Characterizing discrete quantum groups

Question

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.

Answer 1

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.

Answer 2

I think this is an interesting question. I wonder if the OP had any partial results? I give below an argument showing (1)$\implies$(2), but maybe the OP already knew this?

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If (1) holds, then in the operator algebraic setting, we pretty much define¹ discrete quantum groups as being the dual of compact quantum groups. So there is a CQG $\mathbb G$ with $M=L^\infty(\widehat{\mathbb G})$. From duality, we know then that $M$ contains a $\sigma$-weakly dense multiplier Hopf algebra $A = c_{00}(\mathbb G)$, which is just the algebraic direct sum of the full matrix algebras. Here multiplier Hopf algebra is in van Daele's sense. In fact, there is a characterisation of which multiplier Hopf algebras are discrete quantum groups, see van Daele, J. Alg. In particular, the maps $T_1, T_2:A\otimes A\rightarrow A\otimes A$ given by $$ T_1(a\otimes b) = \Delta(a)(1\otimes b), \quad T_2(a\otimes b) = (a\otimes 1)\Delta(b) $$ are bijections. (You need multipliers to make sense of these, and it is an assumption that we even map into $A\otimes A$ and not some multiplier algebra).

Firstly, we have a counit $\epsilon$ on $A$. As $A$ is the sum of matrix algebras, $\epsilon$ must be evaluation of some $1\times 1$ matrix block; let $h$ be the unit of this block, which is the support projection of $\epsilon$. Then $h\in M$, and $\epsilon$ extends boundedly to $M$ as a normal $*$-character, and has the expected properties.

Secondly, if $(p\otimes 1)\geq \Delta(h)$ then set $q=1-p$ so as $(p\otimes 1)\Delta(h)=\Delta(h)$ we see that $(q\otimes 1)\Delta(h)=0$. If $q\not=0$ then from the structure of $M$, we can find a projection $q'\leq q$ which is non-zero and only supported on finitely many matrix blocks. Thus $q'\in A$, and $(q'\otimes 1)\Delta(h)=0$. Thus $T_2(q'\otimes h)=0$ so $q'=0$ contradiction. Hence $p=1$. Similarly use $T_1$ in the $(1\otimes p)\geq \Delta(h)$ case.

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I spent some time messing about with trying to prove the converse, but got nowhere. It of course does work when $M = \ell^\infty(S)$ for some set $S$ (which is then a semigroup, and the conditions imply is a group). One issue I have is that I know of rather few constructions of non-trivial $C^*$ or von Neumann algebraic bialgebras which do not come from (quantum) groups.

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1: Actually, I'd be interested in a direct operator algebra characterisation of discrete quantum groups. All descriptions I know either seem to use duality very directly, or are essentially algebraic.