TITLE: Endomorphisms of the Cuntz algebra
QUESTION [5 upvotes]: 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 $k=n-1$?

REPLY [5 votes]: 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.