Approximate ultrahomogeneity in lattices
Abstract.
We show that for with , the doubly atomless separable Banach lattice is approximately ultrahomogeneous (AUH) over the class of its finitely generated sublattices. The above is not true when . However, for any , is AUH over the finitely generated lattices in the class of bands of lattices.
1. Introduction
In this paper, we explore the homogeneity properties (or lack thereof) of the class of lattices under various conditions.
The following is taken from [6]: A Banach lattice is an abstract lattice if there is a measure space such that can be equipped with an -module and a map such that
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For all and , ,
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For all and , .
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For all ,
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If and are disjoint, , and if , then .
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For all , .
When the abstract space is separable, it has a concrete representation: Suppose and are measure spaces. Denote by the space of Bochner-measurable functions such that the function , with for , is in . The class of bands in lattices, which we denote by , has certain analogous properties to those of spaces, particularly with respect to its isometric theory.
lattices (and their sublattices) have been extensively studied for their model theoretic properties in [6] and [7]. It turns out that while abstract lattices themselves are not axiomatizable, the larger class is axiomatizable with certain properties corresponding to those of spaces. For instance, it is known that the class of atomless lattices is separably categorical, meaning that there exists one unique atomless separable lattice up to lattice isometry. Correspondingly, the class of doubly atomless lattices is also separably categorical; in particular, up to lattice isometry, , which throughout will just be referred to as , is the unique separable doubly atomless lattice (see [7, Proposition 2.6]).
Additionally, when , the lattice isometries of lattices can be characterized in a manner echoing those of linear isometries over spaces (with ). Recall from [1, Ch. 11 Theorem 5.1] that a map is a surjective linear isometry iff , where is a measure-preserving transformation and is related to through Radon-Nikodym derivatives. If we want to be a lattice isometry as well, then we also have positive (and the above characterization will also work for ). In [3] (for the case of ) and [13], a corresponding characterization of linear isometries is found for spaces of the form , for certain and Banach spaces . In particular, for lattices with : given , where is understood as a map from to , any surjective linear isometry is of the form
where is a set isomorphism (see [3] and [13] for definitions) is a measurable function related to via Radon-Nikodym derivatives, and is a Bochner-measurable function from to the space of linear maps from to itself such that for each , is a linear isometry over .
In [11], Raynaud obtained results on linear subspaces of spaces, showing that for , some linearly isomorphically embeds into iff it embeds either to or to . However, when , for , the space isometrically embeds as a lattice in , and for any -convex and -concave Orlicz function , the lattice embeds lattice isomorphically into . Thus, unlike with lattices whose infinite dimensional sublattices are determined up to lattice isometry by the number of atoms, the sublattices of are not so simply classifiable.
In fact, the lattice isometry classes behave more like the linear isometries, at least along the positive cone, as is evident in certain equimeasurability results for lattices. In [11], Raynaud also obtained the following on uniqueness of measures, a variation of a result which will be relevant in this paper: let , and suppose two probability measures and on are such that for all ,
Then . Linde gives an alternate proof of this result in [8].
Various versions and expansions of the above result appear in reference to spaces: for instance, an early result from Rudin generalizes the above to equality of integrals over : ([12]). Assume that with , and suppose that for all ,
Then . An application of this result is a similar condition by which one can show that one collection of measurable functions , with is equimeasurable with another collection By defining and as pushforward measures of and . In the case of spaces, if and are corresponding basic sequences whose pushforward measures satisfy the above for , then they generate isometric Banach spaces. Raynaud’s result shows the converse is true for . A similar result in from [7] holds for under certain conditions, except instead of equimeasurable and , when the ’s and are mutually disjoint and positive and the map generates a lattice isometry, and are equimeasurable.
Recall that a space is approximately ultrahomogeneous (AUH) over a class of finitely generated spaces if for all appropriate embeddings with , for all generated by , and for all , there exists an automorphism such that for each , .
In the Banach space setting, the embeddings are linear embeddings and the class of finitely generated spaces are finite dimensional spaces. In the lattice setting, the appropriate maps are isometric lattice embeddings, and one can either choose finite dimensional or finitely generated lattices.
The equimeasurability results described above can be used to show an approximate ultrahomogeneity of over its finite dimensional linear subspaces only so long as (see [10]). Conversely, the cases where are not AUH over finite dimensional linear subspaces, with counterexamples showing linearly isometric spaces whose corresponding basis elements are not equimeasurability. Alternate methods using continuous Fraïssé Theory have since then been used to give alternate proofs of linear approximate ultrahomogeneity of for (see [5]) as well as lattice homogeneity of for all (see [2], [5]).
This paper is structured as follows: in section 2, we first establish basic notation and give a characterization of finite dimensional lattices. This characterization is used in subsequent sections for establishing both equimeasurability and ultrahomogeneity results later on.
In section 3 we show that when , is AUH over the larger class of finite dimensional (and finitely generated) spaces. This is done by characterizing representations of sublattices in such a way that induces automorphisms over making the homogeneity diagram commute. The results here play a role in subsequent sections as well.
In section 4, we prove that if in addition , is also AUH over the class of its finitely generated sublattices. First, we determine the isometric structure of finite dimensional sublattices of lattices by giving an alternate proof of [7, Proposition 3.2] showing that two sublattices and of , with the ’s and ’s each forming the basis of atoms, are lattice isometric iff and are equimeasurable. The equimeasurability result allows us to reduce a homogeneity diagram involving a finite dimensional sublattice of to one with a finite dimensional lattice, from which, in combination with the results in section 3, the main result follows.
Section 5 considers the case of . Here, we provide a counterexample to equimeasurability in the case that and use this counterexample to show that in such cases, is not AUH over the class of its finite dimensional lattices.
2. Preliminaries
We begin with some basic notation and definitions. Given a measurable set , we let refer to the characteristic function over . For a lattice , let be the unit ball, and be the unit sphere.
For elements in some lattice , use bracket notation to refer to the Banach lattice generated by the elements . In addition, we write without the subscript to denote that the generating elements are also mutually disjoint positive elements in the unit sphere. Throughout, we will also use boldface notation to designate a finite sequence of elements: for instance, for or for some lattice , let . Use the same notation to denote a sequence of functions over corresponding elements: for example, let , or , or . Finally, for any element or tuple of elements in some lattice , let and be the band generated by and in , respectively.
Recall that Bochner integrable functions are the norm limits of simple functions , with , where and are the characteristic functions for and , respectively. One can also consider as a -measurable function such that
Unlike the more familiar lattices, the class of abstract lattices is not itself axiomatizable; however, the slightly more general class of bands in lattices is axiomatizable. Additionally, if is a separable lattice, it is lattice isometric to a lattice of the form
lattices may also contain what are called base disjoint elements. and are base disjoint if . Based on this, we call a base atom if whenever with and base disjoint, then either or . Observe this implies that is an atom in . Alternatively, we call a fiber atom if any disjoint are also base disjoint. Finally, we say that is doubly atomless if it contains neither base atoms nor fiber atoms.
Another representation of involves its finite dimensional subspaces. We say that is an lattice, with if for all disjoint and , there is a finite dimensional of that is -isometric to a finite dimensional space containing such that for each , . Henson and Raynaud proved that in fact, any lattice is a space iff is (see [6]). This equivalence can be used to show the following:
Proposition 2.1.
(Henson, Raynaud) If is a separable lattice, then it is the inductive limit of finite dimensional lattices.
The latter statement is not explicitly in the statement of Lemma in [6], but the proof showing that any lattice is was demonstrated by proving the statement itself.
Throughout this paper, we refer to this class of finite dimensional lattices as . Observe that if , then it is of the form where for , the atoms generate .
Proposition 2.2.
Let be a sublattice of with atoms as described above. Then the following are true:
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There exist disjoint measurable such that for all , ,
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For all and for all , .
Conversely, if is a finite dimensional sublattice of satisfying properties (1) and (2), then is in .
In order to prove this theorem, we first need the following lemma:
Lemma 2.3.
Let , with . suppose are such that
Then the ’s are mutually disjoint.
Proof.
If , then
(1) |
Now observe that for all , , with equality iff either or , so . Combined with the above equality in line (1), since it follows that a.e., so must be disjoint from when .
If , proceed as in the proof for , but with the inequalities reversed, given that in this instance for all .
∎
Remark 2.4.
The above implies that a lattice is base atomless if it contains no bands lattice isometric to some or space. Indeed, if there were a base atom , then any two would have to have -norms multiple to each other, so is lattice isometric to . Resultantly, the band generated by is an space. Similarly, if is a fiber atom, then any is also base disjoint, which implies that the band generated by is an space.
We now conclude with the proof of Proposition 2.2:
Proof of Proposition 2.2.
Observe that for each appropriate pair ,
For notational ease, let . Pick with each . Then, by disjointness of the ’s, for all and all ,
Now since the ’s are isometric to ,
Since the ’s are all positive and , by Lemma 2.3, the ’s are disjoint, that is, the are base disjoint.
For , let be mutually disjoint measurable sets each supporting each for . Then each is supported by . Now we prove (2). Fix , Then using similar computations as above, and since the ’s for fixed generate :
By Minkowski’s inequality, as , equality occurs only when a.e. for all .
To show the converse, it is enough to give the computation:
∎
The following results will allow us to reduce homogeneity diagrams to those in which the atoms of some are mapped by both embeddings to characteristic functions of measurable . In fact, we can further simplify such diagrams to cases where is generated by such ’s which additionally are base-simple, i.e., is a simple function.
Proposition 2.5.
Let and let be an element with full support over . Then there exists a lattice automorphism from to itself such that . Furthermore, can be constructed to bijectively map both simple functions to simple functions and base-simple functions to base-simple functions.
Proof.
The proof is an expansion of the technique used in Lemma 3.3 from [5]. Given a function , define by , and for notation, use . Since has full support, we may assume that for all , . From there, Define by
and the rest of the function definition is a composition, so is a lattice homomorphism. To show it is also an isometry, simply compute the norm, using substitution in the appropriate places:
To show surjectivity, let be a measurable set. Note that any can be expressed as for some , since is an increasing continuous function from 0 to 1, while is continuously increasing from 0 to . Thus there exists such that , implying that ’s image is dense in the band generated by since has full support. Therefore, is also surjective.
Finally, consists of function composition into multiplied by , so if and are simple, then it has a finite image, so if is simple, then the product is also simple, maps simple functions to simple functions, Conversely, if is simple, then is also simple. Thus has a finite image. It follows that itself has a finite image.
Using similar reasoning, if is simple, then whenever is simple, must also be simple, and likewise the converse is true, since by the computation above, . ∎
3. Approximate Ultrahomogeneity of over spaces
In this section, we show that for any , is AUH over .
Let and be sequences of measurable functions and let be a measure in . Then we say that and are equimeasurable if for all -measurable ,
We also say that functions and in are base-equimeasurable if and are equimeasurable.
Lusky’s main proof in [10] of linear approximate ultrahomogeneity in for hinges on the equimeasurability of generating elements for two copies of some in containing . But when , there exist finite dimensional such that two linearly isometric copies of in do not have equimeasurable corresponding basis elements. However, if homogeneity properties are limited to with mutually disjoint basis elements, then is linearly isometric to , and for all , is AUH over all spaces. Note that here, an equimeasurability principle (albeit a trivial one) also applies: Any two copies of into with have (trivially) equimeasurable corresponding basis elements to each other as well.
In the setting, similar results arise, except rather than comparing corresponding basis elements of isometric copies of , equimeasurability results hold in the -norms under similar conditions, with finite dimensional lattices taking on a role like does in spaces.
The following shows that equimeasurability plays a strong role in the approximate ultrahomogeneity of by showing that any automorphism fixing preserves base-equimeasurability for characteristic functions:
Proposition 3.1.
Suppose , and let be a lattice automorphism with . Then there exists a function and a measure preserving transformation over such that for a.e. , is also a measure preserving transformation inducing an isometry over , and for all ,
Furthermore, for all measurable with ’s mutually disjoint, and are base-equimeasurable.
Proof.
By the main result in [13], there exists a strongly measurable function , a set isomorphism over (see [13] for a definition on set isomorphisms), and some related to the radon-Nikodym derivative of such that
and for a.e. , is a linear isometry over . Observe first that sends any characteristic function constant over to characteristic function for some , so since is constant over , we can just refer to it as . Also, since is a lattice isometry, , so is measure preserving. Finally, observe that . Thus, for any simple function constant over with the ’s mutually disjoint, we have , and . Then for all ,
It follows that . We can thus adjust by multiplying by where . Note also that acts as a lattice isometry over when restricted to elements constant over , so by Banach’s theorem in [1], the map can be interpreted as , where is a measure preserving transformation over inducing . By Banach’s theorem again for , this can be interpreted by , with a measure preserving transformation for a.e. . But since , this as well.
It remains to prove equimeasurability. Let , and observe that since for a.e. , is a measure preserving transformation inducing a lattice isometry over , it follows that
While
Thus for each with measurable, since is also a measure preserving transformation,
and we are done.
∎
The following theorem describes a comparable equimeasurability property of certain copies of in for any :
Theorem 3.2.
Let , and suppose that are lattice embeddings with generated by a -indexed collection of atoms with and as described in Proposition 2.2. Suppose also that Then and are base-equimeasurable.
Proof.
Let , and note first that each is of the form for some measurable . Second, with measurable for a.e. , so by Proposition 2.2, for each fixed and each , with almost disjoint. It follows that for each appropriate , , so .
To show equimeasurability, observe that for a.e. , we have iff , and 0 otherwise. Let be a measurable set. Note then that any -indexed sequence is of the form with for some unique , and otherwise. It follows then that for some ,
Since the above holds independent of our choice of , we are done.
∎
Remark 3.3.
The above proof shows much more than base-equimeasurability for copies of lattices in . Indeed, if with , then each atom is in fact base-simple, and where . Furthermore, there exist measurable sets partitioning with such that . Based on this, we can come up with a ”canonical” representation of , with , where
This canonical representation will become relevant in later results.
Having characterized representations of lattice in , we now move towards proving the AUH result. Before the final proof, we use the following perturbation lemma.
Lemma 3.4.
Let be a lattice embedding of a lattice . Then for all , there exists an embedding such that fully supports and .
Proof.
Let . For each , we will construct disjoint from with support in . Let be the elements in disjoint from . Starting with , Observe that can be partitioned by . Let
When , let as well. Now, let be the lattice homomorphism induced by
First, we show that is an embedding. Observe that for each ,
It easily follows that is in fact isometric to , and thus to . Furthermore, for every ,
The above can get arbitrarily small.
Now, if , let , and we are done. Otherwise, let , and observe that fully supports . Observe also that . However, both and are lattice isometric to itself. So there exists an isometric copy of fully supporting . Let be the corresponding basic atoms of this copy, and let . Then for ,
Using similar reasoning as in the definition of , one also gets , so can also arbitrarily approximate .
∎
Observe that the lemma above allows us to reduce the approximate homogeneity question down to cases where the copies of a lattice fully support . Combined with Proposition 2.5, we can further reduce the possible scenarios to cases where for each , for some . It turns out these reductions are sufficient for constructing a lattice automorphism that makes the homogeneity diagram commute as desired:
Theorem 3.5.
Suppose , and for , let be a lattice embedding with and and . Suppose also that each fully supports . Then there exists a lattice automorphism over such that .
Proof.
Let ; by Proposition 2.5, we can assume that for both ’s, we have . For notation’s sake, let . By Proposition 2.2, for each there exist mutually disjoint sets partitioning such that for each , . In addition, for the sets , where , partition . It follows also from the statements in Remark 3.3 that for each and .
To prove the theorem, it is enough to generate lattice automorphisms mapping each band to a corresponding band where and are defined as in Remark 3.3, with .
To this end, we make a modified version of the argument in [7, Proposition 2.6] and adopt the notation in Proposition 2.5: construct lattice isometries from to with
By similar reasoning as in the proof of Proposition 2.5, is a lattice embedding. Surjectivity follows as well. Indeed, since , for a.e. the function matches continuously to with mapped a.e. surjectively to . So ’s image is dense in .
Observe that also preserves the random norm along the base (that is: . Resultantly, the function mapping to is also a lattice automorphism. Indeed, for with , one gets
Now let , and observe that given with , since the ’s are base disjoint, we have
Thus is a lattice automorphism over mapping each to .
Use [5, Lemma 3.3] to construct a lattice isometry such that for each , . By [1, Ch. 11 Theorem 5.1] this isometry is induced by a measure preserving transformation from [0,1] to itself such that . It is easy to show that induces a lattice isometry with . In particular, we have , and , now let , and we are done.
∎
Using the above, we can now show:
Theorem 3.6.
For , the lattice is AUH for the class .
Proof.
Remark 3.7.
Observe that the doubly atomless space is unique among separable spaces that are AUH over . Indeed, this follows from the fact that such a space must be doubly atomless to begin with: let be a one dimensional space generated by atom and suppose is not doubly atomless. Suppose also that is embedded by some into a part of supported by some or band, and on the other hand is embedded by some into with a unit in . Then one cannot almost extend to some lattice embedding with almost commutativity.
One can also expand this approximate ultrahomogeneity to separable sublattices with a weaker condition of almost commutativity in the diagram for generating elements: for any sublattice generated by elements , for any , and for all lattice embedding pairs , there exists a lattice automorphism such that for all , .
Theorem 3.8.
For all , The lattice is AUH for the class of finitely generated lattices.
Proof.
Let , and let be lattice embeddings. We can assume that for each . By Proposition 2.1, is the inductive limit of lattices in . Given , pick a lattice such that for each , there is some such that . Each is an embedding into , so pick an automorphism over such that . Then
∎
We can also expand homogeneity to include not just lattice embeddings but also disjointness preserving linear isometries, that is, if embeddings are not necessarily lattice homomorphisms but preserve disjointness, then there exists a disjointness preserving linear automorphism over satisfying almost commutativity:
Corollary 3.9.
is AUH over finitely generated sublattices in with disjointness preserving embeddings.
4. Approximate Ultrahomogeneity of when
The above results largely focused approximate ultrahomogeneity over lattices. What can be said, however, of sublattices of spaces? The answer to this question is split into two cases: first, the cases where , and the second is when . We address the first case in this section. It turns out that if , then is AUH for the class of its finitely generated sublattices. The argument involves certain equimeasurability properties of copies of fixed finite dimensional lattices in . Throughout, we will refer to the class of sublattices of spaces in as simply , and let be the class of finitely generated sublattices of .
The following result appeared as [7, Proposition 3.2], which is a multi-dimensional version based on Raynaud’s proof for the case of (see [11, lemma 18]). The approach taken here is a multi-dimensional version of the proof of Lemma 2 in [8].
Theorem 4.1.
Let , and suppose are lattice isometric embeddings with . Suppose also that for some . Then and are base-equimeasurable.
Throughout the proof, let be a measure in some interval . To this end, we first show the following:
Lemma 4.2.
Suppose , and are positive finite Borel measures on such that for all with ,
Then .
Proof.
It is equivalent to prove that the signed measure . First, observe that since , and for any , .
Now, we show by induction on polynomial degree that for all , , and for all multivariate polynomials of degree ,
This is true for the base case by assumption. Now assume it is true for and let with . For notational ease, let . Then for each and ,
Now, if and , then
Since in this case, and , the right hand side must also be -integrable. On the other hand, If , then we have
which is also -integrable. So now we apply Lebesgue’s differentiation theorem over to get, for any and for each :
since . A similar argument, deriving over , can be made to show that
One can make linear combinations of the above, which implies line .
Now for fixed , we define a measure on , where for measurable
where . It is sufficient to show that . Observe first that is continuous and injective; indeed, if , then it can be shown that . Thus , implying that . Resultantly, for any Borel is also Borel, hence we will have shown that for any such , as well, so .
Observe that by choice of and and since for all , have
Using simple functions and the definition of , one can show both that for each , we have
and also that
More generally, if , then
So it follows that for all polynomials .
Now if and , since , we then we have
so
Thus for each , . So by [4, Theorem 5.2], is the unique positive measure over with moment values . Since yields the same values, and by **, , it follows that , so .
∎
Now we are ready to prove Theorem 4.1.
Proof.
For simplicity of notation, let and . By definition of , the support of as well as of is the unit interval. Define positive measures by
Now, for any measurable , we have
so for any simple function over ,
Using simple functions to approximate , and given that is in and the support of is the unit interval, it follows that
It is sufficient now to show that for all ,
.
For and , let , and let . By assumption, with , so . Therefore, since each is an embedding, for all ,
Let : then in particular it follows that for all , we have
By Lemma 4.2, we can conclude that , so and are equimeasurable. ∎
Using Theorem 4.1, we can uniquely characterize lattices in in a way that parallels Proposition 2.2.
Theorem 4.3.
Suppose that , and let with . Then the following hold:
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iff there exist mutually disjoint measurable functions , with and such that for each , , and .
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Suppose is a lattice embedding with and Then there exist embeddings extending such that .
Proof.
For part 1, clearly the reverse direction is true. To prove the main direction, we can suppose that fully supports . If not, recall that the band generated by is itself doubly atomless, and hence is lattice isometric to itself. Thus, if under these conditions, there is a sublattices extending as in the statement of the theorem, it will also be the case in general.
By Proposition 2.5, we can also suppose that . Now by assumption, since , then there is an embedding such that each , with . Without loss of generality we may also drop any ’s disjoint from and assume that fully supports . Now is a lattice admitting a canonical representation in as described in Theorem 3.2 and Remark 3.3.
So we can assume that embeds into in such a way that fully supports it and each is both simple and base-simple. Now, use Proposition 2.5 to adjust into an automorphism over such that in a way that preserves both simplicity and base-simplicity. By Theorem 4.1, and are base-equimeasurable. Since the are base-simple, there exist tuples such that for a.e. , there is some such that . By equimeasurability, the same is true for .
Let , and let . Let with defined similarly. Note that each is also base-characteristic, as for some , so for fixed and for any , we must have that and are scalar multiples of each other. Thus for each appropriate pair with , define by . By definition of , for any and any appropriate , and are fiber-disjoint, and . Thus by Proposition 2.2, .
To prove part 2, Observe first that we have already essentially proven part 2 in the case that and . To show the general case, we first assume that for each , maps to . Now, by Theorem 4.1, and are also base-equimeasurable, but by the procedure for part 1, we also know that each is also base-simple. Define as above, and Let . Define similarly and the associated characteristic functions for appropriate pairs such that and . Note first that
Second, observe that by equimeasurability, the eligible pairs are the same for . Let . Clearly , and since the eligible pairs are the same, and are isometric to each other. Let be one of the ’s and let be the expected embedding mapping to , and we are done. ∎
From here, we can now easily extend Theorem 3.5 to lattices in :
Corollary 4.4.
Suppose and suppose are lattice embeddings from with fully supporting . Then there exists a lattice automorphism over such that .
Proof.
Use Theorem 4.3 to generate a lattice containing and lattice embeddings such that . Clearly each fully supports . Now apply Theorem 3.5 to generate an automorphism over with . Clearly as well.
∎
When , using Theorem 4.3, we can show that the same holds with the more general class . However, we can make an even stronger claim by showing that homogeneity holds for any finite dimensional sublattice of . This is done using the following result, which gives a standard way of approximating finite dimensional sublattices of with lattices in .
Lemma 4.5.
Suppose , and let be embeddings with . Then for all , there exists a lattice and embeddings such fully supports and for each , .
Proof.
We can assume each fully supports : given , use Lemma 3.4 to get copies of sufficiently close to each with full support. We then also assume that using Proposition 2.5.
By Theorem 4.1, and are base-equimeasurable. In particular, given any measurable , one has . Now pick an almost disjoint partition of , where each is closed, has relatively non-empty interior, and is of diameter less than . Let . Then by equimeasurability, . For each , pick some , and for each , let
Observe that , and for .
Consider now the lattice . Now, for any linear combination , we have, as in the argument in Proposition 2.5, that
implying that . It follows both that embeds into , implying that it is a lattice, and it is isometric to the lattice generated by the ’s. Let , and define as the maps generated by . Clearly these are lattice embeddings and .
∎
Theorem 4.6.
For all with , the lattice is AUH for the class of finite dimensional sublattices of lattices.
Proof.
In a manner similar to that of Theorem 3.8, we can also extend the AUH property to finitely generated sublattices of as well:
Theorem 4.7.
For all with , The lattice is AUH for the class of its finitely generated lattices.
Proof.
Suppose is finitely generated. Then since is order continuous and separable, it is the inductive limit of finite dimensional lattices as well, so pick a finite dimensional with elements sufficiently approximating the generating elements of , and proceed with the same proof as in Theorem 3.8. ∎
The argument used in Corollary 3.9 can also be used to show:
Corollary 4.8.
For , is disjointness preserving AUH over .
Remark 4.9.
for is AUH over the entire class of its finitely generated sublattices, a property which is equivalent to such a class being a metric Fraïssé class with as its Fraïssé limit. Recall that a class of finitely generated lattices is Fraïssé if it satisfies the following properties:
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(1)
Hereditary Property (HP): is closed under finitely generated sublattices.
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(2)
Joint Embedding Property (JEP): any two lattices in lattice embed into a third in .
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(3)
Continuity Property (CP): any lattice operation symbol are continuous with respect to the Fraïssé pseudo-metric in [2, Definition 2.11].
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(4)
Near Amalgamation Property (NAP): for any lattices , and in with lattice embeddings , and for all , there exists a and embeddings such that
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(5)
Polish Property (PP): The Fraïssé pseudo-metric is separable and complete in (the -structures generated by many elements).
Now clearly the finitely generated sublattices of fulfill the first two properties, and the third follows from the lattice and linear operations having moduli of continuity independent of lattice geometry. In addition, if one can show that the class has the , has some separable which is universal for , and its NAP amalgamate lattices can be chosen so that they are closed under inductive limits, then one can prove that also has the Polish Property (a technique demonstrated in [14, Theorem 4.1] and more generally described in Section 2.5 of [9]). The main difficulty in proving that a class of lattices is a Fraïssé class is in showing that it has the NAP. However, thanks to Theorem 4.7, we have
Corollary 4.10.
has the NAP.
Theorem 4.7 implies an additional collection of AUH Banach lattices to the currently known AUH Banach lattices: namely for ,the Gurarij M-space discovered in [5], and the Gurarij lattice discovered in [14].
However, if one considers classes of finite dimensional Banach spaces with Fraïssé limits using linear instead of lattice embeddings, the only known separable AUH Banach spaces are the Gurarij space and for , and it is currently unknown if there are other Banach spaces that are AUH over its finite dimensional subspaces with linear embeddings. Certain combinations of and are also ruled out for as a potential AUH candidate as discussed in Problem 2.9 of [5]: in particular, when , cannot be linearly AUH.
5. Failure of homogeneity for
Recall that when is embedded into through , then we can achieve almost commutativity for any . However, the automorphism in Theorem 3.6 clearly preserves the equimeasurability of the generating basic atoms of as it fixes .
In this section, we show that the results of Section 4 do not hold when . The first results in this section show that when some is sufficiently close to , the automorphism originally used in the argument of Proposition 2.5 sending to also perturbs selected functions piecewise continuous on their support in a controlled way. Second, Theorem 4.1 does not hold, and thus we cannot infer equimeasurability for arbitrary finite dimensional sublattices of . Finally, we use these results to strengthen the homogeneity property for any lattice assumed to be AUH, and then show that when , does not fulfill this stronger homogeneity property, and thus cannot be AUH.
Lemma 5.1.
Let , and let be such that . Suppose also that for a.e. , where each has finitely many discontinuities. Let , and let fully support . Consider
which is the lattice isometry defined in Proposition 2.5 mapping to .
Then there exists such that if , then for each , we have that .
Proof.
We can assume . Let be a closed set such that for , is continuous and . Pick such that for any , if , then . Now, let be such that , and suppose . Observe that for each , we have . For each , let
Observe that , so it is enough to show that is sufficiently small as well.
To this end, first note that since is being composed with increasing continuous functions in both arguments, each is also the characteristic function of an interval: indeed, we have piecewise continuous with and such that for each , . Also observe that for , we have . In addition, as
Where . The above set up, in combination with the triangle inequality properties of , leads us to the following inequalities:
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•
For all , .
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For all , .
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For all and , .
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For all and , if , then (which implies with the above that ).
We now show that the above implies that . Observe first that for all , if it must be because, but , or vice versa. In either case, it can be shown that either or . Suppose and (a similar proof will work in the case that . Then since , , and ,
It follows then that accounting for both ends of the interval and for , we have . Resultantly,
which can be made arbitrarily small.
∎
Theorem 5.2.
Let and suppose is AUH over its finite dimensional sublattices. Let be lattice embeddings with such that for some . Then for all , there exists an automorphism fixing such that .
Proof.
Assume the above, and pick , where with and intervals such that and for each there is such that .
Since is AUH, there exists an automorphism such that , where satisfies the conditions for and each of the ’s in in Lemma 5.1. Now pick the automorphism over mapping to as defined in Lemma 5.1. It follows that for each , , so . Thus for each ,
Now let to obtain the desired automorphism; then
∎
The above can be used to show that if is AUH and contains for , then we can induce almost commutativity with automorphisms fixing as well. This will allow us to reduce possible automorphisms over to those that in particular fix . The importance of this result is that these particular homomorphisms fixing must always preserve base-equimeasurability for characteristic functions, as shown in Proposition 3.1. Thus a natural approach in disproving that is AUH would involve finding sublattices containing which are lattice isometric but whose generating elements are not base-equimeasurable. The following results do exactly that:
Lemma 5.3.
Lemma 4.2 fails when . In particular, there exists a non-zero measure , with and positive measures such that for all polynomials of degree ,
Remark 5.4.
.
Here we provide another example over the unit interval:
Proof.
Fix such an , and define a polynomial of degree with such that for all , . This can be done by finding a non-trivial in the null set of the size matrix with . Then let . Let and . Clearly and are finite positive Borel measures, but since , . ∎
Lemma 5.5.
Let . Then there exists a two dimensional lattice and lattice embeddings with such that and are not base-equimeasurable.
Proof.
Let be a polynomial of degree at least as defined in Lemma 5.3 such that for all , , and . Let , and let . Note that each , and furthermore that . Additionally, each map is strictly increasing with and . Now we will construct characteristic functions such that the linear map induces an isometry, but and are not base-equimeasurable. From there, we let , and let be the lattice isometry induced by ,
To this end, let
Observe that . Indeed, one can show that the associated push forwards for each have the corresponding equivalence:
So and are not equimeasurable. However, For , , so it follows from the construction of the ’s that
Thus for any , since and are both positive, we have
To conclude the proof, let , and let . Clearly . ∎
Theorem 5.6.
If and , then is not AUH for the class of its finite dimensional sublattices.
Proof.
Fix , and let be the 2-dimensional lattice generated in Lemma 5.5, with embeddings mapping to copies of such that and are not base-equimeasurable. In addition, by assumption . For notational ease, let .
Suppose for the sake of contradiction that is AUH. Pick some measurable and such that
where
By Theorem 5.2, there is some lattice automorphism fixing such that . Let . By Proposition 3.1, preserves base-equimeasurability, so for any measurable ,
By the properties of , we also have . It also follows that
so , but this contradicts the assumption (*). So Theorem 5.2 cannot apply, implying that is not AUH as desired. ∎
Remark 5.7.
For , is the unique lattice that is separably AUH over finitely generated spaces, since up to isometry it is the unique doubly atomless space. In light of Theorem 5.6, this implies that the class of finitely generated sublattices of is not a Fraïssé class as defined in [2], as is the only possible candidate as a Fraïssé limit.
In particular, lacks the NAP. Indeed, otherwise, one can use that NAP with amalgamate lattices and [7, Proposition 2.8] to situate a -Cauchy sequence into a Cauchy-sequence of generating elements in an ambient separable lattice. Thus would also have the Polish Property, implying that is a Fraïssé class. Since the only possible candidate Fraïssé limit space is itself, this would contradict Theorem 5.6.
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