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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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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.

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Tom, I believe $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ is monadic, essentially because all objects in $\mathbf{Vect}$, in particular $\mathbb{C}$, 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$\mathbb{C}$k$ is injective, the functor $(-)^\ast = \hom(-, \mathbb{C})$ 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$\hom(-, \mathbb{C})$, 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 $\mathbb{C}$ k\$ is injective, so we're done.

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

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