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.)

isthe answer for the Banach space version of your question $\endgroup$ – Yemon Choi Aug 16 '12 at 0:57