5.5 Holomorphic zeros and the Poincaré–Lelong formula0C34F7796300

Proof. We prove the first two assertions together by induction on $n$. For $n=1$, every zero of $f$ is isolated: at a zero $a$, the Taylor expansion has a first nonzero term $c(z-a)^ m$, since otherwise $f$ would vanish near $a$ and hence on $\Omega $ by the identity theorem; so $f(z)=(z-a)^ mg(z)$ with $g(a)\neq 0$. Hence the zero set meets every compact subset of $\Omega $ in a finite set, and since $\Omega $ is a countable union of compact sets, it is countable, hence null. Moreover, $f$ does not vanish identically on a disc $D\subset \Omega $, by the identity theorem, which proves the second assertion for $n=1$.

Let $n\geq 2$, and let $P=P'\times D$ be as in the statement; after permuting the coordinates, which preserves Lebesgue measure, we may assume that $j=n$, and we write $w=z_ n$ and $w_0$ for the center of $D$. The functions $c_ k(z')=\partial _ w^ kf(z',w_0)/k!$ are holomorphic on $P'$, and, by the one-variable Taylor expansion of $w\mapsto f(z',w)$ on the disc $D$ centered at $w_0$, this function vanishes identically on $D$ if and only if all $c_ k(z')$ vanish. Since $f$ does not vanish identically on the nonempty open set $P$, by the identity theorem, some $c_ k$ is not identically zero on the polydisc $P'$, and its zero set $N\subset P'$ is null by the first assertion in dimension $n-1$. This proves the second assertion in dimension $n$, since the set in question is contained in $N$. For $z'\notin N$, the zero set of $w\mapsto f(z',w)$ in $D$ is null by the case $n=1$. By Tonelli's theorem, applied to the indicator function of the relatively closed zero set of $f$ in $P$, this zero set is null; and countably many such polydiscs, for instance those with rational centers and radii contained in $\Omega $, cover $\Omega $.

For the last assertion, translate $a$ to the origin and let $P_ m$ be the first nonzero term of the homogeneous expansion of $f$ at $0$; it exists because $f$ does not vanish identically near $0$, and $m\geq 1$ because $f(0)=0$. The set $\{ P_ m\neq 0\} $ is open and dense in $\mathbb C^ n$, because a polynomial vanishing on a nonempty open set is zero: its derivatives at a point $b$ of that set vanish, and the polynomial is the sum of its finite Taylor expansion at $b$. Choose $v_1\in \{ P_ m\neq 0\} $, and inductively $v_ j\in \{ P_ m\neq 0\} $ outside the linear span of $v_1,\ldots ,v_{j-1}$; this is possible because the complement of a proper linear subspace is open and dense, and the intersection of two dense open sets is dense. Then $v_1,\ldots ,v_ n$ is a basis, and $f(tv_ j)=t^ mP_ m(v_ j)+O(t^{m+1})$ by (5.2.2), so $t\mapsto f(tv_ j)$ does not vanish identically near $t=0$. Taking $v_1,\ldots ,v_ n$ as coordinate axes proves the assertion. $\square$

Proof. Since the zeros of $f(0,\cdot )$ are isolated, there is $\rho _0>0$ such that $\{ 0\} \times \{ |w|\leq \rho _0\} \subset U$ and $f(0,w)\neq 0$ for $0<|w|\leq \rho _0$. Let $0<\rho <\rho _0$, and choose $\rho _1<\rho <\rho _2\leq \rho _0$. Then $f(0,w)\neq 0$ on the compact annulus $\{ \rho _1\leq |w|\leq \rho _2\} $. Since $|f(0,\cdot )|$ has a positive minimum on this annulus and $f$ is uniformly continuous on a compact neighborhood of $\{ 0\} \times \{ \rho _1\leq |w|\leq \rho _2\} $ in $U$, there is a polydisc $P'$ centered at $0$, as small as we wish, such that $P'\times \{ |w|\leq \rho _2\} \subset U$ and $f(z',w)\neq 0$ for $z'\in P'$ and $\rho _1\leq |w|\leq \rho _2$. By the argument principle, the number of zeros of $f(z',\cdot )$ in $\{ |w|<\rho \} $, counted with multiplicity, is

a continuous integer-valued function of $z'$ on the connected set $P'$; hence it equals its value $m$ at $z'=0$. Denote these zeros by $\lambda _1(z'),\ldots ,\lambda _ m(z')$, in any order; they lie in $\{ |w|<\rho _1\} $.

By the residue theorem, their power sums are

which are holomorphic functions on $P'$, by differentiation under the integral sign. The elementary symmetric functions $e_ k$ of the roots satisfy Newton's identities

To prove them, put $E(t)=\prod _ l(1+t\lambda _ l)=\sum _ ke_ kt^ k$ and compare the coefficients of $t^{k-1}$ in $E'(t)=E(t)\sum _{j\geq 1}(-1)^{j-1}p_ jt^{j-1}$, which is the expansion of the logarithmic derivative $E'/E=\sum _ l\lambda _ l/(1+t\lambda _ l)$. By induction on $k$, every $e_ k$ is holomorphic on $P'$. Hence $W(z',w)=\prod _ l(w-\lambda _ l(z'))$ is a monic polynomial in $w$ with holomorphic coefficients $a_ k=(-1)^ ke_ k$, and $a_ k(0)=0$ because all $\lambda _ l(0)=0$.

It remains to show that $f/W$ extends to a nowhere vanishing holomorphic function on $P'\times \{ |w|<\rho \} $. For fixed $z'\in P'$, the functions $f(z',\cdot )$ and $W(z',\cdot )$ have the same zeros, with the same multiplicities, in $\{ |w|<\rho _2\} $; so $f/W$ and $W/f$ extend to holomorphic functions of $w$ there. On the annulus $\{ \rho _1<|w|<\rho _2\} $ both quotients are holomorphic in $(z',w)$. The Cauchy integral

is holomorphic in $(z',w)$, by differentiation under the integral sign, and equals $f/W$ by Cauchy's formula in $w$. Defining $\widetilde G$ in the same way from $W/f$, we obtain $G\widetilde G=1$. Thus $g=G$ is holomorphic and nowhere zero, and $f=gW$. $\square$

Proof. Local integrability. Near a point where $f\neq 0$, the function $\log |f|^2$ is smooth. Let $a$ be a zero of $f$. After a translation and a complex-linear change of coordinates, which preserve local integrability, we may assume that $a=0$ and that $f(0,w)$ does not vanish identically in the last variable $w$ (lemma 5.5.1). Then $f(0,\cdot )$ has a zero of some order $m\geq 1$ at $0$, and by theorem 5.5.2, $f=gW$ on a polydisc $P'\times \{ |w|<\rho \} $, where $g$ is a holomorphic unit and $W(z',w)=\prod _{l=1}^ m(w-\lambda _ l(z'))$ with all roots in $\{ |w|<\rho _1\} $ for some $\rho _1<\rho $. Choose $r\in (\rho _1,\rho )$ and replace $P'$ by a smaller concentric polydisc with compact closure in $P'$; since $g$ is continuous and nowhere zero, $\log |g|^2$ is then bounded on $P'\times \{ |w|<r\} $. For $|\lambda |\leq \rho _1$ we have

with $C$ independent of $\lambda $: the positive part of $\log |w-\lambda |$ is at most $\max \{ \log (r+\rho _1),0\} $, and after the translation $\zeta =w-\lambda $ the integral of its negative part is at most $\int _{|\zeta |<1}(-\log |\zeta |)\, \mathrm dV_2(\zeta )$. Since $|\log |W(z',w)||\leq \sum _ l|\log |w-\lambda _ l(z')||$, it follows that $\int _{|w|<r}|\log |W(z',w)||\, \mathrm dV_2(w)\leq mC$ for every $z'\in P'$. (The estimate holds for each fiber separately, so we need not choose the roots as measurable functions of $z'$.) By Tonelli's theorem $\log |W|^2$, and hence $\log |f|^2$, is integrable on $P'\times \{ |w|<r\} $, a neighborhood of $a=0$.

The slice formula. For $n=1$, example 3.4.15 gives $\Delta \log |f|^2=4\pi \sum _ am(a)\delta _ a$ locally, that is, $(\log |f|^2)_{z\bar z}=\pi \sum _ am(a)\delta _ a$; this is (5.5.4) with $\mathbb C^0$ a point. In general, write $P=P'\times D$ with $D$ a disc in the variable $z_ j$. By lemma 5.5.1, the set $N_0$ of points $z'\in P'$ over which $f$ vanishes identically on $D$ is null. By Fubini's theorem,

the inner integral being an integrable function of $z'$. For $z'\notin N_0$, the one-variable case applied to $f(z',\cdot )$ on $D$ shows that the inner integral equals $\pi \sum _ am(z',a)\varphi (z',a)$. This proves (5.5.4); the integrand on the right is measurable and integrable, being equal almost everywhere to the inner integral.

Positivity. By (5.5.4), $(\log |f|^2)_{j\bar j}$ is nonnegative on nonnegative test functions supported in polydiscs contained in $\Omega $, hence on all nonnegative test functions, which are finite sums of such functions by lemma 2.5.2. This holds in every system of complex linear coordinates, since $f$ remains holomorphic after a linear change of coordinates. Applying this after a unitary change of coordinates, as in the matrix criterion of section 5.4, we see that $\sum _{j,k}(\log |f|^2)_{j\bar k}\xi _ j\bar\xi _ k$ is a positive measure for every $\xi \in \mathbb C^ n$. By the matrix criterion, $\mathrm{dd}^{\mathrm{c}}\log |f|^2\geq 0$, and by theorem 5.3.7 its coefficients are measures.

Null sets. Let $\nu =(\log |f|^2)_{j\bar j}$ on $P=P'\times D$, and let $\eta \geq 0$ be a test function on $D$. By (5.5.4), for every nonnegative test function $\psi $ on $P'$,

Both sides define positive Radon measures on $P'$ in $\psi $, the function $h_\eta $ being locally integrable by the identity itself. By the uniqueness in theorem 2.5.6, the image measure $A\mapsto \int _{\pi _ j^{-1}(A)}\eta \, \mathrm d\nu $ of $\eta \nu $ under $\pi _ j$ is therefore $h_\eta \, \mathrm dV_{2n-2}$. Every null set is contained in a Borel null set, by outer regularity of Lebesgue measure, so we may assume that $N$ is a Borel set. Then $\int _{\pi _ j^{-1}(N)}\eta \, \mathrm d\nu =\int _ Nh_\eta \, \mathrm dV_{2n-2}=0$. Letting $\eta $ increase to $1$ on $D$ along a sequence, we obtain $\nu (\pi _ j^{-1}(N))=0$ by monotone convergence.

The canonical representative. The trace of $\mathrm{dd}^{\mathrm{c}}\log |f|^2$ is positive, so by theorem 3.4.12 the class of $\log |f|^2$ has a canonical subharmonic representative $v$, given by the limits of ball averages in (3.4.13). At a point where $f\neq 0$, the function $\log |f|^2$ is continuous, so $v$ agrees with it there. At a zero $a$, there is $C$ with $|f(z)|\leq C|z-a|$ near $a$, since $f$ is smooth and $f(a)=0$; hence the average of $\log |f|^2$ over $B(a,r)$ is at most $\log C^2+2\log r$, which tends to $-\infty $ as $r\to 0$. Thus $v(a)=-\infty =\log |f(a)|^2$, and $v=\log |f|^2$ everywhere. $\square$

Proof. Choose coordinates as in lemma 5.5.1, so that $f$ does not vanish identically on any coordinate axis through $z_0=0$. Fix $j$, and write points as $(z',w)$ with $w=z_ j$. By theorem 5.5.2 in the variable $w$, there is a polydisc $P_ j=P'_ j\times \{ |w|<\rho \} $ in the domain of $f$ on which $f=gW$, where $g$ is a holomorphic unit and $W$ is a monic polynomial in $w$ of some degree $m\geq 1$ whose coefficients are holomorphic on $P'_ j$ and whose roots lie in $\{ |w|<\rho _1\} $ for some $\rho _1<\rho $. We shall find a holomorphic function $h$ on $P'_ j$, not identically zero, such that every zero of $f$ in $P_ j$ lying over $\{ h\neq 0\} $ belongs to $V_{\mathrm{reg}}$. Then $S\cap P_ j\subset \pi _ j^{-1}(N_ j)$ with $N_ j=\{ h=0\} $, which is null by lemma 5.5.1 if $n\geq 2$, and empty if $n=1$, when $P'_ j$ is a point. Any polydisc $P$ centered at $0$ and contained in $P_1\cap \cdots \cap P_ n$ then has the required property.

For $z'\in P'_ j$, let $\lambda _1(z'),\ldots ,\lambda _ m(z')$ be the roots of $W(z',\cdot )$, repeated according to multiplicity, and put $p_ k(z')=\sum _ l\lambda _ l(z')^ k$ for $k\geq 0$. By the residue theorem,

so $p_ k$ is holomorphic on $P'_ j$, by differentiation under the integral sign. Let $s$ be the largest number of distinct roots of $W(z',\cdot )$ for $z'\in P'_ j$, and let $h$ be the determinant of the $s\times s$ matrix $(p_{a+b})_{0\leq a,b\leq s-1}$; it is holomorphic on $P'_ j$. For fixed $z'$, this matrix is $A^{\mathsf T}A$, where $A$ is the $m\times s$ matrix with entries $A_{la}=\lambda _ l^ a$.

To compute $h(z')$, note that the $b$th column of $A^{\mathsf T}A$ is $\sum _ l\lambda _ l^ bv_ l$, where $v_ l=(\lambda _ l^0,\ldots ,\lambda _ l^{s-1})$ is the $l$th row of $A$. Expanding the determinant by multilinearity in the columns, discarding the terms in which some $v_ l$ occurs twice, and grouping the remaining terms according to the set $L$ of indices that occur, we find that each group contributes $(\det A_ L)^2$. Indeed, if $L=\{ l_1<\cdots <l_ s\} $, the terms of this group are indexed by the permutations $\pi $ of $\{ 1,\ldots ,s\} $, the term of $\pi $ taking $\lambda _{l_{\pi (b+1)}}^ bv_{l_{\pi (b+1)}}$ from the $b$th column. Reordering the columns multiplies the determinant by $\operatorname {sign}(\pi )$, so the group contributes $\det A_ L\sum _\pi \operatorname {sign}(\pi )\prod _{b=0}^{s-1}\lambda _{l_{\pi (b+1)}}^ b$, and this sum is the Leibniz expansion of $\det A_ L$. Thus we obtain the Cauchy–Binet formula

where $L$ runs over the $s$-element subsets of $\{ 1,\ldots ,m\} $, $A_ L$ is the square matrix formed by the rows of $A$ with indices in $L$, and the last equality is the Vandermonde formula. If $W(z',\cdot )$ has fewer than $s$ distinct roots, every set $L$ contains two indices with equal roots, and $h(z')=0$. If $W(z',\cdot )$ has exactly $s$ distinct roots $\mu _1,\ldots ,\mu _ s$, of multiplicities $k_1,\ldots ,k_ s$, then the nonzero terms are those of the $k_1\cdots k_ s$ sets $L$ containing exactly one index of each root, and each of them equals $\prod _{c<c'}(\mu _ c-\mu _{c'})^2\neq 0$; hence $h(z')\neq 0$. By the choice of $s$, the function $h$ is not identically zero.

Now let $h(z'_0)\neq 0$, and let $(z'_0,w_0)\in P_ j$ be a zero of $f$. Then $W(z'_0,w_0)=0$, since $g$ is a unit, and $W(z'_0,\cdot )$ has exactly $s$ distinct roots $\mu _1,\ldots ,\mu _ s$, of multiplicities $k_1,\ldots ,k_ s$; say $w_0=\mu _1$. Choose disjoint closed discs $E_ c$ centered at $\mu _ c$ in $\{ |w|<\rho \} $, and a connected neighborhood $Y$ of $z'_0$ in $P'_ j$ such that $W(z',\cdot )$ has no zeros on the boundary circles $\partial E_ c$ for $z'\in Y$; this is possible because $W(z'_0,\cdot )$ has no zeros on these compact circles and $W$ is continuous. By the argument principle, the number of roots of $W(z',\cdot )$ in $E_ c$, counted with multiplicity, is $(2\pi \mathrm i)^{-1}\int _{\partial E_ c}\partial _ wW/W\, \mathrm dw$, a continuous integer-valued function on $Y$, hence equal to $k_ c$. Thus each $E_ c$ contains at least one root of $W(z',\cdot )$. Since $W(z',\cdot )$ has at most $s$ distinct roots, each $E_ c$ contains exactly one, $\mu _ c(z')$, of multiplicity $k_ c$, and since $k_1+\cdots +k_ s=m$, there are no other roots. By the residue theorem,

which is holomorphic on $Y$, and $W(z',w)=\prod _ c(w-\mu _ c(z'))^{k_ c}$ for $z'\in Y$, both sides being monic polynomials in $w$ with the same roots and multiplicities. Near $(z'_0,w_0)$ the function $U=\prod _{c\geq 2}(w-\mu _ c(z'))^{k_ c}$ does not vanish, so $f=(gU)(w-\mu _1(z'))^{k_1}$, where $gU$ is a holomorphic unit and $\mathrm d(w-\mu _1(z'))\neq 0$. That is, $(z'_0,w_0)\in V_{\mathrm{reg}}$. $\square$

Proof. If $X$ has dimension zero, then $f$ has no zeros and both sides vanish. Let $X$ have dimension $n\geq 1$. The set of points near which $f$ vanishes identically is open, and it is closed by the identity theorem, applied in connected coordinate neighborhoods; since $f$ does not vanish identically on any component of $X$, this set is empty. Hence proposition 5.5.3 applies in every connected chart: the function $\log |f|^2$ is locally integrable, and $T=\mathrm{dd}^{\mathrm{c}}\log |f|^2$, given by $T(\varphi )=\int _ X\log |f|^2\, \mathrm{dd}^{\mathrm{c}}\varphi $ independently of coordinates, is a positive current with measure coefficients. It is closed because $\mathrm d\mathrm{dd}^{\mathrm{c}}=\mathrm d\mathrm d\mathrm d^{\mathrm c}=0$. We show that $T$ gives zero mass to $S$ and agrees with $m[V_{\mathrm{reg}}]$ outside $S$.

Step 1: $T$ gives $S$ zero mass. Let $x_0\in S$, and choose coordinates and a polydisc $P$ centered at $x_0$ as in lemma 5.5.5. For each $j$, the last assertion of proposition 5.5.3 shows that the diagonal coefficient $(\log |f|^2)_{j\bar j}$ vanishes on $S\cap P\subset \pi _ j^{-1}(N_ j)$. By the computation of the trace measure in section 5.4, the sum of these diagonal coefficients is a positive multiple of the trace measure of $T$ in these coordinates, which therefore vanishes on $S\cap P$; by theorem 5.3.7 so do all coefficient measures of $T$ in these coordinates. By (5.3.6), the vanishing of the coefficient measures on a Borel set does not depend on the chart. Every compact subset of $S\cap U$, for a chart $U$, is covered by finitely many such polydiscs $P$, so by inner regularity all coefficient measures of $T$ vanish on $S\cap U$.

Step 2: $T=m[V_{\mathrm{reg}}]$ on $X\setminus S$. Since $T$ is defined without reference to coordinates, we may compute it in any holomorphic coordinates. Away from $V$ the function $\log |f|^2$ is pluriharmonic, so $T=0$ there. Near a point $p\in V_{\mathrm{reg}}$ write $f=g\, a^ m$ as above, and use holomorphic coordinates $(t,x')$ with $t=a$. Then $\log |f|^2=\log |g|^2+m\log |t|^2$, where $\log |g|^2$ is pluriharmonic. In one variable, $\partial _ t\partial _{\bar t}\log |t|^2=\pi \delta _0$ by example 3.4.15. Since $\log |t|^2$ does not depend on $x'$, Fubini's theorem shows that its distributional derivatives involving $x'$ or $\bar x'$ vanish and that $\langle (\log |t|^2)_{t\bar t},\psi \rangle =\pi \int \psi (0,x')\, \mathrm dV_{2n-2}(x')$ for test functions $\psi $; thus $\mathrm{dd}^{\mathrm{c}}\log |t|^2=(\mathrm i/2\pi )(\log |t|^2)_{t\bar t}\, \mathrm dt\wedge \mathrm d\bar t$. Let $\varphi $ be a test form of bidegree $(n-1,n-1)$, and let $\varphi _0=c(t,x')\bigwedge _ l(\frac{\mathrm i}2\mathrm dx'_ l\wedge \mathrm d\bar x'_ l)$ be its component without $\mathrm dt$ and $\mathrm d\bar t$, the only one that contributes. Since $(\mathrm i/2\pi )\, \mathrm dt\wedge \mathrm d\bar t\wedge \varphi _0=\pi ^{-1}c\, \mathrm dV_{2n}$, we obtain

that is, $\mathrm{dd}^{\mathrm{c}}\log |t|^2=[\{ t=0\} ]$ in the sense of example 5.3.9, the hypersurface carrying its complex orientation. Therefore $T=m[V_{\mathrm{reg}}]$ near $p$.

Step 3: extension across $S$. Let $U$ be a chart and $K\subset U$ compact. By Step 2, the trace measure of the positive current $m[V_{\mathrm{reg}}]$ on $U\setminus S$ is the restriction of $\mu _ T$, so its mass on $K\setminus S$ is at most $\mu _ T(K)<\infty $. By the coefficient bound of theorem 5.3.7, applied on $U\setminus S$, the coefficient measures of $m[V_{\mathrm{reg}}]$, a positive current of bidegree $(1,1)$, have total variation at most $C_{n,1}\mu _ T(K)$ on $K\setminus S$. Hence they extend by zero on $S$ to complex Radon measures on $U$. Indeed, each coefficient measure $\sigma $ is a complex combination of the positive and negative variations of $\operatorname{Re}\sigma $ and $\operatorname{Im}\sigma $, which are positive Radon measures on $U\setminus S$ bounded by $|\sigma |$; extended by zero on $S$, they become Borel measures on $U$ that are finite on compact sets, hence Radon by proposition 2.2.2. By Step 2 the coefficient measures of $T$ agree with those of $m[V_{\mathrm{reg}}]$ on $U\setminus S$, and by Step 1 they vanish on $S$. Thus the extensions by zero are the coefficient measures of the current $T$ of order zero; we denote this current by $[\operatorname{div}(f)]$. This proves (5.5.7), and $[\operatorname{div}(f)]=T$ is positive and closed. $\square$