What are the algebras for the double dualization monad? - MathOverflow

What are the algebras for the double dualization monad?

2012-08-15T17:03:07Z

Let $k$ be a field, and let $\mathbf{Vect}$ denote the category of vector spaces (possibly infinite-dimensional) over $k$. Taking duals gives a functor $(\ )^*\colon \mathbf{Vect}^{\mathrm{op}} \to \mathbf{Vect}$.

This contravariant functor is self-adjoint on the right, since a linear map $X \to Y^*$ amounts to a bilinear map $X \times Y \to k$, which is essentially the same thing as a bilinear map $Y \times X \to k$, which amounts to a linear map $Y \to X^*$ . It therefore induces a monad $(\ )^{**}$ on $\mathbf{Vect}$.

What are the algebras for this monad?

Remarks

I assume this is known (probably since a long time ago).
The first paper I came across when searching for the answer was Anders Kock, On double dualization monads, Math. Scand. 27 (1970), 151-165. I'm pretty sure it doesn't contain the answer explicitly, but it's possible that it contains some results that would help.
The monad isn't idempotent (that is, the multiplication part of the monad isn't an isomorphism). Indeed, take any infinite-dimensional vector space $X$. Write our monad as $(T, \eta, \mu)$. If $\mu_X$ were an isomorphism then $\eta_{TX}$ would be an isomorphism, since $\mu_X \circ \eta_{TX} = 1$. But $\eta_{TX}$ is the canonical embedding $TX \to (TX)^{**}$, and this is not surjective since $TX$ is not finite-dimensional.
There's another way in which the answer might be somewhat trivial, and that's if $(\ )^*$ is monadic. But it doesn't seem obvious to me that $(\ )^*$ even reflects isomorphisms (which it would have to if it were monadic).
There's a sense in which answering this question amounts to completing the analogy:

sets are to compact Hausdorff spaces as vector spaces are to ?????

Indeed, the codensity monad of the inclusion functor (finite sets) $\hookrightarrow$ (sets) is the ultrafilter monad, whose algebras are the compact Hausdorff spaces. The codensity monad of the inclusion functor (finite-dimensional vector spaces) $\hookrightarrow$ (vector spaces) is the double dualization monad, whose algebras are... what? (Maybe this will help someone to guess what the answer is.)

Answer by Noam Zeilberger for What are the algebras for the double dualization monad?

2012-08-16T11:52:39Z

This is not a direct answer to your question, but are you are familiar with a recent "followup" paper by Kock, Commutative Monads as a Theory of Distributions? There he considers an alternative approach to the theory of distributions starting from a general commutative monad $T$ (with a certain notion of strength), then defining double-dualization with respect to an arbitrary $T$-algebra $B$. He explains that there is a monad morphism from $T$ into any such double-dualization monad $(-\multimap B)\multimap B$, that this morphism may be factored by way of a submonad $(-\multimap B) \multimap^T B$, and states that in certain cases the map $T \Rightarrow (-\multimap B) \multimap^T B$ is an isomorphism.

Answer by Todd Trimble for What are the algebras for the double dualization monad?

2012-08-16T15:10:25Z

Tom, I believe $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ is monadic, essentially because all objects in $\mathbf{Vect}$, in particular $k$ as a module over $k$ as ground field, are injective.

For instance, to check that $(-)^\ast$ reflects isomorphisms, suppose $f: V \to W$ is any linear map. We have two short exact sequences

$$0 \to \ker(f) \to V \to im(f) \to 0$$

$$0 \to im(f) \to W \to coker(f) \to 0$$

Because $k$ is injective, the functor $(-)^\ast = \hom(-, k)$ preserves short exact sequences:

$$0 \to im(f)^\ast \to V^\ast \to \ker(f)^\ast \to 0$$

$$0 \to coker(f)^\ast \to W^\ast \to im(f)^\ast \to 0$$

and if $f^\ast$, the composite $W^\ast \to im(f)^\ast \to V^\ast$, is an isomorphism, then $W^\ast \to im(f)^\ast$ is injective, which forces $coker(f)^\ast = 0$ and therefore $coker(f) = 0$. By a similar argument, $\ker(f) = 0$. Therefore $f$ is an isomorphism.

The remaining hypotheses of Beck's theorem (in the form given in Theorem 2, page 179, of Mac Lane-Moerdijk) are similarly easy to check. Obviously $\mathbf{Vect}^{op}$ has coequalizers of reflexive pairs since $\mathbf{Vect}$ has equalizers. And $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ (which has a left adjoint, as pointed out) preserves coequalizers; this is equivalent to saying that $\hom(-, k)$, as a contravariant functor on $\mathbf{Vect}$, takes equalizers to coequalizers, or takes kernels to cokernels, but that's the same as saying that $k$ is injective, so we're done.

Oh, incidentally, double dualization is not a commutative or monoidal monad, if I recall correctly.

Edit: In a comment below, Tom asks for a more concrete description of $\mathbf{Vect}^{op}$ along the lines of topological algebra. I suspect the way to go is to see $\mathbf{Vect}$ as the Ind-completion (or Ind-cocompletion) of the category of finite-dimensional vector spaces, and therefore $\mathbf{Vect}^{op}$ as the Pro-completion of the opposite category, which is again $\mathbf{Vect}_{fd}$. I think I've seen before a result that this is equivalent to the category of topological $k$-modules which arise as projective limits of (cofiltered diagrams of) finite-dimensional spaces with the discrete topology, or something along similar lines, but I'd have to look this up to be sure. There might be pertinent material in Barr's Springer Lecture Notes on $\ast$-autonomous categories, but again I'm not sure.

Edit 2: Ah, found it. $\mathbf{Vect}^{op}$ is equivalent to the category of linearly compact vector spaces over $k$ and continuous linear maps. See Theorem 3.1 of this paper for example: arxiv.org/pdf/1202.3609. The result is credited to Lefschetz.