Endomorphisms of the Cuntz algebra Consider the Cuntz algebra $\mathcal{O}_n$ with $n \geq 2$ and let $\text{End}(\mathcal{O}_n)$ be the set of all (unital) $\ast$-endomorphisms of $\mathcal{O}_n$. I was wondering if there exists an element $x \in \mathcal{O}_n$ such that the evaluation map $\text{End}(\mathcal{O}_n) \rightarrow \mathcal{O_n},$ $\phi \mapsto \phi (x)$ is injective.
If no, what is the smallest $k \in \mathbb{N}$ for which $x\in \mathcal{O}_n \otimes \mathbb{C}^k$ exists such that the map $\text{End}(\mathcal{O}_n) \rightarrow \mathcal{O}_n \otimes \mathbb{C}^k$ given by $\phi \mapsto (\phi \otimes \mathrm{id}) (x)$ is injective? Is it $k=n-1$?
 A: Although not exactly what the OP has in mind,
there is another interesting characterization of endomorphisms of $\mathcal O_n$ in terms of  single elements.  Namely
there is a one-to-one correspondence between endomorphisms of $\mathcal O_n$ and unitary elements of $\mathcal O_n$ given
as follows:

*

*If $u$ is a unitary element, one defines an endomorphisms $\varphi _u$ by sending each generator $S_i$ to $uS_i$.


*Conversely, given an endomorphism $\varphi $, one defines the unitary element
$$
  u_\varphi  = \sum_{i=1}^n\varphi (S_i)S_i^*.
  $$
It is in fact very easy to show that these correspondences are each other's inverse.
A: This is true: $\mathcal O_n$ is singly generated, i.e. there exists $x\in \mathcal O_n$ such that $C^\ast(x) = \mathcal O_n$. In particular, if $\phi, \psi \colon \mathcal O_n \to B$ are $\ast$-homomorphisms such that $\phi(x) = \psi(x)$, then $\phi = \psi$.
There might be a very direct way of showing this, but here is a proof for $n\geq 2$ ($n\neq \infty$): We will use Theorem 1 from the paper [C. Olsen and W. Zame, Some C*-algebras with a single generator, Trans. Amer. Math. Soc. 215 (1976), 205–217], which states that if $A$ is a unital $C^\ast$-algebra generated by $k(k+1)/2$ elements of which $k(k-1)/2$ are self-adjoint, then $M_k(A)$ is singly generated.
Let $s_1,\dots, s_n \in \mathcal O_n$ be the canonical generators. Then $M_n(\mathcal O_n) \to \mathcal O_n$ given by $(a_{i,j})_{i,j=1}^n \mapsto \sum_{i,j=1}^n s_i a_{i,j} s_j^\ast$ is a $\ast$-isomorphism. In particular, $M_{n^2}(\mathcal O_n)\cong \mathcal O_n$, so it suffices to show that $M_{n^2}(\mathcal O_n)$ is singly generated.
Note that $\mathcal O_n$ is generated by $2n$ self-adjoint elements, namely $s_j + s_j^\ast$ and $i(s_j - s_j^\ast)$ for $j=1,\dots, n$. Taking $A= \mathcal O_n$ and $k=n^2$ in the above theorem (using that $n^2(n^2-1)/2 \geq 2n$ for $n\geq 2$), it follows that $M_{n^2}(\mathcal O_n) \cong \mathcal O_n$ is singly generated.
Note that the theorem above is constructive, so if one wants, one can write down an explicit single geneator of $M_{n^2}(\mathcal O_n)$, and use the isomorphism $M_n(M_n(\mathcal O_n)) \cong M_n(\mathcal O_n) \cong \mathcal O_n$ I described above to express this element in $\mathcal O_n$.
It is also true for $\mathcal O_\infty$, but I lack an elementary proof. You can use much deeper machinery, namely that $\mathcal O_\infty$ is $\mathcal Z$-stable (i.e. $\mathcal O_\infty \otimes \mathcal Z \cong \mathcal O_\infty$ where $\mathcal Z$ is the Jiang-Su algebra). Then it follows from the main result of [Thiel, Hannes; Winter, Wilhelm The generator problem for Z-stable C∗-algebras. Trans. Amer. Math. Soc. 366 (2014), no. 5, 2327–2343], which states that any separable, unital $\mathcal Z$-stable $C^\ast$-algebra is singly generated.
