Theorem 18.46.2318BC0622217

Proof. By lemma 5.37.1,

In the product affine frame, global sections are polynomials of degree at most $m$ in each variable. A global section belongs to $V_ m$ exactly when each of its monomials vanishes to order at least $r_ m$ along $D_0$ and along $D_\infty $, that is, when the exponent $\alpha _0$ of the first variable satisfies $r_ m\leq \alpha _0\leq m-r_ m$; the other exponents range from $0$ to $m$. There is one admissible value of $\alpha _0$ when $m$ is even and two when $m$ is odd. This proves the dimension formula.

To identify the envelope, let $v\in \mathrm{PSH}(X,\theta )$ be a competitor, that is, $v\preceq _{\mathcal I}\psi $ in the multiplier-ideal order of section 5.33 and $v\leq 0$ quasi-everywhere. By theorem 5.37.3, the Lelong numbers of $v$ at the points of $D_0$ and $D_\infty $ are at least $1/2$. Fix $y\in Y$ and restrict $v$ to the line $\mathbb P^1\times \{ y\} $. The restriction is either identically $-\infty $ or $\sigma _0$-plurisubharmonic. In the latter case, its Lelong numbers at the two points over $0$ and $\infty $ are at least those of $v$: the supremum over a ball in the line is at most the supremum over the corresponding ambient ball, and division by $\log r^2<0$ reverses the inequality. By the identification of Lelong numbers with atoms in lemma 5.37.18, the curvature measure of the restriction has atoms of mass at least $1/2$ at these two points. By Stokes' theorem its total mass is one, so these atoms exhaust the measure. The difference between the restriction and $\psi _0$ is therefore harmonic on the compact projective line, hence constant.

It follows that $v(z,y)=\psi _0(z)+c(y)$ off $D_0\cup D_\infty $, where $c(y)=-\infty $ on lines along which $v$ is identically $-\infty $. Restriction to $z=1$ shows that $c(y)=v(1,y)+\log 2$ is a $\theta _ Y$-plurisubharmonic function; it is not identically $-\infty $, since otherwise $v$ would be. Submean values over shrinking balls upgrade the obstacle condition to $v\leq 0$ everywhere. Since $\max \psi _0=-\log 2$, we get $c\leq \log 2$, and hence $v\leq E$. On $D_0\cup D_\infty $ both $v$ and $E$ equal $-\infty $. Conversely, $E\leq 0$ and $E$ differs from $\psi $ by a constant, so $E$ is itself a competitor. This proves (18.46.3).

The norms of the monomials in one factor are beta integrals. With $x=|z|^2/(1+|z|^2)$, they are

Orthogonality under the rotations and integration over the product diagonalize the global norm. Summing over all monomials in one factor gives the constant kernel $m+1$, by the binomial formula. Therefore the full kernel on $Y$ is $(m+1)^ q$. With $C_ k=(2k+1)\binom {2k}{k}$, the selected kernels are

We use the even formula for $k\geq 1$; the odd formula also holds for $k=0$. In the odd case the two monomials in the first factor have equal norms, and the sum of their squared moduli contributes the factor $1+|z|^2$, which gives the second identity. The full kernel on $X$ is $(m+1)^{q+1}$, so evaluation gives $\sup _ X|s|_{h^ m}\leq (m+1)^{(q+1)/2}\| s\| _ m$ for every selected section. This polynomial bound proves the Bernstein–Markov inequality.

The central binomial coefficient is the largest in its row. Hence $4^ k\leq C_ k\leq (2k+1)4^ k$, and $\log C_ k=2k\log 2+O(\log (k+1))$. Taking logarithms in (18.46.5) gives

because $|\psi _0|=\log (2\cosh (\tfrac 12\log |z|^2))$ is bounded by $\log 2+\tfrac 12|\log |z|^2|$. The function $\log |z|^2$ belongs to $L^ p$ for every finite $p$ with respect to smooth volume forms. This proves the convergence of the potentials, and differentiating the $L^1$ limit proves the convergence of the currents. For every $m\geq 2$, both $U_ m$ and $E$ are $-\infty $ on $D_0\cup D_\infty $.

Let $P_ k$ be the probability measure on the first factor with density $C_ k\mathrm e^{2k\psi _0}$ with respect to $\sigma _0$. Under $P_ k$ the angle of $z$ is uniformly distributed, and $x$ has the beta distribution with parameters $(k+1,k+1)$, whose mean is $1/2$ and whose variance is $1/[4(2k+3)]$. Chebyshev's inequality and continuity of the test functions in the angle therefore give $P_ k\rightharpoonup \varsigma $. The measure identities

prove both limits along even and odd degrees and the limit after division by $N_ m$. Since the two limits have different total masses, the $q$-normalized family does not converge. Normalization by the ambient degree, $(q+1)!/m^{q+1}$, instead gives zero in total variation.

Finally, off $D_0\cup D_\infty $ the current in (18.46.1) is $\theta _ Y$, whose rank is $q$. The non-pluripolar product charges neither divisor, so the top product is zero. $\square$