12.6 The Hessian metric and its completion466CEDF9AEBA

Proof. Homogeneity gives $\lambda (s\theta )=s\lambda (\theta )$ for $s>0$. Hence $r=\log \lambda (\theta )$ and $q=\theta /\lambda (\theta )$ define the smooth inverse of $\Phi $. The level set is smooth because the Euler identity gives $D\lambda _\theta (\theta )=\lambda (\theta )>0$.

Differentiating the Euler identity $D\log \lambda _\theta (\theta )=1$ in a direction $\chi \in V$ gives

For directions $\chi ,\eta \in V$, differentiating homogeneity twice also gives

Let $q\in \mathscr S$, let $\xi ,\eta \in T_ q\mathscr S$, and let $a,b,r$ be real numbers. Since $\xi $ and $\eta $ are tangent to a level set of $\lambda $, the first of these identities shows that both are $g_ q$-orthogonal to the radial vector $q$. Using also the second identity and the scaling rule, we obtain

which proves the metric identity.

It remains to prove the distance formula and its consequences. A path $(r(t),q(t))$ in the product has length $\int \sqrt{|r'|^2+|q'|_{g_{\mathscr S}}^2}\, \mathrm{d}t$. The triangle inequality in $\mathbb R^2$ (applied to the integral of the vector $(|r'|,|q'|_{g_{\mathscr S}})$) bounds this below by the square root of the sum of the squared endpoint difference in $r$ and the squared length of the projected path in $\mathscr S$. For the opposite bound, choose a path in $\mathscr S$ of length within $\epsilon $ of the distance of its endpoints, parametrize it with constant speed, and let $r$ vary linearly; then let $\epsilon $ decrease to zero. This proves the distance formula. It identifies Cauchy sequences in $\mathscr C$ with pairs of Cauchy sequences in the two factors; since $\mathbb R$ is complete, the completeness assertion follows. Finally, the length of any path is at least the total change of $\log \lambda $ along it, and this change is infinite when $\lambda $ tends to zero or to infinity. $\square$

Proof. The projection from the blown-up point extends to a morphism $f:X\to \mathbb P^1$ with class $L=H-E$. The incidence map $(\pi ,f):X\to \mathbb P^2\times \mathbb P^1$ is an embedding. If $h$ and $\ell $ are pullbacks of Fubini–Study forms of integral one, then

is positive for $a>b>0$, by the product embedding. Conversely, a positive representative has positive integrals over the exceptional curve and the strict transform of a line through the blown-up point. These integrals are respectively $b$ and $a-b$, proving the cone description. The same curves show that no nonzero linear combination in this two-dimensional space is smoothly $\mathrm{dd}^{\mathrm{c}}$-exact.

The intersection numbers follow from the hyperplane pullback and the normal bundle $\mathcal O_{\mathbb P^1}(-1)$ of the exceptional curve; equivalently $H^2=H\cdot L=1$ and $L^2=0$. Since the classes are closed, integrating the smooth equation gives $\mathcal V_\mu =a^2-b^2$, independently of $\mu $. Direct differentiation of $-\log \sqrt{a^2-b^2}$ gives

Substituting the stated coordinates gives $\mathrm{d}r^2+\mathrm{d}s^2$. The coordinate domain is exactly $\mathbb R\times (0,\infty )$, so the intrinsic distance and completion are those of that Euclidean half-plane. Finally, $H$ has a semipositive pullback representative and is a limit of positive classes, hence is nef. Its square is one, but its integral over $E$ is zero, so it has no positive representative. This proves all assertions about the added boundary. $\square$

Proof. Invariant averages and balancedness. Let $Z_1,\ldots ,Z_ n$ be the invariant holomorphic vector fields dual to $\alpha _1,\ldots ,\alpha _ n$, and set $\Upsilon =\alpha _1\wedge \cdots \wedge \alpha _ n$. This holomorphic top-degree form and its conjugate are closed. For an invariant holomorphic vector field $Z$, evaluation on the invariant frame gives

The second identity also follows from Cartan's formula. Since $h_0^ n$ is a positive constant multiple of the appropriately oriented form $\Upsilon \wedge \bar\Upsilon $, Stokes' theorem on compact $X$ gives

Thus the fields $Z_ j$ and their conjugates have zero divergence with respect to $\mu $. In particular,

Cartan's formula also gives $\mathrm d(\iota _{Z_ j}\Upsilon )=0$. These contractions span the invariant $(n-1,0)$-forms. The form $\theta _ A^{n-1}$ is a constant linear combination of their wedges with conjugates, so $\mathrm d\theta _ A^{n-1}=0$.

The holomorphic fields commute with the antiholomorphic fields, and $\mathrm d\alpha _ j$ has type $(2,0)$. The coefficient matrix of $\mathrm{dd}^{\mathrm{c}}u$ is therefore $(Z_ j\bar Z_ k u)$. For smooth $u$ its entries have $\mu $-mean zero by (12.6.8), so the matrix of $\mu $-means of the coefficients of $\theta _ A+\mathrm{dd}^{\mathrm{c}}u$ is $A$. If $\theta _ A+\mathrm{dd}^{\mathrm{c}}u>0$, averaging the positive coefficient matrix gives $A>0$. The converse follows by taking $u=0$. The same average shows that $V_ W$ contains no nonzero smooth $\mathrm{dd}^{\mathrm{c}}$ direction.

For a distribution $u$, the adjoints of $Z_ j$ and $\bar Z_ k$ with respect to $\mu $ annihilate constants, so the mean of each $Z_ j\bar Z_ k u$ is still zero. The coefficients of a positive current in this global frame form a positive matrix of measures, whose matrix of total masses is semipositive; for $\theta _ A+\mathrm{dd}^{\mathrm{c}}u\geq 0$ this gives $A\geq 0$. Conversely, $\theta _ A$ itself is a positive current when $A\geq 0$. If $A=0$, the positive trace measure of $\mathrm{dd}^{\mathrm{c}}u$ has total mass zero, and polarization then makes the entire current zero. Finally, for a fixed positive $A_+\in W$, the matrices $A+\varepsilon A_+$ with $\varepsilon \downarrow 0$ approximate every semipositive $A\in W$ from the positive cone. This proves the closure assertion.

The quotient and its equality case. Put $Q=H\det A$, so that $\theta _ A^ n=Q\mu $. For every smooth positive Hermitian form $\gamma $, the matrix arithmetic–geometric mean inequality gives

Integration yields $\int _ X\theta _ A\wedge \gamma ^{n-1}\geq Q^{1/n}D_\mu (\gamma )$, which bounds every Gauduchon quotient in definition 12.2.1 from below. The balanced, hence Gauduchon, metric $\gamma =\theta _ A$ has $D_\mu (\theta _ A)=Q^{(n-1)/n}$ and attains equality. This proves the formula for $\lambda _\mu (\theta _ A)$.

For a minimizing Gauduchon metric $\gamma $, the continuous nonnegative difference of the two sides of the pointwise inequality has integral zero. Pointwise equality then forces $\gamma =f\theta _ A$ for a smooth positive function $f$. Set $q=f^{n-1}$ and $T=\theta _ A^{n-1}$. Since $\mathrm dT=0$, the Gauduchon equation becomes $\mathrm{dd}^{\mathrm{c}}q\wedge T=0$, and

The last integrand is nonnegative and vanishes exactly where $\mathrm dq=0$. Thus $f$ is constant. As $n\geq 2$ and

normalization determines $c=\lambda _\mu (\theta _ A)^{-1}$. This proves the unique minimizer and its volume identity.

Differentiating $-\log \lambda _\mu (\theta _ A)=-(\log H+\log \det A)/n$ twice gives (12.6.7). The corresponding squared norm of $U$ is $\| A^{-1/2}UA^{-1/2}\| _{\mathrm F}^2/n$, which is positive for $U\ne 0$; here $\| B\| _{\mathrm F}^2=\operatorname {tr}(B^*B)$ denotes the squared Frobenius norm.

Finally,

For invertible $A$, the vanishing of this form forces every $\mathrm d\alpha _ j=0$, by independence of the conjugate coframe, and the Lie algebra would then be abelian. Since adding $\mathrm{dd}^{\mathrm{c}}u$ does not change $\mathrm d\theta _ A$, this proves the last assertion. $\square$

Proof. By the trace formula, the congruence $M\mapsto P^*MP$ by a fixed invertible matrix $P$ is an isometry. Taking $P=A^{-1/2}$, we may assume that the endpoints are $I$ and $C=A^{-1/2}BA^{-1/2}>0$. The curve $\exp (t\log C)$, $0\leq t\leq 1$, has constant speed $\| \log C\| _{\mathrm F}/\sqrt n$, which gives the upper bound for the distance.

For the reverse inequality, diagonalization gives the convergent resolvent formula for the matrix logarithm and its derivative

On a neighborhood of $M$ where the spectrum is bounded below by a positive constant, differentiation under the integral sign is justified by an integrable bound. In a unitary basis diagonalizing $M$, with eigenvalues $d_ i>0$, the derivative has entries $c_{ij}U_{ij}$, where

Writing $d_ i/d_ j=\mathrm e^{2s}$ gives $\sqrt{d_ id_ j}|c_{ij}|=|s/\sinh s|\leq 1$, including the limiting value at $s=0$. Hence

Applying this along a piecewise $C^1$ path of positive matrices from $I$ to $C$ and integrating, we bound its length below by $\| \log C\| _{\mathrm F}/\sqrt n$. This proves the distance formula. In terms of the original endpoints, the length-minimizing curve is

Its determinant is $(\det A)^{1-t}(\det B)^ t$, so for endpoints with equal determinant it stays in the fixed-determinant hypersurface. $\square$

Proof. Paths in $W$. Every path in $S_ W$ is a path in the hypersurface of all positive matrices with $H\det A=1$, so lemma 12.6.9 gives (12.6.13). To obtain an upper bound by paths inside $W$, put

Since the positive cone is convex and $W$ is linear, $N(t)$ stays in $S_ W$ and joins $A$ to $B$. With $K=C^{-1/2}UC^{-1/2}$, differentiation gives

Indeed $C(t)\geq mI$ and $(\log \lambda (C))'=\operatorname {tr}(C^{-1}U)/n$. Integrating proves (12.6.14), and shows in particular that $S_ W$ is connected.

Put $d=d_ W(A,B)$. Every eigenvalue of $A^{-1/2}BA^{-1/2}$ has a logarithm of absolute value at most $\| \log (A^{-1/2}BA^{-1/2})\| _{\mathrm F}\leq \sqrt n\, d$, by (12.6.13). Hence

Since $\gamma _ A=\theta _ A$ on $S_ W$, this proves (12.6.15).

Cauchy sequences and bounded balls. For a Cauchy sequence $(A_ j)$ in $(S_ W,d_ W)$, comparison with one fixed element of the sequence, using (12.6.17), gives $mI\leq A_ j\leq MI$ with fixed positive $m,M$. Comparing two elements then gives

where $\| \cdot \| _{\mathrm{op}}$ is the operator norm. The coefficient matrices therefore form a Cauchy sequence. Their limit lies in the closed linear space $W$, is positive, and satisfies $H\det A=1$, and the upper bound (12.6.14) proves convergence to that point in $d_ W$. Conversely, coefficient convergence to a positive point gives convergence in $d_ W$ by the same bound. This proves completeness and the Cauchy criterion. The minimizing forms then converge in $C^\infty $, because their coefficients multiply a fixed smooth coframe.

By these estimates, $d_ W$ induces the topology of coefficient convergence. A closed ball of radius $R$ around $A_\star $ is contained in the compact matrix interval

The ball is closed in that interval, hence compact. This proves properness. In particular, distances from a fixed reference point are bounded exactly when the eigenvalues relative to that point are bounded above and away from zero.

The coefficient boundary. Write $\lambda _{\max }$ for the largest eigenvalue. If $A_ j\to A_0$ with $A_0\geq 0$ singular and nonzero, then $\lambda (A_ j)\to 0$ while $\lambda _{\max }(A_ j)$ has a positive limit, so the largest eigenvalue of $\widehat A_ j$ tends to infinity. Applying (12.6.17) to $A_\star ,\widehat A_ j$ gives

The error term is bounded because $A_0\ne 0$. This proves (12.6.16). For comparison, $A_ j=t_ jA_\star $ with $t_ j\downarrow 0$ normalizes to $A_\star $ for every $j$; its coefficient limit, the zero matrix, gives no new point of the completion. $\square$

Proof. The bilinear cross term makes the multiplication associative, with identity zero and inverse $(-z_1,-z_2,-z_3+z_1z_2)$. These operations preserve the Gaussian lattice. Its action by left multiplication is free and properly discontinuous: for a compact set $K\subset G$, elements $\gamma $ with $\gamma K\cap K\ne \varnothing $ belong to the compact set $KK^{-1}$, containing only finitely many lattice elements.

Choose Gaussian integers $\gamma _1,\gamma _2$ that move $z_1,z_2$ into the half-open unit square, and then $\gamma _3$ that moves $z_3+\gamma _1z_2$ into it. This gives a representative in the real six-dimensional unit cube. If two points of the half-open cube are equivalent, then $\gamma _1=\gamma _2=0$ and then $\gamma _3=0$. The cube is therefore a measurable fundamental domain, and $X$ is compact, being the image of its closure.

This coframe is left-invariant, descends to $X$, and has

The density of $h_0^3$ is $6/\pi ^3$ times real Lebesgue volume on the fundamental cube, giving the stated $H$. The coefficient of $\alpha _1\wedge \alpha _2\wedge \bar\alpha _3$ in $\partial \theta _ A$ is $-\mathrm iA_{33}/(2\pi )$, so positivity makes every such background nonclosed.

To see that $X$ admits no Kähler metric at all, suppose that a positive closed $(1,1)$-form $\kappa $ existed, and average its coefficient functions in this coframe against $\mu $. Their mean is a positive matrix $A$. In the coefficients of $\partial \kappa $, derivative terms have mean zero by (12.6.8); the remaining means are exactly the coefficients of $\partial \theta _ A$. Closedness would force them to vanish, contradicting $A_{33}>0$. Thus $X$ is non-Kähler.

The assertions about the volume, the minimizers and completeness follow from proposition 12.6.5, theorem 12.6.12. As $W$ is the full matrix space, the equal-determinant exponential path stays in $S$, and lemma 12.6.9 gives (12.6.19). For $A_\star =aI$, $a=(\pi ^3/6)^{1/3}$, this becomes

The three logarithms sum to zero.

For a concrete degeneration, take $A_\varepsilon =\operatorname {diag}(1,1,\varepsilon )$, $0<\varepsilon \leq 1$. As $\varepsilon \to 0$, the coefficients converge to the nonzero smooth semipositive form $\theta _{\operatorname {diag}(1,1,0)}$. The normalized matrices and their distances, on the other hand, are

Thus, after normalization, the boundary form $\theta _{\operatorname {diag}(1,1,0)}$ lies at infinite distance. $\square$

Proof. Since the forms $\mathrm d\alpha _ j$ are holomorphic, we have

Indeed $\bar\partial (\alpha _ j\wedge \bar\alpha _ k) =-\alpha _ j\wedge \mathrm d\bar\alpha _ k$, and applying $\partial $ and multiplying by $(\mathrm i/(2\pi ))^2$ gives the identity in (12.6.24). Diagonalizing the positive coefficient matrix expresses the result as a sum of nonnegative multiples of $\beta \wedge \bar\beta $ for invariant $(2,0)$-forms $\beta $. In dimension three these are nonnegative $(2,2)$-forms: the wedge product with $\mathrm i\xi \wedge \bar\xi $, for a $(1,0)$-form $\xi $, has the squared modulus of $\xi \wedge \beta $ as coefficient.

By proposition 12.6.5, $\theta _ A$ is balanced, hence Gauduchon. The maximum case of theorem 12.16.3 therefore gives $v_+(\theta _ A)=\int _ X\theta _ A^3=H\det A$. The supremum of $\mathcal V_\nu (\theta _ A)$ over $\nu \in \mathscr M$ equals this upper mass by theorem 12.15.4, and (12.6.6) shows that it is attained at $\mu _0$. Finally, proposition 13.14.15 makes the finiteness of the upper mass a property of the manifold. $\square$

Proof. For $\varepsilon >0$, the same bounded $u$ is $\theta _{A+\varepsilon I}$-plurisubharmonic, so the preceding proposition bounds its mass by $H\det (A+\varepsilon I)$, which tends to zero with $\varepsilon $. By the multilinear construction of bounded products in proposition 5.28.2,

is a polynomial in $\varepsilon $ whose coefficients are mixed products of finite mass. Its total mass therefore converges to $\int _ X(\theta _ A+\mathrm{dd}^{\mathrm{c}}u)^3$ as $\varepsilon \downarrow 0$, so this mass is zero, and positivity makes the measure zero.

For unbounded $u$, we subtract a constant so that $u\leq 0$. Since $\theta _ A\geq 0$, the truncations $u_ k=\max (u,-k)$ are bounded and $\theta _ A$-plurisubharmonic, so their top products vanish by the bounded case. Restricting them to $\{ u>-k\} $ and passing to the limit that defines the non-pluripolar product gives zero as well. $\square$

Proof. The map $p$ is well defined because multiplication by lattice elements adds Gaussian integers to the first two coordinates. It is a submersion, and fixing representatives of those coordinates identifies each fiber with the central quotient $E$. Integrating first in the third coordinate over the fundamental cube proves the pushforward identity.

The closed equation on the base. Write the positive Hermitian matrix in block form and take its Schur complement:

On $B$, let

We apply theorem 14.2.1 to this closed background with the density $Q_ Bf\mu _ B$. Since $\omega _ S$ is closed and the density has total mass $Q_ B$, Stokes' theorem forces the constant in the equation to be one. We obtain a smooth real $v=v_{A,f}$, unique up to an additive constant, such that

If $K$ is the coefficient matrix of $\mathrm{dd}^{\mathrm{c}}v$ in the base coframe, this says $\det (S+K)=(\det S)f$. The form

has coefficient matrix $\left(\begin{smallmatrix}C+K& b\\ b^*& a\end{smallmatrix}\right)$ in the invariant coframe. Since $a>0$ and the Schur complement $S+K$ is positive, $\Omega >0$; its determinant is $a\det (S+K)=(\det A)(f\circ p)$. Consequently $\Omega ^3=(H\det A)\mu _ f$.

This solution form is also balanced. Put $\beta =\mathrm{dd}^{\mathrm{c}}(p^*v)$, which is closed and uses only $\alpha _1,\alpha _2$ and their conjugates. Since $\mathrm d\alpha _3=-\alpha _1\wedge \alpha _2$, every term of $\partial \theta _ A$ contains $\alpha _1\wedge \alpha _2$, and every conjugate term contains $\bar\alpha _1\wedge \bar\alpha _2$. Thus $\mathrm d\theta _ A\wedge \beta =0$. Together with $\mathrm d\theta _ A^2=0$, this proves

Theorem 12.2.3 now gives (12.6.27). Its characterization of the minimizers, the balancedness of $\Omega $ and the homogeneity of degree two of $D_{\mu _ f}$ give the unique normalized minimizer (12.6.28).

All maximizing densities. Let $Z_3=\partial /\partial z_3$ be dual to $\alpha _3$, and set $\tau _ j=(\mathrm i/(2\pi ))\alpha _ j\wedge \bar\alpha _ j$. Directly from (12.6.24),

Since $\theta _ A$ is Gauduchon, the identity (12.16.4) therefore becomes, for every smooth real $u$,

Here $\mathrm du\wedge \mathrm d^{\mathrm c}u=(\mathrm i/(2\pi ))\partial u\wedge \bar\partial u$, and wedging with $\tau _1\wedge \tau _2$ selects its coefficient on $\alpha _3\wedge \bar\alpha _3$; this also accounts for the factor $1/2$.

For an arbitrary $\nu \in \mathscr M$, let $u_\nu $ be the smooth solution of the equation for $\theta _ A$ and $\nu $. Its mass equals $\mathcal V_\nu (\theta _ A)$ by theorem 12.2.3. By (12.6.30), this mass equals $H\det A$ exactly when $Z_3u_\nu =0$. As $u_\nu $ is real, its derivatives in both real directions of each elliptic fiber then vanish, and connectedness of the fibers gives $u_\nu =p^*v$ for a smooth $v$ on $B$. The density of $(\theta _ A+\mathrm{dd}^{\mathrm{c}}u_\nu )^3$ relative to $\mu _0$ is then a function on $B$, so the equation makes $\nu $ fiber invariant. The converse was proved by the construction on the base.

Completion in the background parameters. For fixed $f$, formula (12.6.27) holds on the entire cone of positive matrices, and differentiating the logarithm of its cube root twice gives the stated Hessian metric. Since this is the same metric as in theorem 12.6.12, the matrix-path and coefficient-limit arguments of that proof apply unchanged on each fixed $W$. They give the distance bounds, completeness, properness and the Cauchy criterion; they do not use that the minimizing forms have constant coefficients. $\square$

Proof. Pairing a smooth positive representative $\theta +\mathrm{dd}^{\mathrm{c}}u$ with the closed semipositive form $\kappa ^{m-1}\wedge w$ gives $b>0$, and its integral over the compact fiber $\{ y\} \times C$ is $L(y)>0$, by Stokes' theorem. These are exactly the stated inequalities; conversely, they make $\theta $ itself positive. The same tests applied to a smooth $\mathrm{dd}^{\mathrm{c}}$ direction give $b=0$ and $ca+e=0$ everywhere, and since $a$ is nonconstant, $c=e=0$. Also $\mathrm{d}\theta =c\, \mathrm{d}a\wedge w$, which is not identically zero when $c\ne 0$.

Next we apply theorem 14.2.1 on $B$ to the closed positive background $b\kappa $ and the density $(b^ m/H)\sigma /L$. Since this density has integral $b^ m=\int _ B(b\kappa )^ m$, Stokes' theorem forces the constant in the equation to be one, and we obtain $A$. Put $\Omega =A+Lw$. Since $A^{m+1}=0$ and $(Lw)^2=0$, the binomial expansion gives

The first summand is the pullback of a top-degree form on $B$, and the second is closed. Hence $\gamma $ is balanced, and in particular Gauduchon. For every Gauduchon metric $\eta $ on $X$, Stokes' theorem and the pointwise arithmetic–geometric mean inequality give

For $\eta =\gamma $ the numerator is $nb^ m$, while $D_\mu (\gamma )=(nb^ m/H)^{m/n}H$. The quotient of $\gamma $ therefore attains the lower bound, which proves (12.6.33) and the minimizing property of $\gamma $. $\square$

Proof. Consider an affine variation $\delta =(\delta b,\delta c,\delta e)$ of the coefficients, write $F=\delta L/L$, and note that $\sigma /(HL)=\mathrm{d}\nu _ v$ because $HL=Q(v)(1+va)$. Differentiating under the integral sign gives

where $\mathbb E_{\nu _ v}$ denotes expectation with respect to $\nu _ v$. This is positive unless $\delta L=0$ everywhere, which forces $\delta c=\delta e=0$. Since $\log \lambda _\mu =(\log n+m\log b-\log H)/n$,

is positive definite. On the unit-volume hypersurface, the relation $H(c,e)=Q(v)/e=nb^ m$ gives

To pull back the Hessian, write $t=\mathrm{d}e/e$ and $z=Q'/Q$. Along an affine path of coefficients one has $v''=-2tv'$, so differentiating $\log H=-\log e+\log Q(v)$ twice gives

On $\mathscr S$ we have $t=z\, \mathrm{d}v-m\, \mathrm{d}U$; substituting this and using $n=m+1$ gives (12.6.37). Since $Q'/Q=-\mathbb E_{\nu _ v}\ell _ v$ and $Q''/Q=2\mathbb E_{\nu _ v}\ell _ v^2$, the variance formula follows, and nonconstancy of $a$ makes this variance positive. In the coordinates $(Y,R)$ the metric is Euclidean on the convex set $\mathbb R\times I$, so straight segments prove the distance assertion. $\square$

Proof. At the end $v\to \infty $, where $\epsilon =1/v$, we have

At the end $v\to -1/M$, where $\epsilon =-1/v-M$, we obtain

In both cases $\ell _ v$ is an affine function of $1/(f+\epsilon )$ with coefficients depending only on $\epsilon $, and (12.6.40) expresses $\rho (v)\, \mathrm{d}v^2$ through the variance of $1/(f+\epsilon )$ under the probability $\sigma /[H_ f(f+\epsilon )]$. This probability is $\nu _ v$, the marginal in (12.6.35), because $L$ is a positive multiple of $f+\epsilon $ and positive scaling of $L$ does not change that marginal.

Integration over tubes around the components $Z_ i$, splitting the normal radius at $\epsilon ^{1/k_ i}$, gives

The analogous threshold for $J_ f$ is two. For a finite sum of tube integrals, the smallest ratio governs.

If $r<1$, choose a component attaining $r$, and two normal annuli at fixed multiples of $\epsilon ^{1/k_ i}$, so far apart that the two-sided bounds separate their ranges of $f/\epsilon $. Each annulus has probability bounded below, since its weighted mass and $H_ f$ both have order $\epsilon ^{r-1}$, and the values of $1/(f+\epsilon )$ on the two annuli differ by at least $c\epsilon ^{-1}$. The identity

applied to $T=1/(f+\epsilon )$ gives the lower bound, and $T\leq \epsilon ^{-1}$ gives the upper bound.

If $r=1$, the near annulus has probability at least $c/\log (1/\epsilon )$. In the same tube, the normal radii between $\epsilon ^{1/(2k_ i)}$ and a fixed small radius carry probability bounded below, by the logarithmic integral. There $1/(f+\epsilon )\leq C\epsilon ^{-1/2}$, while $1/(f+\epsilon )\geq c\epsilon ^{-1}$ on the near annulus. The variance identity and the bound $G_ f\leq 2K_ f/(nH_ f)$ prove the claimed logarithmic estimate. In both cases $r<1$ and $r=1$, the integral of $\sqrt{G_ f}$ diverges at $\epsilon =0$.

For $r>1$, $H_ f$ tends to a finite positive integral and $G_ f\leq C K_ f$. For $1<r<3$, $\sqrt{G_ f}$ is therefore bounded by $C\epsilon ^{(r-3)/2}$, whose exponent exceeds $-1$. At $r=3$ the bound is $C\sqrt{\log (1/\epsilon )}$, and for $r>3$ it is a constant. All these bounds are integrable. $\square$

Proof. The completion of the convex product $\mathbb R\times I$ of proposition 12.6.36 is its Euclidean closure. Along a Cauchy sequence, $Y$ converges, so $b$ has a finite positive limit, and $R$ converges to a point of $I$ or to a finite endpoint of $I$. Lemma 12.6.39 decides which endpoints are finite and gives the three alternatives.

On $\mathscr S$ we have $e=Q(v)/(nb^ m)$ and $c=ve$. At the end $v\to \infty $, $vQ(v)\to H_ a$ by monotone convergence, which gives the coefficient limit $\theta _+(b)$. At the end $v\to -1/M$, write $1+va=1-a/M+(v+1/M)a$; dominated convergence gives $Q(v)\to H_-$, and hence the limit $\theta _-(b)$. These are the coefficient limits at all finite ends. The forms are semipositive, hence nef, and they are nonclosed. Increasing their fiber coefficient by $\epsilon >0$ makes them positive, and (12.6.33) shows that the resulting volumes tend to one as $\epsilon \downarrow 0$. Monotonicity bounds the root volume of the limit form by this limit. Conversely, each fixed Gauduchon quotient is continuous in $\epsilon $, so the limit of the root volumes is at most every quotient of the limit form, hence at most its root volume. Thus the variational volume of $\theta _\pm (b)$ is one.

Suppose that some quasi-plurisubharmonic $u$ satisfied

For every smooth nonnegative top-degree form $\eta $ on $B$, its pullback is $\mathrm{dd}^{\mathrm{c}}$-closed for degree reasons, and $\kappa \wedge \eta =0$. Stokes' theorem for currents therefore gives

where $L_\pm $ is the fiber coefficient of $\theta _\pm $. But $L_\pm $ vanishes somewhere, and a nonzero $\eta $ supported where $L_\pm <\delta $ contradicts this inequality. Thus $\theta _\pm $ is not big. $\square$

Proof. Put $L=ca+e$, and suppose that a quasi-plurisubharmonic $u$ satisfies $\theta +\mathrm{dd}^{\mathrm{c}}u\geq \delta h$ for some $\delta \geq 0$. Pairing with the closed semipositive form $\kappa ^{m-1}\wedge w$ and using Stokes' theorem for currents gives $b\geq \delta $. For every smooth nonnegative top-degree form $\eta $ on $B$, its pullback to $X$ is closed for degree reasons, and

If $L<\delta $ at some point, a nonzero form $\eta $ supported in the open set where $L<\delta $ contradicts this inequality. Hence $L\geq \delta $ everywhere. For $\delta =0$ these conditions are exactly $b\geq 0$, $e\geq 0$ and $e+cM\geq 0$; conversely, they make $\theta $ semipositive, and such forms are nef because $\theta +\varepsilon h>0$ for every $\varepsilon >0$. Conversely, if $\theta $ is nef, applying the same tests to smooth positive representatives of $\theta +\varepsilon h$ and letting $\varepsilon $ tend to zero gives these inequalities. For $\delta >0$ the conditions force the three strict inequalities, and conversely these make $\theta $ itself strictly positive. $\square$

Proof. Let $f=a$ for $\theta _+(b)$ and $f=1-a/M$ for $\theta _-(b)$, and let $r>1$ be the corresponding ratio $r_0$ or $r_ M$. The coefficient $L_0$ is a positive constant multiple of $f$. In each tube around an extreme component, the integral of $f^{-p}$ converges exactly when

converges, where $s$ is the normal radius and $s_0>0$ is small. Thus $1/L_0\in L^ p(B,\sigma )$ for every $1<p<r$. The zero set of $L_0$ has volume zero, so the value assigned to the density there is irrelevant. The formulas for $\theta _\pm (b)$ give

We apply theorem 14.19.2 on $B$ to the positive closed background $b\kappa $ and obtain a continuous $\phi $ with $A=b\kappa +\mathrm{dd}^{\mathrm{c}}\phi \geq 0$ and $A^ m=c_0\sigma /L_0$ for a constant $c_0>0$. Since $b\kappa $ is closed, bounded potentials do not change its mass, by theorem 11.1.7; hence $\int _ BA^ m=b^ m$ and $c_0=b^ m/H_0$. Set $u=\phi \circ p_ B$, where $p_ B:X\to B$ is the projection. Then $\theta +\mathrm{dd}^{\mathrm{c}}u=A+L_0w\geq 0$.

The pullback identity of lemma 5.28.4 gives $A^{m+1}=0$ on $X$, and $(L_0w)^2=0$. The multilinear expansion of proposition 5.28.2 therefore gives

The cancellation $L_0(\sigma /L_0)=\sigma $ holds as measures, since the zero set of $L_0$ has volume zero and $A^ m$ is absolutely continuous with density $c_0/L_0$ relative to $\sigma $.

Suppose that some solution $q$, not necessarily pulled back from $B$, were of class $C^{1,1}$ on $X$. Since $\mu /h^ n$ is smooth and strictly positive on compact $X$, proposition 5.28.6 would give $\theta +\mathrm{dd}^{\mathrm{c}}q\geq \delta h$ for some $\delta >0$. This contradicts proposition 12.6.43, because $L_0$ vanishes somewhere. $\square$

Proof. Write $v_ j=c_ j/e_ j$ and $\psi _ j=\phi _ j/b_ j$. On $\mathscr S$ the equation for $\phi _ j$ on $B$ becomes

At the end where $\theta _0=\theta _+(b)$, $v_ j\to \infty $. Set $\epsilon =1/v$ and $H_\epsilon =\int _ B\sigma /(a+\epsilon )$. Then $F_ v=[H_\epsilon (a+\epsilon )]^{-1}$ and $H_\epsilon \to H_ a$. At the end where $\theta _0=\theta _-(b)$, $v_ j\to -1/M$. With $f=1-a/M$ and $t=v+1/M>0$, the denominator of $F_ v$ is $Q(v)(f+ta)$, and $Q(v)\to H_-$. Thus the respective limits of $F_{v_ j}$ are

Choose $1<p<r$, where $r>1$ is the ratio $r_0$ or $r_ M$ of that end. Integration over tubes as in the proof of proposition 12.6.44 gives $a^{-1}\in L^ p(B,\sigma )$, respectively $f^{-1}\in L^ p(B,\sigma )$. Near the first end $F_ v\leq C/a$, and near the second $F_ v\leq C/f$. Since the extreme sets have volume zero, dominated convergence gives $F_{v_ j}\to F_\pm $ in $L^ p(B,\sigma )$, and each limiting density has integral one. Since $\sigma /\kappa ^ m$ is smooth and strictly positive on compact $B$, the densities of the equations also have a common bound in $L^ p(B,\kappa ^ m)$.

We apply theorem 14.19.21 to the fixed background $\kappa $: every subsequence of $(\psi _ j)$ has a uniformly convergent further subsequence. Continuity of bounded products under uniform convergence identifies its limit with a normalized solution of $(\kappa +\mathrm{dd}^{\mathrm{c}}\psi _0)^ m=F_\pm \sigma $. Since $\kappa $ is closed, it satisfies the Guan–Li condition, so 14.25.2 makes this solution unique. Every cluster limit is consequently the same, which proves uniform convergence of the whole sequence. Since $b_ j\to b>0$, we obtain $u_ j\to u_0=(b\psi _0)\circ p_ B$ uniformly.

To prove convergence of the products, note that on each coordinate ball the convergence of the background coefficients provides a common smooth function $q$, a quadratic polynomial, such that all $u_ j+q$ are plurisubharmonic. These corrected potentials are uniformly bounded and converge uniformly. The continuity of bounded products under convergence in capacity (section 5.27) gives weak convergence of all their products, with uniform local mass bounds. Expanding the products as in (5.28.1), and using smooth convergence of the remaining background factors, we obtain the asserted convergence of currents. In top degree, $(\theta _ j+\mathrm{dd}^{\mathrm{c}}u_ j)^ n=\mu $ for all $j$, so the limit solves the endpoint equation.

For uniqueness, let $v=\chi \circ p_ B$ be a bounded solution pulled back from $B$, with $\sup _ B\chi =0$. The current $A_\chi =b\kappa +\mathrm{dd}^{\mathrm{c}}\chi $ on $B$ is positive, as one sees by pairing the pulled-back current $\theta _0+\mathrm{dd}^{\mathrm{c}}v$ with $w$ and with pullbacks of positive forms on $B$. The product expansion gives $nL_0A_\chi ^ m\wedge w=\sigma \wedge w$, which determines $A_\chi ^ m=\sigma /(nL_0)$ off the extreme set. That density already has total mass $H_0/n=b^ m$, whereas mass invariance for the closed background $b\kappa $ gives $\int _ BA_\chi ^ m=b^ m$. Positivity therefore excludes any mass of $A_\chi ^ m$ on the extreme set. Thus $\chi /b$ satisfies the same normalized equation on $B$ as $\psi _0$, and 14.25.2 proves the asserted uniqueness. $\square$

Proof. On $\mathscr S$, $H(c_ j,e_ j)=nb_ j^ m$, so $A_ j^ m=\sigma /(nL_ j)$. Substituting this in (12.6.34) gives the formula for $G_ j$. Stokes' theorem on $B$, for the closed form $\kappa $, gives $\int _ B\kappa \wedge A_ j^{m-1}=b_ j^{m-1}$, which proves the trace identity. The factorization of $L_ j$ uses $\epsilon _ j=e_ j/c_ j$ at the first end and $\epsilon _ j=e_ j/(-c_ j)-M$ at the second, and the coefficient limits give $s_ j\to s_0>0$.

On each tube around an extreme component, $f\asymp \operatorname {dist}(Z_ i)^{k_ i}$, so the integral of $f^{-2}$ is comparable to

where $\varrho $ is normal radius, $\varrho _0>0$ is small and $r_ i=\operatorname {codim}_{\mathbb R}Z_ i/k_ i$. Splitting the integral at $\varrho =\epsilon _ j^{1/k_ i}$ gives order $\epsilon _ j^{r_ i-2}$ for $r_ i<2$, order $\log (1/\epsilon _ j)$ for $r_ i=2$, and a finite limit for $r_ i>2$. The finite sum over components is governed by $r=\min _ i r_ i$. Since $b_ j$ and $s_ j$ have positive finite limits, these estimates prove the three alternatives for $t_ j$.

If $r>2$, then $f^{-2}$ is integrable and dominates $L_ j^{-2}$ up to a fixed constant factor, so $\sigma /L_ j^2\to \sigma /L_0^2$ in total variation. Since $w\wedge w=0$, we have $A_ j^{m-1}\wedge w=(\theta _ j+\mathrm{dd}^{\mathrm{c}}u_ j)^{m-1}\wedge w$, which converges weakly by proposition 12.6.47. This proves $G_ j\rightharpoonup G_0$, and $G_0$ is positive and closed as a weak limit of positive closed forms. Pairing with $h$ gives the limit of the traces, which is positive because of the term $\sigma /(nL_0^2)$. $\square$

Proof. Let $\nu _ j=\sigma /(J_ jL_ j^2)$. These are probability measures on $B$, and

The first coefficient tends to one. The second current is positive with trace $mb_ j^{m-1}/t_ j\to 0$, hence converges weakly to zero. Compactness of probability measures on $B$ and weak continuity of pullback under the submersion $p_ B$ therefore show that every cluster limit is $p_ B^*\nu $ for a cluster limit $\nu $ of $(\nu _ j)$. Away from small tubes around the extreme set, the density $1/L_ j^2$ is uniformly bounded, while $J_ j\to \infty $; that region receives no mass in the limit. The tube integrals for components with ratio greater than $r$ are $o(J_ j)$ by the preceding estimates, so only components with ratio $r$ keep mass.

For a limit $T=p_ B^*\nu $, $\kappa \wedge T=0$ for degree reasons on $B$, while $L_0w\wedge T=0$ because $L_0$ vanishes on the support of $\nu $. Its trace is one because $\int _ Cw=1$. A top-degree measure on $B$ is closed, so its pullback is closed as well. If the support is the single point $p$, every cluster limit of $(\nu _ j)$ is $\delta _ p$; compactness then gives convergence of the whole sequence, and $p_ B^*\delta _ p$ is the current of integration over the fiber $\{ p\} \times C$.

The minimizing quotient is one on $\mathscr S$, and before trace normalization both its numerator and its denominator equal $nb_ j^ m$. Each is multiplied by $t_ j^{-1}$ under the normalization, and the trace estimates give their limits and rates. $\square$

Proof. The extreme sets are the points $[1:0:0]$ and $[0:0:1]$. In affine coordinates, $a$ and $1-a$ vanish there to exactly second order, so both real codimensions are four and both orders are two. To compute the law of $a$ under $\sigma $, polar integration in the chart $Z_0=1$ gives $2\, \mathrm{d}s_1\mathrm{d}s_2/(1+s_1+s_2)^3$, where $s_ j=|z_ j|^2$. The transformation $u_ j=s_ j/(1+s_1+s_2)$ gives the uniform density two on the triangle $u_1,u_2\geq 0$, $u_1+u_2\leq 1$, and $a=u_1/2+u_2$. For $0\leq s\leq 1/2$, the triangle $u_1/2+u_2\leq s$ has area $s^2$, so the law of $a$ has density $4s$ on $[0,1/2]$. With $u_0=1-u_1-u_2$, exchanging $u_0$ and $u_2$ preserves the uniform measure and replaces $a$ by $1-a$; hence the law has density $4(1-s)$ on $[1/2,1]$. Integrating against these two densities gives $H_ a=H_-=4\log 2$ and the formula for $Q$, and the integral definition of $Q$ shows that $Q(0)=1$ is its continuous extension. Theorem 12.6.41 gives the boundary lines and the finite width, with Euclidean coordinates $Y=\sqrt2\log b$ and $R(v)$. $\square$

Proof. The forms descend to $B$, and $\sigma =\eta _1\wedge \cdots \wedge \eta _ m$ is a probability form. Both extreme sets are real hypersurfaces along which $a$, respectively $1-a$, vanishes to second order, so $r_0=r_ M=1/2$. Splitting the period and substituting $t=\tan (\pi u)$ gives

Let $F_ v(u)=[Q(v)(1+v\sin ^2(\pi u))]^{-1}$, which has mean one over a period. Integrating the equation

twice, whose right side has mean zero, gives a smooth periodic potential $\phi _ v$ of $u=u_1$; the first integration constant is chosen so that $\phi _ v'$ has mean zero. In our normalization $\mathrm{dd}^{\mathrm{c}}\phi _ v=\phi _ v''\eta _1/(4\pi )$, hence

Thus the minimizer of theorem 12.6.32, with its cofactor formula (12.6.34), is explicit in this example, without an appeal to theorem 14.2.1. Differentiating $Q$ gives

As $U$ and $v$ range over $\mathbb R$ and $(-1,\infty )$, the coordinates $(Y,R)$ range over all of $\mathbb R^2$. Both boundary faces of the coefficient cone are therefore at infinite distance on the unit-volume hypersurface. $\square$