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18.43 A logarithmic kernel scale in two dimensions9C0BFE5CB0FF

9C0BFE5CB0FF

In the curve models above, the metric weight has a single homogeneous leading term. In two dimensions, a mixed monomial in the weight can compete with the pure monomials and introduce a logarithm into the scale of the kernel. For an explicit metric on $\mathbb P^2$, the relevant global section norm can be computed directly.

Write $\mathrm dA$ for Euclidean area on $\mathbb C$. We first isolate the integral responsible for the logarithm.

Lemma 18.43.1.99A4F9F834BF Set $W(a,b)=ab+a^3+b^3$ on $[0,\infty )^2$. As $t\to \infty $,

\[ \begin{gathered} J(t):=\int _0^\infty \int _0^\infty \mathrm e^{-tW(a,b)}\, \mathrm da\, \mathrm db =\frac{\log t}{3t}+O(t^{-1}), \\ \int _0^\infty \int _0^\infty (a+b)\mathrm e^{-tW(a,b)} \, \mathrm da\, \mathrm db=O(t^{-4/3}). \end{gathered} \]

Proof. Put $A=t^{-2/3}$ and $B=t^{-1/3}$; the logarithm comes from the interval $A\leq a\leq B$. For $a>0$, the substitution $u=tab$ gives

\[ H_ t(a):=\int _0^\infty \mathrm e^{-t(ab+b^3)}\, \mathrm db =\frac1{ta}\int _0^\infty \mathrm e^{-u-u^3/(t^2a^3)}\, \mathrm du. \]

Since $0\leq 1-\mathrm e^{-v}\leq v$ for $v\geq 0$,

\[ 0\leq \frac1{ta}-H_ t(a)\leq \frac6{t^3a^4}. \]

Over $A\leq a\leq B$, the integral of this error is $O(t^{-1})$. Replacing $\mathrm e^{-ta^3}$ by one in $\int _ A^ B\mathrm e^{-ta^3}/(ta)\, \mathrm da$ costs at most $\int _ A^ B a^2\, \mathrm da=O(t^{-1})$ as well. Thus the middle interval contributes

\[ \int _ A^ B\mathrm e^{-ta^3}H_ t(a)\, \mathrm da =\frac1t\log \frac BA+O(t^{-1}) =\frac{\log t}{3t}+O(t^{-1}). \]

For $a<A$, the bound $H_ t(a)\leq \int _0^\infty \mathrm e^{-tb^3}\, \mathrm db=O(t^{-1/3})$ gives a contribution $O(t^{-1})$. For $a>B$, use $H_ t(a)\leq 1/(ta)$ and substitute $v=ta^3$ to obtain the bound

\[ \int _ B^\infty \frac{\mathrm e^{-ta^3}}{ta}\, \mathrm da =\frac1{3t}\int _1^\infty \frac{\mathrm e^{-v}}v\, \mathrm dv =O(t^{-1}). \]

Finally, $aH_ t(a)\leq 1/t$ gives

\[ \int _0^\infty \int _0^\infty a\mathrm e^{-tW(a,b)} \, \mathrm da\, \mathrm db \leq \frac1t\int _0^\infty \mathrm e^{-ta^3}\, \mathrm da =O(t^{-4/3}). \]

Symmetry proves the same bound for $b$. $\square$

On $L=\mathcal O_{\mathbb P^2}(1)$, define a Hermitian metric by

\[ |S|_ h^2([Z])= \frac{|S(Z)|^2}{\bigl (|Z_0|^6+3|Z_0Z_1Z_2|^2+3|Z_1|^6+3|Z_2|^6\bigr )^{1/3}} \qquad (S\text{ a linear form}). \]

The denominator is positive away from zero and homogeneous of degree two in $|Z|$, so the metric is well defined and smooth. Its local weight is one third of the logarithm of a sum of squared absolute values of holomorphic cubic functions, so its curvature is semipositive. In the chart $Z_0\ne 0$, put $z_ j=Z_ j/Z_0$ and $p=[1:0:0]$. The frame $e=Z_0$ has

AAC0258C6A86 18.43.2
\begin{equation} |e|_ h^2=\mathrm e^{-\phi },\qquad \phi (z)=\tfrac 13\log \bigl (1+3W(|z_1|^2,|z_2|^2)\bigr ). \end{equation}

In particular the curvature vanishes at $p$.

Proposition 18.43.3.BDAF8AE299D2 Let $\nu $ be a smooth positive probability volume form on $\mathbb P^2$, invariant under the rotations of $z_1$ and of $z_2$, and write

\[ \nu =\rho (z)\, \mathrm dA(z_1)\, \mathrm dA(z_2),\qquad \rho _0=\rho (0)>0 \]

in the chart above. Give $H^0(\mathbb P^2,L^ m)$ the norm $\int _{\mathbb P^2}|s|_{h^ m}^2\, \mathrm d\nu $, and let $B_ m$ be its kernel density. Then

\[ B_ m(p)=\frac{3}{\pi ^2\rho _0}\frac m{\log m} \left(1+O((\log m)^{-1})\right). \]

For the Fubini–Study probability volume, the leading constant is $3/2$.

Proof. The homogeneous monomials form an orthogonal basis, because the metric and the volume are invariant under the two independent rotations and distinct monomials transform by distinct characters. At $p$, only $Z_0^ m$ has a nonzero value, equal to one in the frame $e^{\otimes m}$. Consequently $B_ m(p)=I_ m^{-1}$, where

\[ I_ m=\int _{\mathbb P^2}|Z_0^ m|_{h^ m}^2\, \mathrm d\nu =\int _{\mathbb C^2}(1+3W(|z_1|^2,|z_2|^2))^{-m/3} \rho (z)\, \mathrm dA(z_1)\, \mathrm dA(z_2). \]

Here the omitted projective line $\{ Z_0=0\} $ has zero volume. By rotational invariance, the first derivatives of $\rho $ vanish at zero, so $\rho (z)=\rho _0+O(|z_1|^2+|z_2|^2)$ near zero. In the variables $a=|z_1|^2$ and $b=|z_2|^2$, polar integration has Jacobian $\pi ^2\, \mathrm da\, \mathrm db$.

Choose a fixed small $\delta >0$. On $W\leq \delta $, set $\ell (W)=\tfrac 13\log (1+3W)$. For constants $c,C>0$ independent of $m$,

\[ cW\leq \ell (W)\leq W,\qquad 0\leq \mathrm e^{-m\ell (W)}-\mathrm e^{-mW} \leq CmW^2\mathrm e^{-cmW} \leq \frac Cm\mathrm e^{-cmW/2}. \]

The first estimate of lemma 18.43.1 bounds the integral of this error in the weight by $O(m^{-2}\log m)$, and the second bounds the error from replacing $\rho $ by $\rho _0$ by $O(m^{-4/3})$. On the complement $W>\delta $, the contribution to $I_ m$ is at most $(1+3\delta )^{-m/3}$, because $\nu $ has mass one, and the corresponding tail of the exponential model is at most $\mathrm e^{-m\delta /2}J(m/2)$. Therefore

\[ I_ m=\pi ^2\rho _0J(m)+O(m^{-4/3})+O(m^{-2}\log m) =\frac{\pi ^2\rho _0}{3m}\log m+O(m^{-1}). \]

Inverting proves the first assertion. The Fubini–Study probability volume has the local expression

\[ \nu =\frac2{\pi ^2}(1+|z_1|^2+|z_2|^2)^{-3} \, \mathrm dA(z_1)\, \mathrm dA(z_2). \]

It has mass one, since $2\int _0^\infty \int _0^\infty (1+a+b)^{-3}\, \mathrm da\, \mathrm db=1$, and its density at zero is $2/\pi ^2$. The leading constant is therefore $3/2$. $\square$

Thus the full kernel, without any selection, already has a logarithmic factor at this fixed curvature zero, and its coefficient depends on the volume density at $p$. The calculation does not give a transition for growing vanishing ideals; that would require control of the norms of the selected monomials on the moving scale, as in the curve case.

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