This question is the outcome of a few naive thoughts, without reading the proof of Gelfand-Neumark theorem.

Given a compact Hausdorff space $X$, the algebra of complex continuous functions on it is enough to capture everything on its space. In fact, by the Gelfand-Neumark theorem, it is enough to consider the commutative C*-algebras instead of considering compact Hausdorff spaces.

The important thing here is that $C*$-algebras have a complex structure. The real structure is not enough. Given the algebra $C(X, \mathbb R) \oplus C(X, \mathbb R)$ of real continuous functions on $X$, the algebra $C(X, \mathbb{C})$ is simply the direct sum $C(X, \mathbb R) \oplus C(X, \mathbb R)$, as a Banach algebra(and this can be given a complex structure, (seeing it as the complexification...)). But to obtain a C*-algebra, we need an additional C*-algebra, and the obvious way, ie, defining $(f + ig)$* $= (f - ig)$ does not work out. More precisely, the C* identity does not hold.

So one cannot weaken (as it stands) the condition in the Gelfand-Neumark theorem that we need the algebra of complex continuous functions on the space $X$, since we do need the C* structure. Of course, this is without an explicit counterexample. Which brings us to:

Qn 1. Please given an example of two non-homeomorphic compact Hausdorff spaces $X$ and $Y$ such that the function algebras $C(X, \mathbb R)$ and $C(Y, \mathbb R)$ are isomorphic(as real Banach algebras)?

(Here I am hoping that such an example exists).

Then again,

Qn 2. From the above it appears that the structure of complex numbers is involved when the algebra of complex functions captures the topology on the space. So how exactly is this happening?

(The vague notions concerning this are something like: the complex plane minus a point contains nontrivial $1$-cycles, so perhaps the continuous maps to the complex plane might perhaps capture all the information in the first homology, etc..)..

Note : Edited in response to the answers. Fixed the concerns of Andrew Stacey, and changed Gelfand-Naimark to Gelfand-Neumark, as suggested by Dmitri Pavlov.

  • 1
    $\begingroup$ If you know a little algebraic geometry, an exercise toward the end of chapter 1 in Atiyah-Macdonald walks you through an attractive proof that a CH space is determined by its ring of real valued functions. What the C* algebra formalism gets you in this context is the "image" of the functor $X \mapsto C(X, \mathbb{C})$ from CH spaces to $\mathbb{C}$-algebras. It is possible to formulate an analogous condition for $\mathbb{R}$-algebras, but the entire theory is more awkward (though more powerful in some ways!). $\endgroup$ – Paul Siegel May 25 '10 at 23:48
  • $\begingroup$ Let me try to justify my last remark. Aside from the awkwardness of the definition, the Bott periodicity theorem for the K-theory of real C* algebras has period 8 rather than period 2 in the complex case, and thus many arguments are much more arduous. But there is also room for more subtlety. For example, the complex structure of C* algebras is at some level responsible for the fact that the Atiyah Singer index theorem is most naturally formulated for even dimensional manifolds. One way to formulate an odd dimensional analogue is to use real C* algebras. This is not without applications. $\endgroup$ – Paul Siegel May 25 '10 at 23:57
  • $\begingroup$ @Paul: That is a nice exercise, but I am confused by "If you know a little algebraic geometry". I guess you mean because of the topology placed on the maximal ideal space? Also, thanks for addressing some differences between the real and complex case. @Akela: In spite of Paul's comments, I must say I'm a little confused by the noncommutative-geometry tag in light of the fact that there is nothing noncommutative in your question. $\endgroup$ – Jonas Meyer May 26 '10 at 2:40
  • 1
    $\begingroup$ @Jonas: In AM, $X$ is recovered from $C(X)$ as the subspace of $Spec(C(X))$ (with the Zariski topology) consisting of maximal ideals. So the exercise requires a casual familiarity with $Spec$, which one usually accumulates via AG. For what it's worth, I support the NCG tag for this question. I don't always work directly with NC C*-algebras in a substantial way, but I usually tell people I work in NCG because I often use $C(X)$ as a proxy for $X$. Maybe I am "mistagging" myself. :) $\endgroup$ – Paul Siegel May 26 '10 at 14:03
  • $\begingroup$ Makes sense. This isn't the first time I've heard of someone working in commutative NCG, but I still appreciate the novelty. $\endgroup$ – Jonas Meyer May 26 '10 at 19:34

Here is a slightly different, perhaps simpler take on showing that $C(X,\mathbb{R})$ determines $X$ if $X$ is compact Hausdorff. For each closed subset $K$ of $X$, define $\mathcal{I}_K$ to be the set of elements of $C(X,\mathbb{R})$ that vanish on $K$. The map $K\mapsto\mathcal{I}_K$ is a bijection from the set of closed subsets of $X$ to the set of closed ideals of $C(X,\mathbb{R})$. Urysohn's lemma and partitions of unity are enough to see this, with no complexification, Gelfand-Neumark, or (explicitly) topologized ideal spaces required. I remember doing this as an exercise in Douglas's Banach algebra techniques in operator theory in the complex setting, but the same proof works in the real setting.

Here are some details in response to a prompt in the comments. (Added later: See Theorem 3.4.1 in Kadison and Ringrose for another proof. Again, the functions are assumed complex-valued there, but you can just ignore that, read $\overline z$ as $z$ and $|z|^2$ as $z^2$, to get the real case.)

I will take it for granted that each $\mathcal{I}_K$ is a closed ideal. This doesn't require that the space is Hausdorff (nor that $K$ is closed). Suppose that $K_1$ and $K_2$ are unequal closed subsets of $X$, and without loss of generality let $x\in K_2\setminus K_1$. Because $X$ is compact Hausdorff and thus normal, Urysohn's lemma yields an $f\in C(X,\mathbb{R})$ such that $f$ vanishes on $K_1$ but $f(x)=1.$ Thus, $f$ is in $\mathcal{I}_{K_1}\setminus\mathcal{I}_{K_2}$, and this shows that $K\mapsto \mathcal{I}_K$ is injective. The work is in showing that it is surjective.

Let $\mathcal{I}$ be a closed ideal in $C(X,\mathbb{R})$, and define $K_\mathcal{I}=\cap_{f\in\mathcal{I}}f^{-1}(0)$, so that $K_\mathcal{I}$ is a closed subset of $X$. Claim: $\mathcal{I}=\mathcal{I}_{K_\mathcal{I}}$.

It is immediate from the definition of $K_\mathcal{I}$ that each element of $\mathcal{I}$ vanishes on $K_\mathcal{I}$, so that $\mathcal{I}\subseteq\mathcal{I}_{K_\mathcal{I}}.$ Let $f$ be an element of $\mathcal{I}_{K_\mathcal{I}}$. Because $\mathcal{I}$ is closed, to show that $f$ is in $\mathcal{I}$ it will suffice to find for each $\epsilon>0$ a $g\in\mathcal{I}$ with $\|f-g\|_\infty<3\epsilon$. Define $U_0=f^{-1}(-\epsilon,\epsilon)$, so $U_0$ is an open set containing $K_\mathcal{I}$. For each $y\in X\setminus U_0$, because $y\notin K_\mathcal{I}$ there is an $f_y\in \mathcal{I}$ such that $f_y(y)\neq0$. Define $$g_y=\frac{f(y)}{f_y(y)}f_y$$ and $U_y=\{x\in X:|g_y(x)-f(x)|<\epsilon\}$. Then $U_y$ is an open set containing $y$. The closed set $X\setminus U_0$ is compact, so there are finitely many points $y_1,\dots,y_n\in X\setminus U_0$ such that $U_{y_1},\ldots,U_{y_n}$ cover $X\setminus U_0$. Relabel: $U_k = U_{y_k}$ and $g_k=g_{y_k}$. Let $\varphi_0,\varphi_1,\ldots,\varphi_n$ be a partition of unity subordinate to the open cover $U_0,U_1,\ldots,U_n$. Finally, define $g=\varphi_1 g_1+\cdots+\varphi_n g_n$. That should do it.

In particular, a closed ideal is maximal if and only if the corresponding closed set is minimal, and because points are closed this means that maximal ideals correspond to points. (Maximal ideals are actually always closed in a Banach algebra, real or complex.)

  • $\begingroup$ To wrap things up, I might as well purchase and wear a donkey costume.. $\endgroup$ – Akela May 25 '10 at 22:09
  • $\begingroup$ I'm not sure how to respond to that, but I hope you don't think I am disparaging your question. $\endgroup$ – Jonas Meyer May 25 '10 at 22:25
  • $\begingroup$ No I didn't mean that. I got thinking into all absurd directions with the wrong premises. I like your answer. This is a complete solution modulo the solution of the exercise. Would you please include a sketch of the proof? Also it may be a good idea to include explicitly that the maximal ideals in C(X,R) are precisely the points of the space. $\endgroup$ – Akela May 25 '10 at 22:30
  • $\begingroup$ I mean I accepted your answer. Since it's an accepted answer, it would be more helpful to any readers when it's complete with more details .. $\endgroup$ – Akela May 25 '10 at 22:30
  • 1
    $\begingroup$ @IntegrateThis: It has been a while and I don't recall exactly what I had in mind, but if my thinking now is correct, my thinking then was convoluted and I didn't need $3$. With $g_0 \equiv 0$, for each $x\in X$, $\begin{align*}|f(x)-g(x)| &= \left|\sum\limits_{k=0}^n \varphi_k(x)(f(x)-g_k(x))\right|\\ &\leq \sum\limits_{k=0}^n\varphi_k(x)|f(x)-g_k(x)|, \end{align*}$, and each nonzero summand has the form $\varphi_k(x)|f(x)-g_k(x)|$ with $|f(x)-g_k(x)|<\epsilon$ (because $x\in U_k$ for such a summand), so the sum is less than $\sum\limits_{k=0}^n \phi_k(x)\epsilon=\epsilon$. Right? $\endgroup$ – Jonas Meyer Jan 29 '19 at 4:20

Qn 1 is trivial because you said "topological spaces" rather than "compact Hausdorff spaces" (or "locally compact Hausdorff" would be okay, I guess). Simply $\lbrace 0,1\rbrace$ with the order topology and $\lbrace 0\rbrace$ will do.

If we refine to "compact Hausdorff spaces" then I take $C(X,\mathbb{R})$ and $C(Y,\mathbb{R})$, complexify, and apply GN to recover $X$ and $Y$, thus I claim that no counterexample exists.

I think that the issue stems from a confusion between the complexification of a real algebra and the underlying real algebra of a complex one. Since I can recover $C(X,\mathbb{C})$ from $C(X,\mathbb{R})$, all the information about the former is captured in the latter. However, since I can find several complex structures on the same real algebra, $C(X,\mathbb{C})_{\mathbb{R}}$ does not contain all the information that is contained in $C(X,\mathbb{C})$. There is a reason why they are called forgetful functors!

So $C(X,\mathbb{R})$ is not the underlying real algebra of $C(X,\mathbb{C})$, but $C(X,\mathbb{C})$ is the complexification of $C(X,\mathbb{R})$.

  • $\begingroup$ Sorry, I unconsciously typed "topological" instead of "compact Hausdorff". How do you get the *-structure after complexification? $\endgroup$ – Akela May 25 '10 at 14:42
  • 1
    $\begingroup$ The next-to-last R should be a C. $\endgroup$ – S. Carnahan May 25 '10 at 15:01
  • $\begingroup$ @Scott: Thanks, fixed. @Akela: since C(X,R)oC = C(X,C), you get it from the same place as normally. $\endgroup$ – Andrew Stacey May 25 '10 at 20:14
  • $\begingroup$ I understand that C(X,R)oC is a complex Banach algebra. What I do not understand is how you gave it the structure of a C*-algebra, as "normal". As I understand Gelfand-Naimark theorem, you need commutative C*-algebras, not commutative Banach algebras. $\endgroup$ – Akela May 25 '10 at 20:21
  • 1
    $\begingroup$ Dear Akela, $C(X,\mathbb R)\otimes_{\mathbb R} \mathbb C$ is isomorphic to $C(X,\mathbb C)$ as a $\mathbb C$-algebra. This is all that is needed to then recover $X$ from the usual Gelfand--Neumark theorem: one takes all maximal ideals, topologizes them via the weak topology, and this is $X$, by Gelfand--Neumark. $\endgroup$ – Emerton May 25 '10 at 20:55

The noncommutative Gelfand-Neumark theorem can be stated and proved for real C*-algebras. See Corollary 4.10 in Johnstone's book “Stone Spaces”.

P.S. “Gelfand-Naimark” theorem is a misnomer. Take a look at the original paper and note how Gelfand and Neumark spell their names. In fact, they consistently use these spellings throughout all of their non-Russian papers.

  • 3
    $\begingroup$ I don't think 'oxymoron' means what you think it means. $\endgroup$ – HJRW May 25 '10 at 22:04
  • $\begingroup$ "Misnomer" instead of "oxymoron" seems apt. However, it is confusing to us ignorant of Russian and the subtleties of its transliteration to Latin characters, because the name is written as "Naĭmark" in some of his other work. (For example, springerlink.com/content/v71158h17p227p39) I believe some choose "Gelfand-Naimark" for simplicity and consistency, but I appreciate your point. $\endgroup$ – Jonas Meyer May 25 '10 at 22:15
  • 1
    $\begingroup$ @Henry: Inconceivable! $\endgroup$ – Yemon Choi May 25 '10 at 22:19
  • $\begingroup$ @Henry: By an oxymoron I meant “a combination of contradictory or incongruous words”. $\endgroup$ – Dmitri Pavlov May 26 '10 at 4:46
  • 2
    $\begingroup$ @Jonas: It's either Gelfand-Neumark if you cite their non-Russian papers, or Gelʹfand-Naĭmark if you cite one of their Russian papers and use the AMS transliteration system. The spelling Naimark is incorrect and should never be used. As for the cited errata, note that the original paper springerlink.com/content/n3m5656p81712676 lists both spellings. $\endgroup$ – Dmitri Pavlov May 26 '10 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.