This is a talk I will give on Monday, December 15th, 2025 at the Seminar on Domain Theory, hosted by the TianYuan Mathematics Center in Kunming, Yunnan, PRC. The organizers are Hui Kou (寇辉, Sichuan University, Chengdu, Sichuan, PRC), Xiaodong Jia (贾晓东, Hunan University, Changsha, Hunan, PRC), and myself.
Abstract
I will give an introduction to barycentric algebras, the plain, the preordered and the (semi)topological. Those include all convex subsets of (plain, preordered, semitopological) cones, but there are a few more. Then I will turn to pointed barycentric algebras, the preordered and the (semi)topological. The leading examples are spaces of probability, and subprobability valuations (or measures).
Slides
The full slides, with animation steps; the shorter presentation, without them.
Videos
- Cones: what we know about them (9:48)
- Barycentric algebras: the plain, the preordered and the (semi)topological (24:32)
- Pointed barycentric algebras: the preordered and the (semi)topological (19:02)
Updates
I have left a few questions open in the slides, and here are a few updates on these questions.
- (February 03rd, 2026) Question 3 is now solved, but leads to more questions.
- On December 17th, 2025, I found a topological barycentric algebras that is not locally convex, even not weakly locally convex. This is inspired by Tychonoff’s example of a non-locally convex topological vector space, lp(N), with 0<p<1. Roughly, we take the subspace of sequences with non-positive values, and give it a quasi-metric that is defined just like the lp metric, except that we replace |x—y| by max(x—y, 0) everywhere.
This non-locally convex topological barycentric algebra is not embeddable. Its free semitopological cone, then, is locally convex. - On February 02nd, 2026, I found a semitopological cone that is not locally convex, even not weakly locally convex. (On February 18th, 2026, I found a bug in the argument, and the following includes the fix.) This is very artificial. You take the set of all maps from [0, 1[ to R+, ordered pointwise, and you give it the finest topology coarser than the Scott topology and making + and . separately continuous such that 2χUn converges to the constant function 1, where Un is the collection of numbers that have a 0 as their n+1st bit when written in binary. Then you can show that fn ≝ (1/n) ∑i=n 2n–1 2χUn does not converge to 1: the set C ≝ ↓{fn | n ∈ N–{0}} ∪ ↓{f∞}, where f∞ ≝ limsup fn, is closed but does not contain 1. Now the complement of C is an open neighborhood of 1 that does not contain any convex neighborhood A of 1: otherwise, U ≝ int(A) would be an open neighborhood of 1, hence would contain 2χUn for n large enough; then A would contain the convex combination fn for n large enough, and that is impossible since A is disjoint from C, and C contains every fn.
This cone is quite probably not topological, only semitopological. Hence the remaining question is: can you find an example of a topological cone that is not (weakly) locally convex?
Also, I still do not know whether the notions ‘weakly locally convex’, ‘locally convex’ and ‘locally linear’ are distinct.
- On December 17th, 2025, I found a topological barycentric algebras that is not locally convex, even not weakly locally convex. This is inspired by Tychonoff’s example of a non-locally convex topological vector space, lp(N), with 0<p<1. Roughly, we take the subspace of sequences with non-positive values, and give it a quasi-metric that is defined just like the lp metric, except that we replace |x—y| by max(x—y, 0) everywhere.
- (March 02nd, 2026) Question 2 is now solved, too, unless I have made a mistake. There is a semitopological pointed barycentric algebra that is not strictly embeddable.
- This is also pretty artificial. We let I ≝ [0, 1] with its metric (or Scott, that does not matter) topology, we give the set N of natural numbers the discrete topology, and we consider the topological cone C ≝ ℒ(I) × ℒ(N), with the product topology, where each component has the Scott topology, or equivalently here, the compact-open topology. For every a ∈ I, let fa be the function from I to R+ that maps every element of ]1–a, 1] to 1, and every other element to 0. Let also 1 be the constant 1 function, and 0 be the zero function. For every n ∈ N, let en be the function from N to R+ that maps n to 1 and every other number to 0. From C, we carve out a T0 topological pointed barycentric algebra B by taking the convex hull of the elements (0, 0), (1, 0) and (fa, en) for all a ∈ I and n ∈ N. Let τ1 be the topology of B. We define a new topology τ2 on B as the finest topology coarser than τ1 that makes the operation (x, a, y) ↦ x +a y separately continuous and such that a.(fa, en) converges to a.(1, 0) when n tends to infinity for every a ∈ I. Let us fix a. One can show that Cb ≝ {(f, e) ∈ B | b.(f, e) ≤ a.(fa, en) for some n ∈ N or (f, e) ≤ a.(1, 0)} is τ2-closed for every b ∈ ]0, 1]. But, given a, b and α such that 0<a, α ≤ b < 1, Cb is not the inverse image of any τ2-closed subset C’ of B under the map α . _: (a/b) . (fa, en) is in Cb for every n ∈ N, so (αa/b) . (fa, en) would have to be in C’, and since C’ is τ2-closed, (αa/b) . (1, 0) would also have to be in C’, and therefore (a/b) . (1, 0) would be in Cb; but that is wrong.
Hence the map α . _ is not full, and therefore Cb is not strictly embeddable.
It is not too hard to show that ↓(f, e) is τ2-closed for every (f, e) ∈ B, hence is the τ2-closure of (f, e) in B; therefore the specialization preordering of B is in the topology τ2 is the usual ordering ≤, and B is T0. - This leads to a new question: can we find a non-strictly embeddable T0 topological pointed barycentric algebra?
- This is also pretty artificial. We let I ≝ [0, 1] with its metric (or Scott, that does not matter) topology, we give the set N of natural numbers the discrete topology, and we consider the topological cone C ≝ ℒ(I) × ℒ(N), with the product topology, where each component has the Scott topology, or equivalently here, the compact-open topology. For every a ∈ I, let fa be the function from I to R+ that maps every element of ]1–a, 1] to 1, and every other element to 0. Let also 1 be the constant 1 function, and 0 be the zero function. For every n ∈ N, let en be the function from N to R+ that maps n to 1 and every other number to 0. From C, we carve out a T0 topological pointed barycentric algebra B by taking the convex hull of the elements (0, 0), (1, 0) and (fa, en) for all a ∈ I and n ∈ N. Let τ1 be the topology of B. We define a new topology τ2 on B as the finest topology coarser than τ1 that makes the operation (x, a, y) ↦ x +a y separately continuous and such that a.(fa, en) converges to a.(1, 0) when n tends to infinity for every a ∈ I. Let us fix a. One can show that Cb ≝ {(f, e) ∈ B | b.(f, e) ≤ a.(fa, en) for some n ∈ N or (f, e) ≤ a.(1, 0)} is τ2-closed for every b ∈ ]0, 1]. But, given a, b and α such that 0<a, α ≤ b < 1, Cb is not the inverse image of any τ2-closed subset C’ of B under the map α . _: (a/b) . (fa, en) is in Cb for every n ∈ N, so (αa/b) . (fa, en) would have to be in C’, and since C’ is τ2-closed, (αa/b) . (1, 0) would also have to be in C’, and therefore (a/b) . (1, 0) would be in Cb; but that is wrong.
