Group rings arrr commutative

If you are familiar with group rings, you might think that the title of this post is false. If G is a nonabelian group, multiplying the basis elements g and h in \mathbb{Z}G can yield gh \neq hg, so we have a problem. In general, if you have a problem that you can’t solve, you should cheat and change it to a solvable one (According to my advisor, this strategy is due to Alexander the Great). Today, we will change the definition of commutative to make things work.

Given a group G, the group ring \mathbb{Z}G is defined to be the free abelian group whose generating basis is given by the elements of G, and with “convolution” multiplication. In coordinates, this means
(\sum_{g \in G} a_g g)(\sum_{h \in G} b_h h) = \sum_{gh=k} a_g b_h k.
These sums make sense, because they have finitely many nonzero summands. In general, we can ask for the coefficients to be elements in other rings, like \mathbb{C}, and these are still called group rings.

What does it mean for a ring to be commutative? Normally, we would say that for any two elements a and b, we have ab = ba. I’d like to phrase this more categorically, without referring to individual elements. Multiplication in a ring A is given by a map A \times A \to A. This is a bilinear operation, in the sense that na*b = a*nb for all integers n, so we can view multiplcation as a linear map m: A \otimes A \to A (Chern apparently liked to say that tensor products replace bilinear maps by linear maps). To say that multiplication is commutative is to say that we have an equality of maps: m \circ T = m, where T: A \otimes A \to A \otimes A is the switching map a \otimes b \mapsto b \otimes a.

This suggests a relatively easy solution to our problem. If we adjust the map T suitably (e.g. using the identity map), we can make any ring commutative. Unfortunately, in solving our problem, we’ve destroyed any meaningful content in the notion of commutativity. We can salvage some meaning by demanding that T satisfy some natural sounding conditions, such as T(1 \otimes a) = a \otimes 1. This conveniently eliminates the identity map from consideration, but we still don’t have any immediate guidance for making a good choice.

I’d like to say that T is more than just a map of abelian groups, i.e., T ought to have some structure that makes it natural in a strong sense. One clue to turning this vague idea into mathematics is the unit compatibility condition I gave above. Since the unit can be viewed as a ring homomorphism \mathbb{Z} \to A, we can say that T should produce isomorphisms like \mathbb{Z} \otimes A \to A \otimes \mathbb{Z} in a manner compatible with the unit maps. One possible first attempt to strengthen this is to demand that T produce isomorphisms B \otimes C \to C \otimes B for all abelian groups in a manner compatible with all abelian group homomorphisms.

We can give this a categorical interpretation. Tensor product takes a pair of abelian groups to an abelian group, and transforms maps in a compatible way. This means tensor product is a functor Ab \times Ab \to Ab, and since it satisfies some conditions like associativity, we say that Ab is a monoidal category. There is also a switch functor \tau: Ab \times Ab \to Ab \times Ab, taking an object (B,C) to (C,B) and also switching homomorphisms. Our demand on T then translates to the assertion that T is a natural isomorphism \otimes \to \otimes \circ \tau.

Unfortunately, there are no natural isomorphisms T that make our group ring commutative in this sense, mostly because the category of abelian groups is really well-behaved. Tensor products commute with colimits, and any abelian group is a colimit of some diagram of copies of the integers. Since T is the switch map whenever one of its inputs is a copy of the integers, T is forced to be the usual switch map on all inputs.

Fortunately, we can view the group ring in a different category, namely, it is a G-graded abelian group. The category of G-graded abelian groups admits a monoidal structure via a graded tensor product that puts the tensor product of a degree g group with a degree h group in degree gh. However, if G is nonabelian, there is no longer an obvious switch transformation, which we need to describe commutativity, since the tensor product in the opposite order would place things in a different degree.

We can try to solve this problem in the following way. Recall that the center of a group or a ring is the set of elements that commute with everything else. We could define the center of a monoidal category to be the subcategory of objects for which the tensor product with anything on the left is isomorphic to the tensor product on the right, and hope the group ring fits in somehow. This category looks nice at first, but since we don’t know what the isomorphisms are, we don’t have a good choice of a natural transformation T.

There is a better version, discovered by Drinfeld, and written up by Majid and Joyal-Street. Given a monoidal category C, one can construct a new category Z(C), called the Drinfeld center. Its objects are pairs (x, \phi), where x is an object in C, and \phi: x \otimes - \to - \otimes x is a natural isomorphism called a braiding, satisfying a compatibility condition:
\phi = (id \otimes \phi) \circ (\phi \otimes id): x \otimes y \otimes z \to y \otimes z \otimes x
(to be precise, I should have inserted some associators). Z(C) then admits a natural monoidal structure induced by the monoidal structure on C, but it also comes with a canonical transformation T, made out of the isomorphisms \phi. With this structure, Z(C) is called a braided monoidal category, and together with the “forget braiding” functor, it is in fact universal with respect to braided monoidal categories with a monoidal functor to C.

Our task is now to find our group ring in some form inside the Drinfeld center of the category of G-graded abelian groups, and hope that it is in some sense commutative. First, we should write down precisely what it means to have a commutative ring in a braided monoidal category. Baez calls these structures r-commutative in his Hochschild homology paper, so if things work out, we could say that group rings are r-commutative.

Given a braided tensor category C, with associator \Psi, commutor
T, and unit structures l and r, a commutative ring is an object A equipped
with a multiplication morphism A \otimes A \to A and a unit
morphism e: 1 \to R, such that:

  • Multiplication with the unit yields the identity: m \circ (e \otimes id) \circ l^{-1} = id, and the corresponding statement on the right.
  • Multiplication can be dragged across the commutor:
    (m \otimes id) \circ (id \otimes T) \circ (T \otimes id) = T \circ (id \otimes m),
    and the reflected version. (I’m omitting the associator here.)
  • Associativity: m \circ (m \otimes id) = m \circ (id \otimes m) \circ \Psi_{R,R,R}.
  • Commutativity: m = m \circ T.

There are diagrammatic ways of representing these axioms, using bits of string on a table, tied together at trivalent vertices. There might even be a Youtube video about it.

Also, we need to understand the center itself. Not every G-graded abelian group admits a braiding transformation. If (A, \phi) lies in the center, we can apply \phi to a copy of the integers in degree g. Since the natural transformation has to respect degree, we find that we have an isomorphism from the part of A in degree h to the part in degree ghg^{-1}. In general, the compatibility condition induces an action of the group ring on A, conjugating degrees.

Objects of the Drinfeld center are then G-equivariant G-graded abelian groups. We can view these as sums of pairs (g,A), where g is a representative of a conjugacy class and A is a representation of the centralizer C_G(g). The G-action on (g,A) is given by conjugating g, and acting on the G-module induced from A. From our choice of commutor notation, the G-action on the grading is from the left, and the braiding is given by (g,A) \otimes (h,B) \mapsto (ghg^{-1},gB) \otimes (g,A).

Now, consider the sum \bigoplus_{[g]} (g,\mathbb{Z}). Under the “forget braiding” functor to G-graded vector spaces, this lands on the group ring. We can pull back the multiplication map to get a copy of the group ring in the center. To check that it is commutative, we compare the two maps g \otimes h \mapsto gh and g \otimes h \mapsto ghg^{-1} \otimes g \mapsto gh. In other words, group rings are commutative because gh = ghg^{-1}g.

I should note that this formalism works if we replace abelian groups with sets. G-graded abelian groups become sets with a map to G, and the center is the category of G-sets with a G-map to G (with action given by conjugation). G lives in this category in a straightforward way, and we can say that G is an abelian group object in this category. One reason I didn’t name this post “All groups are abelian” is that the current title seems slightly less blatantly false.

You might be a bit disappointed if you’ve read this far to find that I’ve just redefined commutativity to hold using a rather tautological-looking conjugation trick. I concede that the punch line is a bit anticlimactic. However, we can think of this example of hidden commutativity as a toy model of a deeper phenomenon, known as transmutation.

There is an old philosophy due to Tannaka that a group is determined by its representation theory, and this was put into a categorical framework by Saavedra-Rivano and later refined by Deligne, Krein, and possibly others. The statement of Tannaka-Krein duality is that any symmetric tensor category with well-behaved duals and a faithful monoidal functor to vector spaces is equivalent to the category of representations of a proalgebraic group that is unique up to isomorphism. We will consider the coordinate ring picture, which is that it is equivalent to the category of comodules of a commutative Hopf algebra. The Hopf algebra is constructed by considering the natural transformations from the faithful functor to itself. Since the functor is linear, we have an additive structure, and composition makes it an algebra. We also have a coproduct arising from the tensor struture on the categories, and the counit and antipode are unique if they exist (this tends to require some completeness assumptions).

In some work in the early 1990s, Majid and Lyubashenko pointed out that one doesn’t need the functor to go to vector spaces. If the functor goes to any symmetric tensor category, you can reconstruct a Hopf algebra object in
that category by the same precedure (assuming existence of suitable colimits). More generally, if you have two braided tensor categories and a faithful braided tensor functor between them, the Hopf algebra of natural transformations has comodule category equivalent to the first category. This is particularly useful in part because we can use the identity functor, so we don’t actually need the second category. Given a braided tensor category, it is the [co]representation category of a commutative Hopf algebra in itself.

This can be applied to familiar examples of braided tensor categories, such as representations of quantum enveloping algebras, and the Drinfeld center that we saw earlier. These particular cases have a nondegeneracy property, known as factorizability: there is a map, called the inverse quantum Killing form (or “braided Fourier transform”), that is an isomorphism between the Hopf algebra and its dual. In particular, if we recast U_q(\mathfrak{g}) as a Hopf algebra in its own category of representations, it becomes transmuted to something commutative and cocommutative, and it is isomorphic to its dual, the “braided coordinate ring”. This is a phenomenon that only happens in the braided world, since sending q to 1 degenerates the two into a universal enveloping algebra and the coordinate ring of an algebraic group, and they are no longer isomorphic. One can take an extreme interpretation of this fact, and claim that braided commutative groups are the natural objects in this picture, and that the algebraic groups that we know and love are degenerate manifestations.

Thanks to David Jordan for introducing me to transmutation in the pre-Talbot seminar. Also, I should point out a neat paper by John Francis and the Davids,that develops Drinfeld centers from the viewpoint of sheaves on loop space, along with other gems.

10 thoughts on “Group rings arrr commutative

  1. Grothendieck proved a very general theorem that seems to make Tannaka’s `philosophy’ precise.

    Let A be any non-zero commutative ring. For a group \Gamma, denote by \mathrm{Rep}_A(\Gamma) the category of finitely presented A-modules with a \Gamma-action. Let u:\Gamma_1\rightarrow \Gamma_2 be a homomorphism of finitely generated groups. The induced map of profinite completions \hat{u}:\hat{\Gamma}_1\rightarrow\hat{\Gamma}_2 is an isomorphism if and only the restriction functor u^*_A : \mathrm{Rep}_A(\Gamma_2)\rightarrow \mathrm{Rep}_A(\hat{\Gamma}_1) is an equivalence of categories.

    So it seems more accurate to say that the profinite completion of a group is determined by its representation theory.

  2. Henry,

    Do you know where he proved this? The statement has a very Grothendieck flavor, but I was unable to find it in SGA 1 or 3.

    If we remove the finite presentation restriction [on modules], we can get a stronger statement. In particular, Higman constructed an infinite finitely presented group with no nontrivial finite quotients, and its regular representation is clearly distinguishable from a sum of trivials. However, expanding to a larger category comes at the cost of having well-behaved duals, and this makes the reconstruction more complicated.

    Incidentally, the statement I gave was for proalgebraic groups, which are a more general setting than profinite groups.

  3. Scott,

    The reference I have for Grothendieck’s theorem is as follows.

    ‘Représentations linéaires et compactification profinie des groupes discrets’, Manuscripta Math. 2 (1970), 375–396.

    I haven’t actually looked at it, though.

    So are you saying that there should be a theorem along the following lines: if a homomorphism u of fg groups induces an isomorphism of representation categories (where we’ve dropped the requirement that the modules be fp) then u induces an isomorphism of proalgebraic completions?

  4. Thanks for the reference. I’m afraid I didn’t explain myself very well, because I was trying to make two points at the same time. The proalgebraic completion of a discrete group is its profinite completion, but the statement in my post also works in a positive dimensional setting. That is the classical Tannakian statement.

    My point in removing the finitely presented hypothesis was that this allows us to distinguish two groups that have the same profinite (or proalgebraic) completion. In particular, the inclusion of the trivial group into Higman’s group induces an isomorphism of their profinite completions, but the restriction functor is not an equivalence if we allow the regular representation into our category.

    I think the Barr-Beck theorem gives us a statement about groups being determined by their representation categories that doesn’t need any finiteness hypotheses on the group, as long as we demand that the representation categories in question contain the regular representation and represent small colimits.

  5. The proalgebraic completion of a discrete group is its profinite completion

    Oh, right – Selberg’s Lemma! What’s the statement for the positive-dimensional case?

    In particular, the inclusion of the trivial group into Higman’s group induces an isomorphism of their profinite completions

    Yes. For what it’s worth, there are much more interesting examples of this sort these days. For instance, Bridson and Grunewald constructed a pair of finitely presented, residually finite groups that aren’t isomorphic but whose profinite completions are isomorphic.

    Anyway, thanks for clarifying, Scott!

  6. Regarding the videos on YouTube, you’re probably thinking of some videos by The Catsters (Eugenia Cheng and Simon Willerton). In particular, there are a bunch about group objects and Hopf algebras which use string diagrams, and another set on string diagrams for monads and adjunctions. Not directly this situation, but after watching those, anyone could figure out what pictures to draw here. :) You can find their channel here: http://www.youtube.com/user/TheCatsters

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