n-group Representation Theory - (part 2: the Poincaré 2-group)

February 14, 2008

n-group Representation Theory - (part 2: the Poincaré 2-group)

Posted by Jeffrey Morton under 2-groups, category theory, higher dimensional algebra, physics, representation theory

It’s been a while since I wrote the last entry, on representation theory of n-groups, partly because I’ve been polishing up a draft of a paper on a different subject. Now that I have it at a plateau where other people are looking at it, I’ll carry on with a more or less concrete description of the situation of a 2-group. For higher values of $n$ , describing things concretely would get very elaborate quite quickly, but interesting things already happen for $n=2$ . In particular, the case that I gave the talk about, a while back, was mostly the Poincaré 2-group, since this is the one Crane, Sheppeard, and Yetter talk about, and probably the one most interesting to physicists. It was first described by John Baez.

So what’s the Poincaré 2-group? To begin with, what’s a 2-group again?

I already said that a 2-group $\mathbb{G}$ is a 2-category with only one object, and all morphisms and 2-morphisms invertible. That’s all very good for summing up the representation theory of $\mathbb{G}$ as I described last time, but it’s sometimes more informative to describe the structure of $\mathbb{G}$ concretely. A good tool for doing this is a crossed module. (A lot more on 2-groups can be found in Baez and Lauda’s HDA V, and there are some more references and information in this page by Ronald Brown, who’s done a lot to popularize crossed modules).

A crossed module has two layers, which correspond to the morphisms and 2-morphisms of $\mathbb{G}$ . These can be represented as $(G,H,\triangleright, \partial)$ , where $G$ is the group of morphisms in $\mathbb{G}$ , $H$ consists of the 2-morphisms ending at the identity of $G$ (a group under horizontal composition).

There has to be an action $\triangleright : G \rightarrow End(H)$ of $G$ on $H$ (morphisms can be composed “horizontally” with 2-morphisms), and a map $\partial : H \rightarrow G$ (which picks out the source of the 2-morphism). The data $(G,H,\triangleright,\partial)$ have to fit together a certain way, which amounts to giving the axioms for a 2-category.

A handy way to remember the conditions is to realize that the action $\triangleright : G \rightarrow End(H)$ and the injection $\partial : H \rightarrow G$ give ways for elements of $G$ to act on each other and for elements of $H$ to act on each other. These amount to doing first $\triangleright$ and then $\partial$ or vice versa, and both of these must amount to conjugation. That is:

$\partial(g \triangleright h) = g (\partial h) g^{-1}$

and

$(\partial h_1) \triangleright h_2 = h_1 h_2 h_2^{-1}$

Both of these are simplified in the case that $\partial$ maps everything in $H$ to the identity of $G$ - in this case, $H$ can be interpreted as the group of 2-automorphisms of the identity 1-morphism of the sole object of $\mathbb{G}$ . In this case, by the Eckmann-Hilton argument (the clearest explanation of which that I know being the one in TWF Week 100) it turns out that $H$ has to be commutative, so the first condition is trivial since $\partial h = 1$ , and the second is trivial since it follows from commutativity. This simpler situation is known as an automorphic 2-group.

In any case, given a 2-group represented as a crossed module, automorphic or not, the collection of all morphisms can be seen as a group in itself - namely the semidirect product $G \ltimes H$ , which is to say $G \times H$ with the multiplication $(g_1,h_1) \cdot (g_2,h_2) = (g_1 g_2 , g_2 \triangleright h_1 h_2)$ . “What?” you may ask, or maybe “Why?”

Maybe a concrete example would help, since we’d like one anyway: the Poincaré 2-group, which is an automorphic 2-group. There are versions of various signatures $(p,q)$ , in which case $G = SO(p,q)$ , and $H = \mathbb{R}^{p+q}$ .

The group $G$ , then, consists of metric-preserving transformations of Minkowski space $R^{p+q}$ with the metric of signature $(p,q)$ - rotations and boosts (if any). The (abelian) group $H$ consists of translations of this space - in fact, being a vector space, it’s just a copy of it. Between them, they cover the basic types of transformation. Thinking of the translations as having a “projection” down to the identity rotation/boost may seem a bit artificial, except insofar as translations “don’t rotate” anything. More obvious is that rotations or boosts act on translations: the same translation can look differently in rotated/boosted coordinate systems - that is, to different observers.

So where does the Poincaré group $SO(p,q) \ltimes \mathbb{R}^{p+q}$ come in? It’s the group of all metric-preserving transformations of Minkowski space, and is built from these two types: but how?

Well, the vector space $H = \mathbb{R}^{p+q}$ is the group of transformations of the identity Lorentz transformation $1 \in G = SO(p,q)$ , since the map $\partial : H \rightarrow G$ is trivial. But suppose that there is another copy of $H$ over each point in $G$ . Then we have the set of points $G \times H$ , but notice that to talk about this as a group, we’d want a way to act on an element $h_1$ of one copy of $H$ over $g_1 \in G$ by another $h_2$ over $g_2$ . The obvious way is to just treat the whole set as a product of groups, but this misses the fundamental relation between $G$ and $H$ , which is that $G$ can act on $H$ , just as morphisms can act on 2-morphisms by “whiskering with the identity”. (Via Google books, here is the description of this in MacLane’s Categories for the Working Mathematician).

Concretely, this is the fact that there is a sensible way for both parts of $(g_1,h_1)$ to affect the $h_2$ , so we can say $(g_2,h_2) \cdot (g_1,h_1) = (g_2 g_1, g_1 h_2 + h_1)$ (using additive notation for translations, since they’re abelian). The point is that the first rotation we do, $g_1$ , changes coordinates, and therefore the definition of the translation $h_2$ .

So that’s the construction of the Poincaré group from the Poincaré 2-group. What would be nice would be to have some clear description of some higher analog of Minkowski space where it makes sense to say the Poincaré 2-group acts as a 2-group. I don’t quite know how to set this up, but if anyone has thoughts, it would be interesting to hear them.

One reason is that, when describing representations of the 2-group, there’s an important role for spaces (or at least sets) with an action of the group $G$ - which raises questions like whether there’s a role for 2-spaces with 2-group actions in representation theory of higher $n$ -groups. Again - I don’t really know the answer to this. However, in Part 3 I’ll describe concretely how this works for 2-groups, and particularly the Poincaré 2-group.

7 Responses to “n-group Representation Theory - (part 2: the Poincaré 2-group)”

John Armstrong Says:
February 14, 2008 at 11:53 pm
Well, how does a 2-group act? (not being rhetorical, actually trying to work this out)

Group actions are pretty easy to understand once you get a categorical view. Any object $X$ in any category whatsoever comes with a group attached to it: the group of automorphisms $\mathrm{Aut}(X)$ . Then a group action is a homomorphism $G\to\mathrm{Aut}(X)$ .

So, any object in a 2-category should come with a 2-group $\mathbb{A}\mathrm{ut}(X)$ . Then a 2-group action should be a 2-group homomorphism $\mathbb{G}\to\mathbb{A}\mathrm{ut}(X)$ . Right?

Minkowski space is an object in the category of semi-Riemannian spaces. The question we need to start with is: what 2-category should we look in to find 2-Minkowski space?
Urs Schreiber Says:
February 15, 2008 at 10:50 am
Once every month, I spend an afternoon wrecking my brain in an attempt to figure out what it might really mean that one of the most important groups in physics, the Poincaré group, happens to be a semidirect product with an abelian factor, and hence secretly a 2-group.

It might just be a coincidence, arranged by the gods of math to trap us in believing there is something to be found here, thus keeping us from finding the real gems.

Or it might not.

Possibly one big hint that it is not just a trap is the Lie 3-algebra of supergravity:

this is a (super) Lie 3-algebra which has the Poincaré Lie algebra in degree 1, has some fermions in degree 1/2 and has u(1) in degree 3.

When you look at how this beast works, you see that it really “wants” to make us pair its fermions with its translation generators to degree 3/2 generators.

But somehow things remain weird. One gets the feeling everything would find a more natural home if we’d pair the translation generators to degree 2-elements. Hence: if we’d make explicit the fact that the Poincaré Lie algebra sitting in that superghravity Lie 3-algebra is itself already secretly a Lie 2-algebra.

I once talked in more detail about that in the entry 2-Palatini.

I am still not entirely sure what all this is trying to tell us, though.

One other tantalizing aspect one sees here is this: passing to Lie n-groups and their Lie n-algebras leads us to the world of graded algebras. The grading being the categorical grading of higher morphisms.

So even though we just categorify and never necessarily superify, things begin to look super. In fact, lots of people think of Lie n-algebras in terms of “supermanifolds” with an odd nilpotent vector field (called “NQ-manifolds”).

That makes me wonder. Is superification maye secretly just another aspect of categorification? Is the fact that the supergravity Lie 3-algebra involves the super-Poincaré Lie 1-algebra related to the fact that we forgot to realize the Poincaré Lie algebra here as a Lie 2-algebra?

This may sound like vain speculation. But actually there are strong indications that something deeper is indeed going on here:

namely we know that the supergravity Lie 3-algebra is actually just one third of a triple, whose other two thirds are two “Chern” Lie 3-algebras which Hisham, Jim and myself describe in L-infinity connections.

It is known in supergravity that all three of these Lie 3-algebras merge to form the Lie 3-algebra which is the true structure Lie 3-algebra of 11-dimensional supergravity. (We talked about that a bit here).

There must be one single Lie 3-algebraic structure which crucially merges the Poincaré Lie algebra with its superification and various Chern-Simons terms. And it involves e8.

Solving this mystery would be important. And would probably shed light on the true meaning of the Poincaré Lie algebra in the context of higher Lie n-algebras.
Jeffrey Morton Says:
February 15, 2008 at 9:32 pm
Urs: That’s a whole lot of fascinating-sounding stuff, most of which is hovering in the vicinity of over my head. I do think this business about treating the categorical levels in an n-group or n-algebra as the grading of a graded algebra sounds like a promising idea. I’ll have to think about that some more, but until I can figure out more of what you said, I’ll attack this question from nearer the beginning.

John: To start with, if there’s to be something called 2-Minkowski space, it should probably be in some 2-category of categories (the analog of a $G$ -set being a $\mathbb{G}$ -category). But more specifically, I’m expecting to see a 2-space, probably in the sense that Toby Bartels uses in his thesis on 2-bundles. Namely, categories internal in $\mathbb{Top}$ , or some other category of spaces. So 2-Minkowski space $\mathbb{M}$ , if there is such a thing, should be a category with a topological space of objects and another one of morphisms, though it might be necessary to internalize it in some more restrictive category (e.g. vector spaces with specified 2-form, etc. etc.)

But its symmetry 2-group ought to be $\mathbb{Aut(M)}$ - assuming the Poincaré 2-group will be precisely the full symmetry 2-group, which of course might not be achievable. In particular, the putative $\mathbb{M}$ should have an $SO(p,q)$ worth of functors $F : \mathbb{M} \rightarrow \mathbb{M}$ , and for each of these an $R^{p+q}$ worth of natural transformations.

Here’s a stab at defining $\mathbb{M}$ . We’re talking about a category internal to semi-Riemannian vector-spaces. Its space of objects is Minkowski space $M$ , and its morphisms are all automorphisms - at any point $x \in M$ , the group $Aut(x)$ is the tangent space to $x$ , which is a group by its vector space addition. So the total space of morphisms is the trivial bundle $M \times M$ .

Then a functor $F : \mathbb{M} \rightarrow \mathbb{M}$ will give a linear map from $M$ to itself respecting the form - i.e. a Lorentz transformation. The morphism maps ought to respect the tangent space structure - it’s not so clear to me that this has to happen, so maybe the definition is off. At any rate if they do, then the morphism maps are completely determined by the same Lorentz transformation that gives the object map.

Then a natural transformation can only go from a functor to itself, since there are only automorphisms to choose from. Then a nat. trans. $\phi$ will be a choice, at each $x \in M$ , of a tangent vector $\phi(x)$ , such that the naturality condition holds - which basically means that choosing a tangent vector at any point determines the choice for any other point on the same mass shell (i.e. taken to it by any Lorentz transformation), and since the map has to be linear, the choice on one mass shell should determine all the others.

So this seemed right at first, but I’m not sure what condition constrains the functors to respect the fact that the bundle is a *tangent* bundle, so there ought to be something more.
Urs Schreiber Says:
February 16, 2008 at 12:17 pm

I do think this business about treating the categorical levels in an n-group or n-algebra as the grading of a graded algebra sounds like a promising idea.

Just in case, I should maybe clarify that this is more than a promising idea. This is a known fact.

As John and Alissa discuss in HDA VI, semistrict Lie n-algebras are the same as n-term L_oo algebras.

These, in turn, are the same (at least when they are finite dimensional) as “quasi free” differential graded algebras, which in degree k have precisely the generators dual to the stuff that the original L-oo algebra had in degree k, which in turn is nothing but the tangents to the space of k-morphisms starting at the identity in the Lie n-group integrating the given Lie n-algebra.

There is a sub-society of mathematical physicists who meet and talk about “cohomological vector fields” and “NQ supermanifolds” and stuff like that all the time. Secretly, they have found the theory of Lie n-algebras in that dual formulation.
Jeffrey Morton Says:
February 16, 2008 at 7:38 pm
Thanks, Urs - I should take another look at HDA VI at some point and clear that stuff up. When I was making that stab at defining 2-Minkowski space above I found I was automatically thinking about tangents to spaces of morphisms, so there’s bound to be some relation.

Plus: mind you - not all known facts are also promising. This fact seems like both.
David Corfield Says:
February 17, 2008 at 1:14 pm
Over here we discovered that the Poincaré was a sub-2-group of the 2-group of symmetries of a certain Baez-Crans 2-vector space.
Jeffrey Morton Says:
February 17, 2008 at 7:59 pm
David: Thanks! That’s great. I did see that it would not be the full symmetry group of what I was defining, since you have to add restrictions to get it. This doesn’t seem to be quite the same thing, since, if I track the conversation over there correctly, over there you’re talking about the general linear group of a BC 2-vector space without any particular structure, so we could say the Poincaré 2-group is the sub-2-group of the whole symmetry 2-group which preserves some structure.

That general linear group does fill in the extra freedom to violate the “tangent space” structure of morphisms, since nothing in the fact of being a BC 2-vector space requires that it be respected. So its objects would be $GL(n_1) \times GL(n_2)$ , or in the special case I was talking about, just $GL(p+q) \times GL(p+q)$ . Here, I tried to put in the semi-Riemannian structure up front, but the way I did it still didn’t seem quite enough to get exactly Poinc. and nothing else. If we add the requirement that the metric form be preserved, then the group of objects would be $SO(p,q) \times SO(p,q)$ …

So maybe to get Poinc. as the full 2-group of symmetries, we need to be in some setting that requires the tangent-space structure of the fibres of morphisms over the base to be preserved. Or maybe one really ought to just deal with the bigger 2-group anyway.

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n-group Representation Theory - (part 2: the Poincaré 2-group)

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