Suppose that A is a unital commutative Banach algebra. It is a nice application of the Shilov

Marianna Stone

Marianna Stone

Answered question

2022-05-24

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?

Answer & Explanation

rideonthebussp

rideonthebussp

Beginner2022-05-25Added 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 = C 2 , endowed with the norm ( x , y ) A 0 = | x | + | y | and the multiplication ( u , v ) ( x , y ) = ( u x , u y + v x ), so A 0 C [ X ] / ( X 2 ), endowed with the 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 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 , C ), where K is the Cantor set ( A 0 would also work). Let A = A 0 × A 1 , with componentwise multiplication, and endow it with the norm ( x , y ) A = max { x A 0 , y 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 } × A 1 are ideals in A, it follows that every unital homomorphism A C is either ψ 0 : ( ( u , v ) , w ) u or ( x , y ) λ ( y ) with λ Δ ( A 1 ). Thus Δ ( A ) Δ ( A 1 ) { ψ 0 }, where ψ 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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