
Let denote the Liouville closure of the field of constants in say transseries.
Question
Before answering the question in the positive, let me introduce an equivalent definition of Liouville extensions. Given two Hardy fields , the extension is a Liouville extension if for all , there are an and nonzero germs with , such that for each , one of the following occurs:
is algeraic over ,
, or
.
Given , we define to be the least such that such a decomposition of length exists. Note that if and only if . Moreover, if in the situation above, then we have for all , which is why I use this presentaion of Liouville extensions.
Proposition
Proof. Let us prove by induction on that for all and , we have . This is immediate when , since then is a constant. Let such that the result holds for germs of rank and let with .
There are nonzero germs with , and where for all , one of the following occurs:
is algeraic over ,
,
Note that for each , we have whence . We distinguish three cases.
Then is algebraic over , whence algeraic over . But is realclosed, so .
Then
Since is closed under integration, it follows that .
As in the previous case . We deduce since is closed under exponential integration that .
This concludes the inductive proof.
Remark
which of course is a problematic inclusion on its own.
Let denote Boshernitzan's Hardy field, i.e. is the intersection of all maximal Hardy fields. In one of your lectures, I asked you if it was known whether every positive infinite germ in has a level in the sense of Rosenlicht/MarkerMiller. I.e. given , is there an with for sufficiently large ?
Proposition
Proof. This can be deduced from a result of Joris in his Transserial Hardy fields paper [1]. Consider the field of gridbased transseries. Let denote the subfield of of transseries. This is an free, Newtonian, Liouvilleclosed Hfield with small derivation. By [1, Theorem 5.12], there is a Hardy field closed under and and an isomorphism
In particular is Hclosed. Let and assume for contradiction that . Let be a maximal Hardy field containing . We have by definition of . Now must be transcendant over , hence also over . This contradicts Boshernitzan's result that each element of is in fact dalgebraic. Thus . In particular, the field embeds into as an ordered exponential field, so each has a level .
As far as I know, the only Hardy field with composition which is known not to have levels in this sense is that which is defined in Adele Padgett's forthcoming thesis [2].