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.