Is there an embedding $v(\mathcal{O}_\infty\otimes \mathcal{K})v^*\to \mathcal{O}_\infty$?

Question

Following situation: If $\mathcal{O}_\infty$ is the Cuntz algebra in infinitely many generators and $\mathcal{K}$ the compact operators on a separable Hilbert space, let $v\in \mathcal{O}_\infty\otimes \mathcal{K}$ be a partial isometry. Questions:

1. Just to make shure: $v(\mathcal{O}_\infty\otimes \mathcal{K})v^*$ is endowed with the multiplication $(vxv^*)(vyv^*)=vxv^*vyv^*$ and the involution $(vxv^*)^*=vx^*v$, where $x,y\in \mathcal{O}_\infty\otimes \mathcal{K}$, a $C^*$-algebra, is it? Or is $v(\mathcal{O}_\infty\otimes \mathcal{K})v^*$ not necessarily closed ?

2. (in this case that $v(\mathcal{O}_\infty\otimes \mathcal{K})v^*$ is a $C^*$-algebra ).. Is there an embedding $v(\mathcal{O}_\infty\otimes \mathcal{K})v^*\to \mathcal{O}_\infty$?

(edit: the following is false, as pointed out in Gabors answer) There is at least a $*$-isomorphism $v(\mathcal{O}_\infty\otimes \mathcal{K})v^*\to \mathcal{O}_\infty\otimes \mathcal{K}$ given by $vnv^*\mapsto n$ where $n \in \mathcal{O}_\infty\otimes \mathcal{K}$. H

owever, I don't know enough about $\mathcal{O}_\infty$ to answer 2 (so, 2. remains my main question).

Thank you.

Answer 1

The answers to both questions is Yes.

For the first one, recall that a partial isometry satisfies the equation $v=vv^*v$ with $p=vv^*$ being a projection (with same range as $v$ if viewed as an operator on a Hilbert space), which easily prompts $$v(\mathcal O_\infty\otimes\mathcal K)v^* = p(\mathcal O_\infty\otimes\mathcal K)p = \{ x\in\mathcal O_\infty\otimes\mathcal K \mid x=pxp \}.$$ A C*-subalgebra of this form is called a corner. If you want to show that it is closed, then assume $x=\lim_{n\to\infty} x_n$ with $x_n=px_np$ for some $x, x_n\in \mathcal O_\infty\otimes\mathcal K$, and you see that then $x=pxp$ must hold by continuity of multiplication.

Note that this first part uses nothing about $\mathcal O_\infty\otimes\mathcal K$ being the C*-algebra in this context, unlike the second part.

For the second part, one needs the fact that $\mathcal O_\infty\otimes\mathcal K$ is a simple and purely infinite C*-algebra. See the papers titled "Simple C*-algebras generated by isometries" and/or "K-theory for certain simple C*-algebras" by Joachim Cuntz if you want to know more. Anyway, from basic results on simple and purely infinite C*-algebras, it is possible to find another partial isometry $w\in\mathcal O_\infty\otimes\mathcal K$ with $w^*w=p$ and $ww^* \in \mathcal O_\infty\otimes e_{11} \cong \mathcal O_\infty$. Then you can check that the assignment $x\mapsto wxw^*$ yields a well-defined embedding from $p(\mathcal O_\infty\otimes\mathcal K)p$ into $\mathcal O_\infty\otimes e_{11} \cong \mathcal O_\infty$.

There is at least a $*$-isomorphism $v(\mathcal O_\infty\otimes\mathcal K)v^* \to \mathcal O_\infty\otimes\mathcal K$ given by $vnv^*\mapsto n$...

Note also that this is false! This map is not even well-defined as a linear map, as you can see when you insert any projection that is orthogonal to $p$.