12.18 Geometric bounds and determinant concentrationAA652CC345D8

Proof. The identity $\mathrm{dd}^{\mathrm{c}}(g^2)=2g\wedge \mathrm{dd}^{\mathrm{c}}g+2\mathrm{d}g\wedge \mathrm d^{\mathrm c}g$ shows that $\mathrm{d}g\wedge \mathrm d^{\mathrm c}g=0$, so theorem 11.1.7 gives $v_-(g)=V_ g$. The quotient in (12.15.5) is monotone and homogeneous in its numerator, so $h\geq cg$ implies $v_-(h)^{1/n}\geq cV_ g^{1/n}$. Theorem 12.9.2 turns these lower-mass bounds into the asserted determinant bounds, with equality when $h=g$.

A Kähler form satisfies both differential identities. On a surface, theorem 11.4.2 gives a metric $g$ with $\mathrm{dd}^{\mathrm{c}}g=0$, and $\mathrm{dd}^{\mathrm{c}}(g^2)=0$ for degree reasons. By compactness, each fixed $h$ satisfies $h\geq cg$ for some $c>0$. $\square$

Proof. Put $H_ i=\bigwedge _{j\ne i}\omega _ j$. Expanding the wedge powers gives $h^{n-1}=(n-1)!\sum _ iH_ i$ and $h\wedge H_ i=\mathrm{d}V$. Thus $T_*=(1/n)\sum _ iH_ i$ is normalized, and the determinant formula gives $k_{T_*}=V^{-1/n}$. The smooth equation for $F=0$ is solved by the zero potential, since $h^ n=V\, \mathrm{d}V$; theorem 12.7.3 therefore gives the maximal value and the uniqueness of the maximizer.

To construct $S_\varepsilon $, choose a nonnegative smooth function $\beta $ compactly supported in the unit disk in $\mathbb C$, of integral one, with $\beta \geq \beta _0>0$ on a disk of radius $r_0\in (0,1)$. On an embedded disk in the first elliptic factor put $b_\varepsilon (z)=\varepsilon ^{-2}\beta (z/\varepsilon )$, extended by zero; it has integral one. For $0<t<1$, set

The form $\gamma _{\varepsilon ,t}$ is strictly positive, and it is closed because $f_{\varepsilon ,t}$ depends only on the first factor, which has real dimension two. Direct expansion gives

Since $\int _ Xf_{\varepsilon ,t}\, \mathrm{d}V=1$, the trace integral is one, so $T_{\varepsilon ,t}\in \mathscr G_ h$, and $\| k_{T_{\varepsilon ,t}}\| _{L^ q}=V^{-1/n}$ because $\alpha q=1$.

The mean-zero function $f_{\varepsilon ,t}-1$ has a smooth real potential $\psi $ on $E$ with $\mathrm{dd}^{\mathrm{c}}\psi =(f_{\varepsilon ,t}-1)\omega _1$; for example, the mean-zero Fourier series solving $\Delta \psi =4\pi (f_{\varepsilon ,t}-1)$ in the area-one coordinates. Consequently $\gamma _{\varepsilon ,t}-ah=\mathrm{dd}^{\mathrm{c}}(a\psi )$. Since $T_*=(ah)^{n-1}$, the closed form $R=\sum _{j=0}^{n-2}\gamma _{\varepsilon ,t}^{n-2-j}\wedge (ah)^ j$ satisfies

This proves all the assertions about classes, including the case $n=2$, where $R=1$.

Jensen's inequality and $f_{\varepsilon ,t}\geq 1-t$ give

where the inequality is strict because $f_{\varepsilon ,t}$ is nonconstant and $0<\alpha <1$. On the product of the disk of radius $r_0\varepsilon $ with the other factors, we have $f_{\varepsilon ,t}\geq t\beta _0\varepsilon ^{-2}$, and this set has volume $\pi r_0^2\varepsilon ^2$. Hence

Now take $t=1/\log (1/\varepsilon )$, which lies in $(0,1)$ by the restriction on $\varepsilon $, and $S_\varepsilon =T_{\varepsilon ,t}$. Then $\mathfrak D_0(S_\varepsilon )$ converges to its maximum, while both lower bounds diverge for $p>q$ because $\alpha p>1$. All constants are independent of $\varepsilon $. $\square$

Proof. In a holomorphic frame put $\lambda =\bigwedge _{j=1}^ n(i\, dz_ j\wedge d\bar z_ j)$, $h=i\sum H_{ij}\, dz_ i\wedge d\bar z_ j$ and $\mathrm dV=v\lambda $. Positive currents have measure coefficients. Choose a positive Radon measure $\nu $ dominating them. For a smooth real form $\xi =i\sum \Xi _{ij}\, dz_ i\wedge d\bar z_ j$, let $Q$ be the positive semidefinite Hermitian coefficient matrix characterized by $\xi \wedge T=\operatorname {tr}(\Xi Q)\nu $. Then

The formula holds on $Q\ne 0$; zero is assigned when $Q=0$. Since $h^ n=n!\det (H)\lambda $, determinant algebra gives

This expression has no $H$. Replacing $\nu $ by $L\nu $, where $L>0$ almost everywhere, replaces $Q$ by $Q/L$, leaving the measure unchanged; a common dominating measure compares any two choices. The original geometric definition ensures compatibility under frame changes. Matrix AM–GM bounds it locally by a constant times a reference trace measure, proving local finiteness. Compactness gives a finite Radon measure. $\square$

Proof. Distributional integration by parts gives $\int h\wedge T=\int \beta \wedge T+\int u\mathrm{dd}^{\mathrm{c}}T =\int \beta \wedge T$. This proves equality of the actual current sets; Lemma 12.18.6 proves equality of their measure sets.

Take a smooth strictly positive real $r$ and put $M=\int r^ n \mathrm dV$. By theorem 14.2.1 there are a smooth potential $w$, a positive form $A=\beta +\mathrm{dd}^{\mathrm{c}}w$ and a positive coefficient in its prescribed-volume equation. The mass-invariance hypothesis and theorem 11.1.7 determine that coefficient, giving

In particular, the total mass of $A$ is $V$. No estimates uniform in a degenerating smooth density are required.

For every $T\in \mathscr G_\beta $, integration by parts gives $\int A\wedge T=1$. In an $A$-unitary frame, matrix AM–GM gives $\det (P_ A)^{1/n}\leq 1$, including singular coefficient densities. Thus as Radon measures

Multiplying by $r$ and using (12.18.10) yields

Replacing a smooth nonnegative $r$ by $r+\eta $ and letting $\eta \downarrow 0$ extends this bound to nonnegative smooth tests. For smooth real or complex $s$, smooth positive approximations $\sqrt{|s|^2+\eta ^2}$ imply $|\int s\, \Gamma _0(T)|\leq V^{-1/n}\| s\| _ n$. Consequently integration against $\Gamma _0(T)$ extends continuously from smooth functions to $L^ n$. Its positive dual representative is $k_ T\in L^{n/(n-1)}$ with the stated bound. Equality on smooth tests identifies the two Radon measures, proving absolute continuity. The integrals of arbitrary $L^ n$ roots are therefore well-defined.

For the reverse inequality with $r>0$ smooth, let $T_ r=A^{n-1}/V$. It is smooth and positive. Write $E=\mathrm{dd}^{\mathrm{c}}w$. For $j\leq n-1$, expansion and closedness of $E$ give

The term with $j=l$ differentiates the constant one. Thus $\mathrm{dd}^{\mathrm{c}}T_ r=0$, and smooth integration by parts gives

Consequently $T_ r\in \mathscr G_\beta $. Its positive root is $\gamma =V^{-1/(n-1)}A$, so

It attains equality in (12.18.11). For a nonnegative $p\in L^ n$, take smooth positive $r_ j\to p$ in $L^ n$. The common dual bound controls $\int (p-r_ j)k_{T_{r_ j}}\mathrm dV$, proving the lower bound in (12.18.9); the upper bound already holds for every current. Apply the same argument to $|p|$ for complex roots. $\square$

Proof. Write $z_ j=x_ j+iy_ j$, $\beta _ j=(i/2)dz_ j\wedge d\bar z_ j=dx_ j\wedge dy_ j$, $\beta =\sum _ j\beta _ j$, $V=n!$, and $\mathrm dV=\beta ^ n/V$. Set

For a function of $x_1$, $\mathrm{dd}^{\mathrm{c}}u=(u''/(4\pi ))\beta _1$. Hence

These are globally periodic smooth positive closed forms with $\varepsilon \leq a_\varepsilon \leq 2-\varepsilon $ and $\omega _\varepsilon \leq 2\beta $. Their minimum relative eigenvalue is $\varepsilon $. They converge smoothly to a semipositive form losing rank on $\{ x_1\in \tfrac 12\mathbb Z\} $. All fixed-order smooth norms of the coefficients and potentials are uniformly bounded. Theorem 12.18.8 proves equality of the entire current data and the exact norm at every $\varepsilon $.

Since $\omega _\varepsilon $ is closed, theorem 11.1.7 gives mass $V$ for every bounded $\omega _\varepsilon $-plurisubharmonic potential. The explicit formula also gives $\int _ X\omega _\varepsilon ^ n=V$, since $a_\varepsilon $ has mean one. Thus both extrema over bounded potentials equal $V$. The global form $dz_1\wedge \cdots \wedge dz_ n$ trivializes the canonical bundle; closed positive metrics are balanced and pluriclosed. $\square$

Proof. With the torus coordinates above set $s=\sin (2\pi x_1)$ and

The first relative Hermitian block is $\left(\begin{smallmatrix}1& \tau s\\ \tau s& 1\end{smallmatrix}\right)$, so positivity follows from $|\tau s|<1$. Moreover

Because $\mathrm{dd}^{\mathrm{c}}s$ is proportional to $dz_1\wedge d\bar z_1$, $\mathrm{dd}^{\mathrm{c}}(s\xi )=0$. The first two-coordinate metric is thus pluriclosed, and its square is top degree on that complex surface. Expanding with the remaining closed curve factors proves

Use the same $u_\varepsilon $ and $a_\varepsilon $ as in Theorem 12.18.13, now with $\varepsilon \leq 1/2$. Then

The first block of $\Omega _\varepsilon -\varepsilon \beta $ has diagonal entries $2(1-\varepsilon )s^2$ and $1-\varepsilon $, and determinant

It is positive semidefinite; the remaining diagonal entries are $1-\varepsilon $. Hence $\Omega _\varepsilon \geq \varepsilon \beta $, with equality in the least eigenvalue at $s=0$. Also $\Omega _\varepsilon \leq 3\beta $, all fixed-order smooth norms are uniformly bounded, and the smooth semipositive limit loses rank at $s=0$. Since $d\mathrm{dd}^{\mathrm{c}}u_\varepsilon =0$, the exterior derivative is the displayed fixed nonzero $dh_\tau $.

The lower-power expansion in the proof of theorem 12.18.8 applies to $\Omega _\varepsilon =h_\tau +\mathrm{dd}^{\mathrm{c}}u_\varepsilon $; it gives $\mathrm{dd}^{\mathrm{c}}(\Omega _\varepsilon ^ j)=0$ for $j\leq n-1$, in particular pluriclosedness. Its volume is $V_\tau $, either by integrating the displayed determinant or by theorem 11.1.7. That theorem also gives mass $V_\tau $ for every bounded $\Omega _\varepsilon $-plurisubharmonic potential, so both mass extrema equal $V_\tau $. Theorem 12.18.8 applied to $h_\tau $ gives equality of the entire current and measure sets and their exact endpoint for every $p\in L^ n$. The balanced trace is fixed as well:

The off-diagonal term has zero trace and the mean of $a_\varepsilon $ equals one. The global holomorphic volume form gives $c_1=0$. The manifold itself is already Kähler, so these nonclosed metrics show that the complete determinant profile and positive bounded-potential mass extrema can stay fixed while the least relative eigenvalue tends to zero. The coexistence problem of 24.4.22 remains open. $\square$

Proof. Put $r=1-\varepsilon $ and

The positive periodic Poisson kernel has mean one and Fourier series

Define

For fixed $\varepsilon >0$ this is smooth. Differentiation gives $w_\varepsilon ''=4\pi (f_\varepsilon -1)$, and hence

It is closed, and $\int A_\varepsilon ^ n=V$. Theorem 12.18.8 and the smooth-root formula give

Thus the displayed power integral attains the upper bound over all currents, not a bound for a selected family. More precisely, $p_\varepsilon =f_\varepsilon ^{1/n}$ has $L^ n$ norm one, and $\int p_\varepsilon k_{T_\varepsilon }\mathrm dV=V^{-1/n}$, so $T_\varepsilon $ attains the original entire-family dual supremum for this varying root. No common fixed density is asserted.

The Fourier series bounds $|w_\varepsilon |\leq \pi /3$ and implies uniform convergence to

This limit is continuous and canonically $\beta $-psh by distributional positivity of $A_0=\beta +\mathrm{dd}^{\mathrm{c}}w_0$. The probability measures $P_ r(x_1)dx_1$ converge weakly to the periodic Dirac mass at $x_1=0$, as follows from their Fourier coefficients $r^{|m|}$. Thus

where the Dirac measure is tensored with Lebesgue measure in the remaining real coordinates. Its bounded local products are genuine Bedford–Taylor products. Locally the first-coordinate current $\beta _1+\mathrm{dd}^{\mathrm{c}}w_0$ is pulled back from one complex coordinate, so its powers of degree at least two vanish in the bounded-product recursion. The remaining factors are smooth and closed. Multilinear expansion therefore contains at most one first-coordinate factor; the same expansion for $A_\varepsilon $ passes directly to the distributional limit. Consequently $T_\varepsilon \to T_0=A_0^{n-1}/V$ weakly. This current is positive, closed and normalized by $\int \beta \wedge T_0=1$.

In its positive coefficient matrix the absolutely continuous part is proportional to the first diagonal matrix below; the singular part on $x_1=0$ is proportional to the second:

Their ranks are respectively one and $n-1$, both less than $n$. With a measure dominating these mutually singular parts, formula (12.18.7) therefore gives $\Gamma _0(T_0)=0$. This is not the Monge–Ampère measure of $w_0$, which has total mass $V$ by theorem 11.1.7.

Away from $x_1=0$, $f_\varepsilon \to 0$ pointwise; its integral against $\mathrm dV$ remains one. Thus the positive power densities are not uniformly integrable: almost-everywhere convergence to zero and uniform integrability would force their fixed positive integrals to vanish. In particular $T\mapsto \int k_ T^{n/(n-1)}\mathrm dV$ is not continuous under these weak current limits, even with uniformly convergent continuous potentials. The intrinsic measures themselves satisfy $\Gamma _0(T_\varepsilon )\to 0$ in total mass; this follows from their uniform $L^{n/(n-1)}$ bound by splitting off arbitrarily small neighborhoods of $x_1=0$. The discontinuity concerns their power integrals, not that last measure limit. $\square$