9.5 Analytic curves and finite maps2BDD13613B99

Proof. Embed $C$ in a local model and apply theorem 9.3.4. The base is a disc $\Delta '=\{ |z|<r\} $, and since the zeros of $\delta $ are isolated, we may take the parametrization polydisc so small that $\delta $ has no zero in $\Delta '\setminus \{ 0\} $. Let $C$ be the representative in this polydisc and $C^*=C\setminus \{ x\} =\pi ^{-1}(\Delta '\setminus \{ 0\} )$. Then $C^*$ is a covering of the punctured disc $\Delta '\setminus \{ 0\} $ with $q$ sheets, and it is connected: if $\delta (0)=0$, it is the covering locus, and if $\delta (0)\neq 0$, then $q=1$, because the fiber over $0$ is $\{ x\} $. We shall show that lifting a circle around the puncture permutes the $q$ points of a fiber cyclically, and then obtain $\gamma $ by lifting the exponential map to $C^*$.

We first recall two facts about coverings. A path in $\Delta '\setminus \{ 0\} $ lifts uniquely to $C^*$ once its initial point is prescribed: subdivide the path into finitely many pieces lying in discs over which the covering is trivial, and lift piece by piece. If two paths with the same endpoints are homotopic with fixed endpoints, their lifts with the same initial point have the same endpoint. Indeed, let $h\colon [0,1]^2\to \Delta '\setminus \{ 0\} $ be the homotopy, with $h(0,s)$ and $h(1,s)$ independent of $s$, and let $e(s)$ be the endpoint of the lift of $h(\cdot ,s)$ with the prescribed initial point. By uniform continuity there is an integer $m\geq 1$ such that $h$ maps each square $Q_{jl}=[(j-1)/m,j/m]\times [(l-1)/m,l/m]$ into a disc $D_{jl}$ over which the covering is trivial. A lift of a path in $D_{jl}$ stays in the sheet over $D_{jl}$ that contains its initial point, because these sheets are open and disjoint.

Fix $l$ and $s,s'\in [(l-1)/m,l/m]$. We show by induction on $j$ that the lifts of $h(\cdot ,s)$ and $h(\cdot ,s')$ over $[(j-1)/m,j/m]$ lie in one sheet over $D_{jl}$. For $j=1$ they start at the same point. If this holds for some $j<m$, the points of the two lifts at the parameter $j/m$ are joined by the lift, inside this sheet, of the path $h(j/m,\cdot )$ between $s$ and $s'$; this lift is a connected set lying over $D_{jl}\cap D_{j+1,l}$, so it lies in one sheet over $D_{j+1,l}$, which then contains both lifts over $[j/m,(j+1)/m]$. For $j=m$ we obtain two points in one sheet over $D_{ml}$ lying over the same point $h(1,s)=h(1,s')$, so $e(s)=e(s')$. Hence $e$ is constant on each interval $[(l-1)/m,l/m]$, and therefore on $[0,1]$.

Next, every loop based at a point $b\in \Delta '\setminus \{ 0\} $ is homotopic with fixed endpoints to the $k$th power of the circle around $0$ through $b$, for some integer $k$. Indeed, lifting the argument continuously along the loop, we write the loop as $\exp $ of a path in the half-plane $H=\{ \zeta :\operatorname{Re}\zeta <\log r\} $ from some $\zeta _0$ to $\zeta _0+2\pi \mathrm ik$; the straight-line homotopy to the segment from $\zeta _0$ to $\zeta _0+2\pi \mathrm ik$ stays in the convex set $H$. Let $\sigma $ be the permutation of the fiber over $b$ obtained by lifting the circle. Lifting any loop at $b$ therefore induces a power of $\sigma $. Since $C^*$ is connected, any two points of the fiber over $b$ are joined by a path in $C^*$, whose image is a loop at $b$; hence the powers of $\sigma $ act transitively on the $q$ points of the fiber, and $\sigma $ is a cycle of length $q$.

The map $\exp \colon H\to \Delta '\setminus \{ 0\} $ lifts to a map $\Phi \colon H\to C^*$ with $\pi \circ \Phi =\exp $: lift along paths from a base point $\zeta _0$, starting at a chosen point of the fiber; since $H$ is convex, the endpoint does not depend on the path. The map $\Phi $ is holomorphic, because locally it is $\exp $ followed by the inverse of $\pi $ on a sheet. The vertical segment from $\zeta $ to $\zeta +2\pi \mathrm i$ is mapped by $\exp $ onto the circle through $\mathrm e^\zeta $; so $\Phi (\zeta +2\pi \mathrm i)=\sigma _\zeta (\Phi (\zeta ))$, where $\sigma _\zeta $ is the permutation of the fiber over $\mathrm e^\zeta $ obtained by lifting this circle. By the preceding paragraph, applied with $b=\mathrm e^\zeta $, it is a cycle of length $q$; hence $\Phi (\zeta +2\pi \mathrm iq)=\Phi (\zeta )$. Therefore

does not depend on the branch of the logarithm; it is holomorphic, and $\pi (\gamma (t))=t^ q$. The coordinates of $\gamma $ are bounded, because $C$ lies in the bounded parametrization polydisc, so by the removable singularity theorem they extend holomorphically to $t=0$. Since $\pi $ is proper, $\gamma (t)$ stays in a compact subset of $C$ as $t\to 0$; hence $\gamma (0)\in C$, and $\gamma (0)=x$ because $\gamma (0)$ lies over $0$.

Let $\Delta =\{ |t|<r^{1/q}\} $. The map $\gamma \colon \Delta \to C$ is surjective: the points of $C$ over $z_0=\mathrm e^{\zeta }\neq 0$ are $\Phi (\zeta +2\pi \mathrm ik)=\gamma (\mathrm e^{(\zeta +2\pi \mathrm ik)/q})$, $0\leq k<q$, because $\sigma _\zeta $ is a $q$-cycle. It is injective: if $\gamma (t_1)=\gamma (t_2)$ with $t_1,t_2\neq 0$, then $t_1^ q=t_2^ q$, so $t_2=t_1\mathrm e^{2\pi \mathrm ik/q}$; writing $q\log t_1=\zeta $, we get $\gamma (t_2)=\Phi (\zeta +2\pi \mathrm ik)=\sigma _\zeta ^ k(\gamma (t_1))$, which forces $k\equiv 0$ modulo $q$. On $\Delta \setminus \{ 0\} $ the map $\gamma $ is a local biholomorphism into the manifold $C^*$, since $\pi \circ \gamma (t)=t^ q$ has nonzero derivative; being bijective, it is a biholomorphism of $\Delta \setminus \{ 0\} $ onto $C\setminus \{ x\} $.

Finally, let $X$ be a normal complex space of dimension one, and let $x\in X$; if $x$ is an isolated point, there is nothing to prove. Otherwise the germ $(X,x)$ has dimension one, and it is irreducible by the definition of normal points; let $\gamma \colon \Delta \to C$ be the map just constructed for a representative $C$ of this germ. Its inverse is holomorphic on the manifold $C\setminus \{ x\} $, which is contained in the regular locus, and it is bounded; so it is weakly holomorphic and, by normality, extends to a holomorphic function $\tau $ near $x$. Then $\tau \circ \gamma =\mathrm{id}$ on $\Delta \setminus \{ 0\} $, hence on $\Delta $, and $\gamma \circ \tau =\mathrm{id}$ on $C\setminus \{ x\} $, hence on $C$, because these maps are continuous and the punctured sets are dense. Thus $\gamma $ is an isomorphism of $\Delta $ onto a neighborhood of $x$, and $x$ is a regular point. $\square$

Proof. The germ of $B\cup X_{\mathrm{sing}}$ is again different from $(X,x)$, since $(X,x)$ is irreducible and $X_{\mathrm{sing}}$ has smaller dimension by corollary 9.3.13(1). Since every analytic germ $(A,x)$ satisfies $V(I_{A,x})=(A,x)$, the ideals of these two germs differ, and we can choose a holomorphic germ $f$ that vanishes on $B\cup X_{\mathrm{sing}}$ but not identically on $X$. Take a local parametrization of $(X,x)$ over a polydisc $\Delta '\subset \mathbb C^ d$, $d\geq 1$, with covering locus $X_\delta $, and for $z'\in \Delta '\setminus \{ \delta =0\} $ let $N(z')$ be the product of the values of $f$ at the $q$ points of $X$ over $z'$. This function is holomorphic and bounded, so it extends holomorphically to $\Delta '$ by lemma 9.2.9. It is not identically zero. Otherwise, choose a small ball in $\Delta '\setminus \{ \delta =0\} $ over which the $q$ points are given by holomorphic sections $s_1,\ldots ,s_ q$; the product of the functions $f\circ s_ i$ would vanish identically there, so one of these functions would vanish on an open set. Then $f$ would vanish on a nonempty open subset of the connected manifold $X_\delta $, hence on $X_\delta $ by the identity theorem, and on $X$ by density, which is not the case.

The function $g=N\delta $ is holomorphic and not identically zero on $\Delta '$. Choose a vector $v\in \mathbb C^ d$ at which the first nonzero homogeneous term of $g$ does not vanish, and $\rho >0$ such that $g(tv)\neq 0$ for $0<|t|<\rho $ and the ball $U=\{ |z'|<\rho |v|\} $ lies in $\Delta '$. Then $Y=X\cap \pi ^{-1}(\{ tv:|t|<\rho \} )$ is an analytic subset of $X\cap \pi ^{-1}(U)$. Over the punctured disc, $Y$ consists of points of $X_\delta $ at which $f\neq 0$, and it is a covering of the punctured disc; over $0$ it consists of the single point $x$. Thus $Y\setminus \{ x\} $ is a complex manifold of dimension one, and by properness of $\pi $ its points accumulate at $x$, so $x$ is not an isolated point of $Y$. Let $(C,x)$ be a branch of $(Y,x)$ of positive dimension. Its regular points near $x$ lie in the one-dimensional manifold $Y\setminus \{ x\} $, so $(C,x)$ has dimension one. Let $\gamma \colon \Delta \to C$ be the parametrization of proposition 9.5.1. For $t\neq 0$, the point $\gamma (t)$ lies in $C\setminus \{ x\} \subset Y\setminus \{ x\} $, hence in $X_\delta \subset X_{\mathrm{reg}}$ and outside $\{ f=0\} \supset B$. $\square$

Proof. Let $X_1$ be a branch of $(X,x)$. If $X_1$ has dimension zero, it is the point $x$. Otherwise, suppose that $f-f(x)$ does not vanish identically on $X_1$, and apply proposition 9.5.2 to $X_1$ and its subgerm $X_1\cap \{ f=f(x)\} $. The holomorphic function $f\circ \gamma $ on the disc has a local maximum of its modulus at $0$, so it is constant by the maximum principle in one variable, and $f(\gamma (t))=f(x)$ for all $t$, contradicting the choice of $\gamma $.

For the second assertion, let $K$ be a compact analytic subset of an open set in $\mathbb C^ N$, and let $K_+$ be the set of its non-isolated points. The isolated points form an open subset of $K$, so $K_+$ is compact. Suppose that $K_+\neq \varnothing $, and let $y$ be a point of $K_+$ where $|z|^2$ attains its maximum on $K_+$. Since $y$ is not isolated, some branch of $(K,y)$ has positive dimension, and its small representatives, being pure-dimensional, consist of non-isolated points of $K$. So proposition 9.5.2, applied to this branch and the subgerm $\{ y\} $, gives a nonconstant holomorphic curve $\gamma $ in $K_+$ with $\gamma (0)=y$. The function $|\gamma (t)|^2$ is subharmonic, with Laplacian $4\sum _ j|\gamma _ j'(t)|^2$, and it has a local maximum at $0$; by the maximum principle (proposition 3.4.7) it is constant near $0$, so its Laplacian vanishes, all $\gamma _ j'$ vanish, and $\gamma $ is constant, a contradiction. Hence $K_+=\varnothing $, and $K$ is compact and discrete, hence finite. $\square$

Proof. Fix $a,b\in B\setminus E$. If $c\in B$ and the segment $[a,c]$ meets $E$ at a point $e=a+s(c-a)$, then $0<s\leq 1$, since $a\notin E$, so $c=a+t(e-a)$ with $t=1/s\geq 1$. We show that the set of such $c$ has $N$-dimensional Lebesgue measure zero. Exhaust $E$ by the compact sets $E_ k=E\cap \overline{B_ k}$, where $B_ k$ are concentric balls increasing to $B$. Since $a\notin E_ k$, the parameter $t$ is bounded on $\{ (e,t):e\in E_ k,\ a+t(e-a)\in B\} $, say by $T_ k$, and the map $(e,t)\mapsto a+t(e-a)$ is Lipschitz on $E_ k\times [1,T_ k]$. Cover $E_ k$ by balls of radii $r_ l$ with $\sum _ lr_ l^{N-1}$ arbitrarily small. The image of each product of such a ball with $[1,T_ k]$ is covered by at most $C_ k/r_ l$ balls of radius $C_ kr_ l$, whose total volume is at most $C'_ kr_ l^{N-1}$. Hence the image of $E_ k\times [1,T_ k]$ is Lebesgue null, and so is the union over $k$. The same holds with $b$ in place of $a$. Choose $c\in B$ outside both null sets. Then the path $[a,c]\cup [c,b]$ lies in the convex set $B$ and does not meet $E$. $\square$

Proof. For $d=0$ the germ is the point $x$, and there is nothing to prove. Let $d\geq 1$, embed a representative $X$ in a local model $X\subset \Omega \subset \mathbb C^ N$ such that $X$ is pure $d$-dimensional and $F$ is the restriction of a holomorphic map on $\Omega $, also denoted $F$, and choose a closed ball $\overline{B(x,\rho )}\subset \Omega $, with $X\cap B(x,\rho )\subset V$, whose intersection with $X\cap F^{-1}(0)$ is $\{ x\} $. The compact set $X\cap \partial B(x,\rho )$ does not meet $F^{-1}(0)$, so $|F|$ has a positive minimum $\eta $ on it (if it is empty, put $\eta =\infty $). Let $B_\varepsilon $ be the ball of radius $\varepsilon <\eta $ about $0\in \mathbb C^ d$, and let $X'=X\cap B(x,\rho )\cap F^{-1}(B_\varepsilon )$. Then $F\colon X'\to B_\varepsilon $ is proper: the inverse image of a compact set $K\subset B_\varepsilon $ is $X\cap \overline{B(x,\rho )}\cap F^{-1}(K)$, which is compact and contains no point of the boundary sphere. Its fibers are compact analytic subsets of the open ball $B(x,\rho )$, hence finite by corollary 9.5.3.

It remains to prove surjectivity. On each connected component of the manifold $X'_{\mathrm{reg}}$, which has dimension $d$, the differential of $F$ has rank $d$ at some point. Otherwise its maximal rank $r$ on some component would be smaller; near a point where the rank is $r$, the rank would be constantly $r$, and by lemma 9.3.14, applied in a chart of $X'_{\mathrm{reg}}$, the fibers near this point would be submanifolds of positive dimension $d-r$, contradicting their finiteness.

Let $C$ be the union of $X'_{\mathrm{sing}}$ and of the set of regular points where the rank of $F$ is less than $d$. In the local model, $C$ is the union of $X'_{\mathrm{sing}}$ and of the common zero set in $X'$ of the minors of size $N$ of the matrix formed by the differentials of local generators of the ideal of $X$ and the rows of $\mathrm dF$; so $C$ is analytic. It contains no branch of $X'$ at any point. Indeed, each branch contains a nonempty open subset of $X'_{\mathrm{reg}}$, consisting of regular points of the branch outside the other branches (corollary 9.3.6), and on each component of $X'_{\mathrm{reg}}$ the points of rank less than $d$ form a proper analytic subset, which is nowhere dense by lemma 9.2.8. Hence $C$ has dimension at most $d-1$ by proposition 9.3.9.

Its image $E=F(C)$ is closed in $B_\varepsilon $ by properness, and it is $\mathcal H^{2d-1}$-null. Indeed, by corollary 9.3.13(2), $C$ is a finite union of complex submanifolds of open subsets of $\Omega $ of dimensions at most $d-1$, on each of which $F$ has rank at most $d-1$, so lemma 9.3.15(1) applies.

Over $B_\varepsilon \setminus E$, the map $F$ is a local biholomorphism at every point of its source, by proposition 5.2.4(1) applied in charts of $X'_{\mathrm{reg}}$, and it is proper; so its image is open and closed in $B_\varepsilon \setminus E$. This image is nonempty: at a regular point of rank $d$, the image of a neighborhood is open, and it is not contained in $E$, which is Lebesgue null because it is $\mathcal H^{2d-1}$-null. Since $B_\varepsilon \setminus E$ is connected by lemma 9.5.4, the image contains $B_\varepsilon \setminus E$, which is dense in $B_\varepsilon $; being closed in $B_\varepsilon $, it equals $B_\varepsilon $. $\square$

Proof. If $d=0$, then $f$ has no zeros. Let $d\geq 1$ and let $x$ be a zero of $f$. We first show that $\dim (\{ f=0\} ,x)=d-1$. For each branch $X_0$ of $(X,x)$, the germ $\{ f=0\} \cap X_0$ is a subgerm of $X_0$ different from it, so $\dim (\{ f=0\} ,x)\leq d-1$ by proposition 9.3.9. Fix a branch $X_0$, and choose linear forms $\ell _1,\ldots ,\ell _{d-1}$ on the ambient space of a local model, vanishing at $x$, inductively such that $\ell _ j$ does not vanish identically on any positive-dimensional branch of the germ at $x$ of $\{ f=0\} \cap X_0\cap \{ \ell _1=\cdots =\ell _{j-1}=0\} $. At each step this excludes finitely many proper linear subspaces, and, by proposition 9.3.9, it lowers the dimension of this germ by at least one as long as it is positive. Since the initial dimension is at most $d-1$, the point $x$ is isolated in the common zero set of $\Phi =(f,\ell _1,\ldots ,\ell _{d-1})$ on $X_0$.

Suppose that $\dim (\{ f=0\} ,x)<d-1$; then $d\geq 2$, and by upper semicontinuity $\dim (\{ f=0\} ,y)\leq d-2$ for all zeros $y$ of $f$ near $x$. By lemma 9.5.5, applied to the germ $(X_0,x)$, which is pure-dimensional, there is a neighborhood $X'_0$ of $x$ in a small representative of $X_0$, on which this bound holds, such that $\Phi $ maps $X'_0$ onto a ball $B\subset \mathbb C^ d$ centered at $0$. Hence $\Phi $ maps $\{ f=0\} \cap X'_0$ onto the set $B\cap \{ w_1=0\} $, which has positive $(2d-2)$-dimensional Lebesgue measure; here $w_1$ is the first coordinate of $\mathbb C^ d$. On the other hand, by corollary 9.3.13(2), $\{ f=0\} \cap X'_0$ is a finite union of complex submanifolds of dimensions at most $d-2$ of open subsets of the ambient space, on each of which $\Phi $ has rank at most $d-2$; by lemma 9.3.15(1), its image under $\Phi $ is $\mathcal H^{2d-2}$-null, hence Lebesgue null in $\{ w_1=0\} \cong \mathbb R^{2d-2}$. This contradiction shows that $\dim (\{ f=0\} ,x)=d-1$.

Now let $V_1$ be a branch of $(\{ f=0\} ,x)$. By corollary 9.3.6, there are regular points $y$ of $V_1$ arbitrarily close to $x$ that do not lie on the other branches. Near such a point, $\{ f=0\} $ coincides with $V_1$, which is a manifold of dimension $\dim V_1$ by purity. The hypotheses hold at $y$, so the first part gives $\dim (\{ f=0\} ,y)=d-1$, and therefore $\dim V_1=d-1$. $\square$