Proof The proof consists of showing that $\Sigma $ is compatible on both charts when $Z_j \in U_j \cap U_{j+1}$ or when $\Sigma (Z_j) \in U_\ell \cap U_{\ell \pm 1}$ . It is a simple computation. Note that for a subdomain $S_j'\subset S_j$ such that $\sigma (S_j')\subset S_{-j}$ , then, in the charts, we have $\Sigma _{-j,j}(Z_j) = \overline {Z_j}$ .

Proof The proof is similar to that found in [Reference Ilyashenko and Yakovenko6]. Let $m(z) = f\circ (\sigma \circ v^{\scriptscriptstyle {1/2}})^{-1}(z) = z + o(z^{2k+1})$ . We see that

$$ \begin{align*} Z_j\circ m(z) &= -{1\over kz^k}(1 + o(z^{2k})) + b\log z + o(z^{2k}) - {ji\pi b\over k}\\ &= Z_j(z) + o(z^k). \end{align*} $$

Since $z^kZ_j(z) \to -{(1/k)}$ when $z\to 0$ , and because $Z_j$ is invertible and $|Z_j(z)|\to \infty $ when $z\to 0$ , it follows that $o(z^k)$ is $O(|Z_j|^{-1})$ when $|Z_j|\to \infty $ . Therefore, $F\circ (\Sigma \circ {T_{\scriptscriptstyle {1/2}}})^{-1}(Z_j) = Z_j + O(|Z_j|^{-1})$ .

Proof The proof makes use of the rigidity of the conformal structure of the doubly punctured sphere $S^2\setminus \{0,\infty \}$ , as in the proof of the uniqueness in the holomorphic case.

Let $\Phi _j$ denote the Fatou coordinate of $f \circ f$ on $U_j$ , that is,

$$ \begin{align*}\Phi_{j} \circ (F \circ F) \circ {(\Phi_{j})}^{-1} = T_1. \end{align*} $$

(We know that it exists, since $f \circ f$ is holomorphic, and that it is unique up to left composition with a translation.) In the space of the Fatou coordinate $W_{j} = \Phi _{j}(Z_j)$ , an orbit $\{ (f\circ f)^{\circ n}(z) \}_n$ corresponds to $\{ W_{j} + n \}_n$ . Note also that $\Phi _j(Z_j) = Z_j + D_j + O(|Z_j|^{-1})$ for some constant $D_j\in \mathbb {C}$ (see [Reference Ilyashenko and Yakovenko6]).

We first note that each $\Phi _j(U_{j})$ contains a vertical strip $B_j$ of width one, by construction of the time coordinate $U_{j}$ and of the Fatou coordinate. We define

$$ \begin{align*}Q_{j} = \Phi_{- j} \circ F\circ {(\Phi_{j})}^{-1} \qquad\textrm{for }-k\leq j\leq k. \end{align*} $$

Then we see that $Q_{j} \circ Q_{-j} = T_1$ , since $\Phi _{j}$ are Fatou coordinates of $F \circ F$ . It follows that $Q_{j}$ commutes with $T_1$ , since

$$ \begin{align*} T_1 \circ Q_{j} &= (Q_{j} \circ Q_{-j}) \circ Q_{j} \\ &= Q_{j} \circ (Q_{-j} \circ Q_{j}) \\ &= Q_{j} \circ T_1. \end{align*} $$

Indeed, $Q_{j}$ represents F in the Fatou coordinates. It is therefore natural that $Q_{j}$ commutes with $T_1$ , which represents $F \circ F$ in the Fatou coordinates.

Because $Q_j$ is the composition of an antiholomorphic germ by a holomorphic diffeomorphism, $Q_j$ is antiholomorphic. In particular, $\Sigma \circ Q_j$ is holomorphic, and $\Sigma \circ Q_{j} - \mathrm {id}$ is $1$ -periodic and holomorphic, so it has a Fourier expansion

$$ \begin{align*}\Sigma\circ Q_{j}(W_{j}) - W_{j} = \sum_{n=-\infty}^\infty c_{n,j} e^{2i\pi n W_{j}}. \end{align*} $$

Moreover, by Lemma 4.4, we have $\Sigma \circ Q_j(W) = W + M_j + O(|W|^{-1})$ , where $M_j\in \mathbb {C}$ is a constant. Therefore, $|\Sigma \circ Q_j - \mathrm {id}|$ is bounded when $|W| \to \infty $ , so we must have $c_{n,j} = 0$ for $n\in \mathbb {Z}^\ast $ .

We conclude that $Q_{j}(W_j) = \overline W_j + \overline {c_{j,0}}$ . Since $Q_j\circ Q_j = T_1$ , it follows that $c_{j,0} = {\tfrac 12} + iy$ . We then adjust all the $\textrm{Im}\kern2pt c_{j,0}$ to 0 by choosing the appropriate Fatou coordinates (that is, composing them with a translation).

The uniqueness comes from a combination of the uniqueness of the Fatou coordinates for the holomorphic $f \circ f$ and having to preserve the constants $c_{j,0} = \frac {1}{2}$ .

Proof Let $(\Psi _{-k},\ldots ,\Psi _{-1}$ , $\Psi _1,\ldots ,\Psi _k)$ be a representative of the analytic invariant of g. Since r is a rotation with an order dividing k, it is an element of $D_{2k}$ . It rotates the coordinate so that the symmetry axis corresponds to the real axis with a repulsive petal on the side of $\mathbb {R}^+$ (this is the case because f is of positive type). The representative of the modulus of g is permuted: $(\Psi _{r(-k)},\ldots ,\Psi _{r(-1)},\Psi _{r(1)},\ldots ,\Psi _{r(k)})$ as in (24); it now has an equivalent representative $(\Psi _{r(-k)}',\ldots ,\Psi _{r(-1)}',\Psi _{r(1)}',\ldots ,\Psi _{r(k)}')$ under relation (23) obtained from the Fatou coordinates of f, so that it satisfies

$$ \begin{align*}\Psi_{r(j)}'\circ \Sigma\circ{T_{\scriptscriptstyle{1/2}}} = \Sigma\circ{T_{\scriptscriptstyle{1/2}}} \circ \Psi_{r(-j)}'. \end{align*} $$

Equation (26) follows from $r^{-1} sr = \sigma $ , where $\sigma $ is the usual complex conjugation. Indeed, on the indices, we have $r(-j) = r\sigma (j)$ and, substituting $\ell = r(j)$ , the previous equation becomes

$$ \begin{align*}\Psi_\ell'\circ \Sigma\circ{T_{\scriptscriptstyle{1/2}}} = \Sigma\circ{T_{\scriptscriptstyle{1/2}}} \circ \Psi_{s(\ell)}'. \\[-36pt] \end{align*} $$

Proof Suppose that g is prenormalized. The dynamics of $g^{-1}$ is reversed; therefore, the dynamics of $g^{-1}$ in the first sector $S_0$ is attractive, but the dynamics in $S_{-1}$ is repulsive. We apply the change of coordinate $y = L_\unicode{x3bb} (z) = \unicode{x3bb} z$ , where $\unicode{x3bb} = e^{i\pi /k}$ . Let $\widetilde g^{-1} = L_\unicode{x3bb} \circ g^{-1}\circ L_\unicode{x3bb} ^{-1}$ .

At the formal level, we apply $(z,t)\mapsto (y,-t)$ to the vector field (5) to obtain

(28) $$ \begin{align} \dot y = w(y) = {y^{k+1}\over 1 - by^k}. \end{align} $$

The formal normal form of $\widetilde g^{-1}$ is the time-1 map $w^1$ of (28). In particular, $\widetilde g^{-1}$ has formal invariant $-b$ . We will denote the sectors of $\widetilde g^{-1}$ in the y-coordinate by $\widetilde S_j$ . Note that $\widetilde S_j = L_\unicode{x3bb} ^{-1}(S_j) = S_{j-1}$ .

The time coordinate on the sector $\widetilde S_j$ is defined as

(29) $$ \begin{align} Y_j(y) = {-1\over ky^k} - b\log y + {ji\pi b\over k}. \end{align} $$

We have the relation $L_{-1}\circ Y_j = Z_{j-1}\circ L_\unicode{x3bb} ^{-1}$ , where the $Z_j$ are the time coordinates of g. Indeed, we see that

$$ \begin{align*}Z_{j-1}(e^{(-i\pi/k)}y) = {1\over k y^k} + b \log y - {i\pi b\over k} - {(j-1)i\pi b\over k} = -Y_j(y). \end{align*} $$

Let $\Phi _j$ be a Fatou coordinate of g in $S_j$ . We know that $\Phi _j\circ Z_j\circ g^{-1}\circ Z_j^{-1} \circ \Phi _j^{-1} = T_{-1}$ . We will show that $L_{-1}\circ \Phi _j\circ L_{-1}$ is a Fatou coordinate of $\widetilde g^{-1}$ on $\widetilde S_{j+1}$ . Indeed, we have

$$ \begin{align*} (L_{-1}\circ \Phi_j\circ L_{-1})\circ Y_{j+1} \circ \widetilde g^{-1}\circ Y_{j+1}^{-1} & {}\circ (L_{-1}\circ \Phi_j\circ L_{-1})^{-1}\\ &= L_{-1}\circ \Phi_j\circ Z_j\circ g^{-1}\circ Z_j^{-1}\circ \Phi_j^{-1}\circ L_{-1}\\ &= L_{-1} \circ T_{-1}\circ L_{-1} = T_1. \end{align*} $$

It follows that the transition functions $\widetilde \Psi _j$ of $\widetilde g^{-1}$ are given by, for $j{\kern-1pt}\not =1,k$ odd,

$$ \begin{align*} \widetilde \Psi_j &= \widetilde \Phi_j\circ T_{\scriptscriptstyle-\mathop{\mathrm{sgn}}\nolimits(j){(i\pi b/ k)}} \circ (\widetilde \Phi_{j - \mathop{\mathrm{sgn}}\nolimits(j)})^{-1}\\ &= L_{-1}\circ \Phi_{j-1} \circ T_{\scriptscriptstyle-\mathop{\mathrm{sgn}}\nolimits(j){(i\pi b/ k)}} \circ \Phi_{j-1-\mathop{\mathrm{sgn}}\nolimits(j-1)}^{-1} \circ L_{-1}\\ &= L_{-1} \circ \Psi_{j-1}^{-1}\circ L_{-1}. \end{align*} $$

The other values of j are done similarly. Note however that for $j=1$ , the equation becomes $\widetilde \Psi _1 = L_{-1}\circ \Psi _{-1}^{-1}\circ L_{-1}$ and that, for $j=k$ , we have $\widetilde \Psi _{-k} = L_{-1}\circ \Psi _k^{-1}\circ L_{-1}$ .

Proof The proof is analogous to that in the holomorphic case, which can be found in [Reference Ilyashenko5, Reference Ilyashenko and Yakovenko6].

(1) Let $f_\ell \colon (\mathbb {C},0) \to (\mathbb {C},0)$ be a germ of an antiholomorphic diffeomorphism with a parabolic fixed point at the origin and let $(k_\ell ,b_\ell ,[\Psi _{1,\ell },\ldots ,\Psi _{k,\ell }])$ be its modulus of classification for $\ell =1,2$ . We can of course suppose that $f_\ell $ is prenormalized.

Suppose first that $f_2(u) = h\circ f_1\circ h^{-1}(u)$ for some germ of an analytic diffeomorphism $u = h(z)$ . The germs $f_1$ and $f_2$ must have the same codimension and formal invariant, since they are topological and formal invariants. For the analytic invariant, first let $F_1$ , $F_2$ and H denote the expressions of $f_1$ , $f_2$ , and h in the time coordinate. If $\Phi _j$ is a Fatou coordinate of $F_2$ on $U_j$ , then $\Phi _j\circ H$ is a Fatou coordinate of $F_1$ . It follows that they have the same transition functions.

Conversely, suppose that $f_1$ and $f_2$ have the same modulus. We can choose a common normalized representative $(\Psi _1,\ldots ,\Psi _k)$ for both classes and Fatou coordinates $\Phi _{j,\ell }$ for $f_\ell $ ( $\ell =1,2$ ) such that the $\Psi _j$ are the transition maps for these Fatou coordinates. Let $\varphi _{j,\ell } = Z_j^{-1}\circ \Phi _{j,\ell }\circ Z_j$ , for $\ell =1,2$ , be the expression of Fatou coordinates in the z-coordinate. Then we define

$$ \begin{align*}h_j(z) = \varphi_{j,1}^{-1}\circ \varphi_{j,2}(z) \end{align*} $$

on $S_j$ . It sends the orbits of $f_2$ to those of $f_1$ on $S_j$ . On the intersection of two consecutive sectors $S_j\cap S_{j+1}$ , we see that

$$ \begin{align*}h_{j+1}\circ h_j^{-1}(z) = \mathrm{id}, \end{align*} $$

so the $h_j$ agree on the intersection. We may define h on $\bigcup _j S_j$ by $h(z) = h_j(z)$ if $z\in S_j$ . It sends a punctured neighbourhood of the origin into another punctured neighbourhood of the origin, so, by the Riemann removable singularity theorem, h extends to a holomorphic diffeomorphism of a neighbourhood of the origin.

Finally, we see that $h\circ f_2 = f_1 \circ h$ , since it sends the orbits of $f_2$ to the orbits of $f_1$ on a whole neighbourhood of the origin.

(2) The proof is in two steps. First we construct an abstract Riemann surface S on which $\Sigma \circ {T_{\scriptscriptstyle {1/2}}}$ is well defined and we prove that this surface has the conformal type of a punctured disk. Then we prove that $\Sigma \circ {T_{\scriptscriptstyle {1/2}}}$ is the germ we are looking for on the disk.

Consider a triple $(k,b, [\Psi _1,\ldots ,\Psi _k])$ . We must find a parabolic germ of an antiholomorphic diffeomorphism with this modulus of classification.

Let $(\Psi _1,\ldots ,\Psi _k)$ be a representative of the analytic invariant and let $\Psi _{-1},\ldots ,\Psi _{-k}$ be the other transition functions obtained from (17). Let $\sigma \circ v^{\scriptscriptstyle {1/2}}$ be the normal form of codimension k with formal invariant b. Let $U_j$ be the charts in the time coordinate of v.

We consider the transition functions defined on those charts. More precisely, $\Psi _1$ is defined on a domain of $U_0$ containing an upper half-plane with its image in $U_1$ ; $\Psi _2$ is defined on a domain of $U_2$ containing a lower half-plane with its image in $U_1$ , and so on. We define the Riemann surface S by

$$ \begin{align*}S = \bigsqcup_{j=-k}^k U_j/\sim, \end{align*} $$

where $\sim $ identifies $W_j\in U_j$ with its image by $\Psi _j$ or $\Psi _{j-1}$ (depending on j). As is done in [Reference Ilyashenko and Yakovenko6], we can build a smooth quasi-conformal mapping $P\colon S\to \mathbb {C}^\ast $ and, from the Ahlfors–Bers theorem (see [Reference Ahlfors1]), find a diffeomorphism $Q\colon D \to (\mathbb {C},0)$ , where $D = P(S)\cup \{0\}$ , so that the composition $H = Q\circ P\colon S\to (\mathbb {C},0)$ is holomorphic. In fact, H is a biholomorphism of S onto some punctured disk of the origin.

By (15), we know that $\Sigma \circ {T_{\scriptscriptstyle {1/2}}}$ is well defined on S. The map $f = H\circ \Sigma \circ {T_{\scriptscriptstyle {1/2}}}\circ H^{-1}$ extends to the origin by $f(0) = 0$ , since it is bounded around $0$ (we can apply the Riemann removable singularity theorem to $f\circ \sigma $ to see this). Lastly, $T_1$ is also well defined on S, so we set $g = H\circ T_1\circ H^{-1}$ . Since $T_1 = (\Sigma \circ {T_{\scriptscriptstyle {1/2}}})\circ (\Sigma \circ {T_{\scriptscriptstyle {1/2}}})$ , it follows that $g = f\circ f$ and, by the chain rule, $|{\partial f/ \partial {\overline z}}|^2 = g'(0)$ . By the holomorphic case, we know that g is a holomorphic parabolic germ of codimension k, so it follows that f is parabolic and of codimension k. The formal invariant of f is b, since it is determined by (19).

It remains only to prove that f is of positive type. The formal symmetry axis of f is the real line, since the $\Psi _1$ and $\Psi _{-1}$ are defined on $U_0$ and $\Psi _j\circ \Sigma \circ {T_{\scriptscriptstyle {1/2}}} = \Sigma \circ {T_{\scriptscriptstyle {1/2}}}\circ \Psi _{-j}$ . Moreover, the petal $S_0$ of f is repulsive, since $H\big |_{U_0}\circ Z_0\colon U_0\to U_0$ is a Fatou coordinate of f and $U_0$ contains a half-plane $\{\Re Z < R\}$ , that is, $S_0$ contains all the backward iterates of f. Therefore, f is of positive type.

We conclude that f is a germ with modulus $(k,b,[\Psi _1,\ldots ,\Psi _k])$ .

Proof The transition functions of the normal form are translations, since the time charts $Z_j$ are Fatou coordinates. Therefore, it follows from the classification theorem (Theorem 6.3).

Questions 7.2. When is a parabolic antiholomorphic germ analytically conjugate to a germ with real coefficients?

Proof (1) That g has a one-parameter family of formal antiholomorphic nth roots on each symmetry axis is a consequence of the fact that g is formally conjugate to the normal form $v^{\scriptscriptstyle 1}$ and Proposition 3.4.

(2) We first prove this for the case k odd. Let

$$ \begin{align*}\boldsymbol{\Psi} = (\Psi_{-k},\ldots,\Psi_{-1},\Psi_1,\ldots,\Psi_k) \end{align*} $$

be a representative of the analytic invariant of g.

First, we suppose that the representative satisfies (33). To realize an antiholomorphic root tangent to the symmetry axis $e^{2i\pi \ell / k}\mathbb {R}$ , we rotate the coordinate by $R_\ell (z) = e^{-{2i\pi \ell /k}}z$ so that this axis is on the real line and the dynamics of $R_\ell \circ g\circ R_\ell ^{-1}$ on the side of the positive real axis is repulsive. Let $g_\ell = R_\ell \circ g\circ R_\ell ^{-1}$ and let r correspond to the permutation of indices defined by $R_\ell $ (see Definition 5.5). The representative is permuted into $(\Psi _{r(-k)},\ldots ,\Psi _{r(-1)},\Psi _{r(1)},\ldots ,\Psi _{r(k)})$ . Also, we have that $r^{-1} s_\ell r = s_0$ , where $s_0(j) = -j$ and $s_\ell $ is the reflection of indices induced by $\sigma _\ell $ . Equation (33) becomes

$$ \begin{align*}\Psi_{r(j)}\circ \Sigma\circ{T_{\scriptscriptstyle{1/2}}} = \Sigma\circ{T_{\scriptscriptstyle{1/2}}} \circ \Psi_{r(-j)}. \end{align*} $$

Now we may repeat the proof of part (2) of Theorem 6.3 to obtain an antiholomorphic germ $f_\ell \colon (\mathbb {C},0)\to (\mathbb {C},0)$ with analytic invariant $[\Psi _{r(1)},\ldots ,\Psi _{r(k)}]$ such that $f_\ell $ is $\Sigma \circ T_{{1/n}}$ in each Fatou coordinate of $g_\ell $ . Since $(\Sigma \circ T_{{1/n}})^{\circ n} = T_1$ in the Fatou coordinates, it follows that $f_\ell ^{\circ n} = g_\ell $ .

Conversely, suppose that g has an nth root $f_\ell $ tangent to the reflection axis $e^{2i\pi \ell /k}\mathbb {R}$ . We rotate the coordinate by $R_\ell $ . Let $f = R_\ell \circ f_\ell \circ R_\ell ^{-1}$ . Now, in every Fatou coordinate of $g_\ell $ , f has the form $\Sigma \circ T_{(1/n)+iy}$ , by Proposition 3.4. By changing Fatou coordinates, we may obtain $\Sigma \circ T_{{1/ n}}$ . Those Fatou coordinates give us a representative $(\Psi _{-k}',\ldots ,\Psi _{-1}',\Psi _1',\ldots , \Psi _k')$ that is equivalent to $\boldsymbol {\Psi }$ under the relations (23) and (24) and that satisfies

$$ \begin{align*}\Psi_{j}'\circ \Sigma\circ{T_{\scriptscriptstyle{1/2}}} = \Sigma\circ{T_{\scriptscriptstyle{1/2}}} \circ \Psi_{-j}'. \end{align*} $$

It follows that $(\Psi _{r^{-1}(-k)}',\ldots ,\Psi _{r^{-1}(-1)}',\Psi _{r^{-1}(1)}',\ldots ,\Psi _{r^{-1}(k)}')$ is a representative equivalent to $\boldsymbol {\Psi }$ under relation (23) which satisfies (33), using the fact that $(r^{-1} s_\ell r)(j) = -j$ .

In the case where k is even, we may have antiholomorphic roots of positive and negative type. Those of positive type are tangent to an axis $e^{i\pi \ell /k}\mathbb {R}$ with $\ell $ even and are done as before. Those of negative type are tangent to an axis $e^{i\pi \ell /k}\mathbb {R}$ with $\ell $ odd. For such a root $f_\ell $ , $f_\ell ^{-1}$ will be a root of positive type of $g^{-1}$ . By Proposition 5.9, the modulus of $g^{-1}$ is

$$ \begin{align*}\let\widetilde\widetilde (k,-b, [\widetilde\Psi_{-k},\ldots,\widetilde\Psi_{-1}, \widetilde\Psi_1, \ldots, \widetilde\Psi_{k}]), \end{align*} $$

where $\widetilde \Psi _j = L_{-1}\circ \Psi _{r_1^{-1}(j)}^{-1}\circ L_{-1}$ and $r_1$ is the rotation of indices induced by $z\mapsto \smash {e^{i\pi /k}} z$ . By the previous two paragraphs, $g^{-1}$ has an nth antiholomorphic root of positive type tangent to $e^{i \pi \ell /k}\mathbb {R}$ if and only if there exists a representative $(\widetilde \Psi _{-k},\ldots ,\widetilde \Psi _{-1}, \widetilde \Psi _1, \ldots , \widetilde \Psi _{k})$ such that

$$ \begin{align*}\widetilde \Psi_j \circ \Sigma\circ T_{{1/ n}} = \Sigma \circ T_{{1/n}} \circ \widetilde \Psi_{s_\ell(j)}. \end{align*} $$

By simplifying $L_{-1}$ and taking the inverse on both sides, we obtain (33).

(3) (i) Suppose that g has two roots $f_1$ and $f_2$ from the same family. In particular, they have the same linear term, so we may suppose that they are tangent to $\sigma $ , modulo conjugating g by a rotation of order k. In the Fatou coordinates, they take the form $\Sigma \circ T_{{(1/2)} + iy_j}$ , for $j=1,2$ , by Proposition 3.4. Since $\Sigma \circ T_{{(1/2)} + iy_j}$ satisfies (15) for $j=1,2$ , and by combining with (18), we see that either $y_1 = y_2$ or the $\Psi _j$ ’s are translations.

(ii) Lastly, consider the dihedral group $D_{2k}$ with its action on the regular k-gon. Recall that we can divide a regular k-gon by its k symmetry axes to form $2k$ sectors, which we associate to the transition functions (see \S5.1 and Figures 13 and 14).

Let $H = \langle s_1,\ldots ,s_m\rangle $ be the subgroup of $D_{2k}$ generated by the reflections along the symmetry axes of $f_{\ell _1},\ldots , f_{\ell _m}$ . Each element of H acts on the modulus by introducing relations of the type

$$ \begin{align*}\Sigma\circ{T_{\scriptscriptstyle{1/2}}}\circ \Psi_j = \Psi_{s_\ell(j)}\circ \Sigma\circ{T_{\scriptscriptstyle{1/2}}}, \end{align*} $$

thus reducing the number of independent $\Psi _j$ (see Figure 18). The orbit by H of a sector represents the transition functions tied together; therefore, the number of independent transition functions corresponds to $|D_{2k}:H|$ . To compute this, we first observe that H must itself be a dihedral group, so that $H = D_{2j}$ for some j (see [Reference Grove and Benson3]). In fact, j is given by

$$ \begin{align*}j = {k\over \gcd(\ell_m - \ell_1,\ldots,\ell_m - \ell_{m-1},k)}, \end{align*} $$

since any rotation $s_ms_p$ has order $k/\gcd (\ell _m - \ell _p,k)$ . It follows that $|D_{2k}:H| = \gcd (\ell _m - \ell _1,\ldots , \ell _m - \ell _{m-1},k)$ .

Figure 18 The orbit of one sector by the subgroup H generated by two reflections (bold axes) in codimension 12.

Proof That f has a unique formal nth root follows from Proposition 3.4.

For the second part, f has an antiholomorphic nth root if and only if the transition functions satisfy $\Psi _{j} \circ \Sigma \circ T_{1/2n} = \Sigma \circ T_{1/2n}\circ \Psi _{-j}$ , since we may realize $\Sigma \circ T_{1/2n}$ on the Riemann surface of $f\circ f$ . We combine this last equation with (15) to obtain $\Psi _{j}\circ T_{(1/2n)-{(1/2)}} = T_{(1/2n)-{(1/2)}}\circ \Psi _{j}$ . To conclude, we note that $\gcd (1-n, 2n)=2$ because n is odd, so there exists $p,q\in \mathbb {Z}$ such that $(1-n)p - 2nq\! =\! 2$ . In other words, we have ${1/n}\! =\! ((1-n)/ 2n)p - q$ . Since $\Psi _{j}$ commutes with $T_{{((1-n)/2n)}p}$ and with $T_q$ (because q is an integer), it follows that $\Psi _{j}$ commutes with $T_{1/n}$ .

Proof Equation (34) is independent of the representative and it is equivalent to a holomorphic parabolic germ having a holomorphic nth root (see [Reference Ilyashenko5]).

Suppose that $g = f\circ f$ and g has a holomorphic nth root. Then the modulus of g and f satisfies (34), so that f has an antiholomorphic nth root.

The converse is direct.

Proof (1) $\Rightarrow $ (2) Let $\gamma $ be the germ of real analytic parametrization of the invariant curve of f. We can extend $\gamma $ on a disk around the origin. Then $\gamma ^{-1}\circ f\circ \gamma $ fixes a germ of the real axis, so its power series has real coefficients.

(2) $\Rightarrow $ (3) We can of course suppose that the power series of f has real coefficients, so that $\sigma \circ f = f\circ \sigma $ . It follows that $\sigma \circ f$ is a holomorphic square root of $g = f\circ f$ . Therefore, we have $\Psi _j\circ {T_{\scriptscriptstyle {1/2}}} = {T_{\scriptscriptstyle {1/2}}} \circ \Psi _j$ .

(3) $\Leftrightarrow $ (4) It follows from equation (15).

(4) $\Rightarrow $ (1) This is the harder part of the proof. The hypothesis implies that $\Sigma $ is well defined on the Riemann surface S constructed in the proof of part (2) of the classification theorem (Theorem 6.3). Let $z = H(W)$ be the coordinate given by $H\colon S\to (\mathbb {C},0)\setminus \{0\}$ , the biholomorphism of S to a punctured neighbourhood of the origin. Let $\sigma '$ and $f'$ be the expressions of $\Sigma $ and $\Sigma \circ {T_{\scriptscriptstyle {1/2}}}$ in this coordinate (note that f and $f'$ are analytically conjugate). Since, on S, $\Sigma $ and $\Sigma \circ {T_{\scriptscriptstyle {1/2}}}$ commute, it follows that $\sigma '$ and $f'$ commute also. So, it remains only to show that $\sigma '$ preserves a germ of real analytic curve at the origin.

We can extend $\sigma '$ by $\sigma '(0) = 0$ by the Riemann removable singularity theorem (this is true for antiholomorphic functions, since we simply apply it to $\sigma '\circ \sigma $ , where $\sigma (z) = {\overline z}$ ). In the charts $U_0$ and $U_k$ , $\Sigma $ fixes the real axis. These two curves carry in the z-coordinate and meet at the origin to form a continuous curve $\gamma $ fixed by $\sigma '$ . In fact, $\gamma $ is a $C^1$ curve, as H can be extended to a $C^1$ diffeomorphism $\widetilde H\colon \widetilde S\to (\mathbb {C},0)$ , where $\widetilde S$ is obtained from S with the point $\infty = H^{-1}(0)$ added.

This curve divides a small disk $D(0,\delta )$ into two connected components A and B. By the Riemann mapping theorem, there exists a biholomorphism $\varphi $ of A to the upper half-plane that sends continuously the boundary of A on the real line; see Figure 19. The image of $\gamma $ corresponds to an interval $[a,b]$ . We can extend $\varphi $ to $A\cup \gamma \cup B$ by

$$ \begin{align*}\widetilde\varphi(z) =\begin{cases} \varphi(z) & \textrm{if } z\in A\cup\gamma,\\ \sigma\circ\varphi\circ\sigma'(z) & \textrm{if } z\in B. \end{cases} \end{align*} $$

This is holomorphic on B, since it is the composition of a holomorphic map with two antiholomorphic maps, and it is continuous on $A\cup \gamma \cup B$ .

Figure 19 Mapping $\varphi $ extended on $A\cup \gamma \cup B$ .

The argument to prove that $\widetilde \varphi $ is holomorphic is similar to that of the Schwarz reflection principle. The idea is to show that the integral of $\widetilde \varphi $ along any triangle in $A\cup \gamma \cup B$ vanishes, and it will then follow from Morera’s theorem. If a simple closed curve is in $A\cup \gamma $ or $B\cup \gamma $ , then it follows from Cauchy’s theorem (we may take a limit of closed curves in A or B converging to the initial one). Then we can divide any triangle in $A\cup \gamma \cup B$ along $\gamma $ to obtain a finite number of closed curves in $A\cup \gamma $ and in $B\cup \gamma $ , as in Figure 20. Thus, $\widetilde \varphi |_{(a,b)}^{-1}$ is a real analytic parametrization of $\gamma $ .

Figure 20 Triangle divided into two closed curves.

Proof (1) $\Rightarrow $ (2) and (4) $\Rightarrow $ (1) are the same as in the previous theorem.

(2) $\Rightarrow $ (3) Suppose that g is in a coordinate such that $g = \sigma \circ g\circ \sigma $ . Let $\Phi _j$ be a Fatou coordinate on $U_j$ for $-k \leq j\leq k$ , where $\Phi _{-k} = \Phi _k$ . Then $\Sigma \circ \Phi _{-j}\circ \Sigma $ is also a Fatou coordinate of g on $U_j$ . By the uniqueness of the Fatou coordinate (for the holomorphic case), there exists a constant $C\in \mathbb {C}$ such that

$$ \begin{align*}\Sigma\circ \Phi_{-j}\circ\Sigma\circ \Phi_j^{-1} = T_C \end{align*} $$

for all j. In particular, for $j=0$ , by taking the inverse and conjugating both sides by $\Sigma $ of the previous equation, we obtain $T_C = T_{-\overline C}$ , so C must be pure imaginary, say $iy$ with $y\in \mathbb {R}$ . For $j>0$ and odd, we conclude with

$$ \begin{align*}\displaystyle\Psi_j = \Phi_j\circ \Phi_{j-1}^{-1} = T_{-iy}\circ \Sigma \circ \Phi_{-j} \circ \Phi_{-j+1}^{-1}\circ \Sigma\circ T_{iy} = T_{-iy}\circ \Sigma \circ \Psi_{-j} \circ \Sigma\circ T_{iy}. \end{align*} $$

The other values of j are done similarly.

(3) $\Leftrightarrow $ (4) It follows from the fact that for any two representatives $(\Psi _{-k},\ldots ,\Psi _{-1}, \Psi _1,\ldots , \Psi _k)$ and $(\Psi _{-k}',\ldots ,\Psi _{-1}',\Psi _1',\ldots ,\Psi _k')$ , there exists $C\in \mathbb {C}$ such that $\Psi _j\circ T_C = T_C \circ \Psi _j'$ . We choose $C = -{iy/2}$ to get a representative satisfying $\Psi _j'\circ \Sigma = \Sigma \circ \Psi _{-j}'$ .

Proof The holomorphic germ realized by the Écalle–Voronin modulus $(1,0,[W + e^{2i\pi W}, W + e^{-2i\pi W}])$ has no antiholomorphic square root, but preserves a germ of real analytic curve.

Proof If m is not the identity or analytically conjugate to $\sigma $ , then m is parabolic. It must have codimension k, since it maps the orbits of g on the orbits of g and the Fatou petals of g on the Fatou petals of g. To see that m has the same formal invariant, we compare degree by degree the power series on both sides of the equation $m\circ g(z) = g\circ m(z)$ .

To see that m has one of the forms of (35), the first part of the proof of Proposition 4.5 applies almost identically to m.

Proof (1) Since g is embeddable, each of its transition functions is a translation, so any $T_t$ will commute with them. Each $T_t$ represents a germ analytically conjugate to $g^t$ for some $t\in \mathbb {C}$ . In the case where m is antiholomorphic, $\Sigma \circ T_t$ is compatible with the transition functions with no restriction on t. It corresponds to $h\circ \sigma \circ v^{t}\circ h^{-1}$ for $t\in \mathbb {C}$ .

(2) Since g is not embeddable, one of the transition functions is not a translation. Suppose that $g_1\in \mathop {\mathrm {Diff}}\nolimits _1(0,\mathbb {C},\overline {\mathbb {C}})$ is holomorphic and commutes with g. By Lemma 7.9, $g_1$ becomes $T_t$ in the Fatou coordinates for some $t\in \mathbb {C}$ . Then $T_t$ must commute with the transition functions. This will only happen if the transition functions expand in the form

(36) $$ \begin{align} \Psi_j(W) = W + (-1)^j{i\pi b\over k} + \sum_{\ell\in\mathbb{Z}^\ast} c_{\ell p,j}e^{2i\pi \ell p W} \end{align} $$

and $t = {a/p}$ for some $a,p\in \mathbb {N}$ . This corresponds to $g_1$ being a holomorphic fractional iterate of order p of g. Let p denote the maximal order of holomorphic roots of g.

Suppose that $f_1\in \mathop {\mathrm {Diff}}\nolimits _1(0,\mathbb {C},\overline {\mathbb {C}})$ is antiholomorphic and commutes with g. We can suppose that $f_1$ is not a Schwarz reflection, since this is covered in Theorem 7.7. Then, by Lemma 7.9, $f_1$ is $\Sigma \circ T_{\scriptscriptstyle t}$ in the Fatou coordinates and it must be compatible with the transition functions. Therefore, $T_{2\Re t}$ commutes with the $\Psi _j$ and, by the previous case, we have $2\Re t = {a / r}$ with $\gcd (a,r) = 1$ and $r\,|\, p$ . We have $\Sigma \circ T_{\scriptscriptstyle t}\circ \Psi _j = \Psi _{-j}\circ \Sigma \circ T_{\scriptscriptstyle t}$ for some representative of the modulus if and only if

(37) $$ \begin{align} \overline{c_{n,j}} = e^{-i\pi an/r}e^{2n\pi y}c_{-n,-j} \end{align} $$

and $t = (a/2r) + iy$ ( $y\in \mathbb {R}$ ), where $c_{n,j}$ are the Fourier coefficients of $\Psi _j$ from (18). Of course, generically there exists no y that satisfy (37), so there are no antiholomorphic germs that commute with g in the generic case. When such a y exists, by changing the $\Psi _j$ by $T_{-{(iy/2)}}\circ \Psi _j\circ T_{{iy/2}}$ , we can suppose that $y=0$ .

Since ${T_{\scriptscriptstyle {1/ p}}}$ commutes with the $\Psi _j$ , we have $c_{n,j} = 0$ if $p{\kern-4pt}\not{\kern0.9pt}|{\kern0.8pt}n$ , so that (37) becomes $\overline {c_{sp,j}} = e^{-i\pi s ap/r} c_{-sp,-j}$ .

If $p/r$ and a are odd, then (37) becomes

$$ \begin{align*}\overline{c_{sp,j}} = \begin{cases} -c_{sp,j}, & s \textrm{ odd,}\\ c_{sp,j}, & s \textrm{ even.} \end{cases} \end{align*} $$

This is precisely the condition for g to have an antiholomorphic root of order $2p$ .

If $p/r$ or a is even, then $e^{-i\pi asp/r} = 1$ for all s, so we obtain the condition necessary for $\Sigma $ to be compatible with the transition functions and then there is a Schwarz reflection in the centralizer of g. In that case, the highest orders of the holomorphic and antiholomorphic iterates coincide.

Proof (1) In the Fatou coordinates, $T_t$ must commute with $\Sigma \circ {T_{\scriptscriptstyle {1/2}}}$ , so t must be real. The rest of the details are exactly as in the proof of Theorem 7.11.

(2) Let $p\in \mathbb {N}$ be the highest order of the holomorphic roots of $f\circ f$ . This number might be one, in which case $Z_f$ contains only iterates of f. Indeed, if $p=1$ , $Z_f$ does not contain a Schwarz reflection, otherwise $f\circ f$ would have the holomorphic root $\sigma \circ f$ , which would contradict that the highest order of the holomorphic root is one. Let us suppose that $p>1$ .

Let $(\Psi _1,\ldots ,\Psi _k)$ be a representative of the analytic invariant of f. Let $\Psi _{-j}$ be determined by (17) for $j=1,\ldots ,k$ . Suppose that the transition functions have a Fourier expansion

$$ \begin{align*}\Psi_j(W) = W + C_j + \sum_{n} c_{n,j}e^{2i\pi nW} \end{align*} $$

for all j, where $\overline {C_{-j}} = C_j = (-1)^{j(i\pi b/k)}$ for $j>0$ . Because the $\Psi _j$ satisfy $\Psi _j\circ \Sigma \circ {T_{\scriptscriptstyle {1/2}}} = \Sigma \circ {T_{\scriptscriptstyle {1/2}}} \circ \Psi _{-j}$ , we obtain

(38) $$ \begin{align} \overline{c_{n,j}} = e^{i\pi n} c_{-n,-j}. \end{align} $$

Since $f\circ f$ has a root of order p, we also have $T_{1/ p}\circ \Psi _j = \Psi _j\circ T_{1/ p}$ , which means that $c_{n,j} = 0$ for all j when $p{\kern-4pt}\not{\kern0.9pt}|{\kern0.8pt}n$ . If we write $n = \ell p$ in (38), it becomes

$$ \begin{align*}\overline{c_{\ell p,j}} = e^{i\pi \ell p} c_{-\ell p,-j} \quad (\ell\in\mathbb{Z}). \end{align*} $$

We have two cases.

If p is odd, then we have

$$ \begin{align*}\overline{c_{\ell p,j}} = \begin{cases} c_{-\ell p,-j} & \textrm{ if }\ell\textrm{ is odd,}\\ -c_{-\ell p,-j} & \textrm{ if }\ell\textrm{ is even.} \end{cases} \end{align*} $$

This corresponds to the necessary condition for f to have an antiholomorphic pth root. Therefore, the maximal order of antiholomorphic root of f is at least p. To see that it is exactly p, we simply note that if $f_1$ is an antiholomorphic root of order $q \geq p$ , then $f_1\circ f_1$ is a holomorphic root of order q of $f\circ f$ and, because p is maximal, we must have $p=q$ . We must also prove that antiholomorphic fractional iterates of $f\circ f$ are fractional iterates of f. Suppose that $f\circ f$ has an antiholomorphic root of order q, where q must be even. Then we have ${q/2}\,|\, p$ . Since p is odd, we know that $q/2$ must be odd. It follows that $(\Sigma \circ T_{\scriptscriptstyle 1/q})^{\circ {q/2}} = \Sigma \circ {T_{\scriptscriptstyle {1/2}}}$ .

If p is even, then we have $\overline {c_{n,j}} = c_{-n,-j}$ , which corresponds to the condition necessary for $\Sigma $ to be compatible with the $\Psi _j$ . In this case, the centralizer of f contains a Schwarz reflection, and it follows that the highest order of antiholomorphic fractional iterates of f and the highest order of holomorphic fractional iterates of $f\circ f$ coincide.