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For a compact quantum group $C_q[G]$, it was shown by Woronowicz that $C_q[G]$ contains a dense Hopf algebra generalising the algebra of representations of $G$. I am interested in the other way around, ie given a Hopf algebra $H$ (say a Drinfeld--Jimbo algebra if it makes things easier) can it always be completed to give a compact quantum group? If so, is this completion unique in anyway, or are there many ways to get a cqg from a Hopf algebra?

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You definitely need to assume some sort of unitartity here. – Noah Snyder Jun 12 '11 at 6:49
up vote 12 down vote accepted

I think this was solved in the paper:

MR1310296 (95m:16029)
Dijkhuizen, Mathijs S.(NL-MATH); Koornwinder, Tom H.(NL-AMST-CS)
CQG algebras: a direct algebraic approach to compact quantum groups. (English summary)
Lett. Math. Phys. 32 (1994), no. 4, 315–330.

They show that a Hopf $*$-algebra $A$ is the maximal Hopf $*$-algebra of a compact quantum group in the Woronowicz sense if (and only if) $A$ is spanned by the matrix coefficients of its finite-dimensional unitary (irreducible) corepresentations. The "only if" part was shown by Woronowicz, of course.

They also show that a Hopf $*$-algebra has this property if and only if it admits a positive definite Haar functional (in the sense Andreas talks about).

You also ask about uniqueness-- this occurs if and only if $A$ is "coamenable" (see various papers by Bedos, Tuset and coauthors). A classical example comes from a discrete group $G$ with Hopf $*$-algebra $\mathbb C[G]$ (under convolution product). Then there is the maximal $C^*$ completion $C^*(G)$ and the completion of $\mathbb C[G]$ acting on $\ell^2(G)$ by the left-regular representation, leading to $C^*_r(G)$. The quotient map $C^*(G) \rightarrow C^*_r(G)$ is injective if and only if $G$ is amenable.

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Do the Drinfeld--Jimbo algebras fit into this category? – John McCarthy Jun 12 '11 at 20:03
I know very little about the more algebraic side of things-- but you might understand, better than I do, the contents of Section 6 of: Or, hopefully someone else will come along with a more definitive answer... – Matthew Daws Jun 12 '11 at 20:07
I do not believe that Drinfel'd-Jimbo algebras fit in (though they are Hopf duals to the algebras which do fit if equipped with a correct $\ast$-structure). – Zoran Skoda Jun 12 '11 at 22:11
They do -- see Chari & Pressley's book Chap. 10-13. They are also amenable as proved by Banica's work. – Makoto Yamashita Jun 12 '11 at 22:47
They do not, I added a remark about this in my comment – Nicola Ciccoli Jun 13 '11 at 7:54

First say we're taking about $*$-Hopf algebras, to start with.

Then, in any case, the answer is no. Any compact quantum group admits a biinvariant Haar measure. On a general Hopf algebra not even the existence of a left integral is granted (Hopf algebra with a left integral, if I remember correctly, are called coFrobenius). Even if left and right integral exists it is not granted that they coincide (again I think it is quite easy to find finite-dimensional examples of this: see van Daele The Haar measure on finite quantum groups, Proc. Amer. Math. Soc. 125, 3489-3500 (1997)).

One has certainly to enlarge the theory to locally compact quantum groups; I'm not completely sure whether one can obtain existence in such a case. I strongly doubt about unicity.

ADDED Drinfel'd-Jimbo quantization, by which I mean quantized universal enveloping algebras, are examples of non commutative Hopf * algebras that do not fall in the CQG setting; in fact they have a real group-like element $K$ and and easy computation shows that any left invariant integral $\phi$ should satisfy $\phi(K^2)=\phi(KK^*)=0$, thus cannot be positive and therefore there is no Haar measure. This should come as no surprise. The quantum duality principle allows to interpretate quantized universal enveloping algebras as quantizationof dual Poisson-Lie groups. Dual Poisson Lie groups of (standard Poisson-Lie) compact groups are not compact. This already shows that the setting of compact quantum groups is too narrow for the whole quantum group thing. If I remember correctly quantized universal enveloping algebras are algebraic quantum groups in the sense of Kustermans - van Daele, i.e. they have a Haar measure on the associated multiplier Hopf * algebra (which is non unital), in perfect agreement with the fact that they quantize non compact unimodular Lie groups.

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The new comment is really useful; thanks for adding it. – Matthew Daws Jun 13 '11 at 8:30

The key is not only the existence and identification of left and right integral, but also that they define a positive linear functional on the $\ast$-algebra, i.e. $$\phi(a^*a)\geq 0, \quad \forall a \in A.$$

This allows to construct a Hilbert space representation via the GNS-construction, and hence yields the desired analytic picture.

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A little side remark, concerning the last part of your question: Marc Rosso describes in

Marc Ross, Algèbres enveloppantes quantifiées, groupes quantiques compacts de matrices et calcul différentiel non commutatif, Duke Math. J. Volume 61, Number 1 (1990), 11-40.

another way to get compact quantum groups from Hopf algebras. He defines inner products on the representation spaces of the deformed enveloping algebras of simple Lie groups (Drinfeld and Jimbo's $U_q(g)$'s), such that they form a concrete monoidal $W^*$ category. Applying Woronowicz' Tannaka-Krein duality he can recover compact quantum groups from this: $C(G_q)$, a deformation of the algebra of continuous functions on the simple Lie group $G$. The Hopf algebras $U_q(g)$ do not sit inside $C(G_q)$, but their restricted duals can be identified with the dense Hopf *-algebra contained in $C(G_q)$.

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