C*-algebra of free monogenic inverse semigroup

Question

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?

Answer 1

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.

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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 $\{0, 1\}^{\mathbb Z}$.