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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.

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2 Answers 2

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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.

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    $\begingroup$ Hi Stefaan, welcome to mathoverflow! Can you suggest any references relating the various definitions of a discrete quantum group, e.g., Van Daele's definition to the von Neumann algebraic notion? $\endgroup$ Commented Jun 8, 2020 at 22:06
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    $\begingroup$ Van Daele's definition of discrete quantum groups can be found in doi.org/10.1006/jabr.1996.0075. In Theorems 5.3 and 5.4 of that paper, it is proven that discrete quantum groups in his sense admit a positive faithful left invariant functional and a positive faithful right invariant functional. Therefore, the von Neumann algebra completion (i.e. replacing the algebraic direct sum of matrix algebras by the $\ell^\infty$ direct sum) is a von Neumann algebraic quantum group. $\endgroup$ Commented Jun 11, 2020 at 16:22
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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?


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.


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.


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.

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  • $\begingroup$ Hi Matt, regarding your footnote/question: are you looking for something which doesn't mention algebraic properties of $L^1({\mathbb G})$? Perhaps Theorem 4.4 of Volker's paper arxiv.org/abs/math/0506493 ? $\endgroup$
    – Yemon Choi
    Commented Jun 5, 2020 at 16:48
  • $\begingroup$ Ah, I think I misunderstood your question, never mind. $\endgroup$
    – Yemon Choi
    Commented Jun 5, 2020 at 16:57
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    $\begingroup$ @YemonChoi I guess Runde's result is not a million miles away from what I had in mind. Thanks for reminding me of it! Although the proof, via Proposition 4.1, is again a duality argument... $\endgroup$ Commented Jun 5, 2020 at 20:11

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