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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

  1. I assume this is known (probably since a long time ago).

  2. 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.

  3. 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.

  4. 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).

  5. 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.)

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My gut feeling is "the dual vector spaces" but maybe this doesn't get all the algebras... –  Yemon Choi Aug 15 '12 at 18:13
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Yemon: dunno. The point about idempotent monads is this theorem: the only algebras for an idempotent monad are the free ones, which in this case are the spaces of the form $X^{**}$ (together with the canonical map $X^{****} \to X^{**}$). So if our monad was idempotent then you'd be right. It's not idempotent, but from that I can't deduce that you're wrong, because I don't know whether the converse of the aforementioned theorem is true. –  Tom Leinster Aug 15 '12 at 18:28
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By the way, when I said dual spaces, I meant $E^*$, equipped withe canonical (co-unit?) map $E^{***} \to E^*$ - IIRC this is the answer for the Banach space version of your question –  Yemon Choi Aug 16 '12 at 0:57
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Here's a sketch of a proof that the algebras are the dual vector spaces. Given an algebra $E^{**}\to E$, topologize $E$ using the quotient topology of the weak* topology on $E^{**}$, and let $F$ be the topological dual of $E$ with respect to this topology. Now you prove that actually $E=F^*$ with the canonical algebra structure. This is plausible since if $E=F^*$, this procedure recovers $F$ from $E$ and its algebra structure. –  Eric Wofsey Aug 16 '12 at 11:29
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For more information on linearly compact vector spaces from a categorical point of view see section 24 of the Bergman and Hausknecht book "Cogroups and Co-rings in Categories of Associative Rings". –  Carl Futia Aug 16 '12 at 17:29
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2 Answers

up vote 14 down vote accepted

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.

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Todd's answer is definitive: the category of algebras is $\mathbf{Vect}^{\mathrm{op}}$. However, I'm left wanting more, since the dual of the category of vector spaces isn't something I readily grasp. (Cf. my remark 5.) Compare the theorem that the catebgory of algebras for the double powerset monad on $\mathbf{Set}$ is $\mathbf{Set}^{\mathrm{op}}$. This is illuminated by the fact that $\mathbf{Set}^{\mathrm{op}}$ is equivalent to the category of complete atomic Boolean algebras. Is there some similar theorem for $\mathbf{Vect}^{\mathrm{op}}$? –  Tom Leinster Aug 16 '12 at 15:55
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@Tom: here is a terrible description. By Pontrjagin duality, $\text{Ab}^{op}$ is the category of compact (Hausdorff) abelian groups. A given field $k$ is a comonoid in this category (with respect to the monoidal product given by dualizing the tensor product; I am not really sure what this looks like) and $\text{Vect}^{op}$ is the category of comodules in $\text{Ab}^{op}$ over this comonoid... –  Qiaochu Yuan Aug 16 '12 at 16:22
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In case someone didn't see it, I added an edit to my answer. –  Todd Trimble Aug 16 '12 at 16:34
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And now a second edit. –  Todd Trimble Aug 16 '12 at 16:50
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This extends even further to a duality between coalgebras and pseudocompact algebras (inverse limits of finite dimensional algebras) if I remember correctly. –  Benjamin Steinberg Aug 16 '12 at 17:54
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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.

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Thanks, Noam. I did see that paper, though I hadn't read the part that you have. If I don't get an answer here, I think my next step is to mail Anders Kock. –  Tom Leinster Aug 16 '12 at 13:29
    
PS - I very much like the idea of a "follow up" paper 42 years after the original. –  Tom Leinster Aug 16 '12 at 13:30
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