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 $\left(p>3\right)$-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

(1)$\cdot :G×G\to G$

that has the usual commuting diagrams defining a group product.

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

(2)$\left(\left(G⋉H\right)⋉J\right)$

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)$\left(\left(g,h\right),j\right)$

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

(4)$\left(\left(g,h\right),j\right)\cdot \left(\left({g}^{\prime },{h}^{\prime }\right),{j}^{\prime }\right)=\left(\left(g,h\right){\cdot }_{1}\left({g}^{\prime },{h}^{\prime }\right),j\phantom{\rule{thinmathspace}{0ex}}{\alpha }_{2}\left(\left(g,h\right)\right)\left({j}^{\prime }\right)\right)=\left(\left({\mathrm{gg}}^{\prime },h{\alpha }_{1}\left(g\right)\left({h}^{\prime }\right)\right),j\phantom{\rule{thinmathspace}{0ex}}{\alpha }_{2}\left(\left(g,h\right)\right)\left({j}^{\prime }\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

First of all it follows that

Proposition 1

we have a complex

(5)$J\stackrel{{t}_{2}}{\to }H\stackrel{{t}_{1}}{\to }G\phantom{\rule{thinmathspace}{0ex}}.$

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)$\left(\left(g,h\right),j\right):\left(g,h\right)\to \left(g,{h}^{\prime }\right)={t}_{2}\left(j\right)\phantom{\rule{thinmathspace}{0ex}}\left(g,h\right)$

in our 3-group goes between two 2-morphisms

(7)$\left(g,h\right):g\to {t}_{1}\left(h\right)\phantom{\rule{thinmathspace}{0ex}}g$

and

(8)$\left(g,{h}^{\prime }\right):g\to {t}_{1}\left({h}^{\prime }\right)g={t}_{1}\left(h\right)g$

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

(9)${t}_{1}\left(h\right)={t}_{1}\left({h}^{\prime }\right)\phantom{\rule{thinmathspace}{0ex}}.$

Thus we have

(10)${t}_{2}\left(j\right)=\left(g,h\prime \right){\cdot }_{1}\left(g,h{\right)}^{-1}=\left(g,h\prime \right){\cdot }_{1}\left({g}^{-1},{\alpha }_{1}\left({g}^{-1}\right)\left({h}^{-1}\right)\right)=\left(1,h\prime {h}^{-1}\right)\cong h\prime {h}^{-1}$

and hence

(11)${t}_{1}\left({t}_{2}\left(j\right)\right)={t}_{1}\left(h\prime {h}^{-1}\right)={t}_{1}\left(h\prime \right){t}_{1}\left({h}^{-1}\right)=1$

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 ${ℝ}^{3}$ 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)$\left(\left(g,h\right),j\right)\circ \left(\left({t}_{1}\left(h\right)g,{h}^{\prime }\right),{j}^{\prime }\right)=\left(\left(g,{h}^{\prime }h\right),{j}^{\prime }\phantom{\rule{thinmathspace}{0ex}}{\alpha }_{2}\left(1,{h}^{\prime }\right)\left(j\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

Proof.

First of all we can identically write

(13)$\left(\left(g,h\right),j\right)\circ \left(\left({t}_{1}\left(h\right)g,{h}^{\prime }\right),{j}^{\prime }\right)=\left(\left(\left(1,1\right),1\right)\cdot \left(\left(g,h\right),j\right)\right)\circ \left(\left(\left(1,{h}^{\prime }\right),{j}^{\prime }\right)\cdot \left(\left({t}_{1}\left(h\right)g,1\right),1\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

Using the exchange law this is rearranged to

(14)$\cdots =\left(\left(\left(1,1\right),1\right)\circ \left(\left(1,{h}^{\prime }\right),{j}^{\prime }\right)\right)\cdot \left(\left(\left(g,h\right),j\right)\circ \left(\left({t}_{1}\left(h\right)g,1\right),1\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

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

(15)$\cdots =\left(\left(1,{h}^{\prime }\right),{j}^{\prime }\right)\cdot \left(\left(g,h\right),j\right)=\left(\left(g,{h}^{\prime }h\right),{j}^{\prime }\phantom{\rule{thinmathspace}{0ex}}{\alpha }_{2}\left(1,{h}^{\prime }\right)\left(j\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

$\square$

This can now be used to show that

Proposition 3

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

Proof.

Use proposition 2 in two ways to compute

(16)$\left(\left(\left(1,1\right),j\right)\circ \left(\left(1,1\right),1\right)\right)\cdot \left(\left(\left(1,1\right),1\right)\circ \left(\left(1,1\right),{j}^{\prime }\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

Performing the product operation as indicated yields

(17)$\cdots =\left(\left(1,1\right),j{j}^{\prime }\right)\phantom{\rule{thinmathspace}{0ex}}.$

Using the exchange law first to reorder the products gives

(18)$\cdots =\left(\left(1,1\right),{j}^{\prime }j\right)\phantom{\rule{thinmathspace}{0ex}}.$

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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