PomeProposition 23.2.14.E5F423FF7707 On the interval where $\eta _ t=\omega +t\kappa _\lambda >0$, the polynomial $V(t)=\int _ X\eta _ t^ n$ has degree at most $n-1$ and
\[ V'(t)=n(n-1)\int _ XQ\wedge \omega \wedge \eta _ t^{n-2}>0. \]
23.2.13 The least eigenvalue of Lee curvature942B6FCFDC1A
Fix a strictly LCK metric $\omega $ with Lee form $\lambda $. The volume along $\omega +t\kappa _\lambda $ increases as long as this form is positive. If the least relative eigenvalue of $\kappa _\lambda $ is a negative constant, the first boundary form has a kernel everywhere and its total volume is zero. This contradiction gives a pointwise positivity criterion.
Set $Q=\lambda \wedge \lambda ^{\mathrm c}\geq 0$. The form $Q$ is nonzero on an open set, since $\lambda $ is not exact and hence is not identically zero.
Proof. Since $\mathrm d\eta _ t=\lambda \wedge \omega $ and $\mathrm d\lambda ^{\mathrm c}=-\kappa _\lambda $, Stokes' theorem gives
\[ 0=\int _ X\mathrm d(\lambda ^{\mathrm c}\wedge \eta _ t^{n-1}) =-\int _ X\kappa _\lambda \wedge \eta _ t^{n-1} +(n-1)\int _ XQ\wedge \omega \wedge \eta _ t^{n-2}. \]
Differentiating the wedge polynomial proves the formula. The integrand on the right is positive wherever $\lambda \ne 0$, including when $n=2$. Its integral is therefore strictly positive. Finally, $\kappa _\lambda $ is closed and de Rham exact, so
\[ \int _ X\kappa _\lambda ^ n =-\int _ X\mathrm d(\lambda ^{\mathrm c}\wedge \kappa _\lambda ^{n-1})=0. \]
This removes the coefficient of degree $n$. $\square$
Theorem 23.2.15.B8583A0E17A6 If the least eigenvalue of $\kappa _\lambda $ relative to $\omega $ is constant on $X$, then it is zero and $\kappa _\lambda $ is semipositive. Its ordinary Bott–Chern class is nonzero and pseudoeffective.
If all relative eigenvalues are constant, the rank is constant and lies between $1$ and $n-1$. In particular, the conclusion holds on a compact quotient $\Gamma \backslash G$ with left-invariant complex structure and strictly LCK tensors.
Proof. Let $r$ be the constant least relative eigenvalue. If $r<0$, put $t_0=-1/r$. Then $\eta _ t>0$ for $0\leq t<t_0$, whereas $\eta _{t_0}$ is semipositive with a kernel at every point. Thus $V(t_0)=0$. The preceding proposition and polynomial continuity instead give $V(t_0)\geq V(0)>0$. This proves $r\geq 0$ and semipositivity.
The nonnegative smooth form $\kappa _\lambda ^ n$ has integral zero by the same proposition, hence vanishes everywhere. Thus the least eigenvalue is zero at every point. Applying the Lee degree (23.2.8) to a conformal Gauduchon representative, as above, proves that the class is nonzero. The form $\kappa _\lambda $ itself represents it positively.
Constant spectrum gives constant rank. The positive pairing (23.2.8) excludes rank zero, and the vanishing top power excludes rank $n$. On the stated compact quotient, the relative matrix is constant in a left-invariant frame, which proves the last assertion. $\square$
The class is also nef: the same representative is bounded below by $-\epsilon h$ for every Hermitian form $h$ and every $\epsilon >0$. Thus proposition 23.2.11 excludes balanced metrics under the least-eigenvalue hypothesis.
The hypothesis concerns the chosen LCK metric. The other eigenvalues may vary, as the following examples show. They also explain why first degeneration of a volume ray need not force its total volume to vanish without the constant-minimum condition.
Example 23.2.16.BBCE95D98A62 On the scalar Hopf manifold $H_ n=(\mathbb C^ n\setminus \{ 0\} )/\langle z\mapsto z/2\rangle $, for every $n\geq 2$, there is a strictly LCK metric with indefinite Lee curvature and a smooth semipositive representative of its ordinary Bott–Chern class. The first boundary reached for increasing $t>0$ in $\omega +t\kappa _\lambda $ has positive total volume.
Proof. Fix $0<\epsilon <1$ and put
\[ \rho =\sum _ j|z_ j|^2,\qquad \Phi =\rho +\epsilon \operatorname {Re}(z_1^2),\qquad \omega =\frac{\mathrm{dd}^{\mathrm{c}}\rho }{\Phi },\qquad \lambda =-\mathrm d\log \Phi . \]
The bounds $(1-\epsilon )\rho \leq \Phi \leq (1+\epsilon )\rho $ give positivity. Since $\mathrm{dd}^{\mathrm{c}}\Phi =\mathrm{dd}^{\mathrm{c}}\rho $ and both $\Phi $ and $\mathrm{dd}^{\mathrm{c}}\rho $ scale by $1/4$ under $z\mapsto z/2$, the metric and Lee form descend and satisfy the LCK equation. Polar coordinates identify the quotient with $S^{2n-1}\times (\mathbb R/(\log 2)\mathbb Z)$, so it is compact and connected. The Lee period from $z$ to $z/2$ is $2\log 2$, which proves strictness.
Here $\kappa _\lambda =\mathrm{dd}^{\mathrm{c}}\log \Phi $. With $F_ i=\bar z_ i+\epsilon z_1\delta _{i1}$, its complex Hessian is
\[ (\log \Phi )_{i\bar j} =\frac{\delta _{ij}}{\Phi } -\frac{F_ i\overline{F_ j}}{\Phi ^2}. \]
Relative to $\omega $, the eigenvalues are $1$ with multiplicity $n-1$ and $h=1-|F|^2/\Phi $. Writing $z_1=x+\mathrm i y$ and $B=\sum _{j>1}|z_ j|^2$, one finds
\[ h=\frac{\epsilon ((1-\epsilon )y^2-(1+\epsilon )x^2)}{(1+\epsilon )x^2+(1-\epsilon )y^2+B}, \qquad -\epsilon \leq h\leq \epsilon . \]
At $(1,0,\ldots ,0)$ the spectrum is $-\epsilon ,1,\ldots ,1$, hence is indefinite. The bounded smooth function $u=\log \rho -\log \Phi $ is deck invariant, and
\[ \kappa _\lambda +\mathrm{dd}^{\mathrm{c}}u=\mathrm{dd}^{\mathrm{c}}\log \rho \geq 0, \]
by the Cauchy–Schwarz inequality for the Hessian of $\log \rho $. The correction also gives the standard Hopf conformal representative: $\mathrm e^{-u}\omega =\mathrm{dd}^{\mathrm{c}}\rho /\rho $.
The first boundary occurs at $t_0=1/\epsilon $. At that value,
\[ 1+h/\epsilon =\frac{2(1-\epsilon )y^2+B}{\Phi }. \]
It vanishes only on the real $z_1$ rays with the remaining coordinates zero. The other relative eigenvalues of $\omega +t_0\kappa _\lambda $ equal $1+1/\epsilon >0$. The boundary form is therefore semipositive and its top power is positive on a nonempty open set, proving positive total volume. $\square$
Example 23.2.17.EB59EA55107F For every $n\geq 2$, the same $H_ n$ carries a family of strictly LCK metrics whose least Lee-curvature eigenvalue is identically zero and whose remaining relative spectrum is nonconstant. Thus constancy of the least eigenvalue is strictly weaker than constancy of the whole spectrum.
Proof. Put $q=|z_1|^2$ and, for $0<\epsilon <1$, define
\[ \Psi =\rho +\epsilon \frac{q^2}{\rho },\qquad \omega _\epsilon =\frac{\mathrm{dd}^{\mathrm{c}}\Psi }{\rho },\qquad \lambda =-\mathrm d\log \rho . \]
For $a=q/\rho \in [0,1]$, direct differentiation gives
\[ \left(\frac{q^2}{\rho }\right)_{i\bar j} =\frac{2}{\rho }(q_ i-a\rho _ i)(q_{\bar j}-a\rho _{\bar j}) +2a\delta _{i1}\delta _{j1}-a^2\delta _{ij}. \]
The first two terms are positive semidefinite and $a^2\leq 1$, so $\mathrm{dd}^{\mathrm{c}}\Psi \geq (1-\epsilon )\mathrm{dd}^{\mathrm{c}}\rho >0$. The potential is smooth on the punctured space and scales by $1/4$ under the deck map. Hence $\omega _\epsilon $ descends, satisfies the LCK equation and has Lee period $2\log 2$.
Its Lee curvature is $\mathrm{dd}^{\mathrm{c}}\log \rho $, which is semipositive of rank $n-1$. The least relative eigenvalue is consequently zero everywhere. At $(1,0,\ldots ,0)$, the Hessian of $q^2/\rho $ is $\operatorname {diag}(1,-1,\ldots ,-1)$, giving the spectrum
\[ 0,(1-\epsilon )^{-1},\ldots ,(1-\epsilon )^{-1}. \]
At $(0,1,0,\ldots ,0)$ the term $q^2/\rho $ vanishes to fourth order, so its Hessian is zero and the spectrum is $0,1,\ldots ,1$. These spectra differ for every $\epsilon >0$. $\square$
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