# Invertible elements of the Hopf algebra quantum $SU(2)$

Let $$SU_q(2)$$ be the (polynomial) Hopf algebra introduced by Woronocicz called the quantum special unitary group. For details see

https://en.wikipedia.org/wiki/Compact_quantum_group

(Note that on the Wikipedia page it is the $$C^*$$-algebra that is discussed, but this question is about the dense Hopf algebra of the $$C^*$$-algebra.)

Does $$SU_q(2)$$ contain any (non-unital) invertible elements?

• This Hopf algebra is unital (because all Hopf algebras are) so... the unit is invertible? Feb 23, 2021 at 9:53
• @MathewDaws: I edited to exclude the case of the unit. Feb 23, 2021 at 13:03

Yes, the nonzero multiples of the identity $$1$$ are the only invertible elements in this algebra. I am sure that someone with more expertise in Hopf algebras than I, can provide a 'high level' proof of the result. The following is a 'low level' direct argument.

For the proof, I am using some notations and basic results from Woronowicz's original paper [W].

For $$\nu \in (0,1) \cup (-1,0)$$, we are considering the unital $$*$$-algebra $$A$$ generated by the elements $$a,b$$ subject to the relations $$a^*a + b^* b = 1 \;\; , \;\; aa^* + \nu^2 b^* b = 1 \;\; , \;\; bb^* = b^*b \;\; , \;\; ab = \nu ba \;\; , \;\; ab^* = \nu b^* a \; .$$ In [W] it is proven that the elements $$a^k b^n (b^*)^m$$ and $$(a^*)^k b^n (b^*)^m$$ form a vector space basis of $$A$$. Define for $$k \geq 0$$, the subspace $$A(k) \subset A$$ as the linear span of $$a^k b^n (b^*)^m$$ with $$n,m \geq 0$$. For $$k \leq 0$$, define $$A(k)$$ as the linear span of $$(a^*)^k b^n (b^*)^m$$ with $$n,m \geq 0$$. One has $$A(k_1) A(k_2) \subset A(k_1 + k_2)$$ for all $$k_1,k_2 \in \mathbb{Z}$$.

Another ingredient is the $$*$$-representation $$\pi$$ of $$A$$ on the Hilbert space $$H = \ell^2(\mathbb{N} \times \mathbb{Z})$$, also defined in [W], and given by $$\pi(a) e_{i,j} = \sqrt{1-\nu^2} e_{i-1,j} \;\; , \;\; \pi(b) e_{i,j} = \nu^n e_{i,j+1} .$$ Write $$B = A(0)$$, so that $$B$$ is the unital and abelian $$*$$-subalgebra of $$A$$ generated by $$b$$. We can view $$B$$ as the algebra of polynomials in two variables. We will use that $$B$$ has no zero divisors and that the multiples of $$1$$ are the only invertible elements in $$B$$.

For every $$k \in \mathbb{N}$$, denote by $$H_k$$ the closed linear span of $$e_{i,j}$$ with $$i \geq k$$, $$j \in \mathbb{Z}$$. We use the following observations: for $$k \geq 0$$, the range of $$\pi(a^k)$$ is dense in $$H$$, while the range of $$\pi((a^*)^k)$$ is dense in $$H_k$$. Also, every $$H_k$$ is an invariant subspace for $$\pi(B)$$ and the resulting representation of $$B$$ on $$H_k$$ is faithful for every $$k \geq 0$$.

Assume that $$x,y \in A$$ and $$xy = 1$$. Write $$x = \sum_n x_n$$ and $$y = \sum_n y_n$$ with $$x_n,y_n \in A(n)$$. Let $$n_0$$, $$m_0$$ be the largest integers with $$x_{n_0} \neq 0$$ and $$y_{m_0} \neq 0$$. Since $$xy = 1$$, we must have that $$n_0 + m_0 \geq 0$$. The component of $$xy$$ in $$A(n_0 + m_0)$$ is given by $$x_{n_0} y_{m_0}$$.

We claim that $$n_0 + m_0 = 0$$. Assume that $$n_0 + m_0 > 0$$. Then, $$x_{n_0} y_{m_0} = 0$$. Using the notation $$a_k = a^k$$ for $$k \geq 0$$ and $$a_k = (a^*)^{-k}$$ for $$k \leq 0$$, we can uniquely write $$x_{n_0} = a_{n_0} P$$ and $$y_{m_0} = Q a_{m_0}$$ with $$P,Q \in B$$. We get that $$\pi(a_{n_0}) \pi(PQ) \pi(a_{m_0}) = 0 .$$ It follows that $$\pi(PQ)$$ is zero on $$H_k$$ for $$k$$ large enough. Hence, $$PQ = 0$$. This forces $$P = Q = 0$$, contradicting our choice of $$n_0$$ and $$m_0$$. So the claim is proven.

Similarly define the smallest integers $$n_1$$, $$m_1$$ such that $$x_{n_1} \neq 0$$ and $$y_{m_1} \neq 0$$. The same reasoning leads to $$n_1 + m_1 = 0$$. It follows that $$n_1 = n_0$$ and $$m_1 = m_0$$. Exchanging the roles of $$x$$ and $$y$$ if necessary, we may assume that $$m_0 \geq 0$$ and $$n_0 = -m_0$$. This means that $$x = (a^*)^{m_0} P$$ and $$y = Q a^{m_0}$$ for some $$P,Q \in B$$. It follows that $$\pi(a^*)^{m_0} \pi(PQ) \pi(a^{m_0}) = 1 .$$ If $$m_0 \geq 1$$, the left hand side has a nontrivial kernel. Thus, $$m_0 = 0$$. It then follows that $$PQ = 1$$, which implies that $$P$$ and $$Q$$ are a multiple of $$1$$. We have proven that $$x$$ and $$y$$ are multiples of $$1$$.