PomeExample 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.
Example 23.2.17EB59EA55107F
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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