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The trigonometric $C^*$-algebra is the universal $C^*$-algebra generated by $\mathcal{G}=\{x,y,z\}$ subject to relations \begin{align}x^2=x=x^*, &\quad y^2=y=y^*\\ [x, z]=y, &\quad [y,z]=-x \\ z^*=-z,& \quad|z|\leq 1\end{align}

Does this universal algebra exist? Does it coincide with a well known $C^*$-algebra? What is its precise structure? How can one compute its $K$-theory?

The motivation and reasons we call it trigonometric $C^*$-algebra: The classical trigonometric coalgebra is inspired by sin and cos. On the other hand if we put $x=\sin t, y=\cos t$, then they satisfy $x'=y, y'=-x$. But how can we represent $d/dt$ as a relation? Instead of differentiation $d/dt$ we use the commutator operation with further assumption that $x,y$ are projections. I learned this from the book by Alain Connes Non commutative Geometry page 20 (The e-version) that if $E$ is a curve in the space of projections of an algebra then $E'=[E,Z]$ where $Z$ is an anti self adjoint element, $E^*=-E$. So the differentiation is considered as commutator provided we work on the space of projections. We add $|Z|\leq 1$ to guaranty the condition of boundedness of generators in every arbitrary representation of generators $\mathcal{G}$

Is the boundedness of $z$ a redundant condition?

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    $\begingroup$ Existence follows immediately from the representable functor theorem. The same theorem also implies that the boundedness of z is necessary since otherwise the relevant functor does not preserve limits and therefore is not representable. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 26, 2023 at 2:48
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    $\begingroup$ You can build the algebra as a quotient of the free product $\mathbb{C}^2 \ast \mathbb{C}^2 \ast C([-1,1])$, where $x$ and $y$ are $(1,0)$ interpreted in the first and second algebras and $z$ is the function $\alpha \mapsto i\alpha : [-1,1] \rightarrow \mathbb{C}$ in the third algebra. You take a quotient to impose your relations $[x,z] = y$ and $[y,z] = -x$. $\endgroup$
    – Robert Furber
    Commented Jan 26, 2023 at 4:09

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