Partial Isometries Satisfying Cuntz-like Relations

Question

I have a situation where I have a family of partial isometries, $S_i$, for $i=0,...,N-1$, on a Hilbert space $\mathcal{H}$ such that the adjoint maps, $S_i^*$ for $i=0,..,N-1$ are also partial isometries on $\mathcal{H}$ which satisfy the following further relations:

(1) $\sum_{i=0}^{N-1} S_iS_i^* = \text{id}_{\mathcal{H}}$

(2) $\sum_{i=0}^{N-1} S_i^*S_i = \text{id}_{\mathcal{H}}$

(3)$S_i^*S_j = S_iS_j^* = 0$ for $i \neq j$

This looks a lot like the relations for the Cuntz algebra except the $S_i$'s are partial isometries, rather than full isometries. It is also different than the Cuntz-Krieger algebra, which is the standard generalization of the Cuntz algebra to partial isometries.

I have a couple questions. Is there a way to represent this scenario as a graph $C^*$ algebra? It doesn't seem to fall under the Cuntz-Krieger graph $C^*$ algebra definition. And more generally, is there any known theory developed for this scenario? Thanks for your help!

Answer 1

I don't know if there is an interpretation in terms of graphs, but the following observation may be helpful: the $C^*$ algebra generated by these operators is the same as the $C^*$ algebra generated by a single unitary operator $U$ and a family of $n$ orthogonal projections $P_1, \dots P_n$ with $\sum P_j = I$. The $U, P_j$ are related to the $S_i$ via the identities \begin{equation} U=\sum_i S_i, \quad P_j = S_jS_j^*. \end{equation} Your conditions (1), (2), (3) together imply that $U$ is unitary, and $P_j$ is a projection since the $S_j$ are partial isometries. It is immediate that $C^*(S_1, \dots S_n)$ contains $C^*(U, P_1, \dots P_n)$; for the converse note that we have (again using the defining relations and the relation $S_i=S_iS_i^*S_i$ for partial isometries) \begin{equation} S_i = S_iS_i^*(\sum_j S_j) = P_i U. \end{equation}

I don't know much in the way of a general theory for the algebras $C^*(U, P_1,\dots P_n)$ except that I vaguely recall some results to the effect that their representation theory is very wild; if I have time later I can try to look them up and come back to edit the answer.

EDIT: I haven't been able to track down the reference I was thinking of, but the above observation can be pushed a little further. Consider the case of two generators, then we have $C^*(U, P,Q)$ with $P+Q=I$, but this is the same as $C^*(U,P)$. So we are considering the case of the universal $C^*$-algebra generated by a unitary and a projection. We can replace the projection as a generator by $V=2P-I$; this $V$ is a unitary satisftying $V^2=I$. It follows that $C^*(U,V)$ is the full group $C^*$-algebra of the free product group $\mathbb Z*\mathbb Z_2$. This group is non-amenable, so the $C^*$-algebra is not nuclear, and also non-simple (this is easy to see anyway, since we have both commutative and noncommutative representations). In the case of more generators I don't think we get a group $C^*$-algebra any more, but it shouldn't be hard to prove that they are still non-nuclear. Thus despite appearances these algebras are in many ways very different from the Cuntz or Cuntz-Krieger algebras.