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Consider the free product of $\mathbb{Z}/2$ with itself with generators $$ \mathbb{Z}/2*\mathbb{Z}/2=\langle u,v\mid u^2=1=v^2\rangle $$ and regard its group $C^*$-algebra $$ C^*(\mathbb{Z}/2*\mathbb{Z}/2)=C^*_r(\mathbb{Z}/2*\mathbb{Z}/2)\subseteq B\Bigl(\ell^2(\mathbb{Z}/2*\mathbb{Z}/2)\Bigr). $$

For $a,b\in\mathbb{C}$, does someone know the norm of $$ \lVert au+bv+1\rVert={?} $$ (As another example I found that $\|au+bv\|=|a|+|b|$.)

In principle one should be able to compute the operator norm of this particular class of elements since they're given as concrete infinite dimensional matrices, but I couldn't figure out how.

Does someone have some idea perhaps?

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    $\begingroup$ I don't have time to write a full answer now, but the explicit isomorphism given in Theorem 1.3 of the C*-algebra generated by two projections should allow you to calculate it very explicitly. This isomorphism identifies $C^*(\mathbb{Z}/2 \ast \mathbb{Z}/2)$ with the C*-subalgebra of $C([0,1],M_2(\mathbb{C}))$ given by those functions that are diagonal at the endpoints. $\endgroup$
    – Tobias Fritz
    Commented May 23, 2022 at 19:41
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    $\begingroup$ Perfect, thanks a lot, that was precisely what we were looking for, thanks a lot! $\endgroup$
    – C-star-W-star
    Commented May 24, 2022 at 0:02
  • $\begingroup$ Where you wrote $\mathbb{Z}/2*\mathbb{Z}/2,$ might you have meant $\mathbb{Z}/2\times\mathbb{Z}/2 \text{?} \qquad$ $\endgroup$
    – Michael Hardy
    Commented May 27, 2022 at 23:49
  • $\begingroup$ @MichaelHardy: I really mean the free product $\mathbb{Z}/2 * \mathbb{Z}/2$, so with non commuting generators. $\endgroup$
    – C-star-W-star
    Commented May 30, 2022 at 15:37
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    $\begingroup$ This group is also called the infinite dihedral group and it is amenable, so reduced is not needed here. Moreover, the group was recently studied by Grigorchuk and Yang: mathnet.ru/php/… $\endgroup$
    – Boris Bilich
    Commented Jun 9, 2022 at 7:38

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