Categorified Gauge Theory and M-Theory

Posted by Urs Schreiber

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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 $p \gt 2$ 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 $(p \gt 3)$ -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.

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

(1)

\cdot : G \times G \to G

that has the usual commuting diagrams defining a group product.

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

(2)

\left( \left(G \ltimes H \right) \ltimes J \right)

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

(3)

((g,h),j)

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

(4)

((g,h),j) \cdot ((g^\prime,h^\prime),j^\prime) = \left( \left(g,h\right) \cdot_1 \left(g^\prime,h^\prime\right), j \, \alpha_2 \left( \left(g,h\right) \right) \left( j^\prime \right) \right) = \left( \left( gg^\prime, h \alpha_1\left(g\right)\left(h^\prime\right) \right), j \, \alpha_2 \left( \left(g,h\right) \right) \left( j^\prime \right) \right) \,.

First of all it follows that

Proposition 1

we have a complex

(5)

J \overset{t_2}{\rightarrow} H \overset{t_1}{\rightarrow} G \,.

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

(6)

\left(\left(g,h\right),j\right) : \left(g,h\right) \rightarrow \left(g,h^\prime\right) = t_2\left(j\right)\, \left(g,h\right)

in our 3-group goes between two 2-morphisms

(7)

(g,h) : g \rightarrow t_1(h)\,g

and

(8)

(g,h^\prime) : g \rightarrow t_1(h^\prime)g = t_1(h)g

which share the same source 1-morphism $g$ and target 1-morphism $t_1(h)g = t_1(h^\prime)(g)$ such that

(9)

t_1(h) = t_1(h^\prime) \,.

Thus we have

(10)

t_2(j) = (g,h')\cdot_1 (g,h)^{-1} = (g,h') \cdot_1 (g^{-1}, \alpha_1(g^{-1})(h^{-1})) = (1, h'h^{-1}) \cong h'h^{-1}

and hence

(11)

t_1\left( t_2 \left( j \right) \right) = t_1\left( h'h^{-1} \right) = t_1\left(h'\right) t_1\left(h^{-1}\right) = 1

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

(12)

\left( \left( g,h \right), j \right) \circ \left( \left( t_1(h)g,h^\prime \right), j^\prime \right) = \left( \left( g,h^\prime h \right), j^\prime \, \alpha_2\left(1,h^\prime\right)\left(j\right) \right) \,.

Proof.

First of all we can identically write

(13)

\left( \left( g,h \right), j \right) \circ \left( \left( t_1(h)g,h^\prime \right), j^\prime \right) = \left( \left( \left( 1,1 \right), 1 \right) \cdot \left( \left( g,h \right), j \right) \right) \circ \left( \left( \left( 1,h^\prime \right), j^\prime \right) \cdot \left( \left( t_1(h)g,1 \right), 1 \right) \right) \,.

Using the exchange law this is rearranged to

(14)

\cdots = \left( \left( \left( 1,1 \right), 1 \right) \circ \left( \left( 1,h^\prime \right), j^\prime \right) \right) \cdot \left( \left( \left( g,h \right), j \right) \circ \left( \left( t_1(h)g,1 \right), 1 \right) \right) \,.

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

(15)

\cdots = \left( \left( 1,h^\prime \right), j^\prime \right) \cdot \left( \left( g,h \right), j \right) = \left( \left( g,h^\prime h \right), j^\prime \, \alpha_2\left(1,h^\prime\right)\left(j\right) \right) \,.

$\square$

This can now be used to show that

Proposition 3

The group $J$ in the product $(G\ltimes H)\ltimes J$ must be abelian.

Proof.

Use proposition 2 in two ways to compute

(16)

\left( \left(\left(1,1\right),j\right) \circ \left(\left(1,1\right),1\right) \right) \cdot \left( \left(\left(1,1\right),1\right) \circ \left(\left(1,1\right),j^{\prime}\right) \right) \,.

Performing the product operation as indicated yields

(17)

\cdots = \left(\left(1,1\right), j j^{\prime}\right) \,.

Using the exchange law first to reorder the products gives

(18)

\cdots = \left(\left(1,1\right), j^{\prime} j\right) \,.

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

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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February 9, 2005