishes, so $[\gamma ,\Lambda _\sigma ]v=\gamma \wedge \Lambda _\sigma v$. We have $\iota _ j(e\wedge \bar\theta _ k)=(-1)^{j-1}\theta _1\wedge \cdots \wedge \widehat{\theta _ j}\wedge \cdots \wedge \theta _ n\wedge \bar\theta _ k$, and $\bar\iota _ j$ maps this to $(-1)^{j+n}\delta _{jk}\, \theta _1\wedge \cdots \wedge \widehat{\theta _ j}\wedge \cdots \wedge \theta _ n$, the factor $\bar\theta _ k$ standing in the $n$th place. Hence $\Lambda _\sigma v=-\mathrm i\sum _ j(-1)^{j+n}v_ j\, \theta _1\wedge \cdots \wedge \widehat{\theta _ j}\wedge \cdots \wedge \theta _ n$. In the product of the $j$th term with $\gamma $ only $\mathrm i\lambda _ j\theta _ j\wedge \bar\theta _ j$ contributes, and $\theta _ j\wedge \bar\theta _ j\wedge \theta _1\wedge \cdots \wedge \widehat{\theta _ j}\wedge \cdots \wedge \theta _ n=(-1)^{n-j}e\wedge \bar\theta _ j$; so $\gamma \wedge \Lambda _\sigma v=\sum _ j\lambda _ jv_ j\, e\wedge \bar\theta _ j$. Since the covectors $e\wedge \bar\theta _ j$ are orthonormal, this gives the formula for $\langle [\gamma ,\Lambda _\sigma ]v,v\rangle _{\sigma ,h}$, and the last assertion follows because the $\lambda _ j$ are nonnegative if $\gamma \geq 0$ and positive if $\gamma $ is positive definite.

(2) We may assume that $\alpha \neq 0$. Since the assertion does not involve $\gamma $, we may choose the $\sigma $-unitary coframe with $\theta _1=\alpha /|\alpha |_\sigma $, by completing this unit covector to an orthonormal basis of the $(1,0)$-covectors. Then $\mathrm i\alpha \wedge \bar\alpha =\mathrm i|\alpha |_\sigma ^2\theta _1\wedge \bar\theta _1$, so (1) gives $\langle [\mathrm i\alpha \wedge \bar\alpha ,\Lambda _\sigma ]v,v\rangle _{\sigma ,h}=|\alpha |_\sigma ^2|v_1|_ h^2$. Writing $u=fe$, we have $\bar\alpha \wedge u=(-1)^ n|\alpha |_\sigma f\, e\wedge \bar\theta _1$, hence $|\langle \bar\alpha \wedge u,v\rangle _{\sigma ,h}|\leq |\alpha |_\sigma |f|_ h|v_1|_ h$, and $|f|_ h=|u|_{\sigma ,h}$.

(3) Write $g=e\wedge \tau =\sum _ j\tau _ j\, e\wedge \bar\theta _ j$ with an $F$-valued $(0,1)$-covector $\tau =\sum _ j\tau _ j\bar\theta _ j$. By (1), $\langle [\gamma ,\Lambda _\sigma ]^{-1}g,g\rangle _{\sigma ,h}=\sum _ j|\tau _ j|_ h^2/\lambda _ j=|\tau |_\gamma ^2$, the squared norm of $\tau $ in the metric induced by $\gamma $ on $(0,1)$-covectors, which does not involve $\sigma $. Moreover $\mathrm dV_\sigma =\sigma ^ n/n!=c_ n\, e\wedge \bar e$ with $c_ n=\mathrm i^ n(-1)^{n(n-1)/2}$. Hence $\langle [\gamma ,\Lambda _\sigma ]^{-1}g,g\rangle _{\sigma ,h}\, \mathrm dV_\sigma =c_ n|\tau |_\gamma ^2\, e\wedge \bar e$. If $\kappa =\mu ^{-1}e$, $\mu \neq 0$, is another nonzero $(n,0)$-covector, then $g=\kappa \wedge \mu \tau $ and $c_ n|\mu \tau |_\gamma ^2\, \kappa \wedge \bar\kappa =c_ n|\tau |_\gamma ^2\, e\wedge \bar e$. Thus the density equals $c_ n|\tau '|_\gamma ^2\, \kappa \wedge \bar\kappa $ whenever $g=\kappa \wedge \tau '$, an expression that does not involve $\sigma $. In the same way, $|u|_{\sigma ,h}^2\, \mathrm dV_\sigma =c_ n|f|_ h^2\, \kappa \wedge \bar\kappa $ for $u=f\kappa $. If $\gamma \geq a\sigma $, all $\lambda _ j$ are at least $a$, and $\sum _ j|\tau _ j|_ h^2/\lambda _ j\leq a^{-1}\sum _ j|\tau _ j|_ h^2=a^{-1}|g|_{\sigma ,h}^2$. Finally, the eigenvalues of $\sigma '$ with respect to $\sigma '$ itself are all equal to one, so $[\sigma ',\Lambda _{\sigma '}]$ is the identity on $(n,1)$-covectors by (1), and the independence just proved gives

if $\sigma '\geq c\sigma $. $\square$