14.29 A real analytic densityC3B76FE21E71

A globally real analytic density can occur in an equation with a smooth nonclosed Hermitian background and distinct smooth normalized solutions. We prescribe it as a polynomial in a real analytic function on projective space. Two symmetric radial perturbations give the required determinant on the middle band; patching the coefficients while preserving their product extends the construction across the projective ends. The two hypersurfaces where the density vanishes quadratically separate regions with different extremal values of the difference of the solutions.

14.29.4 Prescribing the determinantCB8BE6BADB1A

On $U$, one has $t=\sigma (r)$. We first construct two triples of coefficients, positive on the middle band, with product $f_*(\sigma )(\sigma ')^2$. At each endpoint of the band, exactly one of the coefficients of $P$ and $Q$ will vanish.

Put $d_*(r)=f_*(\sigma (r))(\sigma '(r))^2$. Since $a=\sigma (-1)$, $b=\sigma (1)$, define the positive smooth divided differences

and $H_*(r)=k_-(r)k_+(r)\sigma '(r)>0$. The fundamental theorem of calculus gives

These formulas show that $H_*$ is smooth and positive also at the endpoints, without any division by a vanishing function. The symmetry $\sigma (-r)=1-\sigma (r)$ makes $d_*$, and hence $H_*$, even.

Choose the function

All its derivatives vanish at $r=\pm 1$: each derivative is a finite sum of polynomials in $r$ times powers of $(1-r^2)^{-1}$, multiplied by the exponential, and the exponential factor dominates every such power. On $|r|<1$, $q=-2rp/(1-r^2)^2$. Define

Both radicands are strictly positive everywhere, so these formulas define smooth functions on $\mathbb R$, including at the endpoints. They satisfy

The coefficients $x_ b,y_ b$ are nonnegative, positive away from the endpoints, with values $(1,0)$ at $-1$ and $(0,1)$ at $1$. On the open band,

so $x_ b>p$, $y_ b>p$. Consequently both triples

are strictly positive there.

We now verify that the two triples have the same product. Set $D_*=x_ by_ b-p^2$, $E_*=p(x_ b-y_ b)=-rp$. Equations (14.29.8) and the formula for $q$ give

The two products in (14.29.9) therefore equal

For $|r|\geq 1$, $p=q=0$ and the same identity reads $x_ by_ bz_ b=d_*$. Thus both triples have exactly the prescribed product, and at the endpoints exactly one of their entries vanishes.

14.29.11 Extension over the projective endsEA62A44D2D9F

The coefficients just constructed have the right product but need not extend across the projective ends. Away from the zero levels of $f_*(t)$, we interpolate the coefficients of $P$ and $Q$ to those of $f_*(t)^{1/3}\omega _ X$, and determine the coefficient of $R$ by the prescribed product.

Choose a smooth even cutoff $0\leq \eta \leq 1$, equal to one on $|r|\leq 3/2$ and zero on $|r|\geq 2$. On $|r|<3/2$ set $(x,y,z)=(x_ b,y_ b,z_ b)$. On $|r|>5/4$, where $f_*(\sigma (r))>0$, put

These coefficients are smooth and positive on the outer region. On the overlap $5/4<|r|<3/2$, we have $\eta =1$, $p=q=0$ and $x_ by_ bz_ b=d_*$, so the two definitions agree. For $|r|\geq 2$ they equal $\kappa (\sigma (r),\sigma (-r),\sigma '(r))$. Thus

extends smoothly across both omitted projective lines, being equal near them to $f_*(t)^{1/3}\omega _ X$, which is smooth and positive there since $f_*(0)=f_*(1)=a^2b^2>0$. On $X$, the form $\omega _0$ is semipositive, positive away from $r=\pm 1$, and of rank two on these levels.

The function

is constant on each of the intervals $(-\infty ,-1]$ and $[1,\infty )$, and so $F(r)$ extends to a smooth nonconstant function on $X$. By (14.26.5), half of its complex Hessian is $H_0=\mathrm{dd}^{\mathrm{c}}F/2=p(P-Q)+qR$. By (14.29.10) and (14.29.3), the forms $A_0=\omega _0-H_0$, $B_0=\omega _0+H_0$ satisfy

on $U$, and hence on $X$. Their ranks and positivity were determined above, after (14.29.9). The construction is symmetric: $x(-r)=y(r)$, and $z$ is even.

14.29.12 A positive equation background38D1973448C7

As in the first example, an exact correction makes the average form strictly positive without changing the two solution forms. Here the coefficient of $R$ is already positive, and the correction restores the missing coefficient of $P$ or $Q$ at each endpoint. Let $\varepsilon >0$ be a parameter, restricted below.

Let $h(r)=(r^2-1)\eta (r)$, extend it by zero on the tails, and set $\rho =\varepsilon h$. The coefficients of $\omega =\omega _0+\mathrm{dd}^{\mathrm{c}}\rho $ are

On $|r|\leq 3/2$, we have $h'=2r$ and $h''=2$. For $0\leq r\leq 3/2$, $x+2\varepsilon r>0$, since the only zero of $x$ there is at $r=1$, where the correction is positive. On $[-3/2,0]$, $x$ has a positive minimum $m_0$, so positivity holds if $3\varepsilon <m_0$. Reflection proves positivity of the second coefficient, and the third is $z+2\varepsilon >0$. On the compact transition region $3/2\leq |r|\leq 2$, all original coefficients have a common positive lower bound $m_1$. If $M_1$ bounds $|h'|+|h''|$, taking also $\varepsilon <m_1/(M_1+1)$ preserves their positivity there. For $|r|\geq 2$, the background equals $f_*(t)^{1/3}\omega _ X>0$. Hence $\omega $ is strictly positive on $X$.

The correction is exact, so $\mathrm{d}\omega =\mathrm{d}\omega _0$. At $r=0$, where $x=x_ b$ and $y=y_ b$, evenness of $S_*$ gives $x'(0)=-1/2$ and $y'(0)=1/2$. Hence (14.26.11) gives

Indeed, if $\mathrm{d}r\wedge (Q-P)=0$, wedging with $P$ and $\mathrm d^{\mathrm c}r$ would contradict $P\wedge Q\wedge R>0$. Thus the Hermitian background $\omega $ is not closed.

Now put $\widetilde u=-F/2-\rho $ and $\widetilde v=F/2-\rho $. Then $\omega +\mathrm{dd}^{\mathrm{c}}\widetilde u=A_0$, $\omega +\mathrm{dd}^{\mathrm{c}}\widetilde v=B_0$. Subtracting their respective suprema gives $u,v$ with $\sup _ Xu=\sup _ Xv=0$, and (14.29.2) holds. Their difference is nonconstant because $F$ is nonconstant.

14.29.14 Zeros and higher dimensionsE82948C4DB2A

It remains to examine the regularity of the density and the geometry of its zero levels. Taking a product with projective space then preserves the equality of the Monge–Ampère measures and gives examples in every higher dimension.

The function $t$ is real analytic in every affine chart of $X$, being a quotient of real polynomials with positive denominator. Hence $f_*(t)$ is real analytic on $X$. Its two zero levels are regular hypersurfaces because $t=\sigma (r)$, $\sigma '>0$, and $\mathrm{d}r\ne 0$. Each level is connected, since it is the quotient of the product of two three-spheres of fixed nonzero radii by the diagonal action of the circle. Each level is compact and has zero volume. The divided differences in (14.29.5) show near the two levels that

respectively, with smooth strictly positive functions $g_\pm $.

For $n\geq 4$, let $\theta $ be the Fubini–Study form on $Y=\mathbb {P}^{n-3}$, let $\pi _1:X\times Y\to X$ and $\pi _2:X\times Y\to Y$ be the projections, and set, on $X\times Y$,

In the expansion of an $n$-fold product, only the terms with exactly three factors pulled back from $X$ survive. Consequently

and the same holds for $v_ n$. On the two product levels, the solution forms have rank $2+(n-3)=n-1$. Restriction to a slice $X\times \{ y\} $ recovers $\mathrm{d}\omega \ne 0$, so $\omega _ n$ is not closed. Normalization, distinctness and real analyticity are preserved by pullback.

Write $\operatorname {Vol}$ for the volume with respect to $\omega _ X^3$, or $\Omega _ n^ n$ in dimension $n$. The quadratic zeros complete the proof of the sharpness assertion in 14.21.13. In finitely many tubular charts around the zero levels, the density is bounded above and below by positive multiples of the square of a real transverse coordinate, and the volume form has a bounded positive density with respect to coordinate volume. Away from these charts, the density is bounded below by a positive constant. Thus, for $s>0$ and $\alpha >0$,

Thus, for every $0<\alpha <1/2$ and every dimension $n\geq 3$, there are smooth examples of nonuniqueness whose density $f$ satisfies $f^{-\alpha }\in L^1$; this condition is weaker than the hypothesis $f^{-1/2}\in L^1$ of 14.21.13.