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We know they’re isomorphic, but how are they isomorphic?

14 October, 2018

The given information

  1. L \cong L \oplus L, in a very nice way
  2. m \cong m \oplus m, in a very nice way
  3. L\cong m \oplus Q_m, in a nice way
  4. m \cong L \oplus Q_L, in a faintly dodgy way

Wizardry from Warsaw

m \cong L\oplus Q_L \cong (L\oplus L) \oplus Q_L \cong L \oplus (L\oplus Q_L) \cong L \oplus m

L \cong m\oplus Q_m \cong (m\oplus m) \oplus Q_m \cong m\oplus (m\oplus Q_m) \cong m\oplus L

Conclusion: m \cong L\oplus m \cong m\oplus L \cong L.

A nagging question

Just how nice or dodgy is our final isomorphism m\cong L?

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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.

Reminder to self: (non-)isomorphism of two von Neumann algebras and their preduals

13 May, 2018

So… it’s not clear if regular blogging will ever resume here, but in the meantime here is something to just clear the pipes, as it were. Although the results in this post are not new, occasionally I want to refer to them, and I don’t recall seeing an explicit reference in the literature. The proofs given here are not really proper expositions for those who don’t know Banach space theory; hopefully they will provide sufficiently suggestive outlines whose details can be filled in.

Theorem 1 below is something I noticed in 2016, but whose proof I forgot to write down at the time. Having just spent a half-hour trying to (re)construct a proof, it seems worth quickly writing down an argument here so that I can find it more easily. Theorems 2 and 3 are then natural things to point out, to indicate the context for Theorem 1; in both cases I’ve tried to piece together proofs from various bits of the literature.


Let T denote the space of trace-class operators on a separable infinite-dimensional Hilbert space H. Let V = L1([0,1], T) be the space of Bochner-integrable T-valued functions on [0,1]; alternatively we could define V to be the projective tensor product of T with L1.

Theorem 1. T and V are not isomorphic as Banach spaces.

Theorem 2. The dual spaces T* and V* are not isometrically isomorphic as Banach spaces.

Theorem 3. The dual spaces T* and V* are isomorphic as Banach spaces.

Proof of Theorem 1

It is known that the Banach space T has the Radon-Nikodym Property (RNP). I will not define the RNP here, but all we need to know is that it passes to closed subspaces, and that L1 does not have the RNP. Since V contains a (complemented) closed subspace isomorphic to L1, it follows that V does not have the RNP.  \Box

Question: Is there a simpler proof of Theorem 1? Invoking the RNP feels like overkill.

Proof of Theorem 2

Observe that T*=B(H) and V*= L([0,1], B(H)); we denote this second von Neumann algebra by N for sake of brevity. Suppose that B(H) is isometrically isomorphic (as a Banach space) to N. By a theorem of Kadison

R. V. Kadison, Isometries of operator algebras. Annals of Math. 54 (1951), no. 2, 325&ndas;338

this would imply that there is a Jordan *-isomorphism φ from B(H) onto N. Because B(H) is a factor, Corollary 11 of Kadison’s paper implies that φ must either be a *-isomorphism or a *-anti-isomorphism. But this is impossible since N has non-trivial centre, while B(H) has trivial centre.  \Box

Question: Can we obtain a more direct proof by investigating Kadison’s arguments and specializing them to the case of B(H)?

Proof of Theorem 3

This can be deduced from a more general result of Robertson and Wassermann:

A. G. Robertson, S. Wassermann, Completely bounded isomorphisms of injective operator systems. Bull. London Math. Soc. 21 (1989), 285–290.

However, it seems better to sketch a simpler argument for this particular case, which admittedly uses some of the same ideas, specifically, some form of Pelczynski’s decomposition method.

Observe that T*=B(H) and V*= L([0,1], B(H)). It is easy to construct an isomorphism of Banach spaces between L[0,1] and C ⊕ L[0,1]; a minor variant of this gives an isomorphism of Banach spaces between V* and T*V*.

Similarly, there is an obvious isomorphism of Banach spaces between L[0,1] and L[0,1]⊕L[0,1], and a minor variation of this gives an isomorphism of Banach spaces between V* and V*V*.

The final ingredient in this proof is the observation that T* is isomorphic as a Banach space to FB*. To justify this, note that there is a projection from B(L2 ⊗ H) onto L([0,1],B(H)) = V*, and that the former space is isomoprhic to B(H) since L2⊗ H is isomorphic to H.

Putting things together:

T*FV*F ⊕ (V*V*) ≅ (FV*) ⊕ V*T*V*V*

as required.  \Box

on the surface again, with a job to do

1 April, 2018
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Let them have what was under the water. What lived in Venice was still afloat.

—from Venice Drowned by Kim Stanley Robinson —

no one told you when to run

9 March, 2018

I’m going to watch the bluebirds fly over my shoulder

I’m going to watch them pass me by, maybe when I’m older

every year is getting shorter, never seem to find the time;

plans that either come to naught, or half a page of scribbled lines

I slept through the night, I got through to the dawn

2 January, 2018
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update 2018-06-10: replaced first link with better version


Though much is taken, much abides; and though

We are not now that strength which in old days

Moved earth and heaven, that which we are, we are;

One equal temper of heroic hearts,

Made weak by time, but strong enough in will

To strive, to seek, to find, and not to yield.

— from Ulysses by Alfred, Lord Tennyson

I wrote down my dream, I made it this song

It was a pretty good day so far

Hello, 2018.

behind brown and mild eyes

31 December, 2017
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I guess it comes down to a simple choice, really.

Get busy living, or get busy dying.