15.20 Symmetry and the selected projective limit457A6A4C41F4

Proof. The barriers of proposition 14.19.34 give a uniform oscillation bound for $u_\lambda $ and uniform convergence $\mathrm{e}^{\lambda u_\lambda }\to c$. Thus the equation densities are uniformly bounded in $L^ p$ for $0<\lambda \leq 1$ and tend to $cf$ in $L^ p$ as $\lambda \downarrow 0$. Apply theorem 14.19.21 to obtain relative compactness of $a_\lambda $ in $C(X)$ and the asserted equation for every cluster limit.

For each $g\in G$, pullback of the positive-exponent equation shows that $u_\lambda \circ g$ solves the same equation. Its uniqueness gives $u_\lambda \circ g=u_\lambda $. Surjectivity of $g$ preserves the supremum, so $a_\lambda \circ g=a_\lambda $. Uniform convergence preserves these equalities for every $g$. Compactness gives at least one cluster limit as $\lambda \downarrow 0$, yielding the invariant solution. $\square$

Proof. Use the radial coordinate $r$, rank-one forms $P,Q,R$, Fubini–Study form $\omega _ X$, and scalar functions $F,x,y,z$ from 14.26.2, 14.26.7. Set $w=F(r)$, $\omega _0=xP+yQ+zR$, $H=\mathrm{dd}^{\mathrm{c}}w/2$, $A_0=\omega _0-H$ and $B_0=\omega _0+H$ as in 14.26.9. The Hermitian form is $\omega =\omega _0+\mathrm{dd}^{\mathrm{c}}\rho $, with $\rho =\varepsilon (r^2-1)\kappa (r)$ from 14.26.10. Choose the cutoff $\kappa $ even; all its required support and plateau properties are preserved by this choice. Then $\rho $ is even, while $w=F(r)$ is odd. The involution sends $r$ to $-r$, interchanges $P,Q$, and fixes $R$. The coefficient identities $x(-r)=y(r)$ and $z(-r)=z(r)$ show that $\iota ^*\omega _0=\omega _0$. Thus $\iota ^*\omega =\omega $, and $\iota ^*\omega _ X=\omega _ X$.

Since $\iota ^*H=-H$, the involution interchanges $A_0$ and $B_0$. Their cubes are equal, so $f\omega _ X^3$ is invariant. It also interchanges $\widetilde u=-w/2-\rho $ and $\widetilde v=w/2-\rho $. These functions therefore have the same supremum, and the normalized solutions $u,v$ are interchanged. They are distinct, so neither is invariant.

Apply proposition 15.20.1 to the group generated by $\iota $. Here $c=1$, as the two constructed solutions already have that scalar. An invariant cluster solution $a$ cannot equal either $u$ or $v$. Its bounded equation with smooth density gives Hölder regularity by theorem 15.5.1. This gives a third solution. Its regularity is determined in subsection 15.20.3, subsection 15.20.7: it is $C^{1,1}$ but not $C^2$. $\square$

Proof. Reduction to one variable. Block-unitary orbits in $U$ are exactly the level sets of $r$: normalize the sum of the squared norms of the two coordinate blocks, and map the two vectors of the same respective lengths by their unitary groups. Thus invariance gives a function $G$ of one real variable.

On a compact $r$-interval, the radial direction annihilates $P,Q$, while $R$ is strictly positive in that direction. Testing positivity of $\omega _0+\mathrm{dd}^{\mathrm{c}}G(r)$ in this direction gives $G''+z\geq 0$ as a one-dimensional distribution. One may obtain this first by smooth approximation of the scalar distribution and then by testing with nonnegative functions. Adding a twice integrated primitive of $z$ makes $G$ convex. A finite bounded convex function on an interval is locally Lipschitz, and its derivative is locally of bounded variation. The quasi-plurisubharmonic representative is the corresponding continuous representative. The horizontal directions give $x+h\geq 0$, $y-h\geq 0$ almost everywhere. Continuity of $x,y$ and the one-sided limits of a function of bounded variation give these bounds for both one-sided representatives. The radial inequality is precisely $\mathrm{d}h+z\, \mathrm{d}r\geq 0$; in particular jumps of $h$ can only be upward.

The radial form of the measure. We next justify the formula for the measure, including possible jumps. Identify $U$ smoothly with $\mathbb R\times Y$, where $Y=\{ r=0\} $, by real scaling of one block. The forms $P,Q$ are independent of $r$ in this identification. So is $\mathrm d^{\mathrm c}r$: in local coordinates it is the angular differential of the relative complex scale plus fixed base terms. Therefore

where $\nu $ is a fixed smooth positive volume form on the compact orbit $Y$, with the orientation chosen by this equality.

For smooth $G$, the rank-one wedge formula gives the radial coefficient of its Monge–Ampère measure, after dividing by $6\nu $, as

This identity passes to a locally Lipschitz $G$ with the preceding lower Hessian bound. Indeed, convolve $G$ on a slightly larger compact interval. The derivatives $h_\epsilon $ are uniformly bounded and converge to $h$ almost everywhere and in every finite $L^ q$ space. Their second derivatives have a uniform lower bound. Consequently the functions $G_\epsilon (r)$ have a common quasi-plurisubharmonic lower Hessian bound on compact subsets of $U$, and converge uniformly to $G(r)$. A fixed smooth local correction and bounded Bedford–Taylor continuity give convergence of their signed products. On the right-hand side of the formula, the two ordinary terms converge in $L^1$, and $\Phi (r,h_\epsilon )\to \Phi (r,h)$ in $L^1$, giving distributional convergence of its derivative. This proves the same formula for the bounded product of $G$.

Continuity of $h$. After the same division by $6\nu $, the right-hand side $f\omega _ X^3$ has radial coefficient $k(r)\, \mathrm{d}r$. This gives (15.20.6), initially for the almost-everywhere representative of $h$. If $h$ jumped at $r_0$, the atom of its left side would be

The right side has no atom. An upward jump of positive size lies in $[-x(r_0),y(r_0)]$, where the integrand is strictly positive in the interior. Since $x+y>0$ everywhere, such an integral is strictly positive. Thus there are no jumps. A locally bounded-variation function whose jumps all vanish has a continuous representative here: its distributional derivative consists of an absolutely continuous measure and a nonnegative singular measure, so its one-sided representatives coincide at every point. We henceforth use this representative. In particular $G$ is $C^1$ and $G'=h$ everywhere.

All coefficients on the right of (15.20.6) are now continuous, so $\Phi (r,h(r))$ is $C^1$. Where $\partial _ s\Phi =A>0$, the implicit function theorem gives $h\in C^1$ and the stated ordinary differential equation. Smoothness then follows by the smooth coefficients of that equation. The function $k$ is smooth, since the original density and its flat joins are smooth. $\square$

of theorem 15.20.4. We first show that a normalized bounded $K$-invariant solution $u$ coincides with $u_*$, by computing its slope $h$ on the band $|r|\leq 1$ and outside it. Apply lemma 15.20.5. Because $\rho $ and $u$ are invariant under block exchange, $G$ is even, so $h$ is odd and $h(0)=0$.

The slope on the band. On $0<r<1$ we have $x=p>0$ and $y=p+b\geq p>0$. Starting from zero, as long as $-p<h<y$, the equation is $h'=-z_0$. Thus $h(r)=-\int _0^ r z_0(s)\, \mathrm{d}s$. This decreasing function cannot meet the positive upper boundary $y$. Its first meeting with the lower boundary $-p$ occurs at the unique $t_*\in (0,1)$ satisfying

whose existence and uniqueness were proved in proposition 14.26.14.

After that meeting, $h$ must stay on $-p$ until $r=1$. To prove it, suppose a connected component $(a_0,b_0)$ of $\{ h+p>0\} $ lies to its right. Its left endpoint has $a_0\geq t_*$ and $h(a_0)=-p(a_0)<0$. Near that endpoint, $h<0<y$; hence on the component the interior equation applies for as long as $h<0$. It forces $h'=-z_0<0$, so $h$ cannot first reach zero. The equation therefore holds throughout the component. But then

contradicting positivity of $h+p$ immediately to the right of its zero at $a_0$. Thus $h=-p$ on $[t_*,1]$. Oddness determines the entire negative half of the band. The slope on $[-1,1]$ is exactly that of the symmetric explicit solution.

The slope outside the band. Here boundedness of $u$ at the projective ends matters. On $r>1$, $k=xyz>0$. Neither horizontal coefficient can vanish at a finite point of this interval. For the lower one define

Equation (15.20.6) and the smooth lower endpoint show that $\Psi $ is $C^1$. If $h(r_0)=-x(r_0)$, its derivative at that point is $k(r_0)>0$: the $Az$ term vanishes, and the two partial-$r$ terms agree at the common endpoint. This contradicts that the nonnegative $C^1$ function $\Psi $ has an interior minimum zero there. The upper-endpoint integral $\int _{h(r)}^{y(r)}A(r,s)\, \mathrm{d}s$ gives the same contradiction, with derivative $-k(r_0)$, when $h=y$. Thus $-x<h<y$ everywhere on $r>1$, and the smooth ordinary differential equation of the lemma applies.

For $r\geq 2$ the background is Fubini–Study:

Set $a=\sigma +h\in (0,1)$. Its equation is

Hence $a$ is increasing and has a limit in $(0,1]$ at $+\infty $. If that limit were less than one, $h=a-\sigma $ would approach a negative constant and its primitive $G$ would be unbounded below, contrary to boundedness of $u$ and $\rho $. The limit is therefore one. Integrating the equation gives

Taking the limit at infinity gives $C=0$. The scalar function $s^2/2-s^3/3$ is strictly increasing on $(0,1)$, so $a=\sigma $ and $h=0$ for $r\geq 2$. Ordinary uniqueness propagates this zero solution throughout $r>1$, since the coefficients are smooth and the two horizontal coefficients stay positive there. Oddness gives $h=0$ on $r<-1$ as well.

We have proved that $G'$ agrees everywhere with the symmetric explicit slope. Thus $u$ differs from $u_*$ by a constant on $U$. The two quasi-plurisubharmonic functions then agree up to that constant on $X$ by their canonical representatives. Their supremum normalization removes the constant. This proves uniqueness within the bounded $K$-invariant class.

Convergence. Finally proposition 15.20.1 gives uniform precompactness of the normalized positive-exponent family. Every cluster limit solves the zero-exponent equation and is $K$-invariant, because the data are $K$-invariant. The scalar is one for these data. The uniqueness just proved identifies every cluster limit with $u_*$. If convergence of the whole family failed, a sequence staying a fixed positive uniform distance from $u_*$ would have a convergent subsequence, contradicting that identification. This proves uniform convergence and the asserted $C^{1,1}$ regularity of the limit. $\square$

Proof. Uniqueness and invariance of the data imply that $u_\lambda $ is $K$-invariant. Write $u_\lambda +\rho =G_\lambda (r)$ on $U$. The positivity argument in lemma 15.20.5 uses only boundedness and invariance, so $G_\lambda $ is locally Lipschitz and $h_\lambda =G_\lambda '$ is locally of bounded variation, with

Its weak flux equation is the same one, with the prescribed coefficient replaced by

Indeed, the scalar convolution and signed-product argument of that lemma is unchanged, and $k_\lambda $ is exactly the radial coefficient of the right-hand measure. Since $G_\lambda $ is bounded on compact $r$-intervals, $k_\lambda $ is locally bounded. Hence the right side of the flux equation has no atoms. The same strictly positive jump integral eliminates every jump of $h_\lambda $. Thus $h_\lambda $ is continuous and $G_\lambda $ is $C^1$. The coefficient $k_\lambda $ is consequently continuous. Where $A(r,h_\lambda )>0$, inversion of the $C^1$ flux gives

On $-1<r<1$ the coefficient $k_\lambda $ is identically zero. Exchange invariance gives $h_\lambda (0)=0$ and oddness. On the positive half-band, the first-contact and no-departure arguments in the proof of theorem 15.20.4 apply verbatim: the first branch solves $h'=-z_0$, reaches $-p$ at $t_*$, and cannot leave that lower boundary to its right because $(h+p)'=p'-z_0<0$ on every proposed departure interval. Therefore

This is the derivative of $u_*+\rho $. Its primitive differs from $G_\lambda $ by one constant on the connected band. The claimed local $C^{1,1}$ regularity follows from the explicit slope.

At $t_*\in (0,1)$ the two one-sided derivatives of this slope are

These unequal derivatives prevent the scalar primitive from having a second derivative at $t_*$. Through every point of that level set, the submersion $r$ provides a smooth curve parametrized by $r$, transferring the failure to $u_\lambda $. Smooth $\rho $ cannot cancel the jump. Oddness gives the conclusion at $-t_*$, and the identical slope proves it for $u_*$. $\square$