Consider the measures on the circle, $M(\mathbb T)$, endowed with the convolution product which makes it a unital Banach algebra under the total variation norm. Denote by $\Delta$ the maximal ideal space of $M(\mathbb T)$. This space is quite large–recently it was shown that $\Delta$ is non-separable, for example. It also has many copies of $\beta \mathbb N$, yet it is not extremely disconnected, however, in certain sense, it is not too far from being so.

Extremely disconnected spaces do not have injective convergent sequences, again $\beta \mathbb N$ serves a paradigm example.

Does $\Delta$ have any injective convergent sequences?

I have some evidence that it should not be the case but I am unable to make it into a proof. Maybe this follows from some known facts anyway?

  • 2
    $\begingroup$ I think this Ph.D. thesis of J. Taylor might be of use. $\endgroup$
    – Norbert
    Jun 30 '18 at 20:40

Norbert is right (see the paper The Structure of Convolution Measure Algebras). Actually, I had known the result but it didn't occur to me that it would have solved the problem.

The maximal ideal space of $M(\mathbb T)$ contains an analytic disk, which in particular, is a homeomorphic copy of $\mathbb D$, so there are plenty of convergent sequences therein.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.