Convergent Puiseux parametrisation of an irreducible plane branch

Write $f_m$ for the lowest-order part of [F1]. Since $f_m$ is a nonzero homogeneous polynomial, there is a vector $w$ with $f_m(w)\mathrm{\ne }0$; choose such a $w$ and use complex-linear coordinates $(x,y)$ whose $y$-axis is the line $\mathbb{C}w$. Then the one-variable slice satisfies $f(0,\zeta )=f(\zeta w)={\zeta }^m{f}_m(w)+O({\zeta }^{m+1})$, so $f$ is regular in $y$ of order exactly $m$.

By [F3] the germ of step 1.1 is $f=uW$, where $u$ is a unit and $W$ is a Weierstrass polynomial of degree $m$ in $y$ with $W(0,T)=T^m$. Since $W$ is associate to $f$ it is reduced, and it is irreducible in ${\mathcal{O}}_{{\mathbb{C}}^2,0}$: by [F2] the germ $f$ is irreducible, and if $W=W_1{W}_2$ with nonunit factors, then after preparing the two factors with [F3] the product of the resulting Weierstrass polynomials is $W$ up to a unit, so [F11] makes $f$ reducible, a contradiction.

Apply [F4] to the polynomial $W$ of step 2.1: after shrinking $\epsilon$ if necessary there is a disc $D_y$ such that $F=Z(W)\cap (D^{\ast }\times D_y)$, $D^{\ast }=\{x:0<\mathrm{\mid }x\mathrm{\mid }<\epsilon \}$, is a connected $m$-sheeted unramified covering of $D^{\ast }$, all zeros tend to the origin over the base point, and every fibre of $F\to D^{\ast }$ consists of $m$ distinct points.

Put $D_{\epsilon }:=\{\mathrm{\mid }x\mathrm{\mid }<\epsilon \}$, let $N:=\{x\in \mathbb{C}:\mathrm{Im}x=0,\mathrm{Re}x\le 0\}$ be the closed negative real axis, and set $\mathrm{\Omega }:=D_{\epsilon }\setminus N\subseteq D^{\ast }$, an open subset. The map $\phi (w):=w^2$ is a homeomorphism from the convex half-disc $H:=\{w:\mathrm{Re}w>0,\mathrm{\mid }w\mathrm{\mid }<\sqrt{\epsilon }\}$ onto $\mathrm{\Omega }$ with continuous inverse the principal square root: $\mathrm{Re}w>0$ gives $w^2\in \mathrm{\Omega }$, and every $x\in \mathrm{\Omega }$ has principal argument in $(-\pi ,\pi )$ and a square root of modulus $\sqrt{\mathrm{\mid }x\mathrm{\mid }}$ and positive real part. By [F7] the nonempty convex set $H$ is simply connected, so [F8] applied to $\phi$ and to its inverse makes ${\phi }_{\ast }$ an isomorphism of fundamental groups; hence $\mathrm{\Omega }$ is nonempty, path-connected and has trivial fundamental group, that is, simply connected by [F6]. By [F5] the restriction $F_{\mathrm{\Omega }}:=F\cap (\mathrm{\Omega }\times D_y)$ is a covering of $\mathrm{\Omega }$, with $m$ points in every fibre. Every connected component $C$ of $F_{\mathrm{\Omega }}$ is open and maps onto $\mathrm{\Omega }$ as a covering: $p(C)$ is open because $p$ is an open map, and it is closed because over an evenly covered disc a component meeting the preimage of that disc contains a whole sheet; $\mathrm{\Omega }$ connected gives $p(C)=\mathrm{\Omega }$. Since $C$ is connected and $\mathrm{\Omega }$ is simply connected and locally path-connected, [F6] makes $C\to \mathrm{\Omega }$ one-sheeted, so $C$ is the graph of a continuous section ${\alpha }_C:\mathrm{\Omega }\to D_y$. Each such section is holomorphic: at $x_0\in \mathrm{\Omega }$ the point $(x_0,{\alpha }_C(x_0))$ is a simple root of the slice $W(x_0,\cdot )$, the fibre being unramified, so [F12] provides a local holomorphic graph through it, which agrees with ${\alpha }_C$ near $x_0$. Counting the $m$ points of a fibre among the components shows that there are exactly $m$ of them; enumerate them as ${\alpha }_1,\dots ,{\alpha }_m$. Then $F_{\mathrm{\Omega }}={\bigcup }_j\mathrm{graph}({\alpha }_j)$ and, both sides being monic of degree $m$ in $T$ with the same $m$ distinct roots at every $x\in \mathrm{\Omega }$, the identity $W(x,T)={\prod }_{j=1}^m(T-{\alpha }_j(x))$ holds on $\mathrm{\Omega }\times \mathbb{C}$.

Fix $x_1\in N\setminus \{0\}$ and choose a disc $U\subset D^{\ast }$ centered at $x_1$ that meets the slit in an open segment. Since $U$ is convex and simply connected, the restriction of the covering to $U$ is a disjoint union of $m$ holomorphic graphs ${\beta }_1,\dots ,{\beta }_m:U\to D_y$, by [F5], [F6] and [F12]. Put $U^{+}=U\cap \{\mathrm{Im}x>0\}$ and $U^{-}=U\cap \{\mathrm{Im}x<0\}$. These half-discs are connected; on each half every ${\beta }_k$ agrees with one section ${\alpha }_j$. Relabel so that on $U^{+}$, ${\beta }_k={\alpha }_k$. On $U^{-}$ it agrees with a unique ${\alpha }_{{\sigma }_{x_1}(k)}$, and the resulting map ${\sigma }_{x_1}$ is a permutation because the graphs are disjoint and enumerate each fibre. On overlapping discs, the graph continuations agreeing on the upper half-overlap agree throughout the overlap by the identity theorem, so the local permutations agree. Thus ${\sigma }_{x_1}$ is locally constant along the connected slit $N\setminus \{0\}$ and defines a single permutation $\sigma$, with continuation from the upper side to the lower side sending ${\alpha }_k$ to ${\alpha }_{\sigma (k)}$.

$\sigma$ is transitive. Suppose instead that $O\subseteq \{1,\dots ,m\}$ is a $\sigma$-orbit with $1\le \mathrm{\mid }O\mathrm{\mid }<m$, and let $O^{\mathrm{\prime }}$ be its complement, also $\sigma$-invariant. For $x\in \mathrm{\Omega }$ put $W_1(x,T):={\prod }_{j\in O}(T-{\alpha }_j(x))$ and $W_2(x,T):={\prod }_{j\in O^{\mathrm{\prime }}}(T-{\alpha }_j(x))$; their coefficients are holomorphic on $\mathrm{\Omega }$, and $W=W_1{W}_2$ there by step 4.1. Crossing the slit permutes the factors of $W_1$ among themselves by step 5.1, so each coefficient of $W_1$ continues across $N\setminus \{0\}$ to itself; together with the local graphs ${\beta }_k$ of step 5.1 the coefficients glue to holomorphic functions on $D^{\ast }$, and each of them is bounded on $D^{\ast }$, being an elementary symmetric function of $\mathrm{\mid }O\mathrm{\mid }$ roots that all lie in the bounded disc $D_y$. By [F9] the coefficients extend holomorphically across $x=0$. Moreover for every $\eta >0$ all roots of $W_1(x,\cdot )$ satisfy $\mathrm{\mid }y\mathrm{\mid }<\eta$ once $\mathrm{\mid }x\mathrm{\mid }$ is small enough: otherwise there are $x_n\to 0$ and roots $y_n$ of $W_1(x_n,\cdot )$ with $\mathrm{\mid }{y}_n\mathrm{\mid }\ge \eta$, which are zeros of $W$ and contradict the limit property of [F4]. Hence every coefficient of $W_1$ vanishes at $x=0$, by the elementary-symmetric bound for $\mathrm{\mid }O\mathrm{\mid }$ numbers of modulus $<\eta$ on $0<\mathrm{\mid }x\mathrm{\mid }<\rho (\eta )$ and $\eta \to 0$; thus $W_1$ is a Weierstrass polynomial of degree $\mathrm{\mid }O\mathrm{\mid }\ge 1$, and in the same way $W_2$ is a Weierstrass polynomial of degree $m-\mathrm{\mid }O\mathrm{\mid }\ge 1$. On $D^{\ast }$ the monic degree-$m$ polynomials $W$ and $W_1{W}_2$ have the same $m$ distinct roots, hence coincide; at $x=0$ both equal $T^m$, so $W=W_1{W}_2$ on $D_{\epsilon }\times \mathbb{C}$, and [F11] makes $W$ reducible in ${\mathcal{O}}_{{\mathbb{C}}^2,0}$, contradicting step 2.1. Therefore no such $O$ exists and $\sigma$ is transitive.

A transitive permutation of the finite set $\{1,\dots ,m\}$ is a single cycle of length $m$: its orbits are the cycles, and transitivity says that there is exactly one orbit. Hence ${\sigma }^m$ is the identity and ${\sigma }^k(1)=1$ exactly when $m$ divides $k$.

Let $\zeta :=e^{2\pi i\mathrm{/}m}$ and $\delta :={\epsilon }^{1\mathrm{/}m}$, so $\mathrm{\mid }t\mathrm{\mid }<\delta$ implies $\mathrm{\mid }{t}^m\mathrm{\mid }<\epsilon$. Set $S_0:=\{t:0<\mathrm{\mid }t\mathrm{\mid }<\delta ,-\pi \mathrm{/}m<\arg t<\pi \mathrm{/}m\}$ and $S_k:={\zeta }^k{S}_0$ for $k=0,\dots ,m-1$. These are disjoint sectors whose union is the punctured disc minus the $m$ boundary rays, and $t\in S_k$ implies $t^m\in \mathrm{\Omega }$. Define $g(t):={\alpha }_{{\sigma }^k(1)}(t^m)$ on $S_k$. Across each boundary ray the base crosses the slit from its upper side to its lower side, so step 5.1 makes the adjacent definitions the same local implicit-function graph; they glue holomorphically on the punctured disc. For $s\in S_0$ this gives $g({\zeta }^{j}s)={\alpha }_{{\sigma }^j(1)}(s^m)$ for $0\le j<m$, denoted (8.1.1). At a boundary point $t_0$, put $x_0=t_0^m\in N\setminus \{0\}$. The simple roots at $x_0$ have $m$ disjoint local implicit-function graphs by [F12]. Approaching the $m$ boundary parameters ${\zeta }^j{t}_0$ from the side whose base image lies above the slit, their sector labels run through one full $\sigma$-orbit, so they are distinct by step 7.1. Step 5.1 therefore extends $g$ at each boundary parameter as a different local graph; the values $g({\zeta }^j{t}_0)$ are the $m$ distinct roots of $W(x_0,\cdot )$.

The function $g$ is bounded on ${\mathrm{\Delta }}_{\delta }\setminus \{0\}$: every value $g(t)$ is one of the $m$ roots of the monic polynomial $W(t^m,\cdot )$ of degree $m$, and all roots of $W(x,\cdot )$ for $x\in D^{\ast }$ lie in the bounded disc $D_y$ of [F4]. Since $W(t^m,g(t))=0$ for $0<\mathrm{\mid }t\mathrm{\mid }<\delta$ and $g$ is bounded, [F9] extends $g$ holomorphically to ${\mathrm{\Delta }}_{\delta }$; the limit value is $g(0)=0$, because every zero of $W$ tends to the origin as $x=t^m\to 0$ by [F4].

Define $\gamma (t):=(t^m,g(t))$ on ${\mathrm{\Delta }}_{\delta }$. If $\gamma (t_1)=\gamma (t_2)$ and $t_1\mathrm{\ne }0$, then $t_2={\zeta }^j{t}_1$ for some $0\le j<m$. If $t_1$ is on a boundary ray, step 8.1 says that the values $g({\zeta }^j{t}_1)$ for $0\le j<m$ are pairwise distinct, so equality forces $j=0$. Otherwise choose $r$ with $s:={\zeta }^{-r}{t}_1\in S_0$. By (8.1.1), $g(t_1)={\alpha }_{{\sigma }^r(1)}(s^m)$ and $g(t_2)={\alpha }_{{\sigma }^{r+j}(1)}(s^m)$. The $m$ fibre values are distinct by step 3.1, so equality forces ${\sigma }^j(1)=1$; step 7.1 gives $m\mid j$, hence $j=0$. If $t_1=0$ and $\gamma (t_2)=\gamma (0)$, then $t_2^m=0$ and $t_2=0$. Therefore $\gamma$ is injective.

The image of $\gamma$ is a full representative of $X$. For $x\in D_{\epsilon }^{\ast }\setminus N$, choose the unique $s\in S_0$ with $s^m=x$; (8.1.1) and the $m$-cycle $\sigma$ show that the values $g({\zeta }^{j}s)$ enumerate the $m$ roots of $W(x,\cdot )$, so the image of $\gamma$ contains the full fibre. If $x\in N\setminus \{0\}$, choose any $t_0$ with $t_0^m=x$. The parameters ${\zeta }^j{t}_0$ lie on the boundary rays; by step 8.1 their $g$-values are $m$ distinct roots of $W(x,\cdot )$, hence enumerate its full degree-$m$ fibre. At $x=0$, the only point of $Z(W)$ is $(0,0)$ because $W(0,T)=T^m$, and $\gamma (0)=(0,0)$ by step 9.1. Therefore $\gamma ({\mathrm{\Delta }}_{\delta })=Z(W)\cap (D_{\epsilon }\times D_y)$. On a neighbourhood of $0$ the unit $u$ in $f=uW$ does not vanish, so $Z(W)=Z(f)$ there and this image is a full representative of the germ $X$.

The order of $g$ is $0$, or at least $m$: either $g\equiv 0$, or $n:={\mathrm{ord}}_{0}g\ge m$. Indeed, suppose $g$ is not identically zero and $1\le n<m$. Since $\gamma$ takes values in $Z(W)=Z(f)$ near the origin by step 11.1, the holomorphic germ $f\circ \gamma$ vanishes identically. Write $f=f_m+(\text{terms of order}>m)$ as in [F1]; by construction of the $y$-axis in step 1.1 the coefficient of $y^m$ in $f_m$ is $f_m(0,1)=f_m(w)\mathrm{\ne }0$. Substituting the expansion and $g(t)=a_n{t}^n+\cdots$ with $a_n\mathrm{\ne }0$, the monomial $y^m$ of $f_m$ contributes $f_m(0,1)a_n^m{t}^{mn}+O(t^{mn+1})$, while every monomial $x^i{y}^j$ of $f_m$ with $i>0$, $i+j=m$ contributes order $im+jn=mn+i(m-n)>mn$, and every homogeneous part of order $\mathrm{\ell }>m$ contributes order at least $\mathrm{\ell }n>mn$. Hence $f\circ \gamma$ has exact order $mn<\mathrm{\infty }$, contradicting $f\circ \gamma \equiv 0$.

If $g\equiv 0$ or ${\mathrm{ord}}_{0}g>m$, keep the coordinates and put $h:=g$. If ${\mathrm{ord}}_{0}g=m$, write $g(t)=a_m{t}^m+O(t^{m+1})$ with $a_m\mathrm{\ne }0$, let $\mathrm{\Phi }(x,y):=(x,y-a_{m}x)$ be the invertible complex-linear shear, and put $h(t):=g(t)-a_m{t}^m$; in the new coordinates $(x^{\mathrm{\prime }},y^{\mathrm{\prime }}):=\mathrm{\Phi }(x,y)$ the curve $X$ has defining polynomial $W^{\mathrm{\prime }}(x^{\mathrm{\prime }},y^{\mathrm{\prime }}):=W(x^{\mathrm{\prime }},y^{\mathrm{\prime }}+a_m{x}^{\mathrm{\prime }})$, again a Weierstrass polynomial of degree $m$ with $W^{\mathrm{\prime }}(0,T)=T^m$, reduced and irreducible because $\mathrm{\Phi }$ induces an automorphism of ${\mathcal{O}}_{{\mathbb{C}}^2,0}$, and ${\gamma }^{\mathrm{\prime }}(t):=(t^m,h(t))=\mathrm{\Phi }(\gamma (t))$ satisfies $W^{\mathrm{\prime }}({\gamma }^{\mathrm{\prime }}(t))=W(t^m,g(t))=0$. In both cases $h$ is holomorphic on ${\mathrm{\Delta }}_{\delta }$ with $h(t)={\sum }_{k>m}{a}_k{t}^k$, and the map $\gamma (t)=(t^m,h(t))$ is injective by step 10.1 (in the sheared case it is $\mathrm{\Phi }$ composed with the injective $\gamma$, and $\mathrm{\Phi }$ is injective). Moreover, for $x=t_0^m\in D_{\epsilon }^{\ast }$ the fibre of $X$ over $x$ in the final coordinates is $\mathrm{\Phi }$ applied to the old fibre, that is, by step 11.1, the set $\{(x,h({\zeta }^j{t}_0)):j=0,\dots ,m-1\}$, and the image of $\gamma$ is $\mathrm{\Phi }$ of a full representative of $X$, hence again a full representative of $X$.

The exponent $m$ is minimal. Let $\tilde{\gamma }(s)=(s^k,j(s))$, $k\ge 1$, be any parametrisation of the germ $X$ in the coordinates of step 13.1 with $j$ holomorphic near $0$, so that its image contains a full representative of $X$ by step 13.1. Choose a nonzero base value $x_0$ with $\mathrm{\mid }{x}_0\mathrm{\mid }<\epsilon$ small enough that the $m$ points of $X$ over $x_0$ lie in that representative; they are distinct by the fibre description of step 13.1 together with the injectivity of step 10.1. Each of these $m$ points equals $\tilde{\gamma }(s)$ for some $s$ with $s^k=x_0$, and distinct points have distinct parameters, while the monic polynomial $s^k-x_0$ has exactly $k$ roots counted with multiplicity by [F13] and all of them are simple because $x_0\mathrm{\ne }0$; hence there are exactly $k$ solutions. Therefore $m\le k$: the parametrisation of step 13.1 is primitive.

Let $\tilde{\gamma }(t)=(t^m,j(t))$ be any parametrisation of $X$ in the coordinates of step 13.1, defined on $\mathrm{\mid }t\mathrm{\mid }<\eta$ with ${\eta }^m<\epsilon$. For $t_0\in {\mathrm{\Delta }}_{\eta }\setminus \{0\}$ the point $\tilde{\gamma }(t_0)$ lies in $X$ over $x=t_0^m$, so by the fibre description of step 13.1 there is an index $\mathrm{\ell }$ with $j(t_0)=h({\zeta }^{\mathrm{\ell }}{t}_0)$; hence the sets $S_{\mathrm{\ell }}:=\{t\in {\mathrm{\Delta }}_{\eta }\setminus \{0\}:j(t)=h({\zeta }^{\mathrm{\ell }}t)\}$, $\mathrm{\ell }=0,\dots ,m-1$, are closed, cover the punctured disc, and are pairwise disjoint: if $t$ lay in $S_{\mathrm{\ell }}\cap S_{{\mathrm{\ell }}^{\mathrm{\prime }}}$ with $\mathrm{\ell }\mathrm{\ne }{\mathrm{\ell }}^{\mathrm{\prime }}$, then the distinct points ${\zeta }^{\mathrm{\ell }}t,{\zeta }^{{\mathrm{\ell }}^{\mathrm{\prime }}}t$ would satisfy $\gamma ({\zeta }^{\mathrm{\ell }}t)=(t^m,h({\zeta }^{\mathrm{\ell }}t))=(t^m,j(t))=\gamma ({\zeta }^{{\mathrm{\ell }}^{\mathrm{\prime }}}t)$, contradicting step 10.1. The punctured disc is connected by [F14], so [F15] shows that one $S_{\mathrm{\ell }}$ is everything: there is an $m$-th root of unity ${\zeta }_0$ with $j(t)=h({\zeta }_{0}t)$ for all $t\mathrm{\ne }0$ in ${\mathrm{\Delta }}_{\eta }$. Then $\tilde{\gamma }$ is injective: if $\tilde{\gamma }(t_1)=\tilde{\gamma }(t_2)$ with $t_1\mathrm{\ne }0$, then $t_2={\zeta }^r{t}_1$ and $j(t_2)=j(t_1)$ give $h({\zeta }_0{\zeta }^r{t}_1)=h({\zeta }_0{t}_1)$, so $\gamma ({\zeta }_0{\zeta }^r{t}_1)=\gamma ({\zeta }_0{t}_1)$ and step 10.1 forces ${\zeta }^r{t}_1=t_1$, that is $t_2=t_1$; and $t_1=0$ gives $t_2^m=0$, $t_2=0$.

Steps 1.1 and 8.1–13.1 give the invertible complex-linear change of coordinates, the integer $m\ge 1$ and the holomorphic $h(t)={\sum }_{k>m}{a}_k{t}^k$ with injective $\gamma (t)=(t^m,h(t))$ whose image germ is exactly $X$; step 14.1 shows that $m$ is minimal among the exponents of parametrisations $s\mapsto (s^k,j(s))$ of the germ in these coordinates, and step 14.2 shows that every such parametrisation with first component $t\mapsto t^m$ equals $t\mapsto (t^m,h({\zeta }_{0}t))$ for an $m$-th root of unity ${\zeta }_0$ and is injective. This is the asserted convergent Puiseux parametrisation, obtained without any formal-series step.