I'd like to formulate an abstract definition of convex sets: a set $K$ is convex if it is endowed with a ternary operation $K\times[0,1]\times K\to K$, written $(x:t:y)$, satisfying axioms

  • $(x:0:y)=(x:t:x)=x$
  • $(x:t:y)=(y:1-t:x)$
  • $(x:t:(y:\frac ut:z))=((x:\frac{t-u}{1-u}:y):u:z)$

The axioms imply that, for every $x_1,\dots,x_n\in K$ and $t_i\ge0$ with $\sum t_i=1$ the convex combination $\sum t_i x_i\in K$ is well defined.

Examples are the usual convex subsets of real vector spaces, with $(x:t:y)=(1-t)x+ty$, but also trees with $(x:t:y)$ the unique point at ratio $t$ on the geodesic from $x$ to $y$, and more generally $\mathbb R$-trees (geodesic metric spaces in which all triangles are isometric to tripods).

I'm sure this has already been explored, and I'd rather not reinvent the wheel (and I possibly missed some useful axioms), but I couldn't find any reference to such notions.

Natural questions that come to mind are:

  • define an affine map between convex sets $K,L$ as a map $f\colon K\to L$ with $f(x:t:y)=(f(x):t:f(y))$. Topologize then convex sets by making all affine maps to $\mathbb R$ continuous. What can be said of these topological spaces?
  • can every convex set be represented in a vector space? I have in mind the map $K\to\ell^1(K)/\{\delta_{(x:t:y)}=(1-t)\delta_x+t\delta_y\forall x,t,y\}$, though the topologies will probably not match (and I'm not sure I want to close the space I quotient $\ell^1$ by).

Thanks to all! Any kinds of references or comments are welcome.

  • 1
    $\begingroup$ When topologizing, you should use convex maps to $\mathbb{R}$, as you might not have enough affine ones. $\endgroup$
    – Uri Bader
    Commented Oct 27, 2016 at 8:20

2 Answers 2


There has been a bunch of work along these lines, and I think the idea has been rediscovered several times. I suggest looking at the papers of Anna Romanowska, who refers to them as "barycentric algebras", to get an idea of what's known. Her book with Smith, "Modes", covers this as well as generalizations where $t$ is not required to be in $[0,1]$. Here are some slides that cover the basics.

It's known that not all of them are representable as vector spaces. For example, if you mush everything in the interior of $[0,1]$ to a single point, your operations are still well-defined, but you can't embed it in a vector space. The Modes book has a structure theorem.

  • 3
    $\begingroup$ There's also an nlab page: ncatlab.org/nlab/show/convex+space They attribute the characterization of barycentric algebra that can be embedded in a vector space (cancellativity) to a 1949 paper of Stone, "Postulates for the barycentric calculus". $\endgroup$
    – arsmath
    Commented Oct 27, 2016 at 9:57
  • $\begingroup$ In regards to your counterexample, would a requirement that $(x : t_1 : y) \neq (x : t_2 : y)$ unless $x=y$ or $t_1=t_2$, or similar, recover the hope of representation as a vector space? $\endgroup$ Commented Oct 27, 2016 at 19:32
  • $\begingroup$ You need for all $t$ that $(x:t:y) = (x:t:z)$ implies that $y = z$. $\endgroup$
    – arsmath
    Commented Oct 27, 2016 at 19:41

A good (but probably not up to date) reference on several abstractions of the notion of convexity is

Singer, Ivan. Abstract convex analysis. Vol. 20. John Wiley & Sons, 1997.

There you'll find notions like "order convexity" in posets, "metric convexity" for metric spaces, abstract convex combinations, but also notions based on approximation from the outside or based on abstraction of the "convex hull" operation. I am not sure if something along th lines of your ideas is in there, but it is a good read anyway.


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