PomeD85397FA2C2AD85397FA2C2A
The first example places the freedom in the potential inside an open region where the density vanishes. Three independent semipositive forms of rank one reduce equality of the Monge–Ampère measures to equality of products of three scalar coefficients. We construct two semipositive forms with the same cube, and then add the same exact correction to both to obtain a strictly positive background.
In this section and the following ones, $X=\mathbb P(\mathbb C^2\oplus \mathbb C^2)=\mathbb P^3$, with block homogeneous coordinates $[A:B]$, where $A,B\in \mathbb C^2$, and $\omega _ X$ is its Fubini–Study form. Here $A$ and $B$ denote coordinates, not the forms of (14.20.1); the two solution forms will be written $A_0$ and $B_0$.
Theorem 14.26.1.050594389138 On $X=\mathbb P^3$ there exist a smooth Hermitian form $\omega $ with $\mathrm{d}\omega \ne 0$, a smooth nonnegative nonzero function $f$, and distinct smooth functions $u,v\in \mathrm{PSH}(X,\omega )$ such that
The density vanishes on a nonempty open set, and both potentials have minimal singularities.
0CFD28034339On $U=\{ A\ne 0,B\ne 0\} $ put
The logarithms are computed in a local homogeneous lift, but their Hessians do not depend on the lift. The forms $P,Q$ are pullbacks of Fubini–Study forms from the two projective lines $\mathbb P(\mathbb C^2)$. All three forms are semipositive of rank one, and
Indeed, in local coordinates $A=(1,a)$ and $B=\tau (1,b)$, $P$ and $Q$ are positive multiples of $\mathrm{i}\, \mathrm{d}a\wedge \mathrm{d}\bar a$ and $\mathrm{i}\, \mathrm{d}b\wedge \mathrm{d}\bar b$, while
Its $\mathrm{d}\tau $ component is nonzero, so the three rank-one forms correspond to independent directions. In particular, for nonnegative $x,y,z$,
and the form is positive exactly when all three coefficients are positive.
Since $\mathrm{dd}^{\mathrm{c}}r=P-Q$, the chain rule gives, for a smooth function $G:\mathbb R\to \mathbb R$,
1C52D8C53893
14.26.5
Let $\sigma (r)=\mathrm{e}^ r/(1+\mathrm{e}^ r)$. Since $\log (|A|^2+|B|^2)=\log |B|^2+\log (1+\mathrm{e}^ r)$, we have
DE71ED12948B
14.26.6
A function of $r$ that is constant near each end of $\mathbb R$ extends smoothly across $X\setminus U$. A form $xP+yQ+zR$ whose coefficients equal those of (14.26.6) for $|r|\geq 2$ extends smoothly by $\omega _ X$ across $X\setminus U$.
68DF63E92045On the central interval $-1\leq r\leq 1$, the coefficients will make a different rank-one factor vanish in each of the two solution forms. Outside a larger interval they coincide with the Fubini–Study coefficients, so the resulting forms extend across the omitted projective lines. Primes denote derivatives with respect to $r$.
Choose an even smooth function $\beta $ supported in $[-1,1]$, strictly positive on $(-1,1)$, with $\beta '>0$ on $(-1,0)$, and normalized by $\int _{-1}^1\beta =2$; for instance, a positive multiple of $\exp (-1/(1-r^2))$ on $(-1,1)$, extended by zero. Define
and put $p=\beta /2$, $q=\beta '/2$. Then $F$ is smooth and odd, $p$ is even, and $q$ is odd, positive on $(-1,0)$ and negative on $(0,1)$. Both $p$ and $q$ vanish to infinite order at the endpoints.
Take smooth $a\geq 0$ with $a=1$ on $(-\infty ,-2/3]$ and $a=0$ on $[-1/3,\infty )$, and put $b(r)=a(-r)$. Choose an even nonnegative $\chi \in C_ c^\infty ((-1/3,1/3))$ with $\chi (0)>0$. On $[-1,1]$ set
It is smooth: the expression under the square root is positive in the interior, and near each endpoint $z_0=|q|$, where $q$ has a fixed sign. It is positive in the interior and vanishes to infinite order at the endpoints. Also,
6C8664C40E89
14.26.8
The first cancellation holds because $\chi =0$ on the support of $a$ and $q\geq 0$ there; the second follows by reflection.
On $(-\infty ,-1]$ choose smooth functions $\ell ,s$ such that $0\leq \ell \leq 1$, $\ell =0$ on $(-\infty ,-2]$, $\ell =1$ on $[-3/2,-1]$, while $s>0$ for $r<-1$, $s=1$ on $(-\infty ,-3/2]$, and $s$ vanishes to infinite order at $-1$. Such functions are easily obtained from smooth cutoff functions and a function like $\exp (-1/(r+1)^2)$. Define
and for $r\geq 1$ use $x(r)=y(-r)$, $y(r)=x(-r)$, $z(r)=z(-r)$.
The definitions join smoothly: at $-1$ both definitions give the values $(1,0,0)$, and their differences vanish to infinite order there, since $a=1$ and $b=0$ near $-1$. The central definitions are symmetric under reflection, so the same holds at $1$. We have
For $|r|\geq 2$ their values are $(\sigma (r),\sigma (-r),\sigma '(r))$.
DBEDC79E97CDThe cancellation identities now turn the scalar coefficients into two semipositive forms with the same cube. Their average is strictly positive except at the two ends of the central interval.
By the extensions described above, the form and the function
are smooth on $X$. The form is semipositive, and positive except on the hypersurfaces $M_- =\{ r=-1\} $ and $M_+=\{ r=1\} $. Put
For $|r|\geq 1$ we have $H=0$. On $|r|<1$,
All coefficients are nonnegative, and by (14.26.4) and (14.26.8), $A_0^3=B_0^3=0$ on this band. Outside it both cubes equal $\omega _0^3$. Thus $A_0,B_0$ are smooth semipositive forms on $X$ with equal cubes. This common top-degree form vanishes on the closed band $|r|\leq 1$ and is positive on its complement, including near $X\setminus U$.
2D54B6D6C7ADThe forms $A_0,B_0$ already have the required common measure. We obtain a Hermitian background by adding an exact correction $\mathrm{dd}^{\mathrm{c}}\rho $ to their average $\omega _0$ and subtracting $\rho $ from both potentials. The correction must restore positivity at $M_\pm $ while preserving it in the transition regions.
Take $\kappa \in C_ c^\infty ((-2,2))$, equal to one on $[-3/2,3/2]$. For a parameter $\varepsilon >0$ to be chosen, set
By (14.26.5), its coefficients on $U$ are $x+\rho '$, $y-\rho '$ and $z+\rho ''$. On $[-3/2,3/2]$ one has $\rho '=2\varepsilon r$ and $\rho ''=2\varepsilon $, so the last coefficient is positive there. For $r\geq 0$, $x+2\varepsilon r$ is positive, including at $r=0$ since $x(0)=p(0)>0$. On $[-3/2,0]$, the function $x$ has a positive minimum $m$, so $x+2\varepsilon r\geq m-3\varepsilon >0$ for small $\varepsilon $. Reflection gives the corresponding estimate for $y-2\varepsilon r$. On the compact region $3/2\leq |r|\leq 2$, all three original coefficients have a positive minimum and the perturbations are $O(\varepsilon )$. For $|r|\geq 2$ and near $X\setminus U$, the form is exactly $\omega _ X$. Hence $\omega $ is strictly positive on $X$ for sufficiently small $\varepsilon >0$.
The form $\omega $ is not closed. Indeed, since $\mathrm{d}R=-\mathrm{d}r\wedge (P-Q)$,
0711E4CCA0FE
14.26.11
At $r=0$, $x'=y'=0$ and $z=\chi (0)>0$. Hence $\mathrm{d}\omega =z\, \mathrm{d}r\wedge (Q-P)\ne 0$ there. The nonzero $\mathrm{d}\tau $ component used in (14.26.3) also shows that $\mathrm{d}r\wedge P$ and $\mathrm{d}r\wedge Q$ are linearly independent.
of theorem 14.26.1. Set $\widetilde u=-w/2-\rho $ and $\widetilde v=w/2-\rho $. Since $\mathrm{dd}^{\mathrm{c}}w=2H$, the construction gives
Subtracting their respective suprema gives $u,v$. Their Hessians are unchanged, and $v-u=w+\text{constant}$ is nonconstant. The function $f=A_0^3/\omega _ X^3=B_0^3/\omega _ X^3$ is smooth, nonnegative and nonzero, and it vanishes on the nonempty open band $|r|<1$. Both potentials are smooth, hence bounded, so they have minimal singularities: for a Hermitian background, the envelope $V_\omega $ of section 5.40 is zero. $\square$
Every Hermitian background on $\mathbb P^3$ has the bounded mass property of definition 13.14.14: since the Fubini–Study form is closed, Stokes' theorem for bounded potentials shows that all its bounded potentials have the same mass, and proposition 13.14.15 shows that the property does not depend on the metric. This applies to all the backgrounds on $\mathbb P^3$ constructed below.
The linearized equation also explains why the maximum principle argument fails. The form
satisfies $\mathrm{dd}^{\mathrm{c}}w\wedge T=B_0^3-A_0^3=0$, but $T=0$ on $M_\pm $, where $A_0=B_0=\omega _0$ has rank one. Thus the linear equation for $w$ is not elliptic everywhere, even though both potentials and their common density are smooth.
6FDC265145E3The open band where the density vanishes allows more solutions than the two smooth ones constructed above. Writing a further potential as $G(r)-\rho $, with $G$ a function of one real variable, reduces the equation on the band to choosing the slope $G'$ so that one of the three coefficients vanishes. Switching the vanishing coefficient at two suitable levels produces a continuum of $C^{1,1}$ solutions for the same background $\omega $ and density $f$.
Recall that $q=p'$ and $z_0\geq |q|$. The strict inequality $z_0>|p'|$ near zero, where $\chi >0$, gives
381112F03D1E
14.26.13
Proposition 14.26.14.3A73F825D91B For the smooth background $\omega $ and the smooth density $f$ of theorem 14.26.1, the equation
has uncountably many distinct $C^{1,1}$ solutions normalized by $\sup _ Xu=0$. If the cutoff $\kappa $ defining $\rho $ is chosen even, one of them, constructed explicitly, is invariant under the block exchange $\iota [A:B]=[B:A]$.
Proof. Consider pairs $-1<I<0<J<1$ satisfying
We first show that such pairs exist, with $I$ ranging over an interval. Define
We have $K(0,0)=-2p(0)<0$ and $K(0,1)>0$ by (14.26.13). By continuity, there is $\epsilon >0$ such that $K(I,0)<0<K(I,1)$ for every $I\in [-\epsilon ,0]$. For $0<J<1$,
The intermediate value theorem therefore gives a unique $J=J(I)\in (0,1)$ satisfying (14.26.15) for every $I\in [-\epsilon ,0)$; we may assume $\epsilon <1$.
For any pair satisfying (14.26.15), define the slope on $[-1,1]$ by
and extend it by zero outside that interval. The balance identity makes it continuous at $J$, and continuity at $I$ is immediate. Each piece has bounded derivative, so $h_{I,J}$ is Lipschitz on $\mathbb R$; it vanishes to infinite order at $\pm 1$. Let $G_{I,J}$ be a primitive of $h_{I,J}$. It is $C^{1,1}$ and constant near each end of the real line. Consequently $G_{I,J}(r)$ extends to a $C^{1,1}$ function on $X$ that is smooth near $X\setminus U$.
On the middle interval $[I,J]$ the slope lies between $-p$ and $p$. Indeed,
The first difference vanishes at $I$, and the second vanishes at $J$. Since both are nonincreasing, $h_{I,J}\leq p$ and $h_{I,J}\geq -p$ on $[I,J]$.
Put $\widetilde u_{I,J}=G_{I,J}(r)-\rho $. By (14.26.5), almost everywhere on $U$,
2EC8417EFF1E
14.26.16
On $I<r<J$, the first two coefficients are nonnegative because $x=p+a$, $y=p+b$ and $-p\leq h_{I,J}\leq p$, and the third is zero. On $-1<r<I$, we have $h_{I,J}=p$ and $b=0$, since $I<0$; so the second coefficient is zero, and the others are nonnegative because $z_0+p'\geq 0$. On $J<r<1$, we have $h_{I,J}=-p$ and $a=0$; so the first coefficient is zero, and the others are nonnegative because $z_0-p'\geq 0$. Outside the band, $h_{I,J}=h_{I,J}'=0$, so the form equals $\omega _0$.
Formula (14.26.16) has no additional term supported on the hypersurfaces $r=I$ and $r=J$: the function $G_{I,J}$ is $C^{1,1}$, so its first derivative is continuous at both transition points, and its distributional Hessian is its bounded almost-everywhere Hessian. The preceding coefficient inequalities therefore prove $\omega +\mathrm{dd}^{\mathrm{c}}\widetilde u_{I,J}\geq 0$ as a current.
By proposition 5.28.5, its Monge–Ampère measure is the pointwise cube of (14.26.16), computed almost everywhere. By the rank-one identities (14.26.3) and (14.26.4), this cube is zero almost everywhere on the band, since one coefficient vanishes there, and it equals $\omega _0^3$ off the band. These are exactly the values of $f\omega _ X^3$ in the preceding construction. Thus $\widetilde u_{I,J}$ solves the equation, and subtracting its supremum gives a normalized solution $u_{I,J}$.
Different values of $I\in [-\epsilon ,0)$ give different normalized solutions. If $I_1<I_2<0$ and $I_1<r<I_2$, then the slope belonging to $I_2$ is $p(r)$, whereas
because $p'>0$ on $(-1,0)$. The two primitives therefore do not differ by a constant. Since the interval $[-\epsilon ,0)$ is uncountable, this proves the first assertion. None of these slopes equals $p$ or $-p$ on all of $(-1,1)$, so the constructed solutions are also distinct from the two smooth solutions of theorem 14.26.1.
Finally, the function
is negative at zero, positive at one, and strictly increasing on $(0,1)$. Its unique zero $t_*$ gives a symmetric pair $I=-t_*$, $J=t_*$ satisfying (14.26.15). Since $z_0$ and $p$ are even, $h_{I,J}$ is odd: on the middle interval it equals $-\int _0^ r z_0(s)\, \mathrm{d}s$. Choose its primitive even. Since $\iota $ changes $r$ into $-r$, if $\rho $ is even, the potential $G_{I,J}(r)-\rho $ and its normalization are fixed by $\iota $. This proves the last assertion. $\square$
For these symmetric data, the normalized solutions of the equations with a positive exponent converge, as the exponent tends to zero, to the symmetric solution just constructed; see theorem 15.20.4.
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