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Let $(\mathcal{A}, \cup, \cap)$ a lattice (with minimum and maximum elements $\bot$ and $\top$). Let $X\subset \mathcal{A}$ a generator set (a set of minimal cardinality that generate $\mathcal{A}$ i.e. such that $\mathcal{A}$ is the intersection of all sublattices containing $X$).

I consider $\mathbb{R}[\mathcal{A}]$ the real vector space with bases the elements of $\mathcal{A}$, then its elements if of like: $r_1\cdot A_1+\ldots r_n\cdot A_n$ where $r_i$ is a real number and $A_i\in \mathcal{A}$.

If $\mathcal{A}$ is a sublattice of the class $\mathcal{P}(X)$ of subsets of a set $X$, we have (see $[1]$) a exact sequence of real vector spaces:

$0\to I\to \mathbb{R}[\mathcal{A}] \xrightarrow{q} Simp(\mathcal{A})\to 0 $

where $Simp(\mathcal{A})\subset \textbf{R}^X$ is the $\mathbb{R}$-vector subspace of generated of simple functions with support on $\mathcal{A}$ i.e. of the functions of type $r_1\cdot \chi_{A_1}+\ldots r_n\cdot \chi_{A_n}$ where $r_i\in \mathbb{R}, A_i\in \mathcal{A}$ for $1\leq i\leq n$ and $n\in\mathbb{N}$, and $\chi_A(X)=\{0,1\},\ \chi_{A}^{-1}(\{1\})=A$. Where $q$ is the natural map, and $I$ is the real vector subspace generated by the equalities like $\bot\equiv 0$ and:

*) $A+B \equiv A\cup B + A\cap B$ for $A, B\in \mathcal{A}$ .

In fact from $[1]$ (p. 28) a linear map $f: \mathbb{R}[\mathcal{A}]\to V$ has a (linear) extension to $Sim[\mathcal{A}]$ iff it is modular i.e. if it null on the elements of $I$.

From $[2]$ (p.60, 61, 36 (prop. 2.1.2(x)) we have that $I$ is also generated by the equalities like:

**) $A_1+\ldots A_n \equiv B_1+\ldots B_m$ where $A_1,\ldots A_n,\ B_1,\ldots B_m\in \mathcal{A}$ (may be with some repeated elements) such that: $\bigcup_{1\leq i_1<\ldots < i_k\leq n} (A_{i_1}\cap \ldots \cap A_{i_k})= \bigcup_{1\leq i_1<\ldots < i_k\leq n} (B_{i_1}\cap \ldots \cap B_{i_k})$ for each $k\leq max (n, m)$ (If $k>n$ we pose equal to $0$ the first member, analogously if $k>m$).

Let $(*)$ and $I(**)$ the subspaces generated by $(*)$ or $(**)$ respectively, of course $I(*)\subset I(**)$, now if a linear map $f$ as above is null on $I(*)$ then it is modular and then (see $[1]$ or $[2]$) it is null also on $I(**)$, then $I(*)\supset I(**)$.

Now, I want to generalize this sets construction to a general alttice $\mathcal{A}$, then I define:

DEF) $Simp(\mathcal{A}):=\mathbb{R}[\mathcal{A}]/I$

where $I$ is the (real) vectorial subspace generated by $(*)$ or equivalently by $(**)$, and indicate an element of $Simp(\mathcal{A})$ as $[v]$ where $v\in \mathbb{R}[\mathcal{A}]$.

Observation 1):

We define the product on $\mathcal{A}$ as the intersection (or "inf") we have a multiplication on $\mathbb{R}[\mathcal{A}]$ that make this a (associative) algebra. Explicitly $(r_1\cdot A_1+\ldots + r_m\cdot A_m)\cdot (s_1\cdot B_1+\ldots + s_n\cdot B_n)= \sum_{i, j} (r_i\cdot s_j)\cdot\ A_i\cap B_j$ .

We have that $I$ is and ideal of the algebra, in fact if we multiply $(**)$ for a elements $v$ of $\mathbb{R}[\mathcal{A}]$ we get a elements of $I$, for distributivity its enough see this for $v=E$ where $E\in \mathcal{A}$ (a base vector), and easly this multiplication preserve $(**)$.

We call the class $Idmp(Simp(\mathcal{A}))$ of idempotents of $Simp(\mathcal{A})$ its a boolean algebra and we have a lattice morphism $\mathcal{A}\to Idmp(Simp(\mathcal{A}))$. Now I wish define a order on $Simp(\mathcal{A})$ in natural way, and such that the inducted order on $Idmp(Simp(\mathcal{A}))$ is just the one defined above.

I observe that give the elements $A_1,\ldots..A_n\in \mathcal{A}$ I can define (in a standard way) a sequence $C_1,\ldots C_N$ of disjoint elements of $Idmp(Simp(\mathcal{A}))$ such that each $A_i$ is a (disjoint) union of some elements of this sequence. Then each element $u= r_1\cdot [A_1]+\ldots + r_m \cdot [A_m]$ of $Simp(\mathcal{A})$ is writable like $u= r'_1\cdot a_1+\ldots + r'_M \cdot a_M$ where $a_1,\ldots,a_M$ are disjoint elements of $Idmp(Simp(\mathcal{A}))$. In similar way is $u, v\in \mathbb{R}[\mathcal{A}]$ we can write $[u]=r_1\cdot a_1+\ldots r_n\cdot a_n$, $[v]=s_1\cdot a_1+\ldots s_n\cdot a_n$ where $a_1,\ldots a_n$ are disjoint elements of $Idmp(Simp(\mathcal{A}))$.

My idea is to define a order on $Idmp(Simp(\mathcal{A}))$ as $u\leq v$ if given a representation $[u]=r_1\cdot a_1+\ldots r_n\cdot a_n$, $[v]=s_1\cdot a_1+\ldots s_n\cdot a_n$ with $a_1,\ldots, a_n$ disjoint, we have that $r_i \leq s_i\ \leq i\leq n$, but is this well defined?

If this is well defined I can consider the free $\sigma$-boolean algebra (relative to the generator set $X$) as in [3] and call this the "Borel algebra" of $\mathcal{A}$.

Conjecture 1) Is the natural map $\mathcal{A}\to Simp(\mathcal{A})$ injective?

Conjecture 2) If $[u]= r_1\cdot a_1+\ldots + r_m\cdot a_m $, $[v]=s_1\cdot b_1+\ldots + s_n\cdot b_n$ with $a_1\ldots \cdot , a_m$ and $b_1,\ldots \cdot , b_n$ disjoint, is true that $[u]= [v]$ iff $n=m$, $A_i= B_i,\ r_i=s_i\ 1\leq i\leq n$?

Conjecture 3) If I consider the lattice $\mathcal{B}:=Idmp(Simp(\mathcal{A}))$ and do the some construction , is true that $Idmp(Simp(\mathcal{A}))\cong Idmp(Simp(\mathcal{B}))$ ?

I ask: are my conjectures true?

Biblio:

[1]: Measure and Integration: An Advanced Course. Heinz Konig, Springer 2009.

[2]: Theory of Charges: A Study of Finitely Additive Measures Bhaskara Rao, M. Bhaskara Rao. AP

[3] Boolean algebras. Roman Sikorski. Springer, 1960.

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    $\begingroup$ Conjecture 1) is true because you construct $\mathbb{R}[\mathcal{A}]$ as having $\mathcal{A}$ as a basis. So the map $A\in \mathcal{A} \mapsto A \in \mathbb{R}[\mathcal{A}]$ must be injective. $\endgroup$ – Dominic van der Zypen Nov 7 '14 at 7:53
  • $\begingroup$ Dear Dominic, I did a mistake (now corrected), I mean $\mathcal{A}\to Sim(\mathcal{A})$, sorry. $\endgroup$ – Buschi Sergio Nov 7 '14 at 17:06
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    $\begingroup$ Question, is $Sim(\mathcal{A})$ the same (or isomorphic to) $Simp(\mathcal{A})$? Also, can you elaborate what simple functions (from where to where?) and "support on $\mathcal{A}$" means? $\endgroup$ – Dominic van der Zypen Nov 10 '14 at 8:33
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    $\begingroup$ I haven't assimilated all the notation (which is why this is only a comment), but I'd expect things like this to work better for modular lattices than in general. Have you looked at the non-modular 5-element lattice (consisting of top, bottom, and 3 other elements $a,b,c$, where $a,b,c$ are incomparable except that $b<c$)? That would be my first candidate for a counterexample to Conjecture 1. $\endgroup$ – Andreas Blass Jul 29 '16 at 17:04
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I'm still leaving this answer, but I wrote it a long time thinking it was about distributive lattices. I would be very surprise if anything of this sort can be done in the non distributive case.


Let me start by the case where $A$ is a boolean lattice.

Let $X$ be its stone spectrum, i.e. a compact totally disconnected space such that $A$ is exactly the lattice of clopen set of $X$.

I will first prove that $Simp(A)$ is the algebra of locally constant $\mathbb{R}$ valued functions on $X$.

There is obviously a unique map from $Simp(A)$ to this algebra which send the generator corresponding to $a \in A$ to the characteristic functions of the clopen set $a \subset X$.

This map is obviously surjective (any locally constant function on $X$ takes a finite number of value, and each values is taken on a clopen set, so it is a linear combination of characteristic functions).

The hard part is to check that it is injective, the trick is as follow: let $\phi$ be a linear form on $Simp(A)$, i.e. it is exactly a valuation on $A$. One can easily construct a measure on $X$ from a valuation on $A$, and integrating against this measure gives a linear form on the algebra of continuous functions on $X$ and one easily see that the composite of this with the map from $Simp(A)$ to the algebra of functions on $X$ is the linear form we started from (it coincide on the generator). and this prove the injectivity.

In particular, if $A$ is a boolean algebra, then the set of idempotent of $Simp(A)$ is exactly $A$.


We now use $A$ a genral lattice. Let $B$ be the universal boolean lattice generated by $A$ (obtained by freely adding complement to all element of $A$). It is not very hard to see that $Simp(A)$ and $Simp(B)$ have the same universal property ! So $Simp(A)=Simp(B)$.

This answer all the conjecture:

Conjecture $1$ : Yes because $A \rightarrow B$ is well kown to be injective and $B \rightarrow Simp(B) = Simp(A)$ is an injection.

Conjecture $2$: it is not true as stated (or I don't understand the statement), for exemple you can modify the order of the element, and if the $r_i$ are not pairwise disjoint you can regroup some of the terms, but the identification of elements of $Simp(A)$ with functions on the stone spectrum of $B$ allow to obtain result of this kind.

Conjecture $3$: it is true, $Idem(Simp(A))$ is the boolean algebra generated by $A$, hence if you iterate you get the same things.

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The space $Simp(\mathcal{A})$ is defined in the following way:

[...] $Simp(\mathcal{A})\subset \textbf{R}^X$ is the $\mathbb{R}$-vector subspace of generated of simple functions with support on $\mathcal{A}$ i.e. of the functions of type $r_1\cdot \chi_{A_1}+\ldots r_n\cdot \chi_{A_n}$ where $r_i\in \mathbb{R}, A_i\in \mathcal{A}$ for $1\leq i\leq n$ and $n\in\mathbb{N}$, and $\chi_A(X)=\{0,1\},\ \chi_{A}^{-1}(\{1\})=A$. <<

In other words, $Simp(\mathcal{A}) = \{f: \mathcal{A} \to \mathbb{R}: f(A) = 0 \textrm{ for all but finitely many } A\in\mathcal{A}\}$. Addition and scalar multiplication are defined pointwise. The ''natural map'' referred to in conjecture 1 maps $A\in\mathcal{A}$ to the characteristic function $\chi_A$ where $\chi_A$ is defined as above. The map $A\mapsto \chi_A$ is injective: if $A\neq B \in \mathcal{A}$ then $\chi_A(A) = 1$ and $\chi_B(A) = 0$ therefore $\chi_A\neq\chi_B$.

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    $\begingroup$ My apologies for my mistakes on formalisms and for my bad English. Of course this true if $\mathcal{A}$ is a lattice of subset's of a set $X$, in this case we can prove that $Simp(\mathcal{A})=\mathbb{R}[\mathcal{A}]/I $ (where $I$ is the a ideal generated by $(*)$ or $(**)$, see above). I use this relation for define it for a general lattice $\mathcal{A}$, and then define $Simp(\mathcal{A}):=\mathbb{R}[\mathcal{A}]/I $ (where $I$ is defined with the some kind of relations) of course I asking about conjecture's about this generalization. THank you anyway. $\endgroup$ – Buschi Sergio Nov 12 '14 at 20:59
  • $\begingroup$ OK I see. If I understand correctly, the natural map works this way: $A\in\mathcal{A} \mapsto [\chi_A]\in \mathbb{R}[\mathcal{A}]/I$. The natural map is injective if for $A\neq B\in\mathcal{A}$ we have $\chi_A - \chi_B \notin I$, right? $\endgroup$ – Dominic van der Zypen Nov 13 '14 at 7:59
  • $\begingroup$ Yes, I have defined $Simp(\mathcal{A}):=\mathbb{R}[\mathcal{A}]/I$ where $\mathbb{R}[\mathcal{A}]$ is the vector space with base the set of elements of $\mathcal{A}$, this is also a algebra: for distributivity its enough define the multiplication between elements of the base,and let $A\cdot B:= A \cap B$, and $I$ is the $\mathbb{R}$-ideal defined as above. $\endgroup$ – Buschi Sergio Nov 13 '14 at 18:45
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Conjecture 2 is false: Let $X = \{1,2\}$ and $\mathcal{A} = \mathcal{P}(X)$. Moreover set $A_i = \{i\}$ and $B_i = X\setminus \{i\}$ for $i \in X$.

Then $[\chi_{A_1} + \chi_{A_2}] = [\chi_{B_1} + \chi_{B_2}]$ but $A_1 \neq B_1$ and $A_2\neq B_2$.

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