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Let us consider the unital commutative $C^*$-algebra $C[0,1]$. We say $A\subseteq C[0,1]$ forms a C*-subring if it satisfies the following conditions:

1- $A$ is an involutive unital subring (closed under pointwise addtion, multiplication and invoultion operations) whose unit is just the constant function $1$.

2- $A$ is norm-closed.

Q. Any characterization of (the elements of) $A$?

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  • $\begingroup$ So the important point is: no $*$-operation, and no scalar multiplication? $\endgroup$
    – Matthew Daws
    Commented Jun 29, 2020 at 12:03
  • $\begingroup$ Dear Matthew, Thanks. Just no scaler multiplication. I made a correction. $\endgroup$
    – ABB
    Commented Jun 29, 2020 at 12:07
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    $\begingroup$ A remark: the C*-subring generated by x is formed by all functions that are integer valued at 0 and 1 (see Theorem 5 in "What Can Be Approximated by Polynomials with Integer Coefficients" by Le Baron O. Ferguson). $\endgroup$
    – Mateusz Wasilewski
    Commented Jun 29, 2020 at 12:15
  • $\begingroup$ @ Mateusz Wasilewski. The Theorem that you mentioned to says: For a continuous real-valued function $f$ on the unit interval $[0, 1]$ to be uniformly approximable by polynomials with integer coefficients it is necessary and sufficient that it be integer-valued at both 0 and 1. $\endgroup$
    – ABB
    Commented Jun 29, 2020 at 13:01

2 Answers 2

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This is not an answer! It is hoped to be a reasonable conjecture, at best!

Based on Mateusz Wasilewski's comment one can build examples of subrings of $C([0,1])$ by choosing a closed subset $K⊆[0,1]$ and taking $A$ to be the set of functions in $C([0,1])$ taking integer values on $K$. However this is not such a general example for at least two reasons. It fails to include:

  1. sub-C*-algebras of $C([0, 1])$, and

  2. algebras of the form $R\cdot1$, where $R$ is a closed subring of $\mathbb{C}$.

Algebras of form (1) can be quite big (think of space filling curves) so the choice of $[0,1]$ as base space does not seem to bring any relevant simplification and indeed it may actually obfuscate the problem. So it is perhaps a good idea to recast the problem asking instead for the general form of subrings of $C(X)$, where $X$ is a compact topological space.

If $A$ is a subring of $C(X)$, then it is also a subring of the sub-C*-algebra generated by $A$, which in turn is of the form $C(Y)$, where $Y$ is a quotient of $X$. It is easy to see that $A$ separates points of $Y$ so it might make sense to require this condition in the original question.

Regarding (2), I haven't given much thought as to what is the general form of a closed subring of $\mathbb{C}$, such as $\mathbb{Z}+i\mathbb{Z}$ (any ideas?), but at first glance it does not seem to be an entirely trivial question so let's do away with this difficulty by sticking to real valued functions instead. After all, apart from $\mathbb{R}$ itself, the only closed unital subring of $\mathbb{R}$ is $\mathbb{Z}$!

So here is my conjecture:

Let $X$ be a compact Hausdorff topological space and let $A$ be a closed unital subring of $C(X, \mathbb{R})$ (continuous real valued functions on $X$), which separates points of $X$. Then there exists a closed subset $K⊆X$ such that $A$ coincides with the set of functions in $C(X)$ taking integer values on $K$.

Hopefully a clever modification of the proof of the Stone-Weierstrass Theorem can prove this as well!

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    $\begingroup$ It feels like you're on the right track, but I think the ring generated by $f(x) = x$ on $[0,2]$ is a counterexample. Anything in the ring is integer-valued at $x = 0$, $1$, or $2$, but also must have the same parity at $x = 0$ and $x = 2$ ... maybe your conjecture can be refined to accomodate this? $\endgroup$
    – Nik Weaver
    Commented Aug 18, 2020 at 15:55
  • $\begingroup$ Yes, you are right! This immediately brings about a bunch of ideals associated to points in the space and a bunch of homomorphisms between them! So far I cannot see how to organize all of this :-( $\endgroup$
    – Ruy
    Commented Aug 18, 2020 at 16:14
  • $\begingroup$ For instance, I think the discrete unital subrings of $\mathbb{R}^2$ are $\mathbb{Z}^2$ and, for each $n$, the set $\{(a,b): a \equiv b \, ({\rm mod}\, n)\}$. $\endgroup$
    – Nik Weaver
    Commented Aug 18, 2020 at 20:53
  • $\begingroup$ @NikWeaver, what scares me most is the subring of $\mathbb{R}^3$ formed by the integer triples $(x,y,z)$ such that $x\equiv y (\text{mod } p)$, $y\equiv z (\text{mod } q)$, and $z\equiv x (\text{mod } r)$. $\endgroup$
    – Ruy
    Commented Aug 18, 2020 at 22:10
  • $\begingroup$ Hmm. So maybe for every pair of points in $K$ we have such a condition? $\endgroup$
    – Nik Weaver
    Commented Aug 18, 2020 at 22:39
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Let $R$ be a unital subring of $\mathbb{Z}^n$ and choose $n$ points $t_1$, $\ldots$, $t_n$ in $[0,1]$. We can see from Proposition 2.3 of this paper that $R$ can be fairly complicated. Then the set of functions in $C[0,1]$ whose restriction to $\{t_1, \ldots, t_n\}$ belongs to $R + iR$ is an involutive unital subring of $C[0,1]$. I think this shows that it is unreasonable to expect there to be any clean characterization of all involutive unital subrings of $C[0,1]$.

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