Is \mathcal S Equipped with a Convex Structure?

[Fredrik]

Suppose that the set of functions $\{P^a:\mathcal S\rightarrow \mathcal L|\,a\in \mathcal L\}$ has the property that for all $s_1,s_2\in\mathcal S$, $$s_1=s_2\ \Leftrightarrow\ \forall a\in \mathcal L~~ P^a(s_1)=P^a(s_2).$$ Suppose also that the following is true for each positive integer n:

Given $s_1,\dots,s_n\in\mathcal S$ and $c_1,\dots,c_n\in[0,1]$ such that $c_1+\cdots+c_n=1$, there exists $s_0\in\mathcal S$ such that for all $a\in\mathcal L$, $$P^a(s_0)=\sum_{k=1}^n c_k P^a(s_k).$$ My problem is that I would like to think of $\mathcal S$ as some sort of structure rather than just a set (because I would like to know what an "automorphism" of $\mathcal S$ would be). So I'm wondering if the information given above (the existence and properties of the $P^a$ functions) is enough to implicitly define some sort of structure on $\mathcal S$?

I would like to use the notation $\sum_{k=1}^n c_k s_k$ for $s_0$, and really think of this as a convex combination of the $s_k$. Is there perhaps an abstract definition of "convex set" that doesn't even refer to a vector space? Maybe it's called something different, like "convex structure" or "convex space"? If there is such a definition, I think I would just like to show that the functions I've mentioned ensure that $\mathcal S$ is the underlying set of such a structure.

Maybe I should be trying to map $\mathcal S$ bijectively onto a convex subset of some vector space, and take the automorphisms to be vector space automorphisms restricted to that convex subset? Hm, that actually sounds good, but how do I do that, and how do I justify thinking of restrictions of vector space automorphisms as automorphisms of $\mathcal S$?

It appears that the books I'm reading (which are doing almost the same thing that I'm trying to do, but with a slightly different goal) don't really try to address this issue at this stage, and wait until they've made several additional assumptions which finally allows them to identify the members of $\mathcal S$ with probability measures on $\mathcal L$ (which by then has been equipped with a partial order and found to be a bounded orthocomplemented lattice). Maybe I'll have to do something like that too. I'm just wondering if something can be said at this early stage.

 

[micromass]

Hi Frederik!

Let's see if I can say something meaningful here.

Something I would immediately do is construct the function

$$\mathcal{S}\rightarrow \prod_{a\in \mathcal{L}}{\mathcal{L}}=\mathcal{L}^\mathcal{L}:s\rightarrow (P^a(s))_a$$

This function is an injection by the property you mention. As such, $\mathcal{S}$ inherits structure from $\mathcal{L}$ in a natural way.

For example, we can define $c_1s_1+...+c_ns_n$ as

$$(c_1P^a(s_1)+...+c_nP^a(s_n))_a$$

this is not necessarily an element of $\mathcal{S}$, but an element of the larger set $\mathcal{L}^\mathcal{L}$. The property you mention is that $c_1s_1+...+c_ns_n$ is an element of $\mathcal{S}$ if $c_1+...+c_n=1$.

So you could say that $\mathcal{S}$ is a convex subset of the vector space $\mathcal{L}^\mathcal{L}$.

Furthermore, if $\mathcal{L}$ has an order/topology/... then $\mathcal{S}$ also has it by the same trick.



The thing about probability measures is an interesting one. I suppose you could form some kind of "product"

$$<.,.>:\mathcal{L}\times\mathcal{S}\rightarrow \mathcal{L}:(a,s)\rightarrow P^a(s)$$

which tend to remind me of some situations in functional analysis. But I'm not sure if this gives us something interesting...

 

[Fredrik]

Hi micromass. Thank you for your help. I was a bit confused at first when you wanted to bring the set $\mathcal L^\mathcal L$ into the mix, but I see now that the reason is that I messed up the very first line in my post. When I said $\{P^a:\mathcal S\rightarrow \mathcal L|\,a\in \mathcal L\}$, I really meant $\{P^a:\mathcal S\rightarrow[0,1]|\,a\in \mathcal L\}$.

So let's denote the map $s\mapsto P^a(s)$ by $P_s:\mathcal L\rightarrow[0,1]$. (So that $P^a(s)=P_s(a)$). Now $s\mapsto P_s$ is an injective function from $\mathcal S$ into the vector space (or algebra) $[0,1]^\mathcal L$. (Yes, I actually like the mapsto arrow ). The $P_s$ functions already have some of the properties of a probability measure on a lattice: I haven't mentioned that there's a partial order on $\mathcal L$, that $\mathcal L$ has a minimum and a maximum element, denoted by 0 and 1 respectively, and that $P_s(0)=0$ and $P_s(1)=1$.

The idea $s\mapsto P_s$ is so simple that I feel that someone should slap me with a fish. I should have seen it. Well, at least I'm past it now, and can focus on understanding the additional assumptions that will turn $\mathcal L$ into a lattice. Thanks again.