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February 9, 2005

Categorified Gauge Theory and M-Theory

Posted by Urs Schreiber

Point particles are described by nonabelian gauge theory. We can lift this by stringifying/categorifying and realize the point particles as boundaries of strings described by abelian 2-bundles. We can lift this again to M-theory where the strings become membranes attached to 5-branes. By all what is known these membranes are described by abelian 3-bundles with their boundaries coupled to nonabelian 2-bundles.

On the physics side we went up from 0-branes to 1-branes to 2-branes. On the math side we went from 0-categories (sets) to 1-categories to 2-categories, i.e. from groups to 2-groups to 3-groups.

This is curious for the following reason: We know that on the physics side there is no further step. While there do exist p-branes with p>2 these are not fundamental like the F-string and the M2 are, and – more importantly for the above pattern – there is no abelian 3-brane whose boundary would be a nonabelian membrane (as far as I am aware at least).

So what happens when we increase on the math side the dimension ever more? Can we get gauge theories for abelian 15-branes whose boundaries are nonabelian 14-branes?

I think the answer is: ‘NO’. More precisely, I think the following is true:

- 1-gauge theory admits arbitrary gauge groups G

- 2-gauge theory admits gauge groups coming from crossed modules subject to the constraint of vanishing fake curvature

- 3-gauge theories admit gauge groups coming from crossed modules of crossed modules which are fake flat at the level of 2-morphism and abelian at the level of 3-morphisms

- necessarily all higher (p>3 )-gauge theories are abelian at the p-morphism level

This looks interesting, because it sort of solves the above puzzle: The membrane is the endpoint in a chain of ‘stringifications’ and the gauge theory it couples to is an endpoint in a chain of categorifications in the sense that beyond it no nonabelian gauge theory is possible.

Here is a more precise statement of what I am saying together with what I think is the proof for it.

I am considering strict 3-groups. These are strict 2-categories G together with a 2-functor


that has the usual commuting diagrams defining a group product.

I am claiming that this ‘horizontal product’ is the group product in


where G, H and J are groups and in particular J is an abelian group.

Let’s see how this works: Being a (strict) 2-functor the operation is a 1-functor on objects and 1-morphisms of our 3-group, which imposes the structure of a 2-group on these. Similarly 1-morphisms and 2-morphisms must form a 2-group.

This implies that any 2-morphism in our 3-group can be labeled by a triple


with gG, hH and jJ such that the product 2-functor acts as follows:

(4)((g,h),j)((g ,h ),j )=((g,h) 1 (g ,h ),jα 2 ((g,h))(j ))=((gg ,hα 1 (g)(h )),jα 2 ((g,h))(j )).

First of all it follows that

Proposition 1

we have a complex

(5)Jt 2 Ht 1 G.


(First of all let’s change the numbering systems. The objects in a group can be thought of as 1-morphism. Than the morphisms of a 2-group are 2-morphisms and those of a 3-group are 3-morphisms. That’s the numbering I’ll use in the following.)

A 3-morphism

(6)((g,h),j):(g,h)(g,h )=t 2 (j)(g,h)

in our 3-group goes between two 2-morphisms

(7)(g,h):gt 1 (h)g


(8)(g,h ):gt 1 (h )g=t 1 (h)g

which share the same source 1-morphism g and target 1-morphism t 1 (h)g=t 1 (h )(g) such that

(9)t 1 (h)=t 1 (h ).

Thus we have

(10)t 2 (j)=(g,h) 1 (g,h) 1 =(g,h) 1 (g 1 ,α 1 (g 1 )(h 1 ))=(1 ,hh 1 )hh 1

and hence

(11)t 1 (t 2 (j))=t 1 (hh 1 )=t 1 (h)t 1 (h 1 )=1

by the above considerations.

That in fact the third group always has to be abelian is a consequence of one of the exchange laws in the 3-group.

There are three different products in a 3-group, one for each direction in 3 and we should invent some convenient notation to keep track of them.

The ‘horizontal product’ defined by the 2-functor in the 3-group has already been denoted ‘’.

The 2-morphisms in the 3-group form a 2-group and in the context of 2-groups I always denote their composition by ‘’. This could be called the frontward product in the 3-group (but the vertical product with respect to the 2-group).

Finally there is ordinary composition of 3-morphisms. This is the vertical product in the 3-group and luckily I will not need it in the following discussion so that I won’t make up notation for it.

Proposition 2. The action of the frontward product on 3-morphisms is given by

(12)((g,h),j)((t 1 (h)g,h ),j )=((g,h h),j α 2 (1 ,h )(j)).


First of all we can identically write

(13)((g,h),j)((t 1 (h)g,h ),j )=(((1,1 ),1 )((g,h),j))(((1 ,h ),j )((t 1 (h)g,1 ),1 )).

Using the exchange law this is rearranged to

(14)=(((1,1 ),1 )((1 ,h ),j ))(((g,h),j)((t 1 (h)g,1 ),1 )).

Here the only vertical composition left is that with trivial 3-morphisms between trivial 2-morphism which must be

(15)=((1 ,h ),j )((g,h),j)=((g,h h),j α 2 (1 ,h )(j)).

This can now be used to show that

Proposition 3

The group J in the product (GH)J must be abelian.


Use proposition 2 in two ways to compute

(16)(((1,1 ),j)((1,1 ),1 ))(((1,1 ),1 )((1,1 ),j )).

Performing the product operation as indicated yields

(17)=((1,1 ),jj ).

Using the exchange law first to reorder the products gives

(18)=((1,1 ),j j).

It follows thst jj =j j for all j and j .

Of course this is a variation of the theme of the well-known Eckmann-Hilton argument.

Posted at February 9, 2005 4:37 AM UTC

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