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 is the unit ball of the Banach space , and thus an operation in this theory is a formal sum for with the property that . The finiteness of this sum forces it to have only countably many non-zero terms, and thus it factors through an operation on . Hence there is a presentation of this theory with operations in (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.
Of course, one would normally write the right-hand side of the last equation as , but that is not technically an operation in the theory, except as mediated by and . A common choice for is the inverse of , where for this expression to work we take 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
A totally convex space is a pointed convex space. The operations of a convex space are encoded in the operations and the “point” comes from the operation . This functor preserves underlying sets and so has a left adjoint; thus any convex space can be “completed” to a totally convex space.
Clearly, the closed unit ball of any Banach space is a totally convex space.
Bizarrely, the open unit ball of a Banach space is a totally convex space. This is because if a sum, for , lies on the boundary of then every for which must have norm . Thus if the series only contains terms from the interior of , the sum remains in the interior. Hence the open unit ball is a subalgebra of the closed unit ball.
Continuing, the quotient of the closed unit ball by the open unit ball is a totally convex space.
In particular, the three-point space is (assuming excluded middle) a totally convex space. The operations on this space are as follows:
Going back one step, is a totally convex space. It is illuminating to describe this as a coequaliser of free totally convex spaces. Consider a functional which is bounded of norm but does not achieve its norm; for example, let be represented by the sequence . Then for any , . Thus is the coequaliser of the inclusion and the zero map .