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Let denote the Liouville closure of the field of
constants
in say transseries.
Question
closed under composition?
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 non-zero
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 of
is
closed under composition.
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 non-zero 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
real-closed, 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 closed under
composition (instead of
),
provided that
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/Marker-Miller. I.e. given
,
is there an
with
for
sufficiently large
?
Proposition has a level.
Proof. This can be deduced from a result of Joris in
his Transserial Hardy fields paper [1]. Consider
the field of grid-based transseries. Let
denote the subfield of
of
transseries. This is an
-free,
Newtonian, Liouville-closed H-field with small derivation. By [1,
Theorem 5.12], there is a Hardy field
closed
under
and
and an
isomorphism
In particular is H-closed. 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
d-algebraic. 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].