Proposition 12.17.113E19D4A0EA58

Proof. In product coordinates, the squared norm of the standard local frame of $L$ is $(1+|z_1|^2)^ a(1+|z_2|^2)^ b$. Therefore

Dilation changes $r$ by a constant, so $\eta $ descends to $X$. The function $r$ changes by the nonzero constant $\log \rho ^2$ under the generator, so the action is free and properly discontinuous; the unit circle bundle, which is compact, times a fundamental interval for $r$ maps onto $X$, which is therefore compact. Each fiber is $\mathbb C^*/\rho ^{\mathbb Z}$. Its logarithmic coordinate has periods $\log \rho $ and $2\pi \mathrm i$, and the curvature $F=2\pi \mathrm i\Omega $ has real part $0$ and imaginary part $2\pi \Omega $; hence

Theorem 12.17.7 now gives (12.17.12) and the endpoint bound, and excludes pluriclosed metrics.

To prove $H^2(X,\mathbb R)\ne 0$, put $A=\eta +\bar\eta $ and $B_1=\mathrm i(\eta -\bar\eta )$. Then $\mathrm dA=0$ and $\mathrm dB_1=-4\pi \pi ^*\Omega $. Suppose that $\pi ^*(\omega _1-\omega _2)=\mathrm d\zeta $ for a real one-form $\zeta $. Averaging $\zeta $ over the compact group of elliptic translations, we may assume $\zeta $ invariant, so that

for a one-form $\xi $ and functions $f,j$ on $B$. The components of $\mathrm d\zeta $ with one vertical and one horizontal direction must vanish, which gives $\mathrm df=\mathrm dj=0$. Thus

Integrating over the two rulings gives $1=-4\pi ja$ and $-1=-4\pi jb$, which is impossible because $a,b>0$. A Hopf threefold as above is diffeomorphic to $S^5\times S^1$, whose second real cohomology vanishes.

Finally, $\pi ^*\Omega =-\mathrm dB_1/(4\pi )$ is a nonzero semipositive exact form. Its integral against the closed positive square of a balanced metric would vanish by Stokes' theorem and be strictly positive by positivity. Hence no balanced metric exists. $\square$