1.1 Germs and morphisms

Let $\mathbb {K}$ be an integral domain. We let $\widehat {\mathrm {Diff}}(\mathbb {K},0)$ be the group of formal power series of the form

$$ \begin{align*}a_1 x + \sum_{n\geq 2} a_n x^n,\end{align*} $$

where $a_n \in \mathbb {K}$ and $a_1 \in \mathbb {K} \setminus \{0\}$ is invertible. The group operation is that of composition of series, which identifies to that of substitution of variables. In the case where $\mathbb {K}$ is a field, it also identifies to the group of formal germs of diffeomorphisms at the origin.

For $\ell \geq 1$ , we consider the subgroup $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0)$ of $\widehat {\mathrm {Diff}}(\mathbb {K},0)$ formed by the germs that are $\ell $ -tangent to the identity, that is, that are of the form

$$ \begin{align*}x + \sum_{n \geq \ell + 1} a_n x^n.\end{align*} $$

On $\widehat {\mathrm {Diff}}(\mathbb {K},0)$ , there is an obvious homomorphism $\Phi _0$ into the multiplicative subgroup of the invertible elements of $\mathbb {K} \setminus \{0\}$ given by

$$ \begin{align*}\Phi_0 : \sum_{n\geq 1} a_n x^n \mapsto a_1.\end{align*} $$

Similarly, on each subgroup $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0)$ , there is an obvious homomorphism $\Phi _{\ell }$ into the additive group $\mathbb {K}$ , namely,

$$ \begin{align*}\Phi_\ell : x + \sum_{n \geq \ell + 1} a_n x^n \mapsto a_{\ell + 1} .\end{align*} $$

Actually, $\Phi _{\ell }$ is the first of $\ell $ homomorphisms $\Phi _{\ell ,1}, \Phi _{\ell ,2} \ldots , \Phi _{\ell ,\ell }$ defined on $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0)$ by

$$ \begin{align*}\Phi_{\ell,i}: x + \sum_{n \geq \ell + 1} a_n x^n \mapsto a_{\ell+i}\end{align*} $$

(checking that these are homomorphisms is straightforward; see also (3) below).

Less trivially, there is another homomorphism from $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0)$ into $\mathbb {K}$ , namely

(1) $$ \begin{align} \overline{\mathrm{Resad}}_{\ell} \!: x + \sum_{n \geq \ell + 1} a_n x^n \mapsto (\ell + 1) \, a_{\ell + 1}^2 - 2 a_{2\ell + 1}. \end{align} $$

Whenever it makes sense (for instance, when $\mathbb {K}$ is a zero-characteristic field), we let

(2) $$ \begin{align} \mathrm{Resad}_{\ell} \!: x + \sum_{n \geq \ell + 1} a_n x^n \mapsto \frac{\ell + 1}{2} \, a_{\ell + 1}^2 - a_{2\ell + 1}. \end{align} $$

We call both $\overline {\mathrm {Resad}}_{\ell }$ and $\mathrm {Resad}_{\ell }$ the additive residues at order $\ell + 1$ . Checking that they yield group homomorphisms is straightforward: If

$$ \begin{align*}f (x) = x + \sum_{n \geq \ell + 1} a_n x^n \quad \mbox{ and } \quad g (x) = x + \sum_{n \geq \ell + 1} a_n' x^n,\end{align*} $$

then $fg(x)$ equals

$$ \begin{align*}\left[ x + \sum_{m \geq \ell + 1} a_m' x^m \right] + \sum_{n \geq \ell + 1} a_n \left[ x + \sum_{m \geq \ell + 1} a_m' x^m \right]^n,\end{align*} $$

that is,

$$ \begin{align*}x + \sum_{m = \ell + 1}^{2\ell + 1} a_m' x^m + \sum_{n = \ell + 1}^{2\ell + 1} a_n x^n + a_{\ell + 1} (\ell + 1) x^{\ell} a_{\ell + 1}' x^{\ell + 1} + T_{2\ell+2},\end{align*} $$

where $T_{2\ell +2}$ stands for a formal power series in x all of whose terms have order at least $2\ell +2$ . Therefore,

(3) $$ \begin{align} fg (x) = x + \sum_{n=\ell + 1}^{2 \ell} [a_n+a_n'] x^n + [a_{2\ell + 1} + a'_{2\ell + 1} + (\ell + 1) a_{\ell + 1} a_{\ell + 1}'] x^{2\ell + 1} + T_{2\ell+2}, \end{align} $$

and

$$ \begin{align*}&\overline{\mathrm{Resad}}_{\ell} (fg) \\&\quad= (\ell + 1) \, [a_{\ell + 1} + a_{\ell + 1}']^2 - 2 \, [a_{2\ell + 1} + a'_{2\ell + 1} + (\ell + 1) a_{\ell + 1} a_{\ell + 1}'] = \overline{\mathrm{Resad}}_{\ell} (f) + \overline{\mathrm{Resad}}_{\ell} (g). \end{align*} $$

A basis for the cohomology. In the case where $\mathbb {K}$ is the field of real numbers, Fukui proved in [Reference Fukui5] that the $\Phi _{\ell ,i}$ together with $\mathrm {Resad}_{\ell }$ are the generators of the continuous cohomology group of $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0)$ . In [Reference Babenko and Bogatyi1], Babenko and Bogatyi treated the case $\mathbb {K} = \mathbb {Z}$ . In particular, they computed the continuous cohomology $H^1 (\widehat {\mathrm {Diff}}_1 (\mathbb {Z},0))$ : Besides

$$ \begin{align*}\Phi_{1}: x + \sum_{n \geq 2} a_n x^n \mapsto a_2\end{align*} $$

and

$$ \begin{align*}\mathrm{Resad}_1: x + \sum_{n \geq 2} a_n x^n \mapsto a_2^2 - a_3,\end{align*} $$

there are two homomorphisms into $\mathbb {Z}_2$ , namely

$$ \begin{align*}x + \sum_{n \geq 2} a_n x^n \mapsto \frac{a_2 \, ( 1 + a_2 )}{2} + a_3 + a_4 \quad (\mbox{mod } 2)\end{align*} $$

and

$$ \begin{align*}x + \sum_{n \geq 2} a_n x^n \mapsto a_2 a_4 + a_4 + a_6 \quad (\mbox{mod } 2).\end{align*} $$

In [Reference Bogatayaa and Bogatyi2], Bogatayaa and Bogatyi do similar computations for $\mathbb {K} = \mathbb {Z}_p$ , with p a prime number. They show that, for $p \neq 2$ , only $\Phi _{1}$ and $\mathrm {Resad}_1$ survive, but for $p=2$ , the situation is slightly more complicated. See also [Reference Pavez17], where among other things it is shown that $H^1 ( \widehat {\mathrm {Diff}}_{\ell }(\mathbb {Z},0))$ is finitely generated for all values of $\ell $ larger than $1$ .

Genuine diffeomorphisms. For $\mathbb {K}$ being the fields of the real, complex or p-adic numbers, we let $\mathrm {Diff}^{\omega } (\mathbb {K},0)$ be the group of genuine germs of analytic diffeomorphisms fixing the origin, and $\mathrm {Diff}^{\omega }_{\ell } (\mathbb {K},0)$ the subgroup of those elements that are $\ell $ -tangent to the identity. These are subgroups of the corresponding groups $\widehat {\mathrm {Diff}}(\mathbb {K},0)$ and $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0)$ , so the homomorphisms $\Phi _{\ell ,i}$ and $\mathrm {Resad}_{\ell }$ restrict to them.

More interestingly, each of the homomorphisms $\Phi _{\ell ,i}$ and $\mathrm {Resad}_{\ell }$ above involve only finitely many derivatives at the origin (namely, $\ell + i$ and $2\ell + 1$ , respectively). Their definitions hence extend to the subgroup $\mathrm {Diff}^{\infty }_{\ell } (\mathbb {K},0)$ of the group $\mathrm {Diff}^{\infty } (\mathbb {K},0)$ of germs of $C^{\infty }$ diffeomorphisms fixing the origin made of the elements that are $\ell $ -tangent to the identity. Actually, they even extend to the larger group of germs of $C^{2\ell + 1}$ diffeomorphisms that are $\ell $ -tangent to the identity. In this framework, it is a nice exercise to check the additive property of $\mathrm {Resad}_{\ell }$ just by using the Faà di Bruno formula. We will come back to this point in §1.3.

Fukui’s theorem stated above still holds in $\mathrm {Diff}^{\infty }_+ (\mathbb {R},0)$ (where $+$ stands for orientation-preserving maps). We do not know whether a complex or p-adic version of this is also valid.

1.2 Res-it-ad

Elements in $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0) \setminus \widehat {\mathrm {Diff}}_{\ell + 1}(\mathbb {K},0)$ are those whose order of contact to the identity at the origin equals $\ell $ . For simplicity, they will be said to be exactly $\ell $ -tangent to the identity. These correspond to series expansions of the form

$$ \begin{align*}x + \sum_{n \geq \ell + 1} a_n x^n, \quad \mbox{ with } \,\, a_{\ell + 1} \neq 0.\end{align*} $$

A well-known lemma gives a normal form for these elements.

Lemma 1.1. If $\mathbb {K}$ is a zero-characteristic field having a square root of $-1$ , then every $f~\in ~\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0) \setminus \widehat {\mathrm {Diff}}_{\ell + 1}(\mathbb {K},0)$ is conjugate in $\widehat {\mathrm {Diff}} (\mathbb {K},0)$ to a (unique) germ of the (normal) form

(4) $$ \begin{align} x \mapsto x + x^{\ell + 1} + \mu x^{2\ell + 1}. \end{align} $$

An explicit argument showing this in the complex case that applies with no modification to the present case appears in [Reference Voronin22]. The value of $\mu $ is called the residue of f and denoted $\mathrm {Res} (f)$ . As it is very well known (and easy to check), this is invariant under conjugacy. Actually, if f with normal form (4) is conjugate in $\widehat {\mathrm {Diff}}(\mathbb {K},0)$ to a germ of the (reduced) form

$$ \begin{align*}x + x^{\ell + 1} + \sum_{n \geq 2\ell + 1} a_n x^{n},\end{align*} $$

then necessarily $a_{2\ell + 1} = \mathrm {Res} (f)$ .

The residue was introduced as a conjugacy invariant in the complex setting: For a germ of a holomorphic map fixing the origin, one defines

$$ \begin{align*}\mathrm{R}(f) := \frac{1}{2 \pi i} \int_{\gamma} \frac{dz}{z - f(z)},\end{align*} $$

where $\gamma $ is a small, simple and positively oriented loop around the origin. For germs of the form

$$ \begin{align*}f (z) = z + z^{\ell + 1} + \sum_{n \geq 2\ell + 1} a_n z^n,\end{align*} $$

one easily checks that $\mathrm {R}(f) = \mathrm {Res} (f) = a_{2\ell +1}$ .

The iterative residue of $f \in \widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0) \setminus \widehat {\mathrm {Diff}}_{\ell + 1}(\mathbb {K},0)$ , denoted $\mathrm {Resit} (f)$ , is simply the value

$$ \begin{align*}\frac{\ell + 1}{2} - \mathrm{Res} (f).\end{align*} $$

This satisfies the fundamental relation

(5) $$ \begin{align} \mathrm{Resit} (f^n) = \frac{\mathrm{Resit}(f)}{|n|} \end{align} $$

for every integer $n \neq 0$ . It was introduced by Écalle in the context of germs of holomorphic diffeomorphisms [Reference Écalle6]. A proof of the formula above in this setting (due to Écalle) appears for instance in [Reference Milnor13]. Below, we give an algebraic proof that works in a much broader context (in particular, it covers the case of germs of diffeomorphisms of the real line with finite regularity discussed just after Lemma 1.3). As we will see, it just follows from the additive properties of $\Phi _{\ell }$ and $\mathrm {Resad}_{\ell }$ .

Start with a germ of the form

$$ \begin{align*}f(x) = x + x^{\ell + 1} + \mu x^{2\ell + 1} + \sum_{n \geq 2\ell+2} a_n x^n.\end{align*} $$

Using the homomorphisms $\Phi _{\ell }$ and $\mathrm {Resad}_{\ell }$ , we obtain

$$ \begin{align*}f^n (x) = x + n x^{\ell + 1} + \mu_n x^{2\ell + 1} + T_{2\ell+2},\end{align*} $$

with

$$ \begin{align*}\frac{(\ell + 1) \, n^2}{2} - \mu_n = \mathrm{Resad}_{\ell} (f^n) = n \, \mathrm{Resad}_{\ell} (f) = n \, \left[ \frac{(\ell + 1)}{2} - \mu \right] = n \, \mathrm{Resit}(f).\end{align*} $$

Assume $n> 0$ . If we conjugate by $H_{\lambda _n} \! : x \mapsto \lambda x$ with $\lambda _n := \sqrt [\ell ]{n}$ , we obtain

$$ \begin{align*}H_{\lambda_n} f^n H_{\lambda_n}^{-1} (x) = x + x^{\ell + 1} + \frac{\mu_n}{n^2} x^{2\ell + 1} + T_{2\ell+2}'.\end{align*} $$

By definition, this implies that

$$ \begin{align*}\mathrm{Resit} (f^n) = \frac{\ell + 1}{2} - \frac{\mu_n}{n^2} = \frac{1}{n^2} \left[ \frac{(\ell + 1) \, n^2}{2} - \mu_n \right] = \frac{ n \, \mathrm{Resit} (f) }{n^2} = \frac{\mathrm{Resit}(f)}{n}\end{align*} $$

as announced. Computations for $n < 0$ are analogous.

Example 1.2. It follows from (the proof of) Lemma 1.1 (see also Lemma 1.3 below) that, for $f \in \widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0) \setminus \widehat {\mathrm {Diff}}_{\ell + 1}(\mathbb {K},0)$ of the form

$$ \begin{align*}f(x) = x + \sum_{n \geq \ell + 1} a_n \, x^n, \qquad a_{\ell + 1} \neq 0,\end{align*} $$

the value of $\mathrm {Resit} (f)$ is a polynomial function of $a_{\ell + 1},a_{\ell +2}, \ldots , a_{2\ell + 1}$ times some integer power of $a_{\ell + 1}$ . For instance, the reader may readily check that, for $a \neq 0$ ,

$$ \begin{align*}\mathrm{Resit} (x + ax^2+bx^3+ \ldots) = \frac{a^2 - b}{a^2},\end{align*} $$

$$ \begin{align*}\mathrm{Resit} (a+ax^3+bx^4+cx^5+\ldots ) = \frac{3a^3 - b^2 - ac}{2a^3}.\end{align*} $$

See [Reference Nordqvist and Rivera-Letelier16, Theorem 1] for a general result in this direction.

The residues in the real case. Much of the discussion above still makes sense for germs of $C^{2\ell + 1}$ diffeomorphisms. In particular, the morphisms $\Phi _{\ell ,i}$ and $\mathrm {Resad}_{\ell }$ extend to the subgroup $\mathrm {Diff}^{2\ell + 1}_{\ell } (\mathbb {R},0)$ made of those germs that are $\ell $ -tangent to the identity. Moreover, the definition of $\mathrm {Resit}$ still extends to the elements of this subgroup that are exactly $\ell $ -tangent to the identity at the origin because of the next very well-known lemma. (Compare Lemma 1.1.) For the statement, recall that a germ f of diffeomorphisms of the real line fixing the origin is (topologically) contracting (resp. expanding) if $f(x) < x$ (resp. $f(x)>x$ ) holds for all small-enough $x>0$ . In all that follows, the little-o-notation $o (\cdot )$ will refer to values of the involved variable that are positive but very close to zero.

Lemma 1.3. Let f be the germ of a $C^{2\ell + 1}$ diffeomorphism of the real line that is exactly $\ell $ -tangent to the identity at the origin. If f is expanding (resp. contracting), then there exists a germ of polynomial diffeomorphism h such that the conjugate germ $hfh^{-1}$ has a Taylor series expansion at the origin of the (reduced) form

$$ \begin{align*}hfh^{-1} (x) = x \pm x^{\ell + 1} + \mu x^{2\ell + 1} + o (x^{2\ell + 1}),\end{align*} $$

where the sign is positive (resp. negative) in the expanding (resp. contracting) case. Moreover, $\mu $ is uniquely determined up to conjugacy by a germ of $C^{2\ell + 1}$ diffeomorphism.

Proof. By assumption, f has a Taylor series expansion at the origin of the form

$$ \begin{align*}f(x) = x + \sum_{n=\ell + 1}^{2\ell + 1} a_n x^n + o (x^{2\ell + 1}), \quad \mbox{with } a_{\ell + 1} \neq 0.\end{align*} $$

By conjugating f by a homothety, we can arrange that $a_{\ell + 1} = 1$ (resp. $a_{\ell + 1} = -1$ ) in the expanding (resp. contracting) case. Now, by conjugating by a germ of the form $x+\alpha x^2$ , we can make $a_{\ell +2}$ vanish without changing $a_{\ell + 1} = \pm 1$ . Next, we conjugate by a germ of the form $x+\alpha x^3$ to make $a_{\ell +3}$ vanish without changing $a_{\ell + 1} = \pm 1$ and $a_{\ell +2}=0$ . One continues this way up to conjugating by a germ of the form $x+\alpha x^{\ell }$ , which allows killing $a_{2\ell }$ . We leave it to the reader to fill in the details as well as the proof of the uniqueness of $\mu $ (see also [Reference Voronin22]).

In any case above, we let

$$ \begin{align*}\mathrm{Res} (f) := \mu \qquad \mbox{ and } \qquad \mathrm{Resit} (f) = \frac{\ell + 1}{2} - \mu.\end{align*} $$

One readily checks that this definition is coherent with that of the complex analytic case.

1.3 Schwarzian derivatives and $\mathbf {Resad}_{\boldsymbol{\ell} }$

For a germ of a genuine diffeomorphism $f \in \mathrm {Diff}^{\infty }(\mathbb {K},0)$ (where $\mathbb {K}$ is, for example, the field of real numbers), one has

$$ \begin{align*}a_n = \frac{D^n f (0)}{n!}.\end{align*} $$

Thus, if $Df (0) = 1$ , then

$$ \begin{align*}\mathrm{Resad}_1 (f) = a_2^2 - a_3 = \frac{(D^2 f (0))^2}{4} - \frac{D^3f(0)}{6} = \frac{1}{6} \left[ \frac{3}{2} \left( \frac{D^2 f(0)}{Df(0)} \right)^2 - \frac{D^3f (0)}{Df(0)} \right] = \frac{Sf (0)}{6},\end{align*} $$

where $Sf$ denotes the Schwarzian derivative. That $Sf$ appears as a group homomorphism is not surprising. Indeed, in a general setting, it satisfies the cocycle identity

(6) $$ \begin{align} S (fg) = S(g) + S (f) \circ g \cdot Dg^2, \end{align} $$

and restricted to parabolic germs, this becomes

$$ \begin{align*}S(fg)(0) = Sg (0) + Sf (0).\end{align*} $$

In this view, $\mathrm {Resad}_{\ell }$ may be seen as a kind of ‘Schwarzian derivative of higher order’ for parabolic germs. Now, for $f \in \mathrm {Diff}_{\ell }^{\infty } (\mathbb {K},0)$ , we have

$$ \begin{align*}&\mathrm{Resad}_{\ell} (f) \\&\quad= \frac{\ell + 1}{2} \left( \frac{D^i f (0)}{(\ell + 1)\,!} \right)^2 - \frac{D^{2\ell + 1}f(0)}{(2\ell + 1)\,!} = \frac{1}{(2\ell + 1)\, !} \left[ {2\ell + 1 \choose \ell} \frac{(D^{\ell + 1} f(0))^2}{2} - D^{2\ell + 1} f (0) \right] \! .\end{align*} $$

Thus, letting

$$ \begin{align*}S_{\ell + 1} (f) := {2\ell + 1 \choose \ell} \frac{(D^{\ell + 1} f(0))^2}{2} - D^{2\ell + 1} f (0) = (2\ell+1)! \, \mathrm{Resad}_{\ell}(f),\end{align*} $$

we have $S_1 (f) = S(f)$ .

In a slightly broader context, if $f,g$ are two germs of $C^{2\ell + 1}$ diffeomorphisms with all derivatives $D^2, D^3, \ldots ,D^{\ell }$ vanishing at the origin, then, using the Faà di Bruno formula, it is straightforward to check the cocycle relation below:

$$ \begin{align*}S_{\ell} (fg) = S_{\ell} (g) + S_{\ell} (f) \cdot (Dg )^{2\ell}.\end{align*} $$

This is an extension of the additive property of $\mathrm {Resad}_{\ell }$ to the framework of nonparabolic germs.

Remark 1.5. A classical theorem of Szekeres establishes that every expanding or contracting germ of $C^2$ diffeomorphism of the real line is the time-1 map of a flow of germs of $C^1$ diffeomorphisms; moreover, a famous lemma of Kopell establishes that its centralizer in the group of germs of $C^1$ diffeomorphisms coincides with this flow (see [Reference Navas15, chapter 4]). Furthermore, it follows from the work of Yoccoz [Reference Yoccoz23] that, for nonflat germs of class $C^k$ , the flow is made of $C^{k-1}$ diffeomorphisms. Therefore, to each $f \in \mathrm {Diff}^{2\ell + 2}_{\ell } (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 2}_{\ell +1} (\mathbb {R},0)$ we may associate its flow $(f^t)$ made of $C^{2\ell +1}$ germs of diffeomorphisms. These are easily seen to be exactly $\ell $ -flat. Moreover, equality (5) has a natural extension in this setting, namely, for all $t \neq 0$ ,

(7) $$ \begin{align} \mathrm{Resit}(f^t) = \frac{\mathrm{Resit}(f)}{|t|}. \end{align} $$

See Remark 3.1 for a proof.

Remark 1.6. In the real or complex setting, if

$$ \begin{align*}f(x) = x + ax^{\ell + 1} + bx^{2\ell + 1} + \ldots,\end{align*} $$

then, letting $h (x) = x^{\ell }$ , one readily checks that

(8) $$ \begin{align} h f h^{-1} (x) = x + \ell \, a \, x^2 + \left[ \ell \, b + \frac{ \ell \, (\ell-1)}{2} a^2 \right] x^3 + \ldots. \end{align} $$

Moreover, using Lemmas 4.4 and 4.5 below, one can show that $hfh^{-1}$ is of class $C^3$ if f is of class $C^{2\ell +1}$ (yet this is not crucial to define residues, according to Remark 1.4). Now, Equation (8) easily yields

$$ \begin{align*}\mathrm{Resad}_1 (h f h^{-1}) = \ell \,\, \mathrm{Resad}_{\ell} (f) \qquad \mbox{ and } \qquad \mathrm{Resit} (h f h^{-1}) = \frac{\mathrm{Resit}(f)}{\ell}.\end{align*} $$

The value of $\mathrm {Resad}_{\ell }$ hence equals (up to a constant) that of $\mathrm {Resad}_1$ of the conjugate, and the same holds for $\mathrm {Resit}$ . Since $\mathrm {Resad}_1$ is nothing but a sixth of the Schwarzian derivative, this gives even more insight on $\mathrm {Resad}_{\ell }$ as a generalization of the Schwarzian derivative. Compare [Reference Burslem and Wilkinson3], where the Schwarzian derivative naturally arises in the study of conjugacy classes of germs of real-analytic diffeomorphisms.

2.1 A fundamental example: (non-)invariance of residues

Let us consider the germs of

$$ \begin{align*}f(x) = x - x^2 \qquad \mbox{ and } \qquad \, \, g(x) = \frac{x}{1+x} = x - x^2 +x^3 - x^4 + \ldots\end{align*} $$

These have different residues:

$$ \begin{align*}\mathrm{Resit} (f) = \mathrm{Resad}_1 (f) = 1 \neq 0 = \mathrm{Resad}_1 (g) = \mathrm{Resit} (g).\end{align*} $$

Using that $\Phi _{1}(f) = \Phi _{1} (g)$ , one easily concludes that any $C^2$ conjugacy between f and g has to be parabolic (see the first proof below). Now, because of the invariance of $\mathrm {Resad}_1$ (equivalently, of the Schwarzian derivative) under conjugacies by parabolic germs of $C^3$ diffeomorphisms, we conclude that no $C^3$ conjugacy between them can exist. However, Theorems A and B above imply that, actually, these germs are not $C^2$ conjugate, though they are $C^1$ conjugate. Below, we elaborate on these two claims. We do this in two different ways: the former by directly looking at the conjugacy relation and the latter by looking at the associated vector fields.

Proof Sketch of proof that $f,\,g$ are not $C^2$ conjugate.

Assume that h is a germ of $C^2$ diffeomorphism conjugating f and g, that is, $hfh^{-1} = g$ . Writing $h(x) = \lambda \, x + a x^2 + o(x^2)$ for $\lambda = Dh (0) \neq 0$ and $a = D^2h(0)/2$ , equality $hf = gh$ translates to

$$ \begin{align*}\lambda \, f (x) + a \, f(x)^2 + o (x^2) = h(x)- h(x)^2+o(x^2).\end{align*} $$

Thus,

$$ \begin{align*}\lambda \, (x-x^2) + a \, (x-x^2)^2 + o(x^2) = (\lambda x + ax^2) - (\lambda x +ax^2)^2 + o(x^2).\end{align*} $$

By identifying the coefficients of $x^2$ , this yields

$$ \begin{align*}- \lambda + a = a - \lambda ^2,\end{align*} $$

hence $\lambda = \lambda ^2$ and, therefore, $\lambda = 1$ .

Now, knowing that $Dh(0)=1$ , we deduce that there exists $C> 0$ such that

(9) $$ \begin{align} | h(x) - x | \leq C \, x^2 \end{align} $$

for small-enough $x> 0$ . Now, notice that $g^n (x) = \frac {x}{1+nx}.$ In particular, given any fixed $x_0'> 0$ ,

(10) $$ \begin{align} g^n (x_0') \geq \frac{1}{n+D} \end{align} $$

for a very large $D> 0$ (namely, for $D \geq 1/x_0'$ ) and all $n \geq 1$ . In what concerns f, a straightforward induction argument (that we leave to the reader; see also Proposition 3.2) shows that, for all $x_0> 0$ , there exists a (very small) constant $D'> 0$ such that, for all $n \geq 1$ ,

(11) $$ \begin{align} f^n (x_0) \leq \frac{1}{n + D' \log(n)}. \end{align} $$

Now, fix $x_0> 0$ , and let $x_0' := h (x_0)$ . The equality $hf^n (x_0) = g^n h(x_0)$ then yields

$$ \begin{align*}hf^n(x_0) - f^n (x_0) = g^n (x_0') - f^n (x_0).\end{align*} $$

Using (9), (10) and (11), we obtain

$$ \begin{align*}\frac{C}{(n+D'\log(n))^2} \geq C (f^n(x_0))^2 \geq \frac{1}{n+D} - \frac{1}{n + D' \log (n)} = \frac{D' \log (n) - D}{ (n+D) (n + D' \log (n))},\end{align*} $$

which is impossible for a large-enough n.

Proof Sketch of proof that $f,\,g$ are $C^1$ conjugate.

The existence of a $C^1$ conjugacy between diffeomorphisms as f and g above is folklore (see, for instance, [Reference O’Farrell and Roginskaya9, Reference Firmo10]). An argument that goes back to Szekeres [Reference Szekeres20] and Sergeraert [Reference Sergeraert19] proceeds as follows: The equality $hfh^{-1} = g$ implies $hf^nh^{-1} = g^n$ for all $n \geq 1$ , hence $h = g^{-n} h f^n$ . Thus, if h is of class $C^1$ , then

(12) $$ \begin{align} Dh (x) = \frac{Df^n (x)}{Dg^n (g^{-n} h f^n (x))} Dh (f^n (x)) = \frac{Df^n (x)}{Dg^n (h(x))} Dh (f^n (x)). \end{align} $$

It turns out that the sequence of functions

$$ \begin{align*}(x,y) \to A_n (x,y) := \frac{Df^n (x)}{Dg^n (y)}\end{align*} $$

is somewhat well behaved. In particular, it converges to a continuous function A away from the origin. Since $Dh(0)=1$ (see the proof above), Equation (12) translates to

(13) $$ \begin{align} Dh (x) = A (x,h(x)). \end{align} $$

This is an ordinary differential equation in h, a careful analysis of which shows that it has a solution. This allows one to obtain the conjugacy between f and g. See [Reference Firmo10] for the details.

It is worth stressing that both proofs above used not just the conjugacy relation but an iterated version of it:

$$ \begin{align*}hfh^{-1} = g \quad \Rightarrow \quad hf^nh^{-1} = g^n.\end{align*} $$

In this regard, it may be clarifying for the reader to look for direct proofs just using the conjugacy relation in order to detect where things get stuck. The moral is that one really needs to use the underlying dynamics. Now, there are objects that encode such dynamics, namely, the vector fields whose time-1 maps correspond to the given diffeomorphisms. These were proved to exist by Szekeres [Reference Szekeres20] and Sergeraert [Reference Sergeraert19] in a much broader context. Their properties were studied, among others, by Takens in the nonflat $C^{\infty }$ case [Reference Takens21] and later by Yoccoz in finite regularity [Reference Yoccoz23]. We next illustrate how to use them with the example above.

Proof Proof that $f,\,g$ are not $C^2$ conjugate using vector fields.

One proof for this particular case that is close to the preceding one works as follows: Let $X,Y$ be, respectively, the (unique) $C^1$ vector fields associated to $f,g$ (according to Takens and Yoccoz, these must be of class $C^{\infty }$ ; actually, the expression for Y is explicit, as shown below). For x close to 0, one has (see §3.1 for X):

$$ \begin{align*}X (x) = -x^2 - x^3 + o(x^3), \qquad Y(x) = -x^2.\end{align*} $$

Extend $X,\,Y$ to $\mathbb {R}_+ := [0, \infty )$ so that their corresponding flows $(f^t)$ and $(g^t)$ are made of $C^{\infty }$ diffeomorphisms of $\mathbb {R}_+$ and are globally contracting. Fix $x_0> 0$ . The map $\tau _X \!: x\mapsto \int _{x_0}^x\frac 1X$ defines a $C^{\infty }$ diffeomorphism from $\mathbb {R}_+^* := \, (0,\infty )$ to $\mathbb {R}$ satisfying $\tau _X(f^t(x_0))=t$ by definition of the flow so that $\tau _X^{-1}$ is the map $t\mapsto f^t(x_0)$ . Obviously, one has a similar construction for Y and $(g^t)$ . (We refer to [Reference Eynard-Bontemps and Navas8] for further details of this construction.) In particular, using the equalities

$$ \begin{align*}\int_{x_0}^{f^n (x_0)} \frac{dy}{X(y)} \,\, = \tau_X (f^n(x_0)) \,\, = \,\, n \,\, = \,\, \tau_Y (g^n(x_0)) \,\, = \,\, \int_{x_0}^{g^n (x_0)} \frac{dy}{Y(y)},\end{align*} $$

one gets equations for the values of $f^n(x_0), g^n(x_0)$ that easily yield the estimates (10) and (11). Then the very same arguments as in the preceding proof allow to conclude.

Another (more direct) argument of proof works as follows. If h is a $C^2$ diffeomorphism that conjugates f and g, then it must conjugate the associated flows (this is a consequence of the famous Kopell lemma; see, for instance, [Reference Eynard-Bontemps and Navas8, Proposition 2.2]). This means that $X \cdot Dh = Y \circ h.$ Writing $h (x) = \lambda x + a x^2 + o(x^2)$ , with $\lambda \neq 0$ , this gives

$$ \begin{align*}(-x^2 - x^3 + o(x^3)) \cdot (\lambda + 2ax + o(x)) = -h(x)^2 + o(x^3) = - (\lambda x + a x^2 + o(x^2))^2 + o(x^3).\end{align*} $$

By identifying the coefficients of $x^2$ above, we obtain $- \lambda = - \lambda ^2$ , hence $\lambda = 1$ . Next, by identifying the coefficients of $x^3$ , we obtain

$$ \begin{align*}-2a-1 = -2a - \lambda = -2a \lambda = - 2a,\end{align*} $$

which is absurd.

In all what follows, whenever $u,v$ are functions of a variable x which are nonzero for $x> 0$ , we will write $u \sim v$ if they satisfy $\lim _{x \to 0} \frac { u(x) }{ v(x) } = 1$ .

Remark 2.3. Proposition 2.1 gives a very simple criterion of $C^1$ conjugacy for contracting smooth vector fields X and Y which are neither hyperbolic nor infinitely flat at $0$ (or for $C^k$ vector fields which are not k-flat). This criterion is simply to have the same order of flatness at $0$ and the same ‘sign’ or equivalently to satisfy that $X/Y$ has a positive limit at $0$ (though the statement above is much more general since it does not even require the vector fields to be $C^1$ !).

We do not know whether this condition remains sufficient if one considers infinitely flat vector fields. Nevertheless, in this context it is not necessary. For example, $X(x):=e^{-\frac 1{x}}$ and $Y(x):=\frac 12 e^{-\frac 1{2x}}$ are conjugate by a homothety of ratio $2$ , but $Y(x) = \frac 12\sqrt {X(x)}$ , so $X/Y$ goes to $0$ . It is not necessary even if one restricts to conjugacies by germs of parabolic $C^1$ diffeomorphisms. For example, consider the vector field $X(x):= e^{-\frac 1{x^2}}$ and the local (analytical) diffeomorphism $h(x):=\frac {x}{1+x}$ . Then $Y=h^*X$ satisfies

$$ \begin{align*}Y(x) = \frac{(X\circ h)(x)}{Dh(x)}\sim e^{-\frac{(1+x)^2}{x^2}} = X(x) \, e^{-\frac2x-1},\end{align*} $$

so $X/Y$ goes to $+\infty $ .

Proof Proof of the ‘necessity’ in Proposition 2.1.

If h is a germ of $C^1$ diffeomorphisms conjugating X to Y, then, for $x>0$ close to the origin,

$$ \begin{align*}Dh(x) = \frac{(Y\circ h)(x)}{X(x)}\sim \frac{\beta \, (h(x))^s}{\alpha x^r}\sim \frac{\beta \, (Dh(0))^s}{\alpha} x^{s-r}.\end{align*} $$

The last expression must have a nonzero limit at $0$ , which forces $r = s$ . Observe that this part of the argument works for any values of r and s. Moreover, if $r=s=1$ , then, since $Dh(x)$ must tend to $Dh(0)\neq 0$ , we have in addition $\alpha = \beta $ .

Proof Proof of the ‘sufficiency’ in Proposition 2.1.

Again, extend $X,Y$ to $\mathbb {R}_+ = [0, \infty )$ so that they do not vanish outside the origin and their corresponding flows $(f^t)$ and $(g^t)$ are made of $C^1$ diffeomorphisms of $\mathbb {R}_+$ . Multiply them by $-1$ in case they are expanding, fix $x_0> 0$ and consider the maps $\tau _X \!: x \mapsto \int _{x_0}^x\frac 1X$ and $\tau _Y \!: x \mapsto \int _{x_0}^x \frac {1}{Y}$ . The $C^1$ diffeomorphism $h := \tau _Y^{-1}\circ \tau _X$ of $\mathbb {R}_+^* \!= \, (0,\infty )$ conjugates $(f^t)$ to $(g^t)$ and thus sends X to Y. Let us check that, under the assumption $r=s>1$ , the map h extends to a $C^1$ diffeomorphism of $\mathbb {R}_+$ . For $x>0$ near $0$ , one has

$$ \begin{align*}Dh(x) = \frac{(Y\circ h)(x)}{X(x)}\sim \frac{\beta}{\alpha}\left(\frac{h(x)}{x}\right)^{\! r},\end{align*} $$

so it suffices to show that $\frac {h(x)}{x}$ has a limit at $0$ . To do this, first observe that the improper integral $\int _{x_0}^x\frac {dy}{y^r}$ diverges when x goes to $0$ , so

$$ \begin{align*}\tau_X(x) = \int_{x_0}^x\frac1X\sim \int_{x_0}^x \frac{dy}{\alpha y^r}\sim -\alpha' x^{1-r}\end{align*} $$

for some constant $\alpha '>0$ . Similarly, $\tau _Y(x)\sim -\beta ' x^{1-r}$ . It follows that when t goes to $\infty $ , one has $t = \tau _Y(\tau _Y^{-1}(t))\sim -\beta ' (\tau _Y^{-1}(t))^{1-r}$ so that

(14) $$ \begin{align} \tau_Y^{-1}(t)\sim -\beta" t^{\frac1{1-r}}. \end{align} $$

Finally,

$$ \begin{align*}h(x) = \tau_Y^{-1}\tau_X(x)\sim \beta" (\tau_X(x))^{\frac1{1-r}}\sim \beta"' x\end{align*} $$

for some new constant $\beta "'$ , which concludes the proof.

Remark 2.4. The inoffensive relation (14) above is a key step. It does not work for $r=s=1$ . In this case, $x^{1-r}$ must be replaced by $\varphi (x) := \log (x)$ , but the function $\varphi $ does not satisfy

$$ \begin{align*}u(t)\sim v(t) \quad \Longleftrightarrow \quad \varphi(u(t))\sim \varphi(v(t)),\end{align*} $$

whereas the power functions do (as it was indeed used in the line just following Equation (14))!

So far we have discussed the $C^1$ conjugacy between elements in $\mathrm {Diff}^1_+ (\mathbb {R},0)$ that are exactly $\ell $ -tangent to the identity for the same value of $\ell $ via two different methods: One of them is direct, and the other one is based on associated vector fields. It turns out, however, that these approaches are somehow equivalent. Indeed, according to Sergeraert [Reference Sergeraert19], the vector field associated to a contracting germ f of class $C^2$ has an explicit iterative formula:

$$ \begin{align*}X (x) = \lim_{n \to \infty} (f^n)^* (f-id) ( x ) = \lim_{n \to \infty} \frac{f^{n+1}(x)- f^n (x)}{Df^n (x)}\end{align*} $$

(we refer to [Reference Eynard-Bontemps and Navas7] for the details and extensions of this construction). Now, if h is a $C^1$ diffeomorphism that conjugates f to another contracting germ g of $C^2$ diffeomorphism, then, as already mentioned, it must send X to the vector field Y associated to g, that is, $X \cdot Dh = Y \circ h$ . Using the iterative formula above, we obtain

$$ \begin{align*}Dh (x) = \lim_{n \to \infty} \frac{Df^n (x)}{f^{n+1}(x)-f^n(x)} \cdot \frac{g^{n+1}(h(x)) - g^n (h(x))}{Dg^n (h(x))}.\end{align*} $$

Since $hfh^{-1} = g$ , we have

$$ \begin{align*}\frac{g^{n+1}(h(x)) - g^n (h(x))}{f^{n+1}(x) - f^n (x)} = \frac{ h (f^{n+1}(x))- h (f^n (x))}{f^{n+1}(x)-f^n(x)} \longrightarrow Dh (0).\end{align*} $$

As a consequence, if $Dh (0) = 1$ , we have

$$ \begin{align*}Dh (x) = \lim_{n \to \infty} \frac{Df^n(x)}{Dg^n (h(x))},\end{align*} $$

and we thus retrieve the equality $Dh(x) = A (x,h(x))$ from (13).

2.2 Examples of low regular conjugacies that preserve residues

The statement of Theorem A would be empty if all $C^{\ell + 1}$ conjugacies between parabolic germs in $\mathrm {Diff}^{2\ell + 1}_{\ell } (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 1}_{\ell + 1} (\mathbb {R},0)$ were automatically $C^{2\ell + 1}$ . However, this is not at all the case. Below we provide the details of a specific example for which $\ell = 1$ ; we leave the extension to higher order of tangency to the reader.

Actually, our example directly deals with (flat) vector fields: We will exhibit two of them of class $C^3$ that are $C^2$ conjugate but not $C^3$ conjugate. The announced diffeomorphisms will be the time-1 maps of their flows.

Let $X(x) := x^2$ , $h(x) := x+x^2+x^3\log x$ and $Y := h^*X$ . One easily checks that h is the germ of $C^2$ diffeomorphism that is not $C^3$ on any neighborhood of the origin. (This follows from the fact that the function $x\mapsto x^{k+1} \log x$ is of class $C^k$ but not $C^{k+1}$ , which can be easily checked by induction.) Every map that conjugates X to Y equals h up to a member of the flow of X (this immediately follows from [Reference Eynard-Bontemps and Navas8, Lemma 2.6]). As these members are all of class $C^{\infty }$ and h is not $C^3$ , there is no $C^3$ conjugacy from X to Y.

We are hence left to show that Y is of class $C^3$ . To do this, we compute:

$$ \begin{align*} Y(x) =\ & \frac{(X\circ h)(x)}{Dh(x)} \\ =\ & \frac{(x+x^2+x^3\log x)^2}{1+2x+3x^2\log x+ x^2} \\ =\ & x^2 \cdot \frac{1+x^2+x^4(\log x)^2+2x+2x^2\log x+2x^3\log x}{1+2x+3x^2\log x+ x^2}\\ =\ & x^2\left(1+\frac{-x^2\log x+2x^3\log x+x^4(\log x)^2}{1+2x+3x^2\log x+ x^2}\right)\\ =\ & x^2+x^4\log x \cdot \frac{-1+2x+x^2\log x}{1+2x+3x^2\log x+ x^2}. \end{align*} $$

This shows that Y is of the form $Y(x) = x^2 + o( x^4 \log x)$ . Showing that Y is of class $C^3$ (with derivatives equal to those of $x^2$ up to order 3 at the origin) requires some extra computational work based on the fact that the function $u (x) := x^4\log x$ satisfies $D^{(3-n)} u (x) = o(x^{n})$ for $n \in [\![0,3]\!] := \{0,1,2,3\}$ . We leave the details to the reader.

2.3 $C^{2\ell +1}$ germs with vanishing residue that are non $C^{\ell +1}$ conjugate

The aim of this section is to give an example showing that the converse to Theorem A does not hold (for $\ell = 1$ ). More precisely, we will give an example of two expanding germs of $C^3$ diffeomorphisms, both exactly 1-tangent to the identity and with vanishing iterative residue, that are not $C^2$ conjugate. Notice that, by Proposition 2.1, these are $C^1$ conjugate.

Again, our example directly deals with (flat) vector fields. Namely, let $X(x):= x^2$ and $Y(x) := h^* X (x)$ , where h is given on $\mathbb {R}_+^* $ by $h(x) := x + x^2 \, \log (\log x).$ Let f (resp. g) denote the time-1 map of X (resp. Y). We claim that these are (germs of) $C^3$ diffeomorphisms. This is obvious for f (which is actually real-analytic). For g, this follows from the fact that Y is of class $C^3$ . To check this, observe that

$$ \begin{align*}Y(x) = \frac{X(h(x))}{Dh (x)} = \frac{ \big( x+x^2 \log (\log x) \big)^2}{ 1 + 2 x \log(\log x) + \frac{x}{\log x} },\end{align*} $$

hence, skipping a few steps,

(15) $$ \begin{align} Y(x) = x^2 + x^3 \cdot \frac{ x \, (\log (\log x))^2 - \frac{1}{\log x}}{1 + 2 x \log(\log(x)) + \frac{x}{\log(x)} } = x^2 + o(x^3). \end{align} $$

Now, letting

$$ \begin{align*}u(x) := x^3, \qquad v(x):= \frac{ x \, (\log (\log x))^2 - \frac{1}{\log x}}{1 + 2 x \log(\log x) + \frac{x}{\log x} },\end{align*} $$

we have $Y(x) = x^2 + (u v) (x)$ , and

$$ \begin{align*}D^{(3)} (uv) (x) = \sum_{j=0}^3 {3 \choose j} \, D^{(3-j)} u (x) \, D^{(j)} (v)(x) = \sum_{j=0}^3 c_j \, x^j \, D^{(j)} v(x)\end{align*} $$

for certain constants $c_j$ . We are hence reduced to showing that $x^j D^{(j)}v(x)$ tends to $0$ as x goes to the origin, which is straightforward and is left to the reader.

It will follow from §3.1 (see Equation (17)) that both f and g have vanishing iterative residue. They are conjugated by h, which is the germ of a $C^1$ diffeomorphism which is easily seen to be non $C^2$ . As in the previous example, this implies that there is no $C^2$ conjugacy between f and g (despite they are both $C^3$ and have vanishing iterative residue !).

3.1 Residues and vector fields

We begin by recalling Yoccoz’ result and, in particular, by relating the residue to their infinitesimal expression.

Let $\mathbb {K}$ be a field. For each integer $k \geq 1$ , let us consider the group $G_k (\mathbb {K})$ made of the expressions of the form

$$ \begin{align*}\sum_{n=1}^k a_n x^n, \quad \mbox{ with } \, a_1 \neq 0,\end{align*} $$

where the product is just the standard composition but neglecting terms of order larger than k. The group $\widehat {\mathrm {Diff}} (\mathbb {K},0)$ has a natural morphism into each group $G_k (\mathbb {K})$ obtained by truncating series expansions at order k:

$$ \begin{align*}\sum_{n \geq 1} a_n x^n \in \widehat{\mathrm{Diff}} (\mathbb{K},0) \quad \longrightarrow \quad \sum_{n=1}^{k} a_n x^n \in G_k (\mathbb{K}).\end{align*} $$

Notice that each group $G_k (\mathbb {K})$ is solvable and finite-dimensional. Moreover, each subgroup $G_{k,\ell } (\mathbb {K})$ , $\ell < k$ , obtained by truncation as above of elements in $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0)$ is nilpotent. In the case where $\mathbb {K}$ is the field of real numbers, these groups have a well-defined exponential map, and every group element is the time-1 element of a unique flow.

Actually, it is not very hard to explicitly compute this flow for elements $\mathrm {f} \in G_{2\ell + 1,\ell } (\mathbb {K})$ , that is for those of the form

$$ \begin{align*}\mathrm{f} \, (x) = x + \sum_{n=\ell + 1}^{2\ell + 1} a_n x^n.\end{align*} $$

Namely, this is given by

(16) $$ \begin{align} \mathrm{f}^t (x) = x + \sum_{n=\ell + 1}^{2 \ell} t \, a_n \, x^n + \left[ \frac{\ell + 1}{2} (t \, a_{\ell})^2 - t \, \mathrm{Resad}_{\ell} (\mathrm{f}) \right] x^{2\ell + 1}, \end{align} $$

where $\mathrm {Resad}_{\ell }$ is defined as in $\widehat {\mathrm {Diff}_{\ell }}(\mathbb {K},0)$ by Equation (2) (assuming that $\mathbb {K}$ has characteristic different from $2$ ). Checking that this is indeed a flow is straightforward: It only uses the additive properties of the corresponding versions of $\Phi _{\ell ,i}$ (for $1\leq i \leq \ell $ ) and $\mathrm {Resad}_{\ell }$ in $G_{2\ell + 1}(\mathbb {K})$ .

Inspired by the work of Takens [Reference Takens21], in [Reference Yoccoz23, Appendice 3], Yoccoz considered germs f of nonflat $C^k$ diffeomorphisms of the real line fixing the origin. In particular, he proved that there exists a unique (germ of) $C^{k-1}$ vector field X that is k times differentiable at the origin and whose flow $(f^t)$ has time-1 map $f^1 = f$ .

If $k = 2\ell + 1$ , by truncating f at order $2\ell + 1$ , we get an element $\mathrm {f}_{2\ell + 1} \in G_{2\ell + 1} (\mathbb {R})$ , and we can hence consider the flow $( \mathrm {f}^t_{2\ell + 1})$ in $G_{2\ell + 1} (\mathbb {R})$ whose time-1 element is $\mathrm {f}_{2\ell + 1}$ . Let P be the associated polynomial vector field defined by

$$ \begin{align*}P(x) := \frac{d}{dt}_{ |_{t=0}} \mathrm{f}^t_{2\ell + 1} (x).\end{align*} $$

As a key step of his proof, Yoccoz showed a general estimate that implies that

$$ \begin{align*}\lim_{x \to 0} \frac{X(x) - P(x)}{x^{2\ell + 1}} = 0.\end{align*} $$

According to Equation (16), if $f \in \mathrm {Diff}^{2\ell + 1}_{\ell } (\mathbb {R},0)$ writes in the form

$$ \begin{align*}f(x) = x + \sum_{n=\ell + 1}^{2\ell + 1} a_n x^n + o (x^{2\ell + 1}),\end{align*} $$

then the polynomial vector field associated to $\mathrm {f}_{2\ell + 1}$ equals

$$ \begin{align*}P(x) = \frac{d}{dt}_{ |_{t=0}} \mathrm{f}^t_{2\ell + 1} (x) = \sum_{n=\ell + 1}^{2 \ell} a_n x^n - \mathrm{Resad}_{\ell}(f) \, x^{2\ell + 1},\end{align*} $$

and therefore

(17) $$ \begin{align} X (x) = \sum_{n=\ell + 1}^{2\ell} a_n x^n - \mathrm{Resad}_{\ell}(f) \, x^{2\ell + 1} + o(x^{2\ell + 1}). \end{align} $$

This formula in which $\mathrm {Resad}_{\ell }$ explicitly appears will be fundamental for the proof of Theorem A.

Remark 3.1. Formula (7), which holds for every $f \in \mathrm {Diff}^{2\ell + 2}_{\ell } (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 2}_{\ell +1} (\mathbb {R},0)$ and all $t \neq 0$ , follows from Equation (17) above. Indeed, in order to check it, we may assume that f is expanding and already in reduced form, say

$$ \begin{align*}f (x) = x + x^{\ell+1} + \mu x^{2\ell+1} + o(x^{2\ell + 1}).\end{align*} $$

If X is the $C^{2\ell +1}$ vector field associated to f, then the vector field $X_t$ associated to $f^t$ is $t X$ . Since

$$ \begin{align*}X_t (x) = t \, X(x) = t \left[ x^{\ell+1} - \mathrm{Resad}_{\ell}(f) \, x^{2\ell + 1} + o(x^{2\ell + 1}) \right],\end{align*} $$

by Equation (17) we have

$$ \begin{align*}f^t (x) = x + t \, x^{\ell+1} + \mu_t \, x^{2\ell +1} + o(x^{2\ell +1}),\end{align*} $$

where $\mu _t$ is such that $\mathrm {Resad}_{\ell } (f^t) = t \, \mathrm {Resad}_{\ell } (f)$ . Hence,

$$ \begin{align*}\frac{(\ell +1) \, t^2}{2} - \mu_t = t \, \left[ \frac{\ell + 1}{2} - \mu \right].\end{align*} $$

Therefore, for $t> 0$ ,

$$ \begin{align*}\mathrm{Resit} (f^t) \kern1.3pt{=}\kern1.3pt \frac{\ell +1}{2} \kern1.3pt{-}\kern1.3pt \frac{\mu_t}{t^2} \kern1.3pt{=}\kern1.3pt \frac{1}{t^2} \left[ \frac{(\ell +1) \kern1.3pt t^2}{2} \kern1.3pt{-}\kern1.3pt \mu_t \right] \kern1.3pt{=}\kern1.3pt \frac{t}{t^2} \kern1.3pt \left[ \frac{\ell + 1}{2} \kern1.3pt{-}\kern1.3pt \mu \right] \kern1.3pt{=}\kern1.3pt \frac{1}{t} \kern1.3pt \left[ \frac{\ell + 1}{2} \kern1.3pt{-}\kern1.3pt \mu \right] \kern1.3pt{=}\kern1.3pt \frac{\mathrm{Resit} (f)}{t},\end{align*} $$

as announced. The case $t < 0$ is analogous.

It is worth mentioning that the argument above shows that the equality (7) holds for every $f \in \mathrm {Diff}^{2\ell + 1}_{\ell } (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 1}_{\ell +1} (\mathbb {R},0)$ and all $t \neq 0$ for which $f^t$ is a germ of $C^{2\ell +1}$ diffeomorphism (or, at least, has $(2\ell +1)$ derivatives at the origin; see Remark 1.4).

3.2 A first proof of the $C^{\ell + 1}$ conjugacy invariance of $\mathbf {Resit}$

We next proceed to the proof of the $C^{\ell + 1}$ conjugacy invariance of $\mathrm {Resit}$ between germs of $C^{2\ell + 1}$ diffeomorphisms that are exactly $\ell $ -tangent to the identity.

Proof Proof of Theorem A

Let f and g be elements in $\mathrm {Diff}^{2\ell + 1}_{\ell } (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 1}_{\ell + 1} (\mathbb {R},0)$ conjugated by an element $h \in \mathrm {Diff}^{\ell + 1}_+ (\mathbb {R},0)$ . Then they are both contracting or both expanding. We will suppose that the second case holds; the other one follows from it by passing to inverses.

As we want to show the equality $\mathrm {Resit} (f) = \mathrm {Resit} (g)$ , by the definition and Lemma 1.3, we may assume that f and g have Taylor series expansions at the origin of the form

$$ \begin{align*}f(x) = x + x^{\ell + 1} + \mu \, x^{2\ell + 1} + o (x^{2\ell + 1}), \qquad g(x) = x + x^{\ell + 1} + \mu' \, x^{2\ell + 1} + o(x^{2\ell + 1}). \end{align*} $$

Notice that

$$ \begin{align*}\mathrm{Resit} (f) = \mathrm{Resad}_{\ell}(f) = \frac{\ell + 1}{2} - \mu =: R, \qquad \mathrm{Resit} (g) = \mathrm{Resad}_{\ell} (g) = \frac{\ell + 1}{2} - \mu' =: R'.\end{align*} $$

In virtue of Equation (17), the vector fields $X,Y$ associated to $f,g$ , respectively, have the form

$$ \begin{align*}X(x) = x^{\ell + 1} - R \, x^{2\ell + 1} + o(x^{2\ell + 1}), \qquad Y(x) = x^{\ell + 1} - R' \, x^{2\ell + 1} + o(x^{2\ell + 1}).\end{align*} $$

Now, write

$$ \begin{align*}h(x) = \lambda \, x + \sum_{n =2}^{\ell + 1} c_n x^n + o(x^{\ell + 1}).\end{align*} $$

Since h conjugates f to g, it must conjugate X to Y, that is,

(18) $$ \begin{align} X \cdot Dh = Y\circ h. \end{align} $$

We first claim that this implies $\lambda = 1$ . Indeed, the relation writes in a summarized way as

$$ \begin{align*}(x^{\ell + 1} + o (x^{\ell + 1})) \cdot (\lambda + o(1)) = (\lambda \, x + o (x))^{\ell + 1} + o(x^{\ell + 1}).\end{align*} $$

By identification of the coefficients of $x^{\ell + 1}$ , we obtain $\lambda = \lambda ^{\ell + 1}$ . Thus, $\lambda = 1$ , as announced.

We next claim that h must be $\ell $ -tangent to the identity. Indeed, assume otherwise and let $2 \leq p < \ell + 1$ be the smallest index for which $c_p \neq 0$ . Then relation (18) above may be summarized as

$$ \begin{align*}(x^{\ell + 1} + o(x^{\ell + 1})) \cdot (1 + p \, c_p \, x^{p-1} + o(x^{p-1})) = (x + c_p \, x^p + o(x^p))^{\ell + 1} + o(x^{2\ell}).\end{align*} $$

By identifying the coefficients of $x^{\ell +p}$ we obtain $p \, c_p = (\ell + 1) \, c_p$ , which is impossible for $c_p \neq 0$ .

Thus, h writes in the form

$$ \begin{align*}h(x) = x + cx^{\ell + 1} + o(x^{\ell + 1}).\end{align*} $$

Relation (18) then becomes

$$ \begin{align*}&( x^{\ell + 1} - R x^{2\ell + 1} + o (x^{2\ell + 1})) \cdot (1 + (\ell + 1) c x^{\ell} + o(x^{\ell})) \\&\quad= (x + cx^{\ell + 1} + o(x^{\ell + 1}))^{\ell + 1} - R' (x + o(x^{\ell}))^{2\ell + 1} + o(x^{2\ell + 1}).\end{align*} $$

Identification of the coefficients of $x^{2\ell + 1}$ then gives

$$ \begin{align*}(\ell + 1) \, c - R = (\ell + 1) \, c - R'.\end{align*} $$

Therefore, $R=R'$ , as we wanted to show.

3.3 Residues and logarithmic deviations of orbits

We next characterize $\mathrm {Resit}$ for contracting germs $f \in \mathrm {Diff}^{2\ell + 1}_{\ell } (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 1}_{\ell + 1} (\mathbb {R},0)$ in terms of the deviation of orbits from those of the corresponding (parabolic) ramified affine flow of order $\ell $ , namely,

$$ \begin{align*}f^t(x) := \frac{x}{\sqrt[\ell]{1 + t x^{\ell}}} = x - \frac{t}{\ell} x^{\ell + 1} + \frac{1}{2 \, \ell} \left( 1+\frac{1}{\ell} \right) t^2 x^{2\ell + 1} + o (x^{2\ell + 1}).\end{align*} $$

Notice that the time-1 map $f := f^1$ of this flow satisfies $\mathrm {Resit} (f) = 0$ . Moreover, for all $x>0$ ,

$$ \begin{align*}\frac{1}{\sqrt[\ell]{n}} - f^n (x) = \frac{1}{\sqrt[\ell]{n}} - \frac{x}{\sqrt[\ell]{1+nx^{\ell}}}\end{align*} $$

which asymtotically behaves (when n goes to infinity) as

$$ \begin{align*}\frac{n^{\frac{\ell-1}{\ell}}}{\ell \, n \, (1+nx^{\ell})} \sim \frac{1}{\ell \, n \, \sqrt[\ell]{n}},\end{align*} $$

where the first equivalence follows from the equality $u^{\ell } - v^{\ell } = (u-v) (u^{\ell -1} + u^{\ell -2}v + \ldots + v^{\ell -1})$ . In particular, letting $a = 1/\ell $ , we have

$$ \begin{align*}\lim_{n \to \infty} \left[ \frac{ \ell \, n \, \sqrt[\ell]{a \, \ell \, n} }{\log (n)} \left( \frac{1}{ \sqrt[\ell]{a \, \ell \, n}} - f^n (x)\right) \right] = 0.\end{align*} $$

This is a particular case of the general proposition below.

Proposition 3.2. If $\ell \geq 1$ , then, for every contracting germ $f \in \mathrm {Diff}^{2\ell + 1}_{\ell } (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 1}_{\ell + 1} (\mathbb {R},0)$ of the form

$$ \begin{align*}f(x) = x - ax^{\ell + 1} + b x^{2\ell + 1} + o(x^{2\ell + 1}), \qquad a> 0,\end{align*} $$

and all $x_0> 0$ , one has

(19) $$ \begin{align} \mathrm{Resit} (f) = \lim_{n \to \infty} \left[ \frac{a \, \ell^2 \, n^2}{\log (n)} \left( \frac{1}{a \, \ell \, n} -[ f^n (x_0)]^{\ell}\right) \right] = \lim_{n \to \infty} \left[ \frac{ \ell \, n \, \sqrt[\ell]{a \, \ell \, n} }{\log (n)} \left( \frac{1}{ \sqrt[\ell]{a \, \ell \, n}} - f^n (x_0)\right) \right]. \end{align} $$

Proof First proof (using coordinates at infinity).

Consider the map $I (z) = 1/z^{1 / \ell }$ , where $z> 0$ is large enough. Then the conjugate germ at infinity $g := I^{-1} f I$ has an expansion of the form

$$ \begin{align*}g(z) = \left( \frac{1}{\frac{1}{z^{1/\ell}} - \frac{a}{z^{(\ell+1)/\ell}} + \frac{b}{z^{(2\ell+1)/\ell}} + ...} \right)^{\ell} = z \left( \frac{1}{1 - \frac{a}{z} + \frac{b}{z^2} + \ldots} \right)^{\ell}.\end{align*} $$

Hence,

$$ \begin{align*}g(z) = z \left( 1 + \left(\frac{a}{z} - \frac{b}{z^2} - \ldots \right) + \left(\frac{a}{z} - \frac{b}{z^2} - \ldots \right)^2 + \ldots \right)^{\ell} = z \left( 1 + \frac{a}{z} + \frac{a^2 - b}{z^2} + O \left( \frac{1}{z^3} \right) \right)^{\ell},\end{align*} $$

that is

$$ \begin{align*}g(z) = z \left( 1 + \ell \left( \frac{a}{z} + \frac{a^2 - b}{z^2} \right) + \frac{\ell (\ell-1)}{2} \left( \frac{a}{z} \right)^2 + O \left( \frac{1}{z^3}\right) \right).\end{align*} $$

Therefore,

(20) $$ \begin{align} g(z) = z + a\ell + \frac{\ell}{z} \left[ \frac{(\ell + 1) a^2}{2} - b \right] + O \left( \frac{1}{z^2}\right) = z + a\ell + \frac{R \ell}{z} + O \left( \frac{1}{z^2}\right), \end{align} $$

where $R = \mathrm {Resad}_{\ell } (f)$ . We claim that, for each fixed z, the sequence below converges as n goes to infinity:

(21) $$ \begin{align} \varphi_n (z) := g^n (z) - a\ell n - \frac{\ell R \log(n)}{a}. \end{align} $$

Assume this for a while. Then we have

$$ \begin{align*}R = \lim_{n \to \infty} \frac{a}{\ell \log(n)} \big[ g^n(I^{-1}(x_0)) - a \ell n \big],\end{align*} $$

hence

$$ \begin{align*}R = \lim_{n \to \infty} \frac{a}{\ell \log(n)} \big[ I^{-1} f^n (x_0) - a \ell n \big] = \lim_{n \to \infty} \frac{a}{\ell \log(n)} \left[ \frac{1}{f^n (x_0)^{\ell}} - a \ell n \right].\end{align*} $$

This easily implies that

(22) $$ \begin{align} \lim_{n \to \infty} (a \ell n)^{1/\ell} f^n (x_0) = 1. \end{align} $$

Moreover, using the identity $u^{\ell } - v^{\ell } = (u-v) (u^{\ell -1} + u^{\ell -2}v + \ldots + v^{\ell -1})$ , one easily deduces that

$$ \begin{align*}R &= \lim_{n \to \infty} \frac{ a \ell (a \ell n)^{(\ell-1)/\ell}}{\ell \log(n)} \left[ \frac{1}{f^n (x_0)} - (a \ell n)^{1/\ell} \right] \\&= \lim_{n \to \infty} \frac{ a (a \ell n)^{(\ell-1)/\ell}}{\log(n)} \left[ \frac{1}{ (a \ell n)^{1/\ell}} - f^{n}(x_0) \right] \frac{(a \ell n)^{1/\ell}}{f^n(x_0)}.\end{align*} $$

Using Equation (22), one thus concludes

$$ \begin{align*}R = \lim_{n \to \infty} \frac{ a (a \ell n)^{(\ell - 1)/\ell} (a \ell n)^{2/\ell} }{\log(n)} \left[ \frac{1}{ (a \ell n)^{1/\ell}} - f^{n}(x_0) \right] .\end{align*} $$

Therefore,

$$ \begin{align*}\mathrm{Resit} (f) = \frac{R}{a^2} = \lim_{n \to \infty} \frac{ \ell n (a \ell n)^{1/\ell} }{\log(n)} \left[ \frac{1}{ (a \ell n)^{1/\ell}} - f^{n}(x_0) \right],\end{align*} $$

as announced.

It remains to show that the sequence $\varphi _n(z)$ defined by Equation (21) converges as n goes to infinity. First notice that, for all large-enough z,

$$ \begin{align*}z + \frac{a \ell}{2} \leq g(z) \leq z + 2a\ell.\end{align*} $$

Hence, for a fixed z and large-enough n,

$$ \begin{align*}\frac{a\ell n}{2} \leq g^n(z) \leq 2 a\ell n.\end{align*} $$

Using Equation (20) and this last estimate, we obtain

$$ \begin{align*} \varphi_{n+1} (z) - \varphi_n (z) \ &= g^{n+1} (z) -a\ell (n+1) - \frac{R \log(n+1)}{a} - \left[ g^n(z) - a\ell n - \frac{R \log(n)}{a} \right] \\ &= \left[ g^n(z) + a\ell + \frac{R \ell}{g^n(z)} + O \left(\frac{1}{g^n(z)^2} \right) \right] - a\ell - \frac{R }{a} \log \left( \frac{n+1}{n} \right) - g^n(z), \end{align*} $$

hence

(23) $$ \begin{align} \varphi_{n+1} (z) - \varphi_n (z) = \frac{R \ell}{g^n(z)} - \frac{R }{a} \log \left( \frac{n+1}{n} \right) + O \left(\frac{1}{g^n(z)^2} \right) = O \left( \frac{1}{n} \right). \end{align} $$

This estimate is still weak to prove the announced convergence, but reintroducing it in the computations above will give the desired convergence. Indeed, letting $\varphi _0 (z) := z$ , it yields

$$ \begin{align*}|\varphi_n (z) - z| \leq \sum_{i=1}^n |\varphi_i (z) - \varphi_{i-1} (z)| = O (\log(n)).\end{align*} $$

Using the left-hand side equality in Equation (23), the definition of $\varphi _n$ and the latter estimate, we finally obtain

$$ \begin{align*} \varphi_{n+1} (z) - \varphi_n (z) =\ & \frac{R \ell}{g^n(z)} - \frac{R }{a} \log \left( \frac{n+1}{n} \right) + O \left( \frac{1}{n^2} \right) \\ =\ & \frac{R \ell}{a\ell n + \varphi_n (z) + R \log(n) / a} - \frac{R}{an} + O \big( \frac{1}{n^2} \big) \\ =\ & \frac{\varphi_n (z) + R \log(n) / a}{an (a\ell n + \varphi_n (z) + R \log(n) / a)} + O \big( \frac{1}{n^2} \big) \\ =\ & O \left( \frac{\log (n)}{n^2} \right). \end{align*} $$

Since the sum $\sum \frac {\log (n)}{ n^2}$ is finite, this shows that $(\varphi _n(z))$ is a Cauchy sequence, which implies the announced convergence.

Proof Second proof (using vector fields).

We know from §3.1 that the vector field X associated to f has a Taylor series expansion of the form

$$ \begin{align*}X(x) = -a x^{\ell + 1} - R x^{2\ell + 1} + o(x^{2\ell + 1}), \qquad R = \mathrm{Resad}_{\ell} (f).\end{align*} $$

We compute

$$ \begin{align*} n =\ & \int_{x_0}^{f^n(x_0)} \frac{dy}{X(y)} \\ =\ & \int_{x_0}^{f^n(x_0)}\!\! \frac{dy}{-a y^{\ell + 1} - R y^{2\ell + 1}} + \int_{x_0}^{f^n(x_0)} \!\!\left[ \frac{1}{a y^{\ell + 1} + R y^{2\ell + 1}} - \frac{1}{a y^{\ell + 1} + R y^{2\ell + 1} + o(y^{2\ell + 1})} \right] \kern-1pt dy \\ =\ & \int_{x_0}^{f^n(x_0)} \frac{dy}{-a y^{\ell + 1} (1 + R y^{\ell} /a)} + \int_{x_0}^{f^n(x_0)} \left[ \frac{o(y^{2\ell + 1})}{y^{2\ell+2} (a^2 + o(y))} \right] \, dy \\ =\ & \int_{x_0}^{f^n(x_0)} \left[ \frac{1 - Ry^{\ell}/a + o(y^{2\ell-1})}{-a y^{\ell + 1}} \right] \, dy + \int_{x_0}^{f^n(x_0)} o \left( \frac{1}{y} \right) \, dy \\ =\ & \frac{1}{a \ell y^{\ell}}\Big|^{f^n(x_0)}_{x_0} + \frac{R \, \log(y)}{a^2}\Big|^{f^n(x_0)}_{x_0} - \frac1a \int_{x_0}^{f^n (x_0)} o \left( y^{\ell-2} \right) + o (\log (f^n (x_0))) \\ =\ & \frac{1}{a \, \ell \, [f^n (x_0)]^{\ell}} + \frac{R \, \log(f^n (x_0))}{a^2} + C_{x_{0}} + o (\log (f^n(x_0))), \end{align*} $$

where $C_{x_0}$ is a constant that depends only on $x_0$ (and is independent of n). The right-side expression is of the form

$$ \begin{align*}\frac{1}{a \, \ell \, [f^n (x_0)]^{\ell}} + o \left( \frac{1}{[f^n(x_0)]^\ell} \right),\end{align*} $$

which shows that, as n goes to infinity,

(24) $$ \begin{align} [f^n (x_0)]^\ell \sim \frac{1}{a \, \ell \, n}. \end{align} $$

Now, from

$$ \begin{align*} R =\ & \frac{a^2 \, (n-C_{x_0})}{\log (f^n (x_0))} - \frac{a}{\ell \, [f^n(x_0)]^\ell \, \log (f^n(x_0))} + o(1) \\ =\ & \frac{a^2 \, (n-C_{x_0})}{[f^n(x_0)]^\ell \, \log (f^n(x_0))} \left[ [f^n(x_0)]^\ell - \frac{1}{a \, \ell \, (n-C_{x_0})} \right] + o(1), \end{align*} $$

using Equation (24) we obtain

$$ \begin{align*}R = \lim_{n \to \infty} \frac{a^2 \, (n-C_{x_0})}{\frac{1}{a \, \ell \, n} \cdot \frac{1}{\ell}\log (\frac{1}{a \, \ell \, n})} \left[ [f^n(x_0) ]^\ell - \frac{1}{a \, \ell \, (n-C_{x_0})} \right] = \lim_{n \to \infty} \frac{a^3 \, \ell^2 \, n^2}{\log(n)} \left[ \frac{1}{a \, \ell \, n} - [f^n (x_0)]^\ell \right].\end{align*} $$

Therefore,

$$ \begin{align*}\mathrm{Resit}(f) = \frac{R}{a^2} = \lim_{n \to \infty} \frac{a \, \ell^2 \, n^2}{\log(n)} \left[ \frac{1}{a \, \ell \, n} - [ f^n (x_0) ]^\ell \right] = \lim_{n \to \infty} \left[ \frac{ \ell \, n \sqrt[\ell]{a \ell n}}{\log (n)} \left( \frac{1}{ \sqrt[\ell]{a \, \ell \, n}} - f^n (x_0)\right) \right],\end{align*} $$

where the last equality follows from the identity $u^\ell - v^\ell = (u-v) (u^{\ell -1} + u^{\ell -2}v + \ldots + v^{\ell -1})$ .

What follows is a direct consequence of the previous proposition; checking the details is left to the reader. It is worth comparing the case $\ell =1$ of the statement with (11).

Corollary 3.4. If $\ell \geq 1$ , then, for every contracting germ $f \in \mathrm {Diff}^{2\ell + 1}_\ell (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 1}_{\ell + 1} (\mathbb {R},0)$ of the form

$$ \begin{align*}f(x) = x - ax^{\ell + 1} + b x^{2\ell + 1} + o(x^{2\ell + 1})\end{align*} $$

and all $x_0> 0$ , one has

(25) $$ \begin{align} f^n (x_0) = \frac{1}{\sqrt[\ell]{a \ell n}} \left( 1 - \frac{[\mathrm{Resit}(f) + \delta_n] \log (n)}{\ell \, n} \right), \end{align} $$

where $\delta _n$ is a sequence that converges to $0$ as n goes to infinity.

3.4 A second proof of the $C^{\ell + 1}$ conjugacy invariance of $\mathrm {Resit}$

We can give an alternative proof of Theorem A using the estimates of the previous section.

Proof Second proof of Theorem A

Let $f,g$ in $\mathrm {Diff}^{2\ell + 1}_{\ell } (\mathbb {R},0) \setminus \mathrm {Diff}^{2\ell + 1}_{\ell + 1} (\mathbb {R},0)$ be elements conjugated by $h \in \mathrm {Diff}^{\ell + 1}_+ (\mathbb {R},0)$ . In order to show that $\mathrm {Resit} (f)$ and $\mathrm {Resit} (g)$ coincide, we may assume that f and g have Taylor series expansions at the origin of the form

$$ \begin{align*}f(x) = x - x^{\ell + 1} + \mu \, x^{2\ell + 1} + o (x^{2\ell + 1}), \qquad g(x) = x - x^{\ell + 1} + \mu' \, x^{2\ell + 1} + o(x^{2\ell + 1}). \end{align*} $$

The very same arguments of the beginning of the first proof show that h must write in the form

$$ \begin{align*}h(x) = x + cx^{\ell + 1} + o(x^{\ell + 1}).\end{align*} $$

In particular, for a certain constant $C> 0$ ,

(26) $$ \begin{align} |h(x) - x| \leq C \, x^{\ell + 1} \end{align} $$

for all small-enough $x> 0$ . Fix such an $x_0> 0$ . From $hf = gh$ , we obtain $hf^n (x_0) = g^n h(x_0)$ for all n, which yields

$$ \begin{align*}h f^n (x_0) - f^n(x_0) = g^n (h(x_0)) - f^n(x_0)\end{align*} $$

Using Equations (25) and (26), this implies

(27) $$ \begin{align} C \left[ \frac{1}{\sqrt[\ell]{a\ell n}} \left( 1 - \frac{[\mathrm{Resit} (f) + \delta_n] \log (n)}{\ell \, n} \right) \right]^{\ell + 1} \geq \left| \frac{[\mathrm{Resit}(g) - \mathrm{Resit}(f) + \delta_n'] \log (n)}{\ell \, n \, \sqrt[\ell]{a \ell n}} \right| \end{align} $$

for certain sequences $\delta _n,\delta _n'$ converging to $0$ . On the one hand, the left-hand term above is of order $1/ (n \, \sqrt [\ell ]{n})$ . On the other hand, if $\mathrm {Resit}(f) \neq \mathrm {Resit}(g)$ , then the right-hand term is of order

$$ \begin{align*}\frac{|\mathrm{Resit}(g) - \mathrm{Resit}(f)| \, \log(n)}{n \, \sqrt[\ell]{n}}.\end{align*} $$

Therefore, if $\mathrm {Resit}(f) \neq \mathrm {Resit}(g)$ , then inequality (27) is impossible for large-enough n. Thus, $\mathrm {Resit}(f)$ and $\mathrm {Resit}(g)$ must coincide in case of $C^{\ell + 1}$ conjugacy.