Example 6.12.19B10A1F1C864B

Example 6.12.19.B10A1F1C864B Let $c>0$ and $\varphi =c\log |z_1|^2$ on a polydisc in $\mathbb C^ n$. The ideal $(z_1)$ is already principal with smooth zero set, so the identity map is a log resolution, with $D=E=\{ z_1=0\} $, $d=1$ and $A_ X(E)=1$. By theorem 6.12.8, a germ $f$ lies in $\mathcal I(\varphi )$ if and only if $\operatorname {ord}_ Ef+1>c$. Here the log resolution has no exceptional divisors, and the only condition comes from the non-exceptional component $E$; this is why theorem 6.12.8 involves all components of $D$ meeting the fiber, exceptional or not.

For the blowup $\pi $ of the origin in dimension $n\geq 2$, consider the chart $z_1=w_1$, $z_ j=w_1w_ j$ for $j>1$ of proposition 9.6.3. The Jacobian matrix $(\partial z_ i/\partial w_ j)$ is lower triangular with diagonal entries $1,w_1,\ldots ,w_1$, so $J_\pi =w_1^{n-1}$, and the exceptional divisor $E=\{ w_1=0\} $ has $A_ X(E)=n$. Let $f$ be holomorphic near $0$ with homogeneous expansion $f=\sum _{k\geq k_0}P_ k$ and $P_{k_0}\neq 0$. Then $f\circ \pi (w)=\sum _{k\geq k_0}w_1^ kP_ k(1,w_2,\ldots ,w_ n)=w_1^{k_0}g(w)$ for small $w_1$ and bounded $(w_2,\ldots ,w_ n)$, where $g(0,w')=P_{k_0}(1,w')$; this polynomial is not identically zero, since $P_{k_0}(z)=z_1^{k_0}P_{k_0}(1,z_2/z_1,\ldots ,z_ n/z_1)$ for $z_1\neq 0$. Hence $\operatorname {ord}_ Ef=\operatorname {ord}_0f$.

For $\varphi =c\log |z|^2$, this blowup is a log resolution of the maximal ideal, with $D=E$ and $d=1$, by proposition 9.6.3, so theorem 6.12.8 gives $f\in \mathcal I(\varphi )_0$ if and only if $\operatorname {ord}_0f+n>c$. This agrees with a direct computation in polar coordinates: since $|f(z)|\leq C|z|^{k_0}$ near $0$ by (5.2.2), and $\int _{S^{2n-1}}|f(t\omega )|^2\, \mathrm dS(\omega )\geq c't^{2k_0}$ for small $t>0$, with $c'>0$, by (6.8.8), the integral of $|f|^2|z|^{-2c}$ over a small ball $B(0,r)$ is finite if and only if $\int _0^ rt^{2k_0-2c+2n-1}\, \mathrm dt<\infty $, by (3.1.2), that is, if and only if $k_0+n>c$.