Example: traceless C*-algebra universally generated by projections

Question

Are there examples of

1. a non-zero C*-algebra which is
2. universally generated by
3. finitely many projections (not all commuting) together with a unit and plus
4. necessarily satisfying some additional relations such that
5. there remain no traces?

In other words, is there a non-zero quotient $$C^*(Z/2*\ldots*Z/2)\to B\to 0$$ which carries no traces?
So basically the question is about the representation theory of $C^*(Z/2*\ldots*Z/2)$:
For instance, can there arise purely infinite projections?

A particular instance I'm curious about is the following:
I remember vaguely that (under certain circumstances) one can surprisingly build partial isometries out of projections solely, and for example, as Yemon Choi mentioned, even obtain the Cuntz algebra $\mathcal{O}_2$. So a particular question I have in mind is:

How can this happen resp. does someone remember a reference for this?

Basically I'm curious about how interesting such quotients can generally look like.
And also more generally, what kind of classes of groups are there with some traceless quotients and what kind of interesting phenomena can happen there?

(Unfortunately, a more narrowed down question might be stepping on someone's toes, and I really wouldn't want to cause some troubles for my colleagues here. So my big apologies that I can't be more specific here. I hope that is understandable.)

Answer 1

I'm not sure if the following is exactly what the OP was looking for, but it definitely solves the question. The following lemma implies that $\mathcal O_2$ is a quotient of $C^*(\underbrace{(\mathbb Z/2\mathbb Z) * \dots * (\mathbb Z/2\mathbb Z)}_{4\textrm{ times (edited)}} )$.

Lemma (edited): Let $A$ be a unital $C^\ast$-algebra which is generated as a unital $C^\ast$-algebra by $n$ arbitrary elements. Then $M_2(A)$ is generated by $n+2$ self-adjoint unitaries.

Proof (edited): Let $x_1,\dots,x_n$ be arbitrary generators. We may assume $\|x_i\| \leq 1$ for all $i$. Then the dilation $u_i = \begin{pmatrix} \sqrt{1-x_i^*x_i} & x_i^*\\ x_i & -\sqrt{1-x_ix_i^*} \end{pmatrix}$ is a self-adjoint unitary. Let $u_{n+1} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $u_{n+2} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ which are self-adjoint unitaries generating $M_2(\mathbb C) \subseteq M_2(A)$. Let $B$ be the unital $C^\ast$-subalgebra of $M_2(A)$ generated by $u_1,\dots, u_{n+2}$. Then $B$ contains $M_2(\mathbb C)$ and thus also \begin{equation} \begin{pmatrix} 0 & 0 \\ x_i & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} u_i \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \end{equation} for $i=1,\dots, n$. As $A= C^\ast(x_1, \dots, x_n, 1)$ it follows that $B$ contains both $A\oplus 0 \subseteq M_2(A)$ and $M_2(\mathbb C) \subseteq M_2(A)$, and hence $B= M_2(A)$. QED.

If $s_1,s_2\in \mathcal O_2$ are the canonical generators of $\mathcal O_2$. There is an isomorphism $M_2(\mathcal O_2) \to \mathcal O_2$ given by $(z_{i,j})_{i,j=1}^2 \mapsto \sum_{i,j=1}^2 s_i z_{i,j} s_j^\ast$. Thus by the above, $M_2(\mathcal O_2) \cong \mathcal O_2$ is generated by $4$ (edited) self-adjoint unitaries, so $\mathcal O_2$ is a quotient of $C^*(\underbrace{(\mathbb Z/2\mathbb Z) * \dots * (\mathbb Z/2\mathbb Z)}_{4\textrm{ times (edited)}} )$.

Finally, I don't know the smallest number of self-adjoint unitaries generating $\mathcal O_2$. Note also that $\mathcal O_2$ is actually singly generated and so even generated by $2$ self-adjoint elements instead of 2 isometries (see my answer for Endomorphisms of the Cuntz algebra). So while the above got the number down to 4 (edited), I strongly believe it can be reduced further.