Ilyashenko algebras based on definable monomials: the construction (base step)

Let $\H$ be the Hardy field of $\Ranexp$, and let $M$ be a multiplicative $\RR$-subvector space of $\H^{>0}$; I continue to assume in this post that $M$ is a pure scale.

A germ $h \in \H^{>0}$ is small if $h(x) \to 0$ as $x \to +\infty$.

The construction discussed here works for the following type of series in $\Gs{\RR}{M}$: recall that a generalized power series $G \in \Ps{R}{X_0^*, \dots, X_k^*}$ has natural support, if $\supp(G) \cap B$ is finite for every bounded box $B \subseteq \RR^{k+1}$.


A series $F \in \Gs{\RR}{M}$ is an $M$-generalized power series if there exist $k \in \NN$, small $m_0, \dots, m_k \in M$, a generalized power series $G \in \Ps{R}{X_0^*, \dots, X_k^*}$ with natural support and a (not necessarily small) $m \in M$ such that $$F = m G(m_0, \dots, m_k).$$


The $L$-generalized power series are exactly the logarithmic generalized power series.


Show that the support of an $M$-generalized power series is of order type at most $\omega^k$, for some $k \in \NN$.

The construction is based on holomorphic extensions to the following type of domain: for $a \in \RR$, set $$H(a):= \set{z \in \CC:\ \re z > a}.$$


A domain $\Omega \subseteq H(0)$ is a standard power domain if there exist $C > 0$ and $\epsilon \in (0,1)$ such that $$\Omega = \Omega^\epsilon_C := \varphi^\epsilon_C(H(0)),$$ where $\varphi^\epsilon_C:H(-1) \into \CC$ is defined by $$\varphi^\epsilon_C(z):= z + C(1+z)^\epsilon$$ using the standard branch of the logarithm.

standard power domain
The standard power domain $\Omega^\epsilon_C$ with its boundary $\varphi^\epsilon_C(i\RR)$.


A standard power domain $\Omega^\epsilon_C$ is a standard quadratic domain if and only if $\epsilon = \frac12$.

The main reason for working with standard power domain is the following:

Uniqueness Principle

(Lemma 24.37 in Ilyashenko and Yakovenko’s book)

Let $\Omega$ be a standard power domain and $\phi:\Omega \into \CC$ be holomorphic. If $\phi$ is bounded and $\phi\rest{\RR} \prec \exp^{-n}$ for every $n \in \NN$, then $\phi = 0$.

To exploit this Uniqueness Principle, we want to extend the notion of asymptotic expansion to standard power domains. First of all, this requires that the elements of $M$ are well behaved on such domains, in the following sense:


Let $\Omega$ be an unbounded domain.

  1. For maps $\phi,\psi:\Omega \into \CC$, we set $\phi \preceq_\Omega \psi$ if and only if $\left|\frac{\phi}{\psi}\right|$ is bounded.
  2. $M$ is a scale on $\Omega$ if, for every $m \in M$,
    • $m$ has a holomorphic extension $\mm$ on $\Omega$, and
    • if $m \preceq 1$, then $\mm \preceq_\Omega 1$.

For a germ $g \in \C$, I shall use the boldface $\gg$ to denote any holomorphic extension of $g$ on some domain in $\CC$ (if such extension exists).


  1. The germ $x \exp^{-x}$ is bounded, but it’s holomorphic extension $z \exp^{-z}$ is not bounded on any right half-plane. Hence neither $L$ nor $L_i$, for $i \in \NN$, are scales on $H(0)$.
  2. If $C>0$ and $\epsilon \in (0,1)$, there exist $k,K>0$ such that $$k \exp(K|z|^\epsilon) \le |\bexp z| \le \exp(|z|).$$ Hence $L$ is a scale on every standard power domain.

I assume from now on that $M$ is a scale on every standard power domain. This implies, in particular, that every $\RR$-linear combination $h$ of germs in $M$ has a holomorphic extension $\hh$ on every standard power domain.


If $F \in \Gs{\RR}{M}$ has natural support, then every truncation $F_n$, for $n \in M$, has a holomorphic extension $\FF_n$ on every standard power domain.

Under this assumption, I can generalize the notion of asymptotic expansion with monomials in $M$ to standard power domains:


Let $f \in \C$ and $F \in \Gs{\RR}{M}$ have $M$-natural support. The germ $f$ has strong asymptotic expansion $F$ if there exists a standard power domain $\Omega$ such that

  1. $f$ has holomorphic extension $\ff$ on $\Omega$, and
  2. the condition $\ff – \FF_n \prec_\Omega \nn$ holds for each $n \in M$.


  1. The set $\C(M)_{\st}$ of all $f \in \C$ that have a strong asymptotic expansion in $\Gs{\RR}{M}$ is an $\RR$-algebra.
  2. Every $f \in \C(M)_{\st}$ has exactly one asymptotic expansion $T_M(f)$ in $\Gs{\RR}{M}$, and the map $T_M:\C(M)_{\st} \into \Gs{\RR}{M}$ is an $\RR$-algebra homomorphism.

By the Uniqueness Principle, every $f \in \C(M)_{\st}$ with $T_M f = 0$ must be 0; hence $T_M$ is injective. Since every truncation of a series with $M$-natural support is itself a germ in $\C(M)_{\st}$, it follows that $\left(\C(M)_{\st}, M, T_M\right)$ is a qaa algebra.

Indeed, there is more:


A qaa algebra $(\A,M,T)$ is strong if there exists a standard power domain $\Omega$ such that $f$ and each $g_n:= T^{-1}\left((Tf)_n\right)$, for $n \in M$, have holomorphic extensions $\ff$ and $\gg_n$ on $\Omega$ satisfying $$\ff – \gg_n \prec \nn.$$


If $M$ is a pure scale on standard power domains, then the qaa algebra $\left(\C(M)_{\st}, M, T_M\right)$ is strong.

I can now start the construction for $L_0$.

Step 0: Set $$\A_0:= \set{f \in \C(L_0)_{\st}:\ T_{L_0} f \in \Ps{R}{L_0}} \quad\text{and}\quad T_0:= T_{L_0} \rest{\A_0}.$$


The algebra $\A_0$ contains all real analytic functions in the logarithmic chart, that is, if $P \in \Ps{R}{X}$ converges, then $P \circ \exp^{-1} \in \A_0$. As discussed here, the algebra $\A_0 \circ (-\log)$ contains every transition map near a non-resonant, hyperbolic singularity of a planar, real analytic vector field.

Note also that, by definition of (strong) asymptotic expansion and the convention that $T_0f \in \Ps{R}{L_0}$ for $f \in \A_0$, every $f \in \A_0$ has a bounded holomorphic extension on some standard power domain.

Finally, let $\F_0$ be the field of fractions of $\A_0$ and extend $T_0$ accordingly; then $(\F_0,L_0,T_0)$ is a strong qaa field.


Show that $(\F_0, L,T_0)$ is also a strong qaa field.

2 thoughts on “Ilyashenko algebras based on definable monomials: the construction (base step)

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.