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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.

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  • $\begingroup$ [deleted comment caused by not reading properly; sorry] $\endgroup$ – Yemon Choi Jul 1 '15 at 16:15
  • $\begingroup$ @Yemon: Hi Yemon. You’re alright. I didn’t see your deleted comment anyway. :) $\endgroup$ – Transcendental Jul 1 '15 at 18:07
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    $\begingroup$ 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). $\endgroup$ – Jan Jitse Venselaar Jul 2 '15 at 8:51
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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 $|q|<1$.

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  • $\begingroup$ 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? $\endgroup$ – Transcendental Jul 12 '15 at 22:42
  • $\begingroup$ $\alpha$ has kernel, so the last equality doesn't hold. $\endgroup$ – Makoto Yamashita Jul 13 '15 at 0:07
  • $\begingroup$ 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}) $. $\endgroup$ – Transcendental Jul 13 '15 at 23:43

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