### 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 $\cdot $ 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 $g\in G$, $h\in H$ and $j\in J$ such that the product 2-functor acts as follows:

First of all it follows that

**Proposition 1**

we have a complex

*Proof.*

(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

in our 3-group goes between two 2-morphisms

and

which share the same source 1-morphism $g$ and target 1-morphism ${t}_{1}(h)g={t}_{1}({h}^{{\textstyle \prime}})(g)$ such that

Thus we have

and hence

by the above considerations. $\square $

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 ${\mathbb{R}}^{3}$ 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 ‘$\cdot $’.

The 2-morphisms in the 3-group form a 2-group and in the context of 2-groups I always denote their composition by ‘$\circ $’. 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 $\circ $ on 3-morphisms is given by

*Proof.*

First of all we can identically write

Using the exchange law this is rearranged to

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

$\square $

This can now be used to show that

**Proposition 3**

The group $J$ in the product $(G\u22c9H)\u22c9J$ must be abelian.

*Proof*.

Use proposition 2 in two ways to compute

Performing the product operation as indicated yields

Using the exchange law first to reorder the products gives

It follows thst $j{j}^{{\textstyle \prime}}={j}^{{\textstyle \prime}}j$ for all j and ${j}^{{\textstyle \prime}}$. $\square $

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