Skip to content

Non-isomorphisms of some commutative Banach algebras

22 July, 2018

The discussion in the previous post was originally motivated by a particular case of the following general problem:

Given two connected Lie groups {G_1} and {G_2}, when are their Fourier algebras {{\rm A}(G_1)} and {{\rm A}(G_2)} isomorphic (as topological algebras)?

Generally speaking, there is no universal algorithm for deciding if two commutative Banach algebras (CBAs) are isomorphic in the sense above. However, there are various standard tools one can try to use.

  1. Are they both unital / non-unital?
  2. Are they both Jacobson semisimple?
  3. Do they have homeomorphic maximal ideal spaces? Shilov boundaries?
  4. Are they both Arens regular?
  5. Can they be distinguished by cohomological invariants? In particular: are they both (non-)amenable? weakly amenable?

One additional test that is sometimes overlooked is:

  1. are the underlying topological vector spaces of the two CBAs isomorphic?

To use slightly more common phrasing: do the two CBAs have “the same” underlying Banach space?

The aim of this belated sequel is to present a few simple and instructive examples where we can easily distinguish two given CBAs, and then to show how the results mentioned in the previous post allow us to distinguish two Fourier algebras when the other simple tests seem inadequate.

As before, I have not tried to make the arguments here self-contained, but hopefully those who are interested can easily look up the relevant terminology and definitions.


Example 1.

Up to isomorphism, there are exactly two unital, commutative, 2-dimensional {{\mathbb C}}-algebras, corresponding to

\displaystyle  \left\{ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\colon a,b\in {\mathbb C} \right\} \quad\hbox{and}\quad \left\{ \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \colon a,b\in {\mathbb C} \right\}

The first algebra is semisimple but the second is not; so the two algebras cannot be isomorphic.

Example 2.

Consider the following function algebras on the closed unit disc: {C(\overline{\mathbb D})}, the algebra of all continuous complex-valued functions on {\overline{\mathbb D}}; and {\mathcal{A}(\overline{\mathbb D})}, the subalgebra of all {f\in C(\overline{\mathbb D})} which are analytic on the open unit disc. We equip both of these with the usual supremum norm. Both are unital, semisimple, Arens regular Banach algebras, and both have maximal ideal space {\overline{\mathbb D}}. However, the Shilov boundary of {\mathcal{A}(\overline{\mathbb D})} is the unit circle, while that of {C(\overline{\mathbb D})} is the whole of the closed disc. So these Banach algebras cannot be isomorphic.

Example 3.

Take {\mathcal{A}(\overline{\mathbb D})}, as in Example 2, but now consider the subalgebra {{\rm A}_+({\mathbb T})}, which consists of all {f\in \mathcal{A}(\overline{\mathbb D})} whose Taylor series (centred at {0}) converge absolutely on the closed unit disc. In other words, such {f} are of the form {f(z) = \sum_{n=0}^\infty a_n z^n} where {\sum_{n=0}^\infty |a_n| <\infty}. We equip {{\rm A}_+({\mathbb T})} with the obvious {\ell^1}-type norm. Both of these CBAs are unital and semisimple, and both have the same maximal ideal space and Shilov boundary. However there are several ways to show that they are not isomorphic:

  • {\mathcal{A}(\overline{\mathbb D})} is Arens regular, while {{\rm A}_+({\mathbb T})} is not;
  • the underlying Banach spaces of {\mathcal{A}(\overline{\mathbb D})} and {{\rm A}_+({\mathbb T})} are not isomorphic (for instance, the latter space has the Schur property while the former one does not);
  • the automorphism group of {\mathcal{A}(\overline{\mathbb D})} is {{\rm SU}(1,1)} (with usual action on the unit disc via Möbius transformations) while the automorphism group of {{\rm A}_+({\mathbb T})} is just {{\mathbb T}} acting by rotations.

Example 4.

Take

\displaystyle  {\bf aff}= {\mathbb R}\rtimes {\mathbb R}^*_+ \cong \left\{ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \colon a,b\in{\mathbb R}; \; a>0 \right\}

Now consider two groups {G_1= {\mathbb R}^2\times {\bf aff}} and {G_2={\bf aff} \times {\bf aff}}. The Fourier algebras {{\rm A}(G_1)} and {{\rm A}(G_2)} share the following properties:

  • both non-unital (and both have bounded approximate identities);
  • both Jacobson-semisimple;
  • both have maximal ideal spaces homeomorphic to {{\mathbb R}^4}, with the Shilov boundary being the whole maximal ideal space in both cases;
  • both Arens irregular;
  • both fail to be weakly amenable.

I do not know if they can be distinguished by their automorphism groups (recall that we are not assuming automorphisms are isometric). However, we do know that {{\rm A}(G_1)} and {{\rm A}(G_2)} are not isomorphic as Banach spaces (and so in particular they cannot be isomorphic as topological algebras).

Why is this? Well, it is known (I think due to Khalil, but possibly also worked out by Gelfand’s school) that {{\rm A}({\bf aff})} is isomorphic as a Banach space to {{\mathcal S}_1(L^2({\mathbb R}^*_+))\oplus{\mathcal S}_1(L^2({\mathbb R}^*_+))}, where {{\mathcal S}_1(H)} denotes the trace-class operators on a Hilbert space {H}.

We also know that if {H} and {K} are separable infnite-dimensional Hilbert spaces, then {{\mathcal S}_1(H)\oplus{\mathcal S}_1(K)\cong{\mathcal S}_1(\ell_2)} and {{\mathcal S}_1(H\otimes_2 K)\cong{\mathcal S}_1(\ell_2)} at the level of Banach spaces.

Now, by using some abstract operator-algebra/operator-space techniques, one can bootstrap this to show that {{\rm A}(G_1)} is Banach-space-isomorphic to {L^1({\mathbb R}^2)\hat{\otimes} {\mathcal S}_1(\ell_2)\cong L^1([0,1]; {\mathcal S}_1(\ell_2))} while {{\rm A}(G_2)} is Banach-space isomorphic to {{\mathcal S}_1(L^2({\mathbb R}^*_+\times{\mathbb R}^*_+)) \cong {\mathcal S}_1(\ell_2)}. And, as observed in the previous post, these two Banach spaces are not isomorphic.

A final question.

Can we prove that {{\rm A}({\mathbb R}^2\times{\bf aff}\times{\bf aff})} and {{\rm A}({\mathbb R}^4\times{\bf aff})} are not isomorphic as Banach algebras?

Note that both these Fourier algebras have underlying Banach space isomorphic to {L^1([0,1];{\mathcal S}_1(\ell_2))} so that the previous argument does not apply. Moreover, both algebras share the same five properties listed in Example 4.

It is my feeling (backed up by some incomplete private calculations) that we can distinguish these two algebras by looking at the space of alternating {2}-cocycles. To use some old terminology introduced by B. E. Johnson: it seems that the second algebra is {2}-dimensionally weakly amenable, while the first one isn’t. However, to my knowledge this has not been worked out explicitly in the literature.

Advertisements
No comments yet

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: