Let $M$ be a von Neumann algebra, and let $\Delta$ be a unital normal $\*$-homomorphism $M \rightarrow M \mathbin{\bar\otimes} M$ that satisfies the coassociativity condition $(\Delta \mathbin{\bar\otimes} \mathrm{id}) \circ \Delta = (\mathrm{id} \mathbin{\bar\otimes} \Delta ) \circ \Delta$. Assume that $M$ is an $\ell^\infty$-direct sum of finite type I factors. Are the following conditions equivalent?


1. The pair $(M, \Delta)$ is a von Neumann algebraic quantum group, in the sense of Kustermans and Vaes [2, definition 1.1].
2. There exists a unital normal $\*$-homomorphism $\varepsilon\colon M \rightarrow \mathbb{C}$, with support projection $e \in M$, such that:
	* (a) $(\varepsilon \mathbin{\bar\otimes} \mathrm{id}) \circ \Delta = \mathrm{id}$
	* (b) $(\mathrm{id} \mathbin{\bar\otimes} \varepsilon) \circ \Delta = \mathrm{id}$
	* (c) For every projection $p \in M$, if $p \otimes 1 \geq \Delta(e)$, then $p = 1$.
	* (d) For every projection $p \in M$, if $1 \otimes p \geq \Delta(e)$, then $p = 1$.


The question is motivated by quantum predicate logic [3, section 2.6]. I am extending my preprint [1] to include a few more examples, but the example of discrete quantum groups is the one it really needs.


[1] A. Kornell, *Quantum predicate logic with equality*. arXiv:2004.04377


[2] J. Kustermans & S. Vaes, *Locally compact quantum groups in the von Neumann algebraic setting*, Math. Scand. **92** (2003), no. 1.


[3] N. Weaver, *Mathematical Quantization*, Studies in Advanced Mathematics, Chapman & Hall/CRC, 2001.



Characterizing discrete quantum groups
Also the implication (2) $\Rightarrow$ (1) holds and can be proven as follows.


Denote by $\mathcal{C}$ the category of all finite dimensional, nondegenerate $\*$-representations of $M$. The morphisms are the intertwining linear maps. Turn $\mathcal{C}$ in a $C^\*$-tensor category by defining $\pi\_1 \otimes \pi\_2$ to be equal to $(\pi\_1 \otimes \pi\_2) \circ \Delta$. Then $\varepsilon$ is a unit object for $\mathcal{C}$. We identify the set $I$ of minimal central projections of $M$ with the representatives for irreducible objects of $\mathcal{C}$. For $p \in I$, we pick a finite dimensional Hilbert space $H\_p$ such that $Mp = B(H\_p)$.


The main point is to prove that $\mathcal{C}$ is rigid, i.e. that every irreducible object has a conjugate with a solution for the conjugate equations. Let $p \in I$. Hypothesis (c) is saying that the join of all the left support projections of elements of the form $p((\text{id} \otimes \omega)\Delta(e))$ equals $p$. Similarly, the join of all the right support projections of elements of the form $((\omega \otimes \text{id})\Delta(e))p$ equals $p$. It follows that we can pick $q,r \in I$ such that
$$(\Delta(e)(r \otimes p) \otimes 1) (1 \otimes (p \otimes q)\Delta(e)) \neq 0 \; .$$
So we can pick morphisms $V : \mathbb{C} \rightarrow H\_p \otimes H\_q$ and $W : \mathbb{C} \rightarrow H\_r \otimes H\_p$ such that
$$(W^\* \otimes 1) (1 \otimes V) \neq 0 \; .$$
Since this expression defines a morphism between the irreducible objects $r$ and $q$, we conclude that $r=q$ and that $V$ and $W$ may be chosen such that
$$(W^\* \otimes 1) (1 \otimes V) = 1\_q \; .$$
Then
$$(1 \otimes (V^\* \otimes 1)(1 \otimes W)) W = (1 \otimes V^\* \otimes 1)(W \otimes W) = W \; .$$
It follows that $(V^\* \otimes 1)(1 \otimes W)$ is nonzero, thus a multiple of $1\_p$ and hence, equal to $1\_p$. We have proven that $\mathcal{C}$ is rigid.


One could already conclude at this point that $(M,\Delta)$ is a discrete quantum group by making a detour via Woronowicz' Tannaka-Krein theorem.


One can also repeat the proof of the first basic properties of rigid $C^\*$-tensor categories and then directly verify Van Daele's axioms cited above. One first proves Frobenius reciprocity. By definition, $p$ is contained in $q \otimes r$ if and only if $\Delta(p) (q \otimes r) \neq 0$. Using Frobenius reciprocity and denoting by $A \subset M$ the dense $\*$-subalgebra spanned by the $Mp$, $p \in I$, one gets that the linear span of $\Delta(A) (1 \otimes A)$ is contained in the algebraic tensor product $A \otimes\_{\text{alg}} A$. One can thus define Van Daele's map
$$T : A \otimes\_{\text{alg}} A \rightarrow A \otimes\_{\text{alg}} A : T(a \otimes b) = \Delta(a) (1 \otimes b) \; .$$
Given $a \in A$ and $p \in I$, take $V$ as above. Define $b \in A \otimes Mp$ such that
$$\Delta(a)\_{13} (1 \otimes V) = (b \otimes 1)(1 \otimes V) \; .$$
Then
$$(T(b) \otimes 1)(1 \otimes V) = (\Delta \otimes \text{id})\Delta(a) (1 \otimes V) = a \otimes V \; .$$
Therefore, $T(b) = a \otimes p$. It follows that $T$ is surjective. One can reason similarly for $\Delta(A)(A \otimes 1)$ and conclude that Van Daele's definition of discrete quantum groups is satisfied.


