### Why does 2-group theory miss surface holonomies with nonvanishing B+F?

#### Posted by Urs Schreiber

Recall that there are (at least) three roads to well-defined surface holonomies in loop space formalism, but apparently only one road in 2-group theory, namely that called $B+{F}_{A}=0$.

What happens to the other two?

I believe the answer is: One of them is in some sense a special case of the abelian theory and hence not strictly missed. But the other one is missed by 2-group theory because of its inherently finite, non-differential, nature. It could be incorporated if we allowed ‘lattice artifacts’ in 2-group theory that vanish in a ‘continuum limit’ where elementary morphisms become infitesimal, which amounts to considering some sort of ‘*weak*’ 2-group in some sense.

I think so for the following reason:

Consider a ‘tesselation’ of a surface where ‘plaquettes’ are labeled by group elements $h$ and edges are labeled by group elements $g$ as in hep-th/0206130 and hep-th/0309173.

For notational simplicity assume that both labels take values in the same group.

We want to multiply all the $h$ labels together to obtain the full surface holonomy. But the ordering of these will matter. For $h$ labels sitting next to each other vertically they can be just multiplied like beads on a string using the geometrically induced order.

Labels horizontally next to each other have to be inserted into this string of beads. In order to move them around we need some parallel transport, namely that provided by the group elements of the edge labels. Parallel transporting some surface label $h$ from target to source of some edge $g$ gives $gh{g}^{-1}$.

But this zipper must have the special properties that it does not matter *where* beads are inserted, otherwise the surface holonomy would be ill-defined.

Therefore consider a plaquette ${f}_{1}$ with an edge ${g}_{1}$ going along its upper and an edge ${g}_{2}$ along its lower boundary and with a surface element ${f}_{2}^{{\textstyle \prime}}$ to its right (where I use labels as on p. 5 of hep-th/0206130).

Either we move ${f}_{2}^{{\textstyle \prime}}$ to the upper boundary of ${f}_{1}$ by parralel transporting it with ${g}_{1}$, or to its lower boundary by using ${g}_{2}$. The resulting string of beads must be the same in both cases, which means that the crucial condition

must be satisfied.

Here I have used non-standard wording, because I think the zipper-process going on here is instructive, but it can be checked that this equation is equivalent to that found in 2-group literature saying that the order of horizontal and vertical composition must not matter.

So given the group labels of our graph and the above rule to multiply them all up, what one needs to do in order to obtain the most general well-defined surface holonomy is find *all* solutions of $(\star )$

*One* way to solve this equation is that used in 2-group theory, namely that obtained by choosing arbitrary edge labels and setting the surface labels equal to the edge-holonomy around the surface. This yields $B+{F}_{A}=0$ in the continuum.

Of course there is a much more trivial way to solve $(\star )$: If we choose arbitrary edge labels and let the surface labels take values generated by an abelian ideal of the group algebra, then surface labels may be commuted with everything (but not edge labels among themselves) and $(\star )$ is easily seen to be satisfied. This is one of the cases used in hep-th/9710147.

In fact, the authors of that paper assume in addition the curvature of the edge labels to vanish, which is necessary to have a flat connection on loop space but not necessary in order to have a well-defined surface holonomy in this case.

So this case kind of belongs to the abelian theory, but it is maybe noteworthy that the ege labels need not be gauge equivalent to the trivial set of labels at all for this to give a well-defined surface holonomy. In this sense this does have some genuine non-abelian character and deserves to be mentioned as one interesting solution of $(\star )$.

What about the second condition considered in hep-th/9710147, that where the edge labels have vanishing loop holonomy and surface labels are ‘covariantly constant’ with respect to the edge labels?

This does solve $(\star )$, too, but only if we let the size of the surface elements tend to 0!

Namely in the covariantly constant case we can pick one reference plaquette on the surface and the label for any other surface element is defined to be the parallel transport of that reference label to the given position. The path of the parallel transport does not matter (in the limit of small lattice spacing) due to the assumption that loop holonomies vanish (that the 1-form connection is flat).

But then one can see (I don’t bother to spell the obvious but tiresome details out), that the right hand side of $(\star )$ is essentially the same group-valued function as the left hand side, but translated by one lattice unit. In other words, left and right hand side differ by something proportional to our lattice spacing.

This means that in the continuum limit this does give a well defined surface holonomy (as we already know from the loop space point of view), but it also means that strict 2-group theory does not *see* this possibility to get such a unique holonomy for $B+{F}_{A}\ne 0$.

I think that alone is (unless I screwed up somewhere) an interesting fact. But it also makes me speculate:

The small translation between the left and the right hand side of $(\star )$ just discussed can, due to the covariant constancy of the surface labels, also be interpreted as a short parallel transport obtained by adjoining the group element of the edge translating between the two nearby positions of the left and the right hand side.

This means that both sides of this equation *are* equal *up to isomorphism* in this sense, doesn’t it?

Possibly I am wrong, being a category-theory layman, but this sounds like some ‘weak’ notion of category or something like that.

Any ideas?

Posted at August 24, 2004 5:14 PM UTC
## Re: Why does 2-group theory miss surface holonomies with nonvanishing B+F?

The discussion is taking place at sci.physics.research.

John Baez’ comment can be found either in your newsreader or in the PhysicsForums archive of s.p.r. articles.

With one day delay the whole thing is best followed in Google Groups.

Here is my reply:

____________________________

Many thanks for your reply!

You (John Baez) wrote:

True, I like that best at the moment, because it seems to me that a very interesting class of physical applications corresponds to $G=H=U(N)$.

But I do understand that 2-group theory covers the more general case where $G\ne H$. I think pretty much everything I wrote so far with $G=H$ in mind directly translates to this more general case when you think of all my $B$s as $t(B)$s. But I will make that more precise and explicit in the future.

Meanwhile, since I have the feeling that I didn’t express myself well enough, let me try to rephrase the question that I am currently concerned with:

Alvarez, Ferreira & Sánchez-Guillén in hep-th/9710147 found two classes of consistent surface holonomies which happen to have $G=H$ but $B+F\ne 0$. (I am pretty sure that their construction can be straightforwardly generalized to $G\ne H$ in which case it would give consistent surface holonomy for $t(B)+F\ne 0$. But the case with $G=H$ alone already raises the following questions.)

As far as I can see, there are only two possible reactions to this result:

1) It contains a flaw and 2-group theory is right that only $t(B)+F=0$ gives well defined surface holonomy.

2) It is correct. Then there must be a reason why 2-group theory cannot obtain these surface holonomies with $t(B)+F$ nonvanishing.

I argued that the latter is the case (namely that there are more solutions to the 2-associativity condition than just those with $t(B)+F=0$), but if I am wrong about that please let me know where I went astray.

I would like to know

a) if you think there is a 3rd alternative to 1) and 2) above (maybe that somehow Alvarez, Ferreira & Sánchez-Guillén are secretly speaking about a different notion of surface holonomy than 2-group theory does or something like that)

b) or else, if you think that their result is flawed, what you think the mistake is

c) or finally, if you think their result is correct, how you see it fit together with the results of 2-group theory.

I am not sure that examples where $t$ is nontrivial are of help for answering the question that I am concerned with. But maybe that’s precisely my problem.

In any case, thanks for your help!