The set of characters

Question

Let $\mathcal{A}$ be a $C*$ subalgebra of $C_b(R)$, generated by $C_\infty(R)$ (functions that are arbitrarily small outside some compact set) and the function $e^{it}$. I need to determine the set of characters of $\mathcal{A}$, $\hat{\mathcal{A}}$, the set of all unital homomorphisms into complex numbers. I know what $\widehat {C_\infty(R)}$ is, or at least I hope I do. But I am not sure this is the right way to go. Any hints?

Edit: I figured what this object is: $C*$ subalgebra of $C_b(R)$, generated by $C_\infty(R)$ (functions that are arbitrarily small outside some compact set) and the function $e^{it}$. Any element in this subalgebra is a sum of a periodic function and a function that vanishes at infinity. But I still do not know how to proceed from here. Is the set of characters on C* algebra just the set of all extensions of characters on a subalgebra?

Answer 1

Instead of regarding your algebra as a subalgebra of $C_b(\mathbb{R})$, I'll further enlarge the ambient algebra to $L^\infty(\mathbb{R})$. We note that any character on $C_\infty(\mathbb{R})$ can be extended to the algebra $\mathcal{A}$ you're concerned with, which can then be further extended to $L^\infty(\mathbb{R})$. Hence, the spectrum of $\mathcal{A}$ is an intermediate space between the spectrum of $L^\infty(\mathbb{R})$ (which we shall denote by $\Omega$) and the one-point compactification of $\mathbb{R}$.

Note that any element of $L^\infty(\mathbb{R})$ must restrict to a character on $C_\infty(\mathbb{R}) + \mathbb{C}$, whence it must restrict to the evaluation functional at some point in $\mathbb{R} \cup \infty$. For any $\varphi$ which restricts to $r \in \mathbb{R}$, one can show that it must satisfy $\varphi(f) = f(r)$ for all $f \in C_b(\mathbb{R})$. (I assume there is an easier way to prove this, presumably using the fact that $C_b(\mathbb{R})$ is the multiplier algebra of $C_\infty(\mathbb{R})$. But I'm not familiar with those, so the only way I know involves showing characters on $L^\infty(\mathbb{R})$ can be identified with ultrafilters on the Boolean algebra of measurable sets modulo a.e. equivalence. For details, see sections II and III of https://arxiv.org/abs/2306.02086 .) Hence, those characters don't "split" in the spectrum of $\mathcal{A}$. More precisely, if a character on $\mathcal{A}$ restricts to evaluation at $r \in \mathbb{R}$ on $C_\infty(\mathbb{R})$, then it must be evaluation at $r \in \mathbb{R}$ on the entirety of $\mathcal{A}$. So the spectrum of $\mathcal{A}$ is a compactification of $\mathbb{R}$.

We still need to figure out how the point at infinity splits. We note that $C_\infty(\mathbb{R})$ is an ideal in $\mathcal{A}$ and these characters correspond exactly to all characters that restrict to 0 on the said ideal, i.e., they correspond exactly to characters of $\mathcal{A}/C_\infty(\mathbb{R})$. The algebra generated by $e^{it}$ contains all continuous functions with period $2\pi$. Except the zero function, none of these functions is contained in $C_\infty(\mathbb{R})$, whence we see that $\mathcal{A}/C_\infty(\mathbb{R})$ is naturally isomorphic to the algebra generated by $e^{it}$, which is $C(\mathbb{T})$. One can then identify the spectrum of $\mathcal{A}/C_\infty(\mathbb{R})$ with $\mathbb{T}$, so the spectrum of $\mathcal{A}$ contains a copy of $\mathbb{T}$ at infinity.

To summarize, the spectrum of $\mathcal{A}$ can be divided into a copy of $\mathbb{R}$ and a copy of $\mathbb{T}$, both subspaces having their usual topologies. Since $\mathcal{A}$ is separable, the spectrum is second-countable, so to describe the topology I only need to describe convergence of sequences. $\mathbb{T}$ is closed in the spectrum, so a sequence solely in $\mathbb{T}$ will converge or diverge as in $\mathbb{T}$. The same goes for any bounded sequence in $\mathbb{R}$. For an unbounded sequence in $\mathbb{R}$, based on things we have obtained it's relatively easy to show that it only converges iff two conditions are met: 1. the sequence diverges to infinity; 2. modulo $2\pi$, the sequence converges in $\mathbb{T}$. In which case the sequence converges to the said limit in $\mathbb{T}$. Finally, for a sequence with both elements in $\mathbb{R}$ and $\mathbb{T}$: If the sequence only has finitely many elements in $\mathbb{R}$ or $\mathbb{T}$, then it reduces to one of the preceding cases. Otherwise, you can similarly show that it only converges iff two conditions are met: 1. the subsequence of all elements in $\mathbb{R}$ diverges to infinity; 2. reducing all elements in $\mathbb{R}$ modulo $2\pi$ to obtain a sequence purely in $\mathbb{T}$, then this sequence converges in $\mathbb{T}$. In which case the sequence converges, again, to the said limit in $\mathbb{T}$.

(I haven't come up with a more concise way to describe this space. If anyone has an idea as to how to do it, feel free to post in comments or edit my answer.)

Answer 2

1. Consider $C$ be an ${}^*$-algebra (not $C^*$ yet) generated by $C_\infty$ and $\{e^{it}\}.$ $C = C_\infty \oplus P(e^{it}, e^{-it})$ where $P$ is polynomial algebra. Direct sum is viewed in category of vector spaces. Each pair $(z(x), p(x))$ corresponds to function $z(x)+p(x).$ Multiplication in terms of direct sum looks as $(z_0, p_0)(z_1, p_1) = (z_0z_1 + z_0p_1 + z_1p_0, p_0p_1).$
2. Consider norm $\|\cdot\|$ on $C$ from $C_b$ and $\| \cdot \|_\infty$ (resp. $\|\cdot\|_P$) on $C_\infty$ (resp. $P(e^{it}, e^{-it})$) from $C_b.$ Then $\|(z, p)\| \le \| z \|_\infty + \|p\|_P,$ $\|p\|_P \le \|(z, p)\|$ (checking near infinity). $\| z \|_\infty - \|p\|_P \le \|(z, p)\|$ and $\| z \|_\infty + \|p\|_P \le 3\|(z, p)\|.$ Norms are equivalent.
3. By equivalence of norms, completion $\hat{C}$ of $C$ is completion of direct sum with norms $\| \cdot \|_\infty$ and $\|\cdot\|_P.$ $\hat{C}$ is required $C^*$-algebra $\mathcal{A}.$ By Stone–Weierstrass theorem, $\hat{P}(e^{it}, e^{-it})=C(\mathbb{T}),$ $\mathbb{T}$ is $S^1,$ it corresponds to period of $e^{it}.$
4. $\mathcal{A} = C_\infty \oplus C(\mathbb{T}).$ Any character $\psi$ on $\mathcal{A}$ is continuous linear functional on direct sum and $\psi = \psi_\infty + \psi_P$ where $\psi_\infty$ and $\psi_P$ are linear functionals on direct sum's components.
5. As $\psi(XY) = \psi(X)\psi(Y)$ one can deduce that $\psi_\infty$ (resp. $\psi_P$) are multiplicative homomorphisms of $C_\infty$ (resp. $C(\mathbb{T})$). $\psi_\infty$ or $\psi_P$ should be non-zero.

More expansions of definitions follow.

6. $\psi_\infty$ is evaluation at point $\mathbb{R}=\Delta(C_\infty)$ or $0,$ similarly $\psi_P$ is evaluation at point $\mathbb{T}=\Delta(C(\mathbb{T}))$ or $0.$ Checking multiplication from p.1, one obtains more restrictions on $\psi_\infty$ and $\psi_P$ and complete set of characters. $\hat{\mathcal{A}}=\Delta(\mathcal{A})$ is union of $\Delta(C_\infty)$ (evaluations of $z+p=(z, p)\in\hat{C}$ at different points) and of $\Delta(C(\mathbb{T}))$ (evaluations of $C(\mathbb{T})$ component at points of $\mathbb{T}$) confirming David Gao's description.
7. Topology on $\Delta(\mathcal{A})$ is weak. Need to analyze simple neighborhoods $V^{c}_{(z,p), \epsilon}$ of $c\in\Delta(\hat{C})$ where $(z,p) \in \hat{C}, \epsilon > 0:$ $$ V^{c}_{(z,p), \epsilon} = \{c'\in \Delta(\hat{C}): |c'(z,p) - c(z,p)| < \epsilon \} $$ Using convenient set of test-functions one can describe these sets. Interesting neighborhoods are $ V^{c}_{(0,p), \epsilon}, c\in \Delta(C_P),$ these contain $2\pi$-periodic intervals in $\Delta(C_\infty).$ I think, this description of topology corresponds to David Gao's description.