12.17 Curvature pairs and positive lower massC18EABD7D630

Proof. Locally write $u=p-a$, with $p$ bounded plurisubharmonic and $a$ smooth. Expanding in $p$ and $a$ and using multilinearity, we obtain closed currents $C_ r=(\mathrm{dd}^{\mathrm{c}}u)^ r$, not necessarily positive, with $C_0=1$. The recursive definition of bounded products, expanded in the same way, gives

By lemma 5.27.2 and invariance under smooth corrections, the weighted product $u\mathrm{dd}^{\mathrm{c}}u$ does not depend on the local decomposition. On a smaller ball choose smooth bounded plurisubharmonic $p_ j$ decreasing to $p$, and put $u_ j=p_ j-a$. The same lemma gives $u_ j\mathrm{dd}^{\mathrm{c}}u_ j\to u\mathrm{dd}^{\mathrm{c}}u$, while boundedness gives $u_ j^2\to u^2$ in $L^1$. Hence

The smooth currents on the left are positive, so their weak limit $E_ u$ is positive and has order zero.

Now expand the cube. Repeated use of distributional Stokes gives

Since $\mathrm{dd}^{\mathrm{c}}g$ is closed, $\int _ X\mathrm{dd}^{\mathrm{c}}(u^2)\wedge \mathrm{dd}^{\mathrm{c}}g=0$, so the last integral equals $-\int _ XE_ u\wedge \mathrm{dd}^{\mathrm{c}}g$. This proves (12.17.3). If $-\mathrm{dd}^{\mathrm{c}}g\geq 0$, the correction term is nonnegative, and $u=0$ attains the lower bound $\int _ Xg^3$. $\square$

Proof. We first check that the determinant measure in (12.7.1) does not depend on the Hermitian form used to express it. Let $T$ be a positive $(2,2)$-current, and let $\sigma _ h=h\wedge T$ and $P=P_ h$ be its trace measure and coefficient matrix relative to $h$, as in section 12.7. In an $h$-unitary frame let $G$ be the matrix of $g$, and set

Comparing the defining trace pairings shows that $\sigma _ g=g\wedge T$ and $P_ g$ are the trace measure and coefficient matrix of $T$ relative to $g$, with $\operatorname {tr}P_ g=3$. Since $G>0$ and $\operatorname {tr}P=3$, one has $a>0$ even where $P$ is singular. Moreover $\det P_ g=a^{-3}\det G\det P$ and $g^3=\det G\, h^3$, so the factors $a$ and $\det G$ cancel in $\Gamma _0(T)$. The calculation applies to the matrix density on the whole trace measure, including its singular part.

For $T\in \mathscr G_ h$ put $m=\int _ Xg\wedge T$. Then $0<m\leq c^{-1}$ and $U=T/m\in \mathscr G_ g$. By homogeneity, $k_ T=m k_ U$. Applying theorem 12.9.2 to $g$ gives the norm bound, and applying it again to $h$ gives the lower-mass bound. The supremum here is taken over all of $\mathscr G_ h$, including singular and degenerate currents. $\square$

Proof. We suppress pullbacks in the curvature calculation. Comparing types in $\mathrm d\eta =F$ gives $\partial \eta =0$, $\bar\partial \eta =F$, $\partial \bar\eta =\bar F$ and $\bar\partial \bar\eta =0$. Hence

Since $e^2=0$ and the base has dimension two,

The form $g$ is positive, because $\pi ^*\omega $ is positive on directions transverse to the fibers and $e$ is positive on the vertical line. Fiber integration of $g^3$ gives $V$, and lemma 12.17.2, lemma 12.17.4 give the mass identity and the endpoint bound. By compactness, every Hermitian form $h$ satisfies $h\geq cg$ for some $c>0$.

Finally, $S=-\mathrm{dd}^{\mathrm{c}}g$ is smooth, semipositive and not identically zero. If a Hermitian form $b$ were pluriclosed, Stokes' theorem would give

where the last inequality holds because $b$ is strictly positive and $S$ is semipositive and nonzero on an open set. $\square$

Proof. One has $\mathrm d(g^2)=2s\pi ^*\omega \wedge \mathrm de$, and $\mathrm de$ has one component containing $\eta $ and one containing $\bar\eta $. These two vertical types are independent, so $\mathrm d(g^2)=0$ if and only if $\omega \wedge F=0$, which is equivalent to the two stated real identities. A primitive real $(1,1)$-form on a surface has eigenvalues $\lambda ,-\lambda $ in an $\omega $-unitary frame, so its square is $-\lambda ^2\omega ^2$. Applying this to $\alpha $ and to $\beta $ gives the sign of $Q$. In this case $\mathrm{dd}^{\mathrm{c}}g\geq 0$, so (12.17.3) bounds the masses of bounded potentials from above rather than from below, and gives no information on the lower mass. $\square$

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. $\square$

Proof. Every orbit contains a unique point with $|z|=|w|=1$: the required parameter is determined by

These smooth formulas identify $(\mathbb C^2\setminus \{ 0\} )^2$ with $\mathbb C\times (S^3\times S^3)$, so the action is free and proper, and $X_\tau $ is diffeomorphic to $S^3\times S^3$. To obtain holomorphic charts, choose a nonzero coordinate $z_ k$ of $z$ and a branch of its logarithm, and use the action to normalize $z_ k=1$; the functions $z_ l/z_ k$, $l\ne k$, and $w_ jz_ k^{-\tau }$ are then local holomorphic coordinates on the quotient. Changing the branch of the logarithm accounts for the periods along the fibers. Thus $X_\tau $ is a compact complex threefold.

The invariant form $\eta $ vanishes on the generator $z\partial _ z+\tau w\partial _ w$ of the action, so it descends. On an elliptic fiber it equals $\mathrm d\chi $, where $\chi =\tau \log v_1-\log v_2$ in terms of fiber coordinates $(v_1z,v_2w)$; its period lattice is generated by $2\pi \mathrm i\tau $ and $-2\pi \mathrm i$. Hence $A_ E=8\pi ^2B_0$. Direct differentiation gives

Since $\int _ B(\omega _1+\omega _2)^2=2$, theorem 12.17.7 gives the lower mass, the endpoint bound and the absence of pluriclosed metrics.

Moreover $\mathrm d\operatorname {Re}\eta =-2\pi B_0\pi ^*\omega _1$. This nonzero semipositive exact form excludes every balanced metric by the Stokes argument at the end of the preceding proof, and hence also every Kähler metric. The quotient is simply connected, so the Lee form of a hypothetical LCK metric would be exact, equal to $\mathrm df$; multiplying the metric by $\mathrm e^{-f}$ would then give a Kähler metric, a contradiction. In particular $X_\tau $ carries no Vaisman metric, that is, no LCK metric with parallel Lee form. $\square$