A Question about an irreducible ultra-power II,

Question

Let $E$ be an irreducible Banach $A$-module, for a Banach algebra $A$. One can easily show that for an ultra filter $\mathcal U$, $(E)_\mathcal U$ is a Banach $(A)_\mathcal U$-module. Is it possible to find a countably incomplete ultra filter $\mathcal U$ such that $(E)_\mathcal U$ becomes an irreducible Banach $(A)_\mathcal U$-module?

Answer 1

This is really an extended comment, at least at the moment.

As you mean "algebraically irreducible", for any $x_0\in E$ the orbit of $x_0$ by $A$ is exactly $E$, and so:

• $I := \{ a\in A : a\cdot x_0 = 0 \}$ is a maximal modular left ideal in $A$.
• To be a little more precise, the modular unit is any $u\in A$ with $u\cdot x_0=x_0$.
• Then $E$ is isomorphic to $A/I$.
• So let's assume that actually $E = A/I$.

It's not so hard to check that $\newcommand{\mc}{\mathcal}(A/I)_{\mc U} = (A)_{\mc U} / (I)_{\mc U}$ for the obvious isomorphism. We have that $u$ is the modular unit for $(I)_{\mc U}$. So the question reduces to asking if $(I)_{\mc U}$ is maximal.

I think of this as a sort of "uniformity" question, for the following reason. For each $a\in A$, the map $A \rightarrow A/I; b\mapsto ba+I$ is surjective, and so there is $b$ with $ba - u \in I$. (Then for any $c\in A$ we have that $cba +I = cu + I = c + I$, and $\|cb\| \leq \|c\|\|b\|$, which is important below).

So, does the modular unit always "uniformly factorise" is this way?

Suppose there is $K>0$ so that if $\|a+I\|=1$ there is $b$ with $\|b\|\leq K$ and $ba-u\in I$. Then for any non-zero $x\in (A)_{\mc U}/(I)_{\mc U}$ we have $x = (a_n) + (I)_{\mc U}$ for some bounded family $(a_n)$ where also $\|a_n+I\|$ is bounded below. So there is a bounded family $(b_n)$ with $(b_na_n) + (I)_{\mc U} = u + (I)_{\mc U}$. Then for any $c = (c_n)\in (A)_{\mc U}$ we have that $c+(I)_{\mc U} = (c_nu) + (I)_{\mc U} = (c_n b_n)(a_n) + (I)_{\mc U}$. It follows that $(I)_{\mc U}$ is maximal.

If there is $K>0$ so that if $\|a+I\|=1, \|c+I\|\leq 1,\epsilon>0$ there is $b$ with $\|b\|\leq K$ and $\|ba-c+I\|<\epsilon$. Then there is $b'$ with $\|b'\|\leq K\epsilon$ and $\|b'a - (c-ba) + I\|<\epsilon^2$. We iterate, and so find $b_0$ with $\|b_0\|\leq K'$ and $b_0a + I = c+I$. So, suppose there is no such $K$, so we can find $\|a_n+I\|=1, \|c_n+I\|\leq 1$ so that $\|b_na_n-c_n+I\|<1/n \implies \|b_n\|\geq n$. Towards a contradiction, if $(I)_{\mc U}$ is maximal, $(A)_{\mc U} (a_n) + (I)_{\mc U} = (A)_{\mc U}$, and so there is $(b_n)$ bounded with $\lim_{n\rightarrow\mc U} \|b_na_n - c_n+I\|=0$. This is a contradiction.

So does the modular unit always "factorise uniformly" in this way??