5
$\begingroup$

Gelfand duality tells us that the category of compact Hausdorff spaces (with continuous maps as morphisms) is contravariantly equivalent to the category of commutative, unital $C^\ast$-algebras (with $\ast$-algebra maps over $\mathbb C$ as morphisms). The latter category can be defined in $L_{\omega_1,\omega}$ and is easily seen to be locally presentable, and in particular it is complete and cocomplete.

Recall that a topological space $X$ is called $\kappa$-Lindelöf (or $\kappa$-compact) if every open cover of $X$ admits a subcover of cardinality $<\kappa$. So compact = $\aleph_0$-Lindelöf.

Question 1: Let $\kappa$ be a cardinal. Is the opposite of the category of $\kappa$-Lindelöf Hausdorff spaces complete and cocomplete?

Question 2: Is it locally presentable? If so, is there a reasonable "algebraic" presentation of the category?

Notes:

  • Tychonoff's theorem is what tells us that the category of compact Hausdorff spaces has products / that the category of commutative unital $C^\ast$-algebras has coproducts. The analog of Tychonoff's theorem for $\kappa$-Lindelöf spaces (and I believe equivalently for $\kappa$-Lindelöf Hausdorff spaces) if and only if $\kappa$ is a strongly compact cardinal. So the above questions trivially have negative answers unless $\kappa$ is strongly compact. Thus we should assume in the above that $\kappa$ is a strongly compact cardinal, and we are squarely in the realm of studying large cardinals.

  • Gelfand duality can be thought of as being analogous to the duality between affine schemes and commutative rings, or to Stone duality between profinite sets -- a "geometric" category is dual to a category which is "algebraic", and hence locally presentable.

$\endgroup$
2
  • $\begingroup$ How exactly does the fact that κ-Lindelöf spaces are not closed under products in topological spaces imply that the category of κ-Lindelöf spaces is not locally presentable? $\endgroup$ Oct 7 at 17:59
  • $\begingroup$ Exactly. The inclusion functor doesn't have to be product-preserving. Anyway, very interesting question. $\endgroup$ Oct 8 at 0:42

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.