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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?

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

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