Let us take the right shift operator $S$ acting on the Hilbert space $l^2(\mathbb{N})$. Consider the C*-algebra generated the operator

$ \begin{pmatrix} S & 0 \\ 0 & S^* \end{pmatrix} $ acting on $l^2(\mathbb{N})\oplus l^2(\mathbb{N}).$

This is, I guess, called C*-algebra of free monogenic inverse semigroup.

My question is, is this C*-algebra a universal C*-algebra for some generators and relations?


Letting $T=\pmatrix{S & 0 \cr 0 & S^*}$, you can show that the map $$ \pi:n\in {\mathbb Z} \mapsto \left\{\matrix{T^n, & \hbox{if } n\geq 0, \cr (T^*)^{-n},& \hbox{otherwise.}}\right. $$ is a partial representation of the group ${\mathbb Z}$, so your algebra is a quotient of the partial group C*-algebra $C^*_{par}(\mathbb Z)$. Moreover your algebra may be described as the crossed product of its diagonal subalgebra by a partial action of the group $\mathbb Z$, and in this sense it is a universal C*-algebra.

Elaborating a bit more on my answer, the crucial point is to analyze the spectrum of the subalgebra generated by the set $$ \big \{π(n)π(n)^*: n∈ {\mathbb Z}\big \}, $$ which turns out to be homeomorphic to the two-point compactification of ${\mathbb Z}$, namely ${\mathbb Z}∪\{∞, -∞\}$.

In the case of the monogenic inverse semi-group, whose C*-algebra is the same as I called $C^*_{par}({\mathbb Z})$, and is much bigger than the one you described, the spectrum of that abelian subalgebra turns out to be the Cantor set $\{0, 1\}^{\mathbb Z}$.

| cite | improve this answer | |
  • $\begingroup$ would you please explain how it can be viewed as partial crossed product and is it full crossed product so that it has universal property? $\endgroup$ – SiOn Oct 30 '17 at 23:19
  • $\begingroup$ The partial crossed product fact follows from (4.21) in [Circle actions on C*-algebras, partial automorphisms and a generalized Pimsner-Voiculescu exact sequence, J. Funct. Analysis, 122 (1994), 361-401]. You might also want to look at [Partial Dynamical Systems, Fell Bundles and Applications, Mathematical Surveys and Monographs, American Mathematical Society, volume 224, 321 pp., 2017] for a more modern treatment of partial actions. $\endgroup$ – Ruy Oct 31 '17 at 11:50

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.