Picard and Brauer 2-Groups

April 5, 2006

Posted by Urs Schreiber

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Picard and Brauer groups of representation categories of vertex algebras encode symmetries of full CFTs ( $\to$ , $\to$ ). In the CFT context the elements of the Picard group are called simple currents (e.g. Int. J.Mod.Phys. A 5 (1990) 2903).

It is rarely explicitly admitted that the Picard group is really a 2-group ( $\to$ ), and so is the Brauer group.

Even better, both fit into a certain 3-group. All this has been well known to a handful of experts, apparently ( $\to$ ) - which of course does not stop me from enjoying rediscovering this from my personal point of view.

Namely, this 3-group is closely related to 2-dimensional field theory, and in particular to CFT. The fact that it is an $n$ -group for precisely $n=3$ is closely related to the holography-like phenomenon that 2D CFT can be described in terms of 3D TFT.

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Before talking about the big 3-group, let me describe how the Picard group and the Brauer group look like as 2-groups. This is nicely described in

Enrico M. Vitale
A Picard-Brauer exact sequence of categorical groups
pdf.

The Picard 2-group $\mathrm{Pic}(C)$ of a monoidal category $C$ is the 2-group of invertible objects of $C$ .

There is a better way to say this: let $\Sigma(C)$ be the suspension of the monoidal category $C$ , which is the (possibly weak) 2-category with a single object $\bullet$ , 1-morphisms the objects of $C$ and 2-morphisms the morphsims of $C$ . Then

The Picard 2-group of $C$ is the automorphism 2-group of the single object $\bullet$ in the suspension of $C$ :

(1) $\mathrm{Pic}(C) = \mathrm{Aut}_{\Sigma(C)}(\bullet) \,.$

What is ordinarily called the Picard group is the decategorification of $\mathrm{Pic}(C)$ , namely the group $\pi_0(\mathrm{Pic}(C))$ of isomorphism classes of objects of $\mathrm{Pic}(C))$ .

A special subclass of Picard 2-groups are the Brauer 2-groups.

Consider the bicategory $\mathrm{BiMod}(C)$ of bimodules of algebra objects internal to $C$ . Its objects are algebras, its 1-morphisms are bimodules and its 2-morphisms are bimodule homomorphisms.

When $C$ is not just monoidal, but braided monoidal (experts will take this as a first hint of a secret 3-categorical structure) we get a monoidal structure - now on the bicategory $\mathrm{BiMod}(C)$ (the periodic table of $n$ -categories is at work here).

Taking 2-isomorphism classes in $\mathrm{BiMod}(C)$ , thus obtaining a monoidal category called $\mathrm{cl}(\mathrm{BiMod}(C))$ , we can again form a Picard 2-group. This is the Brauer 2-group:

The Brauer 2-group $\mathrm{B}(C)$ is the Picard 2-group of 2-isomorphism classes of bimodules internal to $C$ .

(2) $\mathrm{B}(C) := \mathrm{Pic}(\mathrm{cl}(\mathrm{BiMod}(C))) \,.$

The fact that I used one suspension in the definition of the Picard 2-group, and one taking of isomorphism classes in the definition of the Brauer 2-group all indicates that the 2-group nature of Picard and Brauer is still not the full truth.

The need to go to three dimensions is nicely seen in the the way 2D CFT correlators are written in terms of Wilson graphs decorated in a modular tensor category ( $\to$ ).

Take a worldsheet, pick a dual triangulation, label the resulting graph in some tensor category. One quickly finds (see section 2.3 of $\to$ for an illustration) that one ultimately needs to pass some of the graph’s edges “beneath” or “above” each other - to “braid” them. Visually, this requires a third dimension. But this is not just heuristics, there is a precise theorem behind this:

Every braided monoidal category $C$ has a double suspension $\Sigma(\Sigma(C))$ , i.e. a 3-category with a single object and a single 1-morphism.

(See for instance page xx here.)

Now realize that $C$ can be thought of as sitting embedded inside its own 2-category of internal bimodules

(3)

C \to \mathrm{BiMod}(C) \,,

where each object of $C$ is regarded as a bimodule over the trivial algebra constituted by the tensor unit in $C$ .

Hence, by simply generalizing to arbitrary bimodules the way that $C$ sits inside a 3-category, we get the full 3-catgegoy

(4)

\Sigma(\mathrm{BiMod}(C))

(recall that we are assuming $C$ to be braided monoidal for this to make sense!).

This is the 3-category John Baez describes in TWF 209.

(And, guess what, I am once again running out of time…)

Posted at April 5, 2006 6:44 PM UTC

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Re: Picard and Brauer 2-Groups

I wonder: is this 3-category equivalent to a strict one? The unit group, the Picard group, and the Brauer group of A are the 0th, 1st, and 2nd cohomology groups of the sheaf O^* on Spec A. It’s interesting that the same groups appear in this 3-category upside-down. Behind the cohomology interpretation there is of course a chain complex. From one point of view or another a chain complex is some kind of *strict* infinity-category, right?

Posted by: D. on April 6, 2006 4:28 AM | Permalink | Reply to this

Re: Picard and Brauer 2-Groups

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is this 3-category equivalent to a strict one?

I am not sure, but I would expect this to be the case. Isn’t every monoidal category equivalent to its strictification? That should give strictness at the 2-group level.

But I’ll try to find out the answer to this question.

The unit group, the Picard group, and the Brauer group of A are the 0th, 1st, and 2nd cohomology groups of the sheaf $O^*$ on $\mathrm{Spec} A$ .

That’s intersting, didn’t know this. Could you be so kind and point me to a reference/textbook?

Behind the cohomology interpretation there is of course a chain complex.

A strict 3-group should be the same as a 3-term chain complex of groups, with some additional structure. I called this a “2-crossed module”, see this entry, which has become section 10.7 of this. (But there are possibly better accounts of 3-groups.)

From one point of view or another a chain complex is some kind of strict infinity-category, right?

I know this is true at least for the case where there is certain extra structure around, like if we are talking about categories internal to the category of groups or vector spaces. But there may be extra conditions and extra structure on a complex in order to define an internal strict $n$ -category in these cases.

I don’t know about the general situation. But this are certainly a couple of very interesting questions you raise.

Posted by: urs on April 6, 2006 3:48 PM | Permalink | Reply to this

Re: Picard and Brauer 2-Groups

Someone named D. writes:

I wonder: is this 3-category equivalent to a strict one?

Good question; I feel like an idiot for not having pondered it. I don’t know the answer.

The unit group, the Picard group, and the Brauer group of A are the 0th, 1st, and 2nd cohomology groups of the sheaf O^* on Spec A.

That’s interesting; I didn’t know that about the Brauer group, though I knew it for the Picard and unit groups! Is there a pleasant reference for this fact, or is it better to figure this out oneself?

Behind the cohomology interpretation there is of course a chain complex. From one point of view or another a chain complex is some kind of *strict* infinity-category, right?

Right: it’s a rigorous theorem that degree-n chain complexes give strict n-categories. In fact, degree-n complexes of abelian groups are precisely the abelian group objects in the category of strict n-categories. This is also true for n = infinity.

Of course, this doesn’t actually prove that for any commutative ring R, the monoidal 2-category Alg(R) of R-algebras, bimodules and bimodule homomorphisms is equivalent to a strict one.

A weaker statement would be that the core of Alg(R) is equivalent to a strict one. This is the 3-group consisting of weakly invertible R-algebras, bimodules and bimodule homomorphisms. (Since people enjoy jargon they call these “Azumaya algebras”, “Morita equivalences” and “bimodule isomorphisms”.)

This weaker statement could be disproved by calculating the “Whitehead product” in this 3-group and finding that it’s nontrivial.

The “Whitehead product” is an operation that takes two elements of the second homotopy group of a space, and spits out an element of the third homotopy group. A 3-group is the same as a connected space with trivial homotopy groups above the third, so the Whitehead product also exists for any 3-group. In the case at hand, it would be an operation that takes two elements of the Picard group of R and spits out an element of the unit group of R.

If someone knows such an operation, it’s probably the Whitehead product. If it is, the core of Alg(R) cannot be strictified. If that’s true, Alg(R) cannot be strictified either.

If nobody knows such an operation, either they’re being dumb, or the core of Alg(R) can be strictified, or there is some other obstruction to strictifying 3-groups that I’m forgetting.

(Btw, the Whitehead product is the reason the 2-sphere has a nontrivial third homotopy group. In this case we can see it as the obstruction to making a certain braided 2-group into a symmetric one. Topologically, the Whitehead product is closely related to braiding. This stuff is what led Joyal and Tierney to realize that not every 3-category could be strictified!)

Posted by: jcbaez on April 6, 2006 5:54 PM | Permalink | Reply to this

chain complexes

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it’s a rigorous theorem that degree- $n$ chain complexes give strict n-categories

Ok, the question was about chain complexes (of abelian groups, presumeably), while in my reply I was really thinking of more general exact sequences.

So abelian groups internal to $\mathrm{nCat}$ are a special case of strict $n$ -groups. General strict $n$ -groups should come from exact $n$ -term sequences of groups, equipped with actions of the $i$ th group on the $(i+1)$ th group, subject to some conditions (defining something like an “ $(n-1)$ -crossed module” of groups).

If all groups in the sequence are abelian, the extra structure and properties should become trivial and we are left just with a chain complex (= a chain complex of abelian groups).

Posted by: urs on April 6, 2006 7:26 PM | Permalink | Reply to this

Re: Picard and Brauer 2-Groups

About the Brauer group: I think you can find the cohomology interpretation in Milne’s book on etale cohomology. An Azumaya algebra is something that looks locally like a matrix algebra. For some reason I can’t think of now, the transition functions for such a thing should live in PGL(n), so Azumaya algebras are classified by H^1(PGL(n)). There is a connecting homomorphism to H^2(O^*) since O^* is the center of GL(n); this is injective by “theorem 90” and surjective…usually? always? I don’t know.

(I don’t know if the algebraic geometry will be of any use to either of you in the details though, but I’m glad you think the question it motivates is interesting.)

I don’t know of any operation pic x pic -> units, and I don’t think there is one. You’re asking for something that takes two line bundles and gives a number; this is impossible to do analytically on something like a Riemann surface.

Posted by: D. on April 6, 2006 10:42 PM | Permalink | Reply to this

Re: Brauer group Br(X)

(No fancy markup, have no time)

Must say something on the Brauer group:

The coboundary map in nonabelian cohomology (H^1(PGL(n)), here) from H^1 to H^2 maps the Brauer group into the torsion part of H^2(X,O*) (called Br’(X) = cohomological Brauer group). There is a conjecture by Grothendieck that this is true for schemes. It is not so, but is probably true for separated schemes (i.e. without doubled points - the `simple’ counter-example is two double cones (i.e. light-cone shape) glued together along their affine parts, giving a (double) cone with a doubled vertex - will post reference later when I have time)

It is true (topologically) for finite CW complexes (and hence smooth finite dim mflds), complex tori, complex curves (trivially), K3 surfaces, affine schemes and a *short* list of other objects (taking holomorphic/complex analytic coefficients where necessary). It is not known even for surfaces in general, though it is true for separated, *mumble* surfaces (I cannot remember the condition, but it something nice).

The link to 0- 1- and 2-dimensional cohomology is (IMHO) due to the fact higher cohomology can be defined by suspending the coefficients an appropriate number of times and taking H^1.

Must dash,

Posted by: David Roberts on April 7, 2006 6:40 AM | Permalink | Reply to this

Re: Picard and Brauer 2-Groups

“Azumaya algebras are classified by H^1(PGL(n)). There is a connecting homomorphism to H^2(O^*) since O^* is the center of GL(n); this is injective by “theorem 90” and surjective…usually? always? I don’t know.”

I think I wrote this too hastily; all the right ingredients are there but put together it’s sort of gibberish.

Posted by: D. on April 7, 2006 6:47 AM | Permalink | Reply to this

Re: Picard and Brauer 2-Groups

Many thanks to everybody for the intersting input! I am about to catch a train to Vietri so can’t reply in detail for the next week. But I’ll think about this stuff.

Posted by: urs on April 7, 2006 9:58 AM | Permalink | Reply to this

Re:Br(X) again

I meant: the coboundary/connecting map in cohomology induces an injective map of Br(X) to the torsion subgroup of second cohomology with O* coeffs. Sujectivity of this map is the conjecture

A. Grothendieck, Le groupe de Brauer I, II, III, in Dix exposes sur la cohomologie des schemas

He wonders in part I.2 whether a result similar to Serre’s on the case of a finite CW complex holds in a general locally ringed topos (Ouch!)

Some references with some summary of known results(not necessarily the best- I’ve only just gotten into this)

THE BRAUER GROUP OF ANALYTIC K3 SURFACES, D. Huybrechts and S. Schroer, math.AG/0305101

TOPOLOGICAL METHODS FOR COMPLEX-ANALYTIC
BRAUER GROUPS, S. Schroer, math.AG/0405223

for a couple of recent ones

Posted by: David Roberts on April 8, 2006 7:16 AM | Permalink | Reply to this

Re: Pic x Pic \to units

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I don’t know of any operation pic x pic - $\rangle$ units, and I don’t think there is one. You’re asking for something that takes two line bundles and gives a number; this is impossible to do analytically on something like a Riemann surface.

This is a stupid question, but how about the interpretation of the Picard group in terms of divisors? I realise divisors on a Riemann surface are “points”, so we can’t go for intersection number or anything like we could on a complex surface.

Comments?

Posted by: David Roberts on April 11, 2006 6:21 AM | Permalink | Reply to this

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April 5, 2006