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?



Invertible elements of the Hopf algebra quantum $SU(2)$
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]](https://mathscinet.ams.org/mathscinet-getitem?mr=890482).


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 $1$.


