A while ago, I was reading Majid's book Foundations of quantum group theory, and Section 9.4 has a rather fascinating description of a Tannaka-Krein reconstruction result for quantum groups. In particular, there seems to be the claim that if $H$ is a quasi-triangulated quasi-Hopf algebra, then the braided endomorphisms of the identity functor on (a suitably large category of) $H$-comodules form a Hopf algebra object $H'$ in $H$-comodules, in a way that identifies $H'$-comodules with $H$-comodules. Furthermore, $H'$ is commutative and cocommutative with respect to the braided structure on the category of $H$-comodules, and under some nondegeneracy assumptions, it is self-dual. This is called "transmutation" because $H'$ appears to have nicer properties than $H$ (although it may live in a strange category). Some examples are given, e.g., $U_q(g)$ and quantum doubles of finite groups. Unfortunately, the arguments in the proof are given in a diagrammatic language that I was unable to fathom.

Why does this result seem problematic?

The first problem comes from reasoning by analogy. If I want to describe a Hopf algebra object in a monoidal category, I need some kind of commutor transformation $V \otimes W \to W \otimes V$ to even express the compatibility between multiplication and comultiplication, e.g., that comultiplication is an algebra map. In operad language, I need (something resembling) an E[2]-structure on the category to describe compatible E[1]-algebra and E[1]-coalgebra structures. If you think of the spaces in the E[k] operad as configurations of points in $\mathbb{R}^k$, this is roughly saying that you need two dimensions to describe compatible one-dimensional operations. In the above case, the category of $H$-comodules has an E[2]-structure, but I'm supposed to get compatible E[2]-algebra and E[2]-coalgebra structures. Naively, I would expect an E[4]-category to be necessary to make sense of this, but I was unable to wrestle with this successfully.

The second problem comes from a construction I've heard people call Koszul duality, or maybe just Bar and coBar. If we are working in an E[n]-category for n sufficiently large (like infinity, for the symmetric case), then there is a "Bar" operation that takes Hopf algebras with compatible E[m+1]-algebra and E[k]-coalgebra structures, and produces Hopf algebras with compatible E[m]-algebra and E[k+1]-coalgebra structures. There is a "coBar" operation that does the reverse, and under some conditions that I don't understand, composing coBar with Bar (or vice versa) is weakly equivalent to the identity functor. In the above case, I could try to apply Bar to $H'$, but the result cannot have an E[3]-coalgebra structure, since E[3] doesn't act on the category. Applying Bar then coBar implies the coalgebra structure on $H'$ is a priori only E[1], and applying coBar then Bar implies the algebra structure on $H'$ is a priori only E[1]. It is conceivable (in my brain) that the E[2]-structures could somehow appear spontaneously, but that seems a little bizarre.


Am I talking nonsense, or is there a real problem here? (or both?)

  • $\begingroup$ Naturally, you are welcome to replace every appearance of "Hopf" with "bi" if it makes you more comfortable. $\endgroup$
    – S. Carnahan
    Jan 31, 2010 at 19:54
  • $\begingroup$ I don't know the operad language enough to be able to make a stab at answering your questions. I do know that Majid has some notions of "braided Hopf algebra" that make sense in a braided monoidal category. They are somewhat surprising, because they are fairly asymmetric. This is more of a problem for Majid's "braided Lie algebra"s. $\endgroup$ Jan 31, 2010 at 21:24

1 Answer 1


Scott, I believe the source of your confusion is that Majid doesn't claim that the braided Hopf algebra he constructs is both braided commutative and braided co-commutative in C. Just as in the usual case, the Hopf algebra one constructs is braided co-commutative (like U(g)) or braided commutative (like O(G)) if you work dually, but not both. I think your arithmetic about E[n]-operads is correct as to why that would be unusual.

Can you point where in Chapter 9.4 is it claimed that the resulting Hopf algebra is both commutative and co-commutative? For instance in my copy on page 481, he explains U(C) is braided co-commutative but doesn't mention commutative anywhere. On page 477 of my copy, he says "One can use the term 'braided group' more strictly to apply to braided-Hopf algebras which are 'braided-commutative' or 'braided-cocommutative' in some sense." (keyword "or").

By the way, in the Section 3 of http://arxiv.org/abs/0908.3013 is an exposition (not original) reconstructing the algebra A (transmutation of O_q(G)), which uses language probably more familiar to you. Also there is a Remark 3.3 (explained to us by P. Etingof) which gives a more concise description of A (and can be used to derive its key properties like braided-commutativity) by considering module categories and internal homs.

(apologies in advance if i'm incorrect; I read that book awhile ago and opened it up briefly to address your question)

  • $\begingroup$ Thanks David. I had been unable to extract precise claims from his discussion. This gets rid of the E[4], but even a braided commutative Hopf algebra would require an E[3] category to express compatibility with comultiplication, assuming my heuristic arguments hold water. $\endgroup$
    – S. Carnahan
    Jan 31, 2010 at 22:10
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    $\begingroup$ Yes, while it's a very interesting and useful book, it is sometimes difficult to translate the funny language and certainly the diagrammatic proofs, which go beyond just tangles. In the cases pertaining to U_q(g) and O_q(G), there are calculations that only use tangle diagrams, without. Also, what he calls "covariantization" and "transmutation" are called "twist by cocycle" in works by others (e.g. Donin, Mudrov). There is another paper you might want to read, in which he constructs the "CoEnd" algebra A(C) in a more mathematician-friendly way. I'll try to find it for you this week. $\endgroup$ Jan 31, 2010 at 22:49
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    $\begingroup$ This isn't the one I was thinking of, but here is a whole book discussing Hopf algebras in braided categories, and in particular, the so-called "CoEnd" algebra, which is constructed in a nice way from any abelian braided tensor category. <springerlink.com/content/613044426433577h/…> In case the link fails, the book is called "Non-Semisimple Topological Quantum Field Theories for 3-Manifolds with Corners" and Chapter 5 is likely what you're interested in. $\endgroup$ Feb 2, 2010 at 3:06
  • $\begingroup$ Okay, I'm reasonably convinced that this is the sort of compatibility that only works in 1-categories, because there aren't an non-identity higher morphisms to foul things up. $\endgroup$
    – S. Carnahan
    Sep 9, 2013 at 15:48

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