totally convex space

Totally convex spaces

Totally convex spaces


Essentially, the totally convex spaces are what you get if you try to build an algebraic theory of Banach spaces.


Abstract definition

A totally convex space is a module for the monad on Set which sends a set XX to the closed unit ball of the Banach space 1(X)\ell^1(X).

This is an example of an algebraic theory (in the strong sense that it is bounded monadic) but not finitary, hence not a Lawvere theory. The free totally convex space on XX is the unit ball of the Banach space 1(X)\ell^1(X), and thus an operation in this theory is a formal sum xXa xx\sum_{x \in X} a_x x for a xa_x \in \mathbb{R} with the property that xX|a x|1\sum_{x \in X} {|a_x|} \le 1. The finiteness of this sum forces it to have only countably many non-zero terms, and thus it factors through an operation on \mathbb{N}. Hence there is a presentation of this theory with operations in 1()\ell^1(\mathbb{N}) (which is why the theory is bounded monadic). The corresponding identities are simply the substitution rules – namely that substituting a sum into another works as expected – and the reordering rules.

Concrete definition

A totally convex space is a set XX equipped with, for each infinite sequence (a 1,a 2,)(a_1, a_2, \ldots) of real numbers such that i|a i|1\sum_i {|a_i|} \leq 1, an operation from X X^{\mathbb{N}} to XX, written (x 1,x 2,) ia ix i(x_1,x_2,\ldots) \mapsto \sum_i a_i x_i, such that:

  1. Reordering: If σ\sigma is any permutation of \mathbb{N}, then ia ix i= ia σ(i)x σ(i)\sum_i a_i x_i = \sum_i a_{\sigma(i)} x_{\sigma(i)};
  2. Nullary substitution: If δ i\delta_i is the Kronecker delta at 00 (so δ i=0\delta_i = 0 normally, but δ 0=1\delta_0 = 1), then iδ ix i=x 0\sum_i \delta_i x_i = x_0;
  3. Binary substitution: If the functions π,ρ:\pi,\rho\colon \mathbb{N} \to \mathbb{N} express \mathbb{N} as its own product with itself (it is enough to pick one pair of functions and state this axiom only for it), then ia i( jb i,jx i,j)= ka π(k)b π(k),ρ(k)x π(k),ρ(k)\sum_i a_i (\sum_j b_{i,j} x_{i,j}) = \sum_k a_{\pi(k)} b_{\pi(k),\rho(k)} x_{\pi(k),\rho(k)}.

Of course, one would normally write the right-hand side of the last equation as i,ja ib i,jx i,j\sum_{i,j} a_i b_{i,j} x_{i,j}, but that is not technically an operation in the theory, except as mediated by π\pi and ρ\rho. A common choice for (π,ρ)(\pi,\rho) is the inverse of (i,j)(i+j+12)+j(i,j) \mapsto \big({i + j + 1 \atop 2}\big) + j, where for this expression to work we take 00 to be a natural number.

I haven't actually checked that this list is complete; but it's what I get if I take Andrew at his word that we need only substitution and reordering rules. I wouldn't be terribly surprised if nullary substitution is redundant, but right now I don't see how. —Toby


  1. A totally convex space is a pointed convex space. The operations of a convex space are encoded in the operations (r,1r,0,0,0,)(r, 1 - r, 0, 0, 0, \ldots) and the “point” comes from the operation (0,0,)(0, 0, \ldots). This functor preserves underlying sets and so has a left adjoint; thus any convex space can be “completed” to a totally convex space.

  2. The operations for this theory are commutative, hence the category of totally convex spaces is a closed symmetric monoidal category.


  1. Clearly, the closed unit ball BXB X of any Banach space XX is a totally convex space.

  2. Bizarrely, the open unit ball of a Banach space is a totally convex space. This is because if a sum, xBXa xx\sum_{x \in B X} a_x x for x|a x|1\sum_x {|a_x|} \leq 1, lies on the boundary of BXB X then every xx for which a x0a_x \ne 0 must have norm 11. Thus if the series only contains terms from the interior of BXB X, the sum remains in the interior. Hence the open unit ball is a subalgebra? of the closed unit ball.

  3. Continuing, the quotient of the closed unit ball by the open unit ball is a totally convex space.

  4. In particular, the three-point space {1,0,1}\{-1,0,1\} is (assuming excluded middle) a totally convex space. The operations on this space are as follows:

    a jϵ j={1 ϵ j=sign(a j),|a j|=1 1 ϵ j=sign(a j),|a j|=1 0 otherwise \sum a_j \epsilon_j = \begin{cases} 1 & \epsilon_j = \operatorname{sign}(a_j), \sum {|a_j|} = 1 \\ -1 & \epsilon_j = -\operatorname{sign}(a_j), \sum {|a_j|} = 1 \\ 0 & \;\text{otherwise} \end{cases}
  5. Going back one step, (1,1)(-1,1) is a totally convex space. It is illuminating to describe this as a coequaliser of free totally convex spaces. Consider a functional f: 1()f \colon \ell^1(\mathbb{N}) \to \mathbb{R} which is bounded of norm 11 but does not achieve its norm; for example, let ff be represented by the sequence (12,23,34,)(\frac{1}{2},\frac{2}{3},\frac{3}{4},\ldots). Then for any xB 1()x \in B\ell^1(\mathbb{N}), f(x)(1,1)f(x) \in (-1,1). Thus (1,1)(-1,1) is the coequaliser of the inclusion kerf 1()\ker f \to \ell^1(\mathbb{N}) and the zero map kerf 1()\ker f \to \ell^1(\mathbb{N}).

Last revised on January 29, 2010 at 16:52:39. See the history of this page for a list of all contributions to it.