Can we show that idempotents in $A$ are linearly dense if the spectum of $A$ is totally disconnected?

Marianna Stone
2022-05-24
Answered

Suppose that $A$ is a unital commutative Banach algebra. It is a nice application of the Shilov idempotent theorem that if the spectrum of $A$ is totally disconnected, then $A$ is regular.

Can we show that idempotents in $A$ are linearly dense if the spectum of $A$ is totally disconnected?

Can we show that idempotents in $A$ are linearly dense if the spectum of $A$ is totally disconnected?

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rideonthebussp

Answered 2022-05-25
Author has **10** answers

A totally disconnected spectrum doesn't imply a dense span of the idempotents.

Let's look first at the silly example ${A}_{0}={\mathbb{C}}^{2}$, endowed with the norm $\Vert (x,y){\Vert}_{{A}_{0}}=|x|+|y|$ and the multiplication $(u,v)\cdot (x,y)=(ux,uy+vx)$, so ${A}_{0}\cong \mathbb{C}[X]/({X}^{2})$, endowed with the ${\ell}^{1}$-norm. One verifies that this is a unital commutative Banach algebra, and its only idempotent elements are ${e}_{0}=(1,0)$ and $0$, so the linear span of the idempotents is not dense. The spectrum of ${A}_{0}$ is a singleton (the only unital algebra homomorphism to $\mathbb{C}$ is the first coordinate projection), so it's totally disconnected (and connected).

We use that to construct a less silly example, in which the spectrum is totally disconnected and not connected. Let ${A}_{1}$ be a unital commutative Banach algebra with (nonempty) totally disconnected spectrum, for example ${A}_{1}=C(K,\mathbb{C})$, where $K$ is the Cantor set (${A}_{0}$ would also work). Let $A={A}_{0}\times {A}_{1}$, with componentwise multiplication, and endow it with the norm $\Vert (x,y){\Vert}_{A}=max\{\Vert x{\Vert}_{{A}_{0}},\Vert y{\Vert}_{{A}_{1}}\}$. That makes $A$ a unital commutative Banach algebra, and the idempotents in $A$ are the pairs $(x,y)$ where $x$ is an idempotent in ${A}_{0}$ and $y$ an idempotent in ${A}_{1}$. So the span of idempotents is not dense, $((0,1),0)$ does not belong to the closure of that span. Since $\{0\}\times {A}_{1}$ are ideals in $A$, it follows that every unital homomorphism $A\to \mathbb{C}$ is either ${\psi}_{0}:((u,v),w)\mapsto u$ or $(x,y)\mapsto \lambda (y)$ with $\lambda \in \mathrm{\Delta}({A}_{1})$. Thus $\mathrm{\Delta}(A)\cong \mathrm{\Delta}({A}_{1})\cup \{{\psi}_{0}\}$, where ${\psi}_{0}$ is an isolated point, so totally disconnected.

It would be interesting to investigate whether the presence of nilpotent elements is the only thing that can make the span of idempotents non-dense if the spectrum is totally disconnected.

Let's look first at the silly example ${A}_{0}={\mathbb{C}}^{2}$, endowed with the norm $\Vert (x,y){\Vert}_{{A}_{0}}=|x|+|y|$ and the multiplication $(u,v)\cdot (x,y)=(ux,uy+vx)$, so ${A}_{0}\cong \mathbb{C}[X]/({X}^{2})$, endowed with the ${\ell}^{1}$-norm. One verifies that this is a unital commutative Banach algebra, and its only idempotent elements are ${e}_{0}=(1,0)$ and $0$, so the linear span of the idempotents is not dense. The spectrum of ${A}_{0}$ is a singleton (the only unital algebra homomorphism to $\mathbb{C}$ is the first coordinate projection), so it's totally disconnected (and connected).

We use that to construct a less silly example, in which the spectrum is totally disconnected and not connected. Let ${A}_{1}$ be a unital commutative Banach algebra with (nonempty) totally disconnected spectrum, for example ${A}_{1}=C(K,\mathbb{C})$, where $K$ is the Cantor set (${A}_{0}$ would also work). Let $A={A}_{0}\times {A}_{1}$, with componentwise multiplication, and endow it with the norm $\Vert (x,y){\Vert}_{A}=max\{\Vert x{\Vert}_{{A}_{0}},\Vert y{\Vert}_{{A}_{1}}\}$. That makes $A$ a unital commutative Banach algebra, and the idempotents in $A$ are the pairs $(x,y)$ where $x$ is an idempotent in ${A}_{0}$ and $y$ an idempotent in ${A}_{1}$. So the span of idempotents is not dense, $((0,1),0)$ does not belong to the closure of that span. Since $\{0\}\times {A}_{1}$ are ideals in $A$, it follows that every unital homomorphism $A\to \mathbb{C}$ is either ${\psi}_{0}:((u,v),w)\mapsto u$ or $(x,y)\mapsto \lambda (y)$ with $\lambda \in \mathrm{\Delta}({A}_{1})$. Thus $\mathrm{\Delta}(A)\cong \mathrm{\Delta}({A}_{1})\cup \{{\psi}_{0}\}$, where ${\psi}_{0}$ is an isolated point, so totally disconnected.

It would be interesting to investigate whether the presence of nilpotent elements is the only thing that can make the span of idempotents non-dense if the spectrum is totally disconnected.

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