5.37 Multiplier ideals and logarithmic poles5BB7B3EA169D

Proof. The bounded remainder changes the weight only by bounded positive factors. Expand a nonzero holomorphic germ $g(z,w)$ in homogeneous polynomials in $w$, and let $a$ be the smallest degree whose term does not vanish identically. On smaller product neighborhoods, $|g(z,w)|\leq C|w|^ a$. Hence $a+k>tc$ suffices for weighted integrability, since the corresponding radial integral in $\rho =|w|$ is

Conversely, let $z$ be a point near the origin at which the term of degree $a$ does not vanish identically in $w$. Integration of $|g(z,w)|^2$ over the spheres $|w|=\rho $ gives a positive leading term of order $\rho ^{2a}$, so the radial integral in $w$ diverges when $a+k\leq tc$, logarithmically in the case of equality. Such points $z$ form a nonempty open set in every sufficiently small neighborhood, and Fubini's theorem shows that $g$ is not weighted-integrable near the origin. For $d=0$ this is the radial calculation at a point. Thus $g$ is weighted-integrable exactly when $a\geq \lfloor tc\rfloor -k+1$, that is, when $g$ lies in the power of $\mathcal I_ Y$ given in the statement. $\square$

Proof. Choose disjoint coordinate balls centered at the points $p_ j$ and smooth cutoffs $\chi _ j$ supported in them and equal to one near their centers. The function $\chi _ j\log |z|^2$, extended by zero, is plurisubharmonic near its pole and smooth elsewhere, so its complex Hessian is bounded below by $-C_ j\omega $. Set $u=\sum _ ja_ j\chi _ j\log |z|^2$. These lower bounds give the asserted current inequality, and the remaining properties follow from the disjointness of the supports. $\square$

Proof. Fix a coordinate neighborhood $U\subset X$ and a modification $\pi :Y\to U$ with smooth source. At every $y\in Y$, the elementary rules for the Lelong numbers of maxima, sums and positive multiples give

These rules are unaffected by smooth local corrections. Since $u\sim _{\mathcal I}v$ and $u'\sim _{\mathcal I}v'$, the first item of theorem 5.37.3 shows that the right-hand sides do not change when $u,u'$ are replaced by $v,v'$. The converse direction of the same item, applied on each coordinate neighborhood, gives the three asserted equivalences.

For the cancellation rule, suppose that $u+h\sim _{\mathcal I}v+h$. By the first item of theorem 5.37.3, the pullbacks $(u+h)\circ \pi $ and $(v+h)\circ \pi $ have the same Lelong numbers. The common summand $h\circ \pi $ is quasi-plurisubharmonic, so its Lelong numbers are finite, and subtracting them shows that $u\circ \pi $ and $v\circ \pi $ have the same Lelong numbers. Hence $u\sim _{\mathcal I}v$ by the same item. The reverse implication is the addition rule proved above. $\square$

Proof. Fix a point $x$ and add a smooth correction to make $q$ plurisubharmonic on a bounded pseudoconvex coordinate neighborhood. Form the weighted Bergman space of proposition 5.35.1 at weight $t$. If every member vanished at $x$, the Cauchy estimates, together with the local upper bound for $q$, would give

near $x$. Its Bergman potential would consequently have Lelong number at least $1/t$, contradicting (5.35.3). There is therefore a weighted square-integrable holomorphic function nonzero at $x$. On a smaller neighborhood it is bounded away from zero, so its weighted norm proves integrability of $\mathrm e^{-tq}$. Smooth corrections do not change this conclusion. Thus $q\sim _{\mathcal I}0$; the addition rule in lemma 5.37.4 gives $v+q\sim _{\mathcal I}v$, which is precisely the asserted equality at every weight. $\square$

Proof. Work in product coordinates $(z,\zeta )$ with $p(z,\zeta )=z$. Write $\alpha $ for a multi-index in the fiber variables and $\mathrm{d}\lambda _\zeta $ for their Euclidean volume. A germ in the ideal on the right is a finite sum of bounded holomorphic multiples of pullbacks of germs that are square integrable against $\mathrm{e}^{-tu(z)}$. Fubini's theorem shows that it is weighted-integrable on a sufficiently small product neighborhood.

Conversely, let $F(z,\zeta )$ be a weighted-integrable holomorphic germ and expand it in the fiber variables,

The coefficient estimate for a holomorphic function on a polydisk gives, on smaller product neighborhoods,

Multiplying by $\mathrm{e}^{-tu(z)}$ and integrating in $z$ shows that each coefficient germ $f_\alpha $ belongs to $\mathcal I(tu)$. Let $J$ be the extension of this ideal to the local ring $R$ of holomorphic germs on the product. Every finite Taylor sum of $F$ belongs to $J$, and the remainder of total fiber degree at least $N$ belongs to $(\zeta )^ N$. Thus the class of $F$ in $R/J$ lies in $(\zeta )^ N(R/J)$ for every $N$. By the Krull intersection theorem for the finitely generated module $R/J$ over the Noetherian local ring $R$, the intersection of these submodules is zero. Hence $F\in J$. Smooth corrections and smooth positive volume densities do not change the integrability conditions. $\square$

Proof. On the curve the coefficient is $1/2$, so the multiplier ideal is the zeroth power of the maximal ideal. Upstairs $v\circ f=\log |z_0|^2$ has coefficient one along the smooth divisor $\{ z_0=0\} $. Apply lemma 5.37.1 in codimension one to get its first ideal power. $\square$

Proof. Step 1: a twisted $\bar\partial $-estimate. Suppose first that $\Omega $ is bounded, that the weights are smooth on a neighborhood of $\overline\Omega $ with $S<1$ there, and that $\Phi =\log S$ is strictly plurisubharmonic. Set $\chi =\phi _1+\phi _2+\psi $ and $A=(\Phi _{j\bar k})$. For a $(0,1)$-form $b$, write $|b|_ A^2$ for its squared norm with respect to the dual of this positive Hermitian matrix. We claim that every $\bar\partial $-closed $(0,1)$-form $b$ for which the right-hand side below is finite admits a solution $u$ of

To prove this, define for $x<0$

The inequality $-x>1-\mathrm e^ x$ gives $\eta >0$. Also $\eta '= -[1-(1-x)\mathrm e^ x]/(1-\mathrm e^ x)^2<0$, whereas $T'>0$, so $\lambda >0$. Direct differentiation gives

Use the weight $h=\mathrm e^{-\chi -T(\Phi )}$ on the trivial line bundle and evaluate the twisting functions at $\Phi $. The twisted curvature is

Let $\Lambda $ denote contraction by the Euclidean Kähler form and $\bar\partial _ h^*$ the weighted adjoint. For a compactly supported $(n,1)$-form $v$, the twisted Bochner identity and the weighted Cauchy–Schwarz inequality give

More explicitly, the cross term is bounded by $\lambda |\bar\partial _ h^*v|^2+\lambda ^{-1}|(\bar\partial \eta )^*v|^2$; the latter term is exactly the commutator contribution of $\mathrm i\lambda ^{-1}\partial \eta \wedge \bar\partial \eta $. The terms of type $(n+1,1)$ vanish, and the remaining connection norm has a favorable sign.

The Hilbert space existence argument applied to (5.37.13) therefore gives

To see the coefficient, project a test form to $\ker \bar\partial $. On this kernel the second term of (5.37.13) vanishes, and the Cauchy–Schwarz inequality in the curvature norm bounds the functional $v\mapsto \langle b,v\rangle _ h$ by $(\int |b|_ A^2h)^{1/2}\| \sqrt{\eta +\lambda }\, \bar\partial _ h^*v\| _ h$. The Riesz representation theorem gives a solution with the asserted norm. The twisting functions are bounded on $\Omega $, since $\Phi $ is smooth and bounded away from both $-\infty $ and $0$. The argument first uses a complete Kähler metric; approximating the Euclidean metric by the Euclidean metric plus small positive multiples of a complete metric gives the Euclidean statement. Under the identification with canonical-bundle-valued forms, the $(n,0)$ norm integral and the curvature-dual $(n,1)$ integral have the same coordinate expressions as above. Since $h=\mathrm e^{-\chi }(1-\mathrm e^\Phi )$, the identities for $T,\eta ,\lambda $ now give (5.37.12).

Step 2: division for smooth weights. Put $v_ i=\mathrm e^{\phi _ i}/S$ and $\ell =(v_1f,v_2f)$. Equip $\mathbb C^2$ with the diagonal metric $(\mathrm e^{-\phi _1},\mathrm e^{-\phi _2})$. Then $\ell $ is orthogonal to the kernel line of the sum map, generated by $(1,-1)$, and $|\ell |^2=|f|^2/S$. Write $\delta =\phi _1-\phi _2$. The Hessian formula gives

Thus $v_1v_2|\bar\partial \delta |_ A^2\leq 1$ and, for $b=\bar\partial (v_1f)$,

Solve (5.37.12) for this $b$ and set $F=\ell -u(1,-1)$. Since $v_1+v_2=1$, the vector $F=(F_1,F_2)$ is holomorphic and its components sum to $f$. The squared norm of $(1,-1)$ is $\mathrm e^{-\phi _1}+\mathrm e^{-\phi _2}=S\mathrm e^{-\phi _1-\phi _2}$. Orthogonality and (5.37.12) consequently give

This proves (5.37.11) for smooth weights.

Step 3: general weights. Exhaust $\Omega $ by bounded pseudoconvex domains with compact closures in $\Omega $. On each of them, convolve the weights on a slightly larger neighborhood and add a sufficiently small positive multiple of $|z|^2$. These smooth plurisubharmonic weights dominate the original weights, converge to them almost everywhere, preserve $S<1$ on the chosen closure, and make $\Phi $ strictly plurisubharmonic. The corresponding integrals $J$ are bounded by the original one. On every compact subset the weights have common upper bounds, so (5.37.11) gives uniform unweighted local $L^2$ bounds for both components. The holomorphic mean-value inequality and a diagonal extraction give holomorphic limits. Almost-everywhere convergence of the weights and Fatou's lemma yield (5.37.11) on $\Omega $, and $F_1+F_2=f$ passes to the limit. $\square$

Proof. For $t>0$, apply lemma 5.37.10 with

Their exponential sum is $S=(1+t\mathrm e^{p\phi })/(1+t)<1$. The integral $J$ of the lemma is finite, since $\| f\| _\psi ^2\leq I$ and $S\geq 1/(1+t)$. The first component $F_1$ belongs to $\Gamma (\Omega ,\mathcal I(p\phi +\psi ))$, so the second component $F_2=f-F_1$ is admissible for $c$, and the norm estimate gives

Set $t=x^ p$ and integrate over $x>0$. With $\kappa _ k=\int _0^\infty (1+x^ p)^{-k}\, \mathrm{d}x$, Tonelli's theorem and the change of variable $x\mapsto x\mathrm e^\phi $ give $c\kappa _1\leq I\kappa _2$; the pole set of $\phi $ is Lebesgue null. Since $p>1$, both integrals are finite, and integrating the derivative of $x/(1+x^ p)$ over $(0,\infty )$ gives $\kappa _2=(1-1/p)\kappa _1$. The argument does not require the infimum to be attained. $\square$

Proof. Set $\psi _ i=\psi +\sum _{j>i}\phi _ j$, $a_ i=1-1/p_ i$, and $f_0=f$. For arbitrary $\varepsilon _ i>0$, proposition 5.37.14 successively gives holomorphic $f_ i$ such that

Since $\psi _ i\leq \psi $, these ideals are contained in $\mathcal I(p_ i\phi _ i+\psi )$. As $f-f_ r$ is the sum of the differences $f_{i-1}-f_ i$, the function $f_ r$ is admissible for $c$. Moreover, $\psi _ i+\phi _ i=\psi _{i-1}$ for $i\geq 1$ and $\psi _ r=\psi $, so iteration gives

Letting every $\varepsilon _ i$ tend to zero proves (5.37.16).

For sharpness when $r\leq n$, take the unit polydisk $\Omega =\Delta ^ n$, $f=1$, $\psi =0$, and $\phi _ i=p_ i^{-1}\log |z_ i|^2$ for $i\leq r$. By lemma 5.37.1, $\mathcal I(p_ i\phi _ i)=(z_ i)$. Every admissible $g$ equals one on $Y=\{ z_1=\cdots =z_ r=0\} $, hence $g(0)=1$. Orthogonality of the monomials on the polydisk gives $\| g\| _0^2\geq \pi ^ n$, with equality for $g=1$. Direct radial integration gives $I=\pi ^ n\prod _ i p_ i/(p_ i-1)$, so equality holds in (5.37.16). $\square$

Proof. With the normalization of $\mathrm{dd}^{\mathrm{c}}$ fixed in the conventions, $\mu =\mathrm{dd}^{\mathrm{c}}u$ is a positive Radon measure and $a=\mu (\{ x\} )=\nu (u,x)$. After removal of this atom, the distribution $u-a\log |z-x|^2$ has positive complex Hessian, so it has a subharmonic representative $v$, and $\mathrm{dd}^{\mathrm{c}}v$ has no atom at $x$. Since subharmonic functions that agree almost everywhere agree everywhere,

We claim that, for each $t>0$, the function $\mathrm e^{-tv}$ is integrable on some neighborhood of $x$. Indeed, since $\mathrm{dd}^{\mathrm{c}}v$ has no atom at $x$, there is a disk $D=D(x,R)\Subset \Omega $ with $M=(\mathrm{dd}^{\mathrm{c}}v)(D)<1/t$. On $D$ write

where $h$ is harmonic: the integral is locally integrable by Fubini's theorem and has complex Hessian $\mathrm{dd}^{\mathrm{c}}v$ on $D$, so it differs from $v$ by a harmonic distribution. If $M=0$, the claim follows from harmonicity. For $M>0$, Jensen's inequality for the probability measure $(\mathrm{dd}^{\mathrm{c}}v)/M$ gives

Since $tM<1$, the integral of $|z-y|^{-2tM}$ over $z\in D(x,R/2)$ is bounded uniformly in $y\in D$. The harmonic function $h$ is bounded below on this smaller disk, so Tonelli's theorem proves the claim. (Truncating the logarithmic kernel justifies Jensen's inequality at its poles.)

A nonzero holomorphic germ of order $k$ at $x$ is a unit times $(z-x)^ k$. Since $v$ is locally bounded above, weighted integrability requires $k+1>a$; equality is excluded by the logarithmic divergence of the radial integral. Conversely, if $k+1>a$, choose $b>1$ such that $b(a-k)<1$. Then $|z-x|^{2k-2a}$ belongs locally to $L^ b$, and $\mathrm e^{-v}$ belongs locally to $L^{b/(b-1)}$ by the claim, so Hölder's inequality proves integrability. Thus the germ belongs to $\mathcal I(u)_ x$ exactly when $k\geq \lfloor a\rfloor $.

The points $y$ with $\nu (u,y)\geq 1$ form a locally finite set, since each of them carries an atom of mass at least one of the locally finite measure $\mathrm{dd}^{\mathrm{c}}u$. Near any point, therefore, the ideal sheaf is generated by a single power of a coordinate, which proves the last assertion. $\square$

Proof. For $r=1$ there is nothing to prove, so let $r\geq 2$. Put $J_ i=\mathcal I(u_ i)$ and $J=\sum _ iJ_ i$. By lemma 5.37.18, these sheaves are locally free of rank one. At each point, the order of $J$ is the smallest of the orders of the $J_ i$, and a local generator of an ideal $J_ i$ attaining this order also generates $J$. Consequently the sum map $\bigoplus _ iJ_ i\to J$ is locally split, and its kernel $E$ is a holomorphic vector bundle of rank $r-1$.

Fix $f\in \Gamma (\Omega ,J)$. Gluing local holomorphic lifts of $f$ with a smooth partition of unity gives a smooth section $s$ of $\bigoplus _ iJ_ i$ whose image is $f$. Since $f$ is holomorphic, $\beta =\bar\partial s$ is an $E$-valued $(0,1)$-form. It suffices to solve $\bar\partial v=\beta $ for a section $v$ of $E$, for then $s-v$ is a holomorphic lift of $f$.

We solve this equation using a bundle metric of positive curvature. Choose a smooth strictly subharmonic exhaustion function $\rho \geq 1$ of $\Omega $. The Kähler metric

is complete: it dominates $2\mathrm i\partial \rho \wedge \bar\partial \rho $, so $\rho $ has bounded gradient with respect to it, and therefore every curve leaving all compact subsets has infinite length.

Equip $E$ with a smooth Hermitian metric $h$. Choose a smooth increasing convex function $\chi $ on $[1,+\infty )$ so that the metric $h_\chi =h\mathrm e^{-\chi (\rho )}$ has

These two requirements can be imposed successively on the compact sublevel sets of $\rho $: increase $\chi '$ so that $\chi '(\rho )$ times the positive Hessian of $\rho $ dominates the negative part of the curvature, and increase $\chi $ so that the integral over each shell $\{ m\leq \rho <m+1\} $ is at most $2^{-m}$. Increasing the slopes preserves the inequalities already achieved.

Use the global frame $\mathrm{d}z$ of the canonical bundle $K_\Omega $ to identify $E$-valued $(0,q)$-forms with $(E\otimes K_\Omega ^{-1})$-valued $(1,q)$-forms, and give $K_\Omega ^{-1}$ its flat coordinate metric. Its curvature is zero, and $\omega _*$ dominates the Euclidean metric, so the identified form $\beta $ remains $L^2$. The Bochner–Kodaira inequality for compactly supported forms $w$ of type $(1,1)$ gives

Here we used that there are no $(1,2)$-forms. By completeness, cutoffs and smoothing extend this inequality to the domain of the adjoint. Since $\beta $ is $L^2$, the Cauchy–Schwarz inequality bounds the functional $w\mapsto \langle w,\beta \rangle $ by $\| \beta \| \| \bar\partial ^*w\| $, and the Riesz representation theorem gives a distributional solution $v$ of $\bar\partial v=\beta $. Thus $s-v$ is distributionally holomorphic, hence holomorphic in every bundle chart, and it is the required global lift. No global weighted norm condition is imposed on its components. $\square$

Proof. Put $J_ i=\mathcal I(p_ i\phi _ i+\psi )$ and $J=\sum _ iJ_ i$. At a point $x$, write $A=\nu (\psi ,x)$ and $B_ i=\nu (\phi _ i,x)$. By lemma 5.37.18, the order of $J_ x$ is $\min _ i\lfloor A+p_ iB_ i\rfloor $.

Suppose first that $s<1$, and put $P=1/s>1$. The numbers $P/p_ i$ sum to one, so

Taking integer parts and using lemma 5.37.18 gives $\mathcal I(P\sum _ i\phi _ i+\psi )\subseteq J$. By lemma 5.37.20, every global section of this smaller ideal sheaf lies in $\sum _ i\Gamma (\Omega ,J_ i)$, so every remainder admissible for the one-weight problem is admissible for $c$. Proposition 5.37.14 with $\phi =\sum _ i\phi _ i$ and $p=P$ therefore gives $c\leq (1-1/P)I=(1-s)I$.

Now let $s\geq 1$. Since $I<+\infty $, we have $f\in \Gamma (\Omega ,\mathcal I(\psi +\sum _ i\phi _ i))$. If $f\equiv 0$ there is nothing to prove; otherwise let $k$ be its order at $x$. Then $k+1>A+\sum _ iB_ i$. If its germ did not belong to $J_ x$, we would have $k+1\leq A+p_ iB_ i$ for every $i$. With $L=k+1-A>0$, this would give $\sum _ iB_ i\geq sL\geq L$, a contradiction. Hence $f\in \Gamma (\Omega ,J)$, and lemma 5.37.20 shows that $g=0$ is admissible.

For sharpness when $s<1$, take the unit disk, $\psi =0$, $f=1$, and $\phi _ i=p_ i^{-1}\log |z|^2$. Then $J_ i=(z)$ for all $i$, so an admissible $g$ has $g(0)=1$. Orthogonality of the monomials on the disk gives $c=\pi $, attained by $g=1$, whereas $I=\pi /(1-s)$. If $s\geq 1$, keep these weights and take $f=z^ q$ with an integer $q\geq 0$ satisfying $q+1>s$. Then $I=\pi /(q+1-s)>0$ and $c=0$, so the constant $0$ cannot be replaced by a negative number. $\square$

Proof. Choose distinct real numbers $t_1,\ldots ,t_ r$ and unit linear forms

Any $n$ of these forms are linearly independent, since the determinant of their unnormalized coefficients is a nonzero Vandermonde determinant. Put $\phi _ i=p^{-1}\log |\ell _ i|^2$. Since the coefficient vectors of the $\ell _ i$ have norm one, $|\ell _ i(z)|<1$ on $B$, so the $\phi _ i$ are negative plurisubharmonic functions. By lemma 5.37.1, $\mathcal I(p\phi _ i)=(\ell _ i)$.

At a nonzero point, at most $n-1$ of the forms vanish. These forms are linearly independent and can be used as part of a holomorphic coordinate system, while the other forms are bounded away from zero on a small neighborhood of the point. Hence

is locally integrable off the origin, because each factor $|\ell _ i|^{-2/p}$ has exponent $2/p<2$ (note that $p>r/n>1$). By compactness, $W$ is integrable on every closed annulus. Set $s=r/p$ and let $\mathrm{d}\sigma $ denote surface measure on the unit sphere. Homogeneity and polar integration over such an annulus show that

Tonelli then gives

since $s<n$. This takes care of the common zero at the origin as well as of the individual hyperplanes.

Every global section of $(\ell _ i)$ vanishes at the origin. Thus every admissible $g$ in (5.37.26) satisfies $g(0)=1$. For each $z\in B$, the holomorphic function $\lambda \mapsto g(\lambda z)$ is defined on a disk of radius greater than one, and the submean inequality for the subharmonic function $|g(\lambda z)|^2$ gives

Integrating in $z$ and using the rotation invariance of $B$ proves $\int _ B|g|^2\, \mathrm{d}V\geq \operatorname {Vol}(B)=\pi ^ n/n!$, also when the integral is infinite. The constant $g=1$ is admissible and attains this bound. $\square$