# A question on an argument in Woronowicz’s paper on the compact quantum group ${\text{SU}_{q}}(2)$

Let $q \in [0,1)$. The compact quantum group ${\text{SU}_{q}}(2)$ is defined to be the universal unital $C^{*}$-algebra that is generated by two elements $\alpha$ and $\beta$ satisfying the following five relations: \begin{align} \alpha^{*} \alpha + \beta^{*} \beta & = 1, \\ \alpha \alpha^{*} + q^{2} \beta \beta^{*} & = 1, \\ \beta^{*} \beta & = \beta \beta^{*}, \\ \alpha \beta & = q \beta \alpha, \\ \alpha \beta^{*} & = q \beta^{*} \alpha. \end{align} In his paper Twisted $\text{SU}(2)$ Group. An Example of a Non-Commutative Differential Calculus, Woronowicz proves that there is a $*$-isomorphism between ${\text{SU}_{q}}(2)$ and ${\text{SU}_{0}}(2)$. Part of his proof proceeds as follows. He begins with the claim that \begin{align} \sigma(\alpha^{*} \alpha) & \subseteq \left\{ 0,1 - q^{2},1 - q^{4},1 - q^{6},\ldots,1 \right\} ~ \text{and} \\ \sigma(\beta^{*} \beta) & \subseteq \left\{ 0,\ldots,q^{6},q^{4},q^{2},1 \right\}. \end{align} Next, he chooses arbitrary continuous functions $f,g: [0,1] \to \Bbb{R}$ such that:

• $f(0) = 0$ and $f(t) = 1$ for all $t \in \left[ 1 - q^{2},1 \right]$.
• $g(1) = 1$ and $g(t) = 0$ for all $t \in \left[ 0,q^{2} \right]$.

Letting $a = \alpha ~ f(\alpha^{*} \alpha)$ and $b = \beta ~ g(\beta^{*} \beta)$, he then says that $(a,b)$ is a generating pair for ${\text{SU}_{0}}(2)$, i.e., \begin{align} a^{*} a + b^{*} b & = 1, \\ a a^{*} & = 1, \\ b^{*} b & = b b^{*}, \\ a b & = 0, \\ a b^{*} & = 0. \end{align}

The problem is, none of this seems to work. For example, let us try to verify that $a a^{*} = 1$. Observe that \begin{align} a a^{*} & = [\alpha ~ f(\alpha^{*} \alpha)] [\alpha ~ f(\alpha^{*} \alpha)]^{*} \\ & = [\alpha ~ f(\alpha^{*} \alpha)] [f(\alpha^{*} \alpha)^{*} ~ \alpha^{*}] \\ & = [\alpha ~ f(\alpha^{*} \alpha)] \left[ \overline{f}(\alpha^{*} \alpha) ~ \alpha^{*} \right] \qquad (\text{By the continuous functional calculus.}) \\ & = [\alpha ~ f(\alpha^{*} \alpha)] [f(\alpha^{*} \alpha) ~ \alpha^{*}] \qquad \left( \text{As $\overline{f} = f$.} \right) \\ & = [f(\alpha \alpha^{*}) ~ \alpha] [\alpha^{*} ~ f(\alpha \alpha^{*})] \qquad (\text{As $\alpha ~ p(\alpha^{*} \alpha) = p(\alpha \alpha^{*}) ~ \alpha$ for every polynomial $p$.}) \\ & = [f(\alpha \alpha^{*})] (\alpha \alpha^{*}) [f(\alpha \alpha^{*})] \\ & = h(\alpha \alpha^{*}), \end{align} where $h: [0,1] \to \Bbb{R}$ is defined by $h(t) \stackrel{\text{df}}{=} t [f(t)]^{2}$ for all $t \in [0,1]$. However, $$\sigma(\alpha \alpha^{*}) \cup \{ 0 \} = \sigma(\alpha^{*} \alpha) \cup \{ 0 \},$$ so $h|_{\sigma(\alpha \alpha^{*})}$ is the identity function on $\sigma(\alpha \alpha^{*})$, which yields $h(\alpha \alpha^{*}) = \alpha \alpha^{*}$ instead of $h(\alpha \alpha^{*}) = 1$.

Question. For those who are familiar with this topic, am I hallucinating, or is something not right here?

Note: Woronowicz’s argument is given on Page 179, in Section A2, of his paper.

• [deleted comment caused by not reading properly; sorry] – Yemon Choi Jul 1 '15 at 16:15
• @Yemon: Hi Yemon. You’re alright. I didn’t see your deleted comment anyway. :) – Transcendental Jul 1 '15 at 18:07
• Maybe there is a typo in the definition of the functions $f$ and $g$? (Not that I'm able to find a correct definition right now...). I did check that the inverses ($\tilde{T}_H^1$ and $\tilde{T}_H^2$) are maps from $SU_0(2)$ to $SU_q(2)$ however (modulo some concerns I have about some of the manipulations with the infinite sums, at one point I need that $\sum_{n=1}^\infty ({\alpha^*}^{n-1} \alpha^{n-1} - {\alpha^*}^n \alpha^n) = 1$, which is a bit shady. This can probably be fixed if you do the calculation cleverer). – Jan Jitse Venselaar Jul 2 '15 at 8:51

As Jan suggests, there's a typo in the definition of $f$. It should read $$f(t) = \begin{cases}\frac{1}{\sqrt{t}} & (t \ge 1 - q^2)\\ 0 & (t = 0).\end{cases}$$ Since $\alpha \alpha^* \ge 1-q^2$ (this follows from usual realization as in p. 123), this will imply that $a^*$ is an isometry. And $\alpha^* \alpha + \beta^* \beta = 1$ implies that the eigenspace of $\beta^* \beta$ for the eigenvalue $1$ is precisely the kernel of $\alpha$, which implies $a^* a + b^* b = 1$.
Morally, at $q=0$ you have the Toeplitz algebra as the 'quantum limit' of the deformation quantization of the $2$-dimensional symplectic leaves $$\left\{\begin{pmatrix}\bar{z} \lambda & \sqrt{1-|z|^2}\lambda\\-\sqrt{1-|z|^2}\bar{\lambda}&z\bar{\lambda}\end{pmatrix} \mid |z| < 1 \right\} \quad (|\lambda|=1)$$ inside $SU(2)$ together with boundary $U(1)$, and $C(SU_q(2))$ is an elaborate patching of these. The $q$-disc algebra generated by an operator $Z_q$ satisfying $\|Z_q\|=1$ and $1-Z_q^* Z_q = q^2 (1-Z_qZ_q^*)$ interpolates the Toeplitz algebra and the usual function algebra on the closed disc, but these algebras are in fact isomorphic to each other for $|q|<1$.
• Hi Makoto. Thanks for your response! I understand that there’s surely a typo error in the definition of $f$, but if we use $f$ as defined in your response, then we have \begin{align} a^{\ast} a & = [\alpha ~ f(\alpha^{\ast} \alpha)]^{\ast} [\alpha ~ f(\alpha^{\ast} \alpha)] \\ & = [f(\alpha^{\ast} \alpha) ~ \alpha^{\ast}] [\alpha ~ f(\alpha^{\ast} \alpha)] \\ & = [f(\alpha^{\ast} \alpha)] (\alpha^{\ast} \alpha) [f(\alpha^{\ast} \alpha)] \\ & = 1, \end{align} which means that $a$ is unitary. This forces $b^{*} b$, hence $b$, to be $0$. Is this contradictory? – Transcendental Jul 12 '15 at 22:42
• $\alpha$ has kernel, so the last equality doesn't hold. – Makoto Yamashita Jul 13 '15 at 0:07
• Thank you, Makoto. Yes, the last equality doesn’t hold because $0 \in \sigma(\alpha^{\ast} \alpha)$ and $0 [f(0)]^{2} = 0 \neq 1$. This doesn’t happen to $\alpha \alpha^{\ast}$ because $\alpha \alpha^{\ast}$ is invertible by the second relation, so $0 \notin \sigma(\alpha \alpha^{\ast})$, which implies that $t [f(t)]^{2} = 1$ for all $t \in \sigma(\alpha \alpha^{\ast})$. – Transcendental Jul 13 '15 at 23:43