Self-dual Complete Category

Question

Is there an example of a self-dual complete category that is not a partially-ordered set?

Answer 1

The category of suplattices (complete join-semilattices) is complete and self-dual. It is complete because it is monadic over Set (the monad is the covariant powerset monad). The duality $Sup^{\mathrm{op}}\to Sup$ sends a suplattice $X$ to the opposite poset $X^{\mathrm{op}}$, which is again a suplattice since every suplattice is also an inflattice (define $\bigwedge S = \bigvee \{ x | x \le S\}$). It sends a sup-preserving map $f:X\to Y$ to $(f^\ast)^{\mathrm{op}}:Y^{\mathrm{op}}\to X^{\mathrm{op}}$, where $f^\ast:Y\to X$ is the right adjoint of $f$—this exists by the adjoint functor theorem for posets, and preserves infs since it is a right adjoint.

Answer 2

Mike has already answered the question, but I thought I would add some assorted remarks on some of the discussion above.

Mike gave an example of what is called a $\ast$-autonomous category, as first defined by Michael Barr in Springer LNM 752 (1979). The field has been very heavily investigated, partly because $\ast$-autonomous categories are the categorical models of Girard's celebrated linear logic. In brief, a $\ast$-autonomous category is a symmetric monoidal closed category $V$ equipped with a "dualizing object" $D$, which means, letting $\hom$ denote the internal hom of $V$, that the canonical natural transformation

$$A \to \hom(\hom(A, D), D)$$

is a natural isomorphism. This implies in particular that

$$\hom(-, D): V^{op} \to V$$

is an equivalence, so that $V$ is self-dual. Complete $\ast$-autonomous categories are certainly of considerable interest, and many examples are known. Some (including Mike's example) are given in this nLab article.

There are also various constructions which allow one to embed sufficiently nice instances of duality, such as the Pontryagin duality alluded to by Martin, into complete $\ast$-autonomous categories. One of the most potent general constructions is called the Chu construction, which takes an input any symmetric monoidal closed category with pullbacks and a preassigned object $d$, and produces as output a $\ast$-autonomous category $Chu(C, d)$ together with a coreflective (in particular full) symmetric monoidal embedding

$$i: C \to Chu(C, d)$$

whose dualizing object is manufactured from $d$ in a canonical way. If $C$ is complete and cocomplete, then so is $Chu(C, d)$. For a sample theorem which shows how certain nice topological dualities embed into Chu-type dualities, including the case of Pontryagin duality, see this paper by Barr.

Incidentally, with regard to Martin's answer, it is indeed the case that the category $LCHAb$ of locally compact Hausdorff abelian groups is not complete. We know for example that the topological product $\mathbb{R}^{\mathbb{N}}$ of countably many copies of $\mathbb{R}$ is not locally compact; it remains to see that there is no such product in $LCHAb$ (i.e., even if we try to strengthen the product topology on the set $\mathbb{R}^\mathbb{N}$ to some locally compact topology in some way, there still exists no solution to the universal mapping problem). If there were, then using the universal property, we could show that the scalar product $\mathbb{R} \times \mathbb{R}^\mathbb{N} \to \mathbb{R}^\mathbb{N}$ is continuous, making $\mathbb{R}^\mathbb{N}$ a Hausdorff topological vector space. But it is well-known that a locally compact Hausdorff TVS over the real numbers is finite-dimensional (reference; see theorem 4.3 on page 5). As arsmath suggests, this implies that the category of locally compact Hausdorff spaces cannot be complete (if it were, then so would be the category of abelian group objects, which we just proved is not the case).

Answer 3

If $LCA$ denotes the category of locally compact topological abelian groups, then Pontryagin duality implies that $LCA$ is self-dual. To see that this category is complete, note that the forgetful functor to abelian groups creates limits.

EDIT: The category is not complete. See the answer by Todd. I won't delete my answer because Todd refers to it.