Are commutative C*-algebras really dual to locally compact Hausdorff spaces?

Several online sources (e.g. Wikipedia, the nLab) assert that the Gelfand representation defines a contravariant equivalence from the category of (non-unital) commutative $C^{\ast}$-algebras to the category of locally compact Hausdorff (LCH) spaces. This seems wrong to me.

The naive choice is to take all continuous maps between LCH spaces. This doesn't work. For example, the constant map $\mathbb{R} \to \bullet$ does not come from a morphism $\mathbb{C} \to C_0(\mathbb{R})$, the problem being that composing with the map $\bullet \to \mathbb{C}$ sending $\bullet$ to $1$ gives a function on $\mathbb{R}$ which doesn't vanish at infinity. It is necessary for us to restrict our attention to proper maps.

But this still doesn't work. If $A, B$ are any commutative $C^{\ast}$-algebras we can consider the morphism $$A \ni a \mapsto (a, 0) \in A \times B.$$

This morphism does not define a map on Gelfand spectra; if $\lambda : A \times B \to \mathbb{C}$ is a character factoring through the projection $A \times B \to B$, then composing with the above morphism gives the zero map $A \to \mathbb{C}$. This contradicts the nLab's claim that taking Gelfand spectra gives a functor into locally compact Hausdorff spaces (if one requires that the morphisms are defined everywhere on the latter category).

The correct statement appears to be that commutative $C^{\ast}$-algebras are contravariantly equivalent to the category $\text{CHaus}_{\bullet}$ of pointed compact Hausdorff spaces; the functor takes an algebra to the Gelfand spectrum of its unitization (we adjoin a unit whether or not the algebra already had one). There is an inclusion of the category of LCH spaces and proper maps into this category but it is not an equivalence because maps $(C, \bullet) \to (D, \bullet)$ in $\text{CHaus}_{\bullet}$ may send points other than the distinguished point of $C$ to the distinguished point of $D$.

So do sources mean something else when they claim the equivalence with locally compact Hausdorff spaces?

• No, you certainly don't have a straighforward duality as you argue correctly (it is asserted in many places and repeated without thinking it through). Note that you don't just get any *-homomorphism if you have a proper map $f\colon X \to Y$: an exhaustion of $Y$ by compact sets yields an exhaustion of $X$ by compact sets, so the corresponding *-homomorphism should send approximate unities in $C_0(Y)$ to approximate unities in $C_0(X)$ and if I remember correctly the converse is also true.
– t.b.
Jul 15, 2012 at 4:50
• This MO thread lists a few correspondences and I think Martin Brandenburg wrote a careful account of some aspects in German somewhere a few years ago here
– t.b.
Jul 15, 2012 at 5:00
• Note that in the MO thread @t.b. links to, Matt mentions another POV espoused by e.g. Woronowicz, where one adjusts the morphisms on the C^* side to be non-degenerate *-homs into the multiplier algebra of the target object. As non-proper maps are IMHO natural (the covering map from R to T, for instance) this seems philosophically more appealing to me
– user16299
Jul 15, 2012 at 10:24
• To point to an elementary exposition of the failure of Gelfand duality for locally compact Hausdorff spaces, see math.univ-lille1.fr/~dellambr/exercise_C_algebras.pdf Jul 25, 2013 at 5:24
• Thanks for the alert. That was maybe my bad. It seems the actual content secton in the nLab was right, but the Idea-section was wrong. I have fixed it now. Nov 12, 2014 at 17:52

It is true that the category of locally compact Hausdorff spaces is equivalent to the category of commutative $C^*$-algebras . . . with appropriately chosen morphisms.

Let $A$ and $B$ be commutative $C^*$-algebras. Then, a morphism from $A$ to $B$ is defined to be a nondegenerate homomorphism of $^*$-algebras from $\phi :A\rightarrow M(B)$, where $M(B)$ is the multiplier algebra of $B$. Here, nondegeneracy means that the the span of $\left\{ \pi (a)b:a\in A,b\in M(B)\right\}$ is dense in $M(B)$. Note that you need a bit of machinery to even make this into a category because it is not obvious a priori that composition makes sense. Nevertheless, it does work out. Proposition 1 on pg. 11 and Theorem 2 on pg. 12 of Superstrings, Geometry, Topology, and $C^*$-algebras (in fact this chapter is on the arXiv) respectively show that this forms a category and that the dual of this category is equivalent to the category of locally compact Hausdorff spaces.

• You mean the category of locally compact Hausdorff spaces with all continuous maps, I suppose? May 15, 2015 at 16:28
• That's correct. Alternatively, you can take locally compact spaces with continuous proper maps, and this category will be co-equivalent to the category of commutative C*-algebras whose morphisms are non-degenerate continuous homomorphisms (of $^*$-algebras). Either way, you're somehow not taking the 'natural' choice of morphisms. May 15, 2015 at 17:00
• Since writing this answer, I've since come across the 'correct' generalization: The category of completely $T_2$ quasi-topological spaces (with quasi-continuous maps) is co-equivalent to the category of commutative unital locally $C^*$-algebras (with continuous homomorphisms), the co-equivalence being given by $X\mapsto \mathrm{Hom}_{\mathsf{QTop}}(X,\mathbb{C})$. This is Theorem 2.7 of the paper Inverse Limits of $C^*$-algebras by N. C. Phillips (jot.theta.ro/jot/archive/1988-019-001/1988-019-001-010.pdf). May 15, 2015 at 17:15

So, for the sake of having an answer written down to this question: as t.b. says in the comments, we simply don't have this duality as stated.

The following categories are contravariantly equivalent:

• locally compact Hausdorff spaces with proper continuous maps
• commutative C$^*$-algebras with non-degenerate $*$-homomorphisms

Here, a $*$-homomorphism $f : A \to B$ is non-degenerate if the following equivalent conditions are satisfied:

1. The ideal generated by the set-theoretic image of $f$ is dense in $B$.
2. For every approximative unit $(u_i)$ in $A$ its image $f(u_i)$ is an approximative unit in $B$.
3. For some approximative unit $(u_i)$ in $A$ its image $f(u_i)$ is an approximative unit in $B$.

I don't think that multiplier algebras are necessary ...

• Yes, this is mentioned by t.b. in the comments. May 15, 2015 at 16:23
• Sure, I just wanted to make this into an answer, so that it becomes more visible. May 15, 2015 at 16:29

It was recognised in the 50's that if one wants a duality theory for more general spaces than compact ones, then one has to go beyond the category of Banach spaces. In the context of a linear duality for locally compact spaces (generalising the Riesz representation theorem), R.C. Buck introduced the strict topology. This was later extended to the completely regular case by several authors---for example, by using the techniques of mixed topologies and Saks spaces which had been developed by the polish school. In this context, Gelfand-Naimark duality can be also extended and in the book Saks spaces and applications to functional analysis, a class of so-called Saks algebras (see eg these notes) was identified as the dual to the category of locally compact spaces. In the language of category theory, this provides a concrete representation of the opposite category to that of locally compact spaces (with continuous mappings as morphisms) extending the celebrated duality between compact spaces and commutative $C^*$-algebras with unit.

The above negative response means that in order to get a duality theory for locally compact spaces one has to leave the categories of Banach spaces or algebras. This problem has long been recognised and addressed, initially by Beurling, Herz and Buck in the context of harmonic analysis (spectral synthesis) and the Riesz representation theorem. The appropriate topology on the space of bounded, continuous functions was called the strict topology. In the sixties, it was extended to the case of completely regular spaces by several authors. A systematic approach to these topics can be found in the book "Saks spaces and Applications to Functional Analysis".

• Sorry. Answered twice without realising.
– jbc
Nov 5, 2012 at 20:16