Real Gelfand dual

Question

By Gelfand duality, a hausdorff locally compact space $X$ is homeomorphic to the maximal ideals of the complex-valued continuous functions over $X$ that vanish at infinity.

Does this still hold if we consider instead real-valued continuous functions that vanish at infinity? I suspect there are more maximal ideals than points in this case, but I cannot find examples of such ideals.

Are we assuming the ideals in question are modular (or just closed)? (Which always holds when $X$ is compact, say.) If so, the result still holds for real-valued functions, and it is just an easy consequence of the Stone-Weierstrass theorem. (If anything, the proof is even easier than the complex case, since you don’t need to prove maximal ideals are self-adjoint.) — David Gao, Mar 14 '25 at 09:22
@DavidGao. Thanks. Then which parts of Gelfand duality require complex numbers? That's actually my true question, I'll accept an answer about it. — V. Semeria, Mar 14 '25 at 09:40
Depending on what you meant by Gelfand duality - if you meant the maximal ideal space of $C_0(X)$ is $X$, then this statement is true in either the real or complex case. If you meant a commutative $C^\ast$-algebra is of the form $C_0(X)$ where $X$ is its spectrum, then this needs complex numbers - there are commutative real $C^\ast$-algebras that are not $C_0(X)$. An easy example is $\mathbb{C}$ regarded as a real $C^\ast$-algebra. — David Gao, Mar 14 '25 at 09:53
(Also, just as an additional remark, the modularity/closedness assumption is actually not necessary - every maximal ideal of $C_0(X)$ is automatically closed. Though the proof is somewhat technical.) — David Gao, Mar 14 '25 at 10:04
@DavidGao I tried to prove the maximal ideal part with the Stone-Weierstrass theorem, but failed. It would require to prove the maximal ideal is not dense, but I did not find it easier than trying to find a vanishing point directly. — V. Semeria, Mar 15 '25 at 12:43
I’m not sure what you meant by “trying to find a vanishing point directly”, but you have to show maximal ideals are not dense regardless. A dense ideal (say, $c_{00} \subset c_0$) vanishes nowhere, after all. And Stone-Weierstrass is effectively the converse of this, that an ideal that vanishes nowhere must be dense. That’s why my first comment says assuming the maximal ideal in question is closed, since then it cannot be dense and thus must vanish somewhere. Modularity is also sufficient, since maximal ideals that are modular must be closed, which is a standard exercise. — David Gao, Mar 15 '25 at 21:20
Again, maximal ideals of $C_0(X)$ are actually always closed, but that is a harder argument and I was not implying that follows from Stone-Weierstrass. Stone-Weierstrass just tells you that a closed proper ideal must vanish somewhere, that’s all. — David Gao, Mar 15 '25 at 21:22
@DavidGao Do you have a proof of this harder argument that maximal ideals are closed? Again I would accept it as an answer — V. Semeria, Mar 15 '25 at 22:09

score 4 · Accepted Answer · answered Mar 16 '25 at 02:50

Here is an argument proving that every maximal ideal in $C_0(X)$ is closed. This applies to both the real and complex cases. The argument is adapted from Proposition 3.1 in this paper by Dales.

Let $A = C_0(X)$, $I \subset A$ be a maximal ideal. Clearly, $I$ is either closed or dense, so we may assume to the contrary that it is dense. It is standard exercise to show that maximal ideals in unital Banach algebras (real or complex) are closed, so $X$ is non-compact. Let $\tilde{A} = C(X \cup \{\infty\})$ be the unitization of $A$. Note that any ideal in $A$ is also an ideal in $\tilde{A}$. Pick $a \in A \setminus I$. Every element of $C_0(X)$ can be written as the product of two elements, so we may write $a = bc$ for some $b, c \in A$.

Now, let $J = a\tilde{A} + I$. Since $J$ is an ideal that contains $I$ and also $a \in A \setminus I$, by maximality of $I$, $J = A$. Thus, in particular $b - ad \in I$ for some $d \in \tilde{A}$. On the other hand, $I$ is dense in $A$, so we may pick $e \in A$ with $\|e\| < \frac{1}{\|d\|}$ s.t. $c - e \in I$. Thus, $a - be = bc - be \in I$, so,

$$b(1 - ed) = b - bed \in I$$

as $b - ad \in I$ and $a - be \in I$. But $\|ed\| \leq \|e\|\|d\| < 1$, so $1 - ed$ is invertible in $\tilde{A}$. But then,

$$a = bc = cb(1 - ed)(1 - ed)^{-1} \in I$$

as $b(1 - ed) \in I$. This contradicts our assumption that $a \in A \setminus I$, so $I$ cannot be dense. $\square$

Once we know that maximal ideals are closed, the remainder is actually easier in the real case than in the complex case. Indeed, now just apply Stone-Weierstrass to see that, if an ideal $I \subset C_0(X)$ vanished nowhere, then it must be dense. Thus, a maximal ideal must vanish at some point $p \in X$. So $I \subset \{f \in C_0(X): f(p) = 0\}$. The latter set is a proper ideal, whence $I$ must equal the latter set. (In the complex case, this is harder, as one needs to show maximal ideals are also self-adjoint before being able to apply Stone-Weierstrass.)

Thanks a lot. I consider this is another justification of algebraic geometry. Here we handle the usual spaces (the locally compact ones) and their residual field is R (not C). When we switch from Gelfand to the Zariski topology, we get the algebraic schemes. I suppose this is also the reason why schemes have their own ad hoc definition of compactness, since the usual compactness is exhausted here. — V. Semeria, Mar 16 '25 at 09:09
By the way, you can simplify your proof and remove the Alexandrov compactification altogether. Instead, compute in the bigger ring of continuous functions X -> R that do not necessarily vanish at infinity. Divide by twice the norm of d in the definition of e. Then the image of 1 - ed is included in [1/2, 3/2]. So c / (1 - ed) does vanish at infinity thanks to c, and you get the same contradiction. — V. Semeria, Mar 16 '25 at 16:21

Real Gelfand dual

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