Take the 2-minute tour ×

There is a fact that I should have learned a long time ago, but never did; I was reminded that I did not know the answer by Qiaochu's excellent series of posts, the most recent of which is this one.

Let $X$ be a topological space. I can think of at least four rings of continuous functions to attach to $X$:

  1. $C(X)$ is the ring of continuous functions to $\mathbb R$.
  2. $C_b(X)$ is the ring of bounded functions to $\mathbb R$.
  3. $C_0(X)$ is the ring of continuous functions that "vanish at infinity" in the following sense: $f\in C_0(X)$ iff for every $\epsilon>0$, there is a compact subset $K \subseteq X$ such that $\left|f(x)\right| < \epsilon$ when $x \not\in K$.
  4. The ring $C_c(X)$ of functions with compact support.

Option 2. is not very good. For example, it cannot distinguish between a space and its Stone-Cech compactification. My question is what kinds of spaces are distinguished by the others.

For example, I learned from this question that MaxSpec of $C(X)$ is the Stone-Cech compactification of $X$. But it seems that $C(X)$ knows a bit more, in that $C(X)$ has residue fields that are not $\mathbb R$ if $X$ is not compact?

On the other hand, my memory from my C*-algebras class (which was a few years ago and to which I didn't attend well), is that for I think locally-compact Hausdorff spaces $C_0(X)$ knows precisely the space $X$?

So, MOers, what are the exact statements? And if I believe that the correct notion of "space" is "algebra of observables", are there good arguments for preferring one of these algebras (or one I haven't listed) over the others?

share|improve this question
Have you tried looking Gilman and Jerison's "Rings of Continuous Functions"? The problem is that LCHff is atypically nice as a subclass of general topological spaces, and so intuition gained there can be misleading. –  Yemon Choi Nov 29 '09 at 1:10
Also, the ring of functions with compact support could be denoted by $C_c$, or by $C_{00}$ in some contexts, but $C_0$ just seems wrong... –  Yemon Choi Nov 29 '09 at 1:11
Wikipedia confirms that MaxSpec C_0 is homeomorphic to X for X locally compact Hausdorff. I don't know that there is a good answer in general. –  Qiaochu Yuan Nov 29 '09 at 3:00
Another vote for Gillman and Jerison. A great book. When you are ready to leave the locally compact enclave and go out into the completely regular world, try it! Yes, I put it last month in the thread "Examples of great mathematcal writing" mathoverflow.net/questions/358/… –  Gerald Edgar Nov 29 '09 at 11:22
@Gerald I have to confess I've never so much as opened the book - but from talking to some former colleagues it sounded like a seminal text on the subject. –  Yemon Choi Nov 29 '09 at 13:12

4 Answers 4

up vote 10 down vote accepted

I work with infinite dimensional manifolds so am extremely distrustful of anything that requires some sort of compactness condition. Most of the time, it's just too restrictive.

Consider a really nice simple space: the coproduct of a countably infinite number of lines, $\sum_{\mathbb{N}} \mathbb{R}$ (coproduct taken in the category of locally convex topological vector spaces). This has the property that any compact subset is contained in a finite subspace. However, any neighbourhood of the origin has to be absorbing (the union of the scalar multiples of it is the whole space) so there aren't any non-zero continuous functions with compact support. That defenestrates option 4.

Particularly simple functions on an infinite dimensional vector space are the cylinder functions. These are important in measure theory on such spaces. A cylinder function has the property that it factors through a projection to a finite dimensional vector space. Such functions can be continuous and can be bounded, but (apart from the zero function) never vanish at infinity and never have compact support. Thus option 3 joins option 4 in the flowerbed.

As for option 2, I have no particular qualms about it except that it's not stable under partitions-of-unity. Assuming that I have such, then any continuous function can be written as a sum of bounded functions so when doing standard p-of-1 constructions I have to assume that my starting family is uniformly bounded (if that's the right term).

Well, I just remembered one qualm about option 2: if I go up the scale of differentiability then it gets increasingly hard to justify global bounds on the derivatives. I have a memory of John Roe telling me of some result that he'd proved which was to do with bounding all derivatives of a smooth function in some fashion. I don't recall the exact conditions, but the conclusion was that the only functions that satisfied them were trigonometric.

As others have said, if you are really only interested in (locally) compact spaces then the other options have meaning (functionally Hausdorff - points separated by functions - is assumed). But then the title of your question should have been: "Which is the correct ring of functions for a Locally Compact Hausdorff Space?".

share|improve this answer
@Andrew - most of the non locally compact spaces that I deal with on a day to day basis are infinite dimensional topological vector spaces as you point out. But in this case, wouldn't the "correct" ring of functions be the continuous linear maps? What are some other examples? –  Dave Penneys Nov 30 '09 at 17:35
@Dave - not if you're a differential topologist! I use infinite dimensional LCTVS's as models for my infinite dimensional manifolds so I'm particularly interested in their rings of smooth functions. –  Loop Space Nov 30 '09 at 19:11
+1 for "defenestrates". You have consistently excellent mathematics and choice of verbs. –  Steven Gubkin Mar 9 '10 at 1:16
+1 for 'defestrates' from me as well, and also for 'joins option 4 in the flowerbed', which sounds not just like defenestration to me, but something Agatha Christie-like. –  David Roberts Oct 24 '12 at 8:46

As an operator algebraist, I think the space of continuous functions to $\mathbb{C}$ which vanish at infinity is my preferred choice. Let me tell you why.

One of the basic ideas of noncommutative topology/geometry (and probably algebraic geometry, but I don't know much about that) is that we can trade the space for algebras of functions on that space. This is afforded by the Gelfand transform. The spectrum of a commutative $C^\ast$-algebra is the space of characters, i.e., $\ast$-algebra homomorphisms to $\mathbb{C}$.

  • If $X$ is compact Hausdorff, then the spectrum of $C(X)$ is $X$.
  • If $X$ is locally compact Hausdorff, but not compact, the spectrum of the non-unital $C^\ast$-algebra $C_0(X)$ is $X$. The spectrum of the unitalization of $C_0(X)$ ($C_0(X)\oplus \mathbb{C}$) is the one point compactification of $X$. The spectrum of the unital $C^\ast$-algebra $C_b(X)$ is $\beta X$, the Stone-Cech compactification of $X$. One should note that $C_b(X)$ is also the multiplier algebra of $C_0(X)$.
  • If $X$ is compact, but not Hausdorff, then $C(X)$ corresponds to some type of "Hausdorffization" of $X$.

Actually $C(X)$ and $C_0(X)$ are the same if $X$ is compact, but you want to denote it $C(X)$ to emphasize the fact that the algebra is already unital. Otherwise, when you add a unit, you take the one point compactification of a compact space which adds an extra point, which is not what you want.

Now let's suppose you have some additional structure, like $X$ is a compact manifold. Then you probably want the $C^\infty$-functions on $X$. However, these can be recovered from $C(X)$ as those operators whose iterated commutator with the Dirac operator is bounded. This inspired the notion of a spectral triple.

EDIT: In my haste to answer this question, I made some mistakes in the earlier answer as pointed out by @Jonas.

share|improve this answer
This answer is standard and has the ring of truth to it. It suggests that lots of bad things inevitably happen if X is not locally compact. But are there good arguments for that? –  Greg Kuperberg Nov 29 '09 at 5:40
@Jonas - thank you. don't know what i was thinking... –  Dave Penneys Nov 29 '09 at 6:13
@Jonas - you have every reason to harp away. somehow my brain is not working clearly right now. i edited again. i think i need to get off of MO right now and get some sleep. hopefully i haven't made any more ridiculous errors. please point them out if you see them. –  Dave Penneys Nov 29 '09 at 6:45
There isn't one. An unbounded element of C(X) has unbounded spectrum, but the spectrum of an element x of a Banach algebra is always contained in the closed disk of radius ||x|| centered at 0. –  Jonas Meyer Nov 29 '09 at 7:02
Well, that shows that there isn't even a Banach algebra norm on C(X) in general, but I just thought of something else. For commutative C*-algebras, the norm of an element is equal to its spectral radius, so the only possible candidate for a C*-norm on C(X) is the sup norm. –  Jonas Meyer Nov 29 '09 at 7:16

I thought I'd record the misgivings in my comment above as an actual "answer". I think that one needs to be clearer about just what class of topological spaces one is considering. As I said above, even if one starts by restricting attention to Hausdorff spaces, the LCHff class is atypically nice (and the CHff case extraordinarily atypically nice, as passing categorists might attest). Saying that function spaces are dual to topological spaces is a great maxim, but like all maxims needs to be wielded with a modicum of care and not always as broadly as salespeople would have you do...

In this kind of broad generality, my first instinct would be to see what Gillman and Jerison's book "Rings of continuous functions" has to say.

As several people have said above, when X is not compact then one needs to decide how interested one is at "behaviour at infinity", and choose the most appropriate algebra of functions to reflect this. If you really don't care much, then $C_0(X)$ seems natural although as I said above this might only be good for LCHff X, and there are plenty of quite interesting topological spaces which are not locally compact...

I rather suspect that when $X$ is Hausdorff, $\sigma$-compact but non-compact, then $C(X)$ should have a natural Frechet algebra structure, and there might be some work done on this class of examples.

By the way, if $X$ is a metric space then there is a case that one should be looking at the algebra of Lipschitz functions on X (Nik Weaver has championed this viewpoint in the past). However, since this is not a $C^*$-algebra, it might not meet the standards of Proper Functional Analysis and should perhaps be consigned to the dustbin of history (or not, depending on your point of view).

share|improve this answer

If one can distill one central fact from the answers above and elsewhere, then it is that if one wants a good duality theory for algebras of continous functions, one has to move on from Banach spaces or algebras. This fact was recognised at least 50 years ago, and a suitable generalisation was discovered---the so-called strict topology on the space of bounded, continuous functions on a, say, completely regular space. Despite the eminence of its discoverers and advocates (Beurling, Herz, Buck), it seems to have become forgotten lore. A lucid argument on its behalf can be found in the 1960 Transactions paper by Herz, "The spectral theory of bounded functions" (as the title suggests, the motivation for its introduction comes from Harmonic Analysis). The original version used weighted seminorms and was valid for locally compact spaces. It was subsequently extended to completely regular spaces by several authors. The topology can be defined most succinctly as the finest locally convex one which agrees with compact convergence on the unit ball of $C^b(X)$. It is a complete locally convex spaces, its compact sets and convergent sequences can be simply characterised (the latter are the uniformly bounded sequences which converge uniformly on compacta). Its dual is the space of bounded Radon measures on $X$, the natural version of the Stone-Weierstra\ss theorem holds for it, its spectrum is naturally identifiable with $X$ and so one has a version of Gelfand-Naimark theory. One can also characterise the topological properties of $X$ in terms of those of $C^b(X)$, all in principle, many in practice. Importantly, one can characterise local compactness of $X$ in terms of the property of the algebra $C^b(X)$. A full account can be found in the book "Saks Spaces and Applications to Functional Analysis".

share|improve this answer
Here's a link to the book: books.google.com.au/… - thanks for this answer, it looks interesting! –  David Roberts Oct 24 '12 at 8:48

Your Answer


