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$\newcommand{\id}{\mathrm{id}}$Let $H$ be a Hilbert space, and $X \in B(H)$ a proper isometry (i.e. $X^{\star}X = \id$ and $XX^{\star} \neq \id$). Coburn's theorem states that ${\rm C}^{\star}(X)$, the ${\rm C}^{\star}$-algebra generated by $X$, is the Toeplitz algebra.

We are interested in an extension of Coburn's theorem. Consider $Y \in B(H)$ such that $Y^{\star}Y = \id+p$, with $p \in B(H)$ a nonzero projection (so $YY^{\star} \neq \id$, see comments).

Question: Is ${\rm C}^{\star}(Y)$ always the Toeplitz algebra? If not, is it still a ${\rm C}^{\star}$-algebra of type ${\rm I}$?

Motivation: I'm mainly interested in a shift on a trivalent directed tree with one parent and one or two children for each vertex, such that the associated operator is non-normal; it is a generalization of the operator $S^{\star}$ of this post. These questions were inspired by the Collatz graph, in particular, the connected component of an eventual counter-example of Collatz Conjecture going to infinity by the Collatz map iteration (such a connected component would be a trivalent directed tree). The Collatz map induces a shift operator which generates an operator algebra (${\rm C}^{\star}$ or ${\rm W}^{\star}$). I wonder whether this operator algebra is interesting in its own right or for an application to number theory.

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  • $\begingroup$ Those relations have finite-dimensional representations, so the answer to the first question is "no". $\endgroup$
    – user85913
    Commented Dec 22, 2017 at 8:53
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    $\begingroup$ Note that there is nothing stopping $YY^\ast=1+p=Y^\ast Y$ (as long as $p\neq 0$), so you could for example let $Y$ be the positive square root of $1+p$. For an extreme example, take $H=\mathbb C$, $p=1$ and $y=\sqrt{2}$. $\endgroup$
    – user85913
    Commented Dec 22, 2017 at 13:04
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    $\begingroup$ Note that $YY^\ast\neq 1$ actually follows from $Y^\ast Y= 1+p$ (as long as $p\neq 0$), because if $YY^\ast$ is a projection then so is $Y^\ast Y$. Secondly: the operator $Y$ has polar decomposition $Y=T|Y|$ where $T$ is an isometry (and $|Y|=\sqrt{1+p}$). Then $C^*(Y)=C^*(T,p)$. Conversely, if $T$ is any isometry and $p$ is any projection, then $Y=T\sqrt{1+p}$ satisfies $Y^*Y=1+p$. So you're thinking about the universal $C^*$-algebra generated by an isometry and a projection, i.e. the free product $\mathcal T\ast \mathbb C$ where $\mathcal T$ is the Toeplitz algebra. $\endgroup$
    – user85913
    Commented Dec 22, 2017 at 13:56
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    $\begingroup$ I think such algebras are discussed in arxiv.org/pdf/1111.4392.pdf. They have a different $K$-theory than the Toeplitz algebra. $\endgroup$
    – hänsel
    Commented Dec 22, 2017 at 17:00
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    $\begingroup$ $YY^*\neq1$ also follows from $Y^*Y=1+p$ (as long as $p\neq0$) by the C*-identity $\Vert YY^*\Vert = \Vert Y^*Y\Vert = \Vert Y\Vert^2$ $\endgroup$
    – Ruy
    Commented Dec 24, 2017 at 13:53

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