totally convex space

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}).

Revised on January 29, 2010 16:52:39 by Toby Bartels (