## August 24, 2004

### 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 ${f}_{2}^{\prime }$ to its right (where I use labels as on p. 5 of hep-th/0206130).

Either we move ${f}_{2}^{\prime }$ 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

(1)$\left(\star \right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{f}_{1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{g}_{1}{f}_{2}^{\prime }{g}_{1}^{-1}={g}_{2}{f}_{2}^{\prime }{g}_{2}^{-1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{f}_{1}$

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 $\left(\star \right)$

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 $\left(\star \right)$: 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 $\left(\star \right)$ 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 $\left(\star \right)$.

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 $\left(\star \right)$, 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 $\left(\star \right)$ 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 $\left(\star \right)$ 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

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## 11 Comments & 0 Trackbacks

### 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:

When we think about connections in this context, we see the natural equation is not

(1)$B=-F$

(which makes no sense!) but instead

(2)$t\left(B\right)=-F$

where $F$ is a $g$-valued 2-form and $B$ is an $h$-valued 2-form.

We can then consider various special cases.

At one extreme we can have $G$ trivial and $H$ abelian; then the above equation is vacuous. This is what happens in 2-form electromagnetism.

At another extreme we can have $G=H$ and $t$ the identity; then you get $B=-F$. That’s the case you seem to like best.

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

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 $t\left(B\right)$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 $t\left(B\right)+F\ne 0$. 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 $t\left(B\right)+F=0$ 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 $t\left(B\right)+F$ nonvanishing.

I argued that the latter is the case (namely that there are more solutions to the 2-associativity condition than just those with $t\left(B\right)+F=0$), 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.

But there are lots of intermediate cases. Maybe you want some concrete examples? I can manufacture examples, but not very interesting ones if $G$ and $H$ are required to be compact Lie groups, because the Lie algebra of these is always semisimple + abelian, and the options for homomorphisms dt are severely limited.

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!

Posted by: Urs Schreiber on August 25, 2004 10:31 AM | Permalink | PGP Sig | Reply to this

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

The `other’ class is one for which F=0, i.e. A is a flat connection, right? Then B is in an Abelian ideal according to Alvarez et al. (This may not be the only solution of the zero curvature condition, but this is the one they consider.)

I think this fits in the 2-group construction if we take this equation to hold only in the Abelian ideal, and F=0 outside of that ideal.

I do have a related question … what do (local) gauge transformations do for this `other’ class of loop space connections? Their transformation properties may tell you how this class is different.

Posted by: Amitabha on August 25, 2004 12:18 PM | Permalink | Reply to this

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

The ‘other’ class is one for which $F=0$, i.e. $A$ is a flat connection, right? Then $B$ is in an Abelian ideal according to Alvarez et al. (This may not be the only solution of the zero curvature condition, but this is the one they consider.)

This is what these authors consider, because that’s what gives a flat connection on loop space, and that’s what they are looking for. But not every connection on loop space which gives well-defined surface holonomy needs to be flat. The case of abelian $B$ and vanishing $A$ is the simplest example of that. And the case where $B$ is in an abelian ideal and $A$ is arbitrary is the next-simplest example.

That even for $A$ arbitrary a $B$ in an abelian ideal is sufficient to give consistent surface holonomy can be seen either in higher algebra or in loop space formalism. (This is probably obvious to you, but I’ll spell it out here just for the record, because it serves to emphasize the point I want to make.)

In higher-dimensional algebra the condition to be satisfied is the 2-associativity (e.g. p. 8 of Baez’ hep-th/0206130)

(1)$\left({f}_{1}\circ {f}_{1}^{\prime }\right)\cdot \left({f}_{2}\circ {f}_{2}^{\prime }\right)=\left({f}_{1}\cdot {f}_{2}\right)\circ \left({f}_{1}^{\prime }{f}_{2}^{\prime }\right)$

where ‘$\circ$’ is the vertical and ‘$\cdot$’ is the horizontal product. In the case where both groups are the same $G=H$ we have

(2)${f}_{1}\circ {f}_{1}^{\prime }={f}_{1}^{\prime }{f}_{1}$
(3)${f}_{1}\cdot {f}_{2}={f}_{1}\phantom{\rule{thinmathspace}{0ex}}{g}_{1}{f}_{2}{g}_{1}^{-1}$

where simple juxtaposition denotes the ordinary group product.

Insetrting these explicit expression in the above shows that this is equivalent to

(4)$\cdots ⇔{f}_{1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{g}_{1}{f}_{2}^{\prime }{g}_{1}^{-1}={g}_{2}{f}_{2}^{\prime }{g}_{2}^{-1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{f}_{1}$

where all products are ordinary products in the group.

This equation is solved in particular for

(5)${f}_{1}={g}_{2}{g}_{1}^{-1}\phantom{\rule{thinmathspace}{0ex}}.$

This is the standard $B+F=0$ solution.

But there are other solutions, too: If $B$ is in an abelian ideal this means that the surface labels are in the center of the group and commute with everything. In this case we have, for arbitrary edge labels $g$

(6)${f}_{1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{g}_{1}{f}_{2}^{\prime }{g}_{1}^{-1}={f}_{1}{f}_{2}^{\prime }{g}_{1}{g}_{1}^{\prime }={f}_{1}{f}_{2}^{\prime }$

and similarly on the right hand side of the above condition. So one further solution of the 2-associativity condition is the case where $B$ is in an abelian ideal and $A$ is arbitrary.

To quickly see the same result in loop space formalism it suffices to note that for $B$ in an abelian ideal we have

(7)${\int }_{0}^{2\pi }d\sigma \phantom{\rule{thinmathspace}{0ex}}{W}_{A}\left(\sigma \right){B}_{\mu \nu }\left(\sigma \right){W}_{A}^{-1}\left(\sigma \right){X}^{\prime \mu }\left(\sigma \right){\mathrm{dX}}^{\nu }\left(\sigma \right)={\int }_{0}^{2\pi }d\sigma \phantom{\rule{thinmathspace}{0ex}}{B}_{\mu \nu }\left(\sigma \right){X}^{\prime \mu }\left(\sigma \right){\mathrm{dX}}^{\nu }\left(\sigma \right)\phantom{\rule{thinmathspace}{0ex}},$

which leaves us with the same expression as for abelian $B$ and vanishing $A$.

So this does give a well-defined surface holonomy, even though one could argue that it essentially reduces to the case where $B$ is abelian and $A$ vanishes. That’s true, but still it is one non-abelian and nontrivial solution of the 2-associativity condition.

But I think the second class of solutions by Alvarez et al., that using covariantly constant $B$ and flat $A$ is more interesting. In particular this is interesting because it gives a valid surface holonomy but does not solve the above consistency condition of 2-group theory at all! It solves it only up to terms which vanish when all the group elements involved there are close to the identity (i.e. in the continuum limit).

I am in the process of writing out the above 2-associativity condition in differential form, i.e. substituting

(8)$f\to 1+{ϵ}^{2}G$
(9)$g\to 1+ϵA$

and expanding in $ϵ$. This should give the differential 2-associativity condition which I believe should be used in 2-Lie-algebras and which should be solved by the above class of solutions which covariantly constant $B$ and flat $A$.

I think this fits in the 2-group construction if we take this equation to hold only in the Abelian ideal, and $F=0$ outside of that ideal.

Hm, I have thought about this, but I don’t seem to see exactly what you have in mind. Maybe you are talking about what John Baez has mentioned a minute ago on sci.physics.research, namely

Of course it’s possible to have $G=H$ but $t:H\to G$ not equal to the identity! That would be one possible explanation of their work, since this can give $B+\mathrm{dt}\left(F\right)=0$ but $B+F\ne 0$.

But it seems to me that this does not solve the problem either. Of course I could let $t$ be the projection on the center of the group. But since, as I have just tried to discuss, $F$ can be really absolutely arbitrary if $B$ is in an abelian ideal, even this does not seem to help. Or does it?

I do have a related question… what do (local) gauge transformations do for this ‘other’ class of loop space connections? Their transformation properties may tell you how this class is different.

If you are talking about local gauge transformations on loop space then I fully agree that this is a crucial issue. As I have tried to discuss in hep-th/0407122 I think that local gauge transformations on loop space preserve the form

(10)$𝒜={\int }_{0}^{2\pi }d\sigma \phantom{\rule{thinmathspace}{0ex}}{W}_{A}\left(\sigma \right)B\left(\sigma \right){W}_{A}^{-1}\left(\sigma \right)$

of a loop space connection $𝒜$ in terms of a 1-form $A$ and a 2-form $B$ if and only if $B+{F}_{A}=0$. Otherwise after a gauge transformation we will in general be left with an expression that involves for instance two different 1-forms or integrals with two insertions in between Wilson lines ${W}_{A}$ instead of just one.

Therefore even if we take the two extra classes of solutions by Alvarez,Ferreira & Sánchez-Guillén into account, we know that they fit in the scheme of surface holonomy defined using precisely one 1-form and one 2-form only in a special gauge! This is in contrast to the solutions with $B+{F}_{A}=0$. For these the gauge transformations respect this class.

So the solutions of Alvarez et al. are in a sense really living in a theory more general that 2-group theory. They are easily written down in loop space formalism, because loop space 1-form connections can be cooked up from all kinds of ingredients.

So in any case these solutions are borderline with respect to 2-group theory. Still, I think it would be nice to see what is exactly going on.

Posted by: Urs Schreiber on August 25, 2004 2:35 PM | Permalink | PGP Sig | Reply to this

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

I wrote:

I am in the process of writing out the above 2-associativity condition in differential form

Unfortunately I didn’t find the time today and now I have to call it quits for tonight, but it is easy to see what is roughly going on:

I had discussed that the 2-associativity condition (the exchange law) is equivalent (when $G=H$ for simplicity) to

(1)${f}_{1}{g}_{1}{f}_{2}^{\prime }{g}_{1}^{-1}={g}_{2}{f}_{2}^{\prime }{g}_{2}^{-1}{f}_{1}\phantom{\rule{thinmathspace}{0ex}}.$

It is maybe more convenient to rewrite this equivalently by multiplying from the left with ${g}_{2}^{-1}$ and from the right with ${g}_{1}$. This gives

(2)$\cdots ⇔\left({g}_{2}^{-1}{f}_{1}{g}_{1}\right){f}_{2}^{\prime }={f}_{2}^{\prime }\left({g}_{2}^{-1}{f}_{1}{g}_{1}\right)$

where I have put brackets just to emphasize the structure of the result. Namely the terms in brackets are precisely expression (3.22) in hep-th/0309173

(3)${g}_{2}^{-1}{f}_{1}{g}_{1}\approx {e}^{{a}^{2}\left({B}_{1}+{F}_{1}\right)}$

for the left of the two plaquettes, where $a$ is the ‘lattice constant’ as in Girelli&Pfieffer’s paper. Writing ${B}_{2}$ for the value of $B$ in the right plaquette the 2-associativity law hence says that

(4)$\left[{e}^{{a}^{2}\left({B}_{1}+{F}_{1}\right)},{e}^{{a}^{2}{B}_{2}}\right]\approx 0$

where one must take care to expand consistently in powers of $a$, a simple task that unfortunately I don’t have the time for right now, because I am in a hell of a hurry.

But one thing is clear: This condition is solved for $B+F=0$, because then the left part of the commutator becomes just the unit, but does indeed potentially have more solutions. In fact, using BCH again it is pretty obvious that $B$ with values in an abelian ideal does solve this, too, indeed, as we already know. Remains to be seen if we can read off the case with $F=0$ and $B$ covariantly constant. Seems like that requires a more careful analysis.

More tomorrow! :-)

Posted by: Urs Schreiber on August 25, 2004 8:39 PM | Permalink | PGP Sig | Reply to this

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

John Baez’ reply to my above message can be found here. My reply to that in turn is the following:

__________________________________

John Baez wrote:

By the way, there are closely related ways to build a 2-group, that are easy to mix up. It would be unfortunate to think you were working with one when you were really working with the other! I’m not accusing you of doing this, but I just think I should warn you of the danger, since I’m gotten confused myself plenty of times. Here they are:

1) Take $G=H$, let $t:H\to G$ be the identity homomorphism, and let $\alpha$ be the action of $G$ on $H$ be via conjugation:

(1)$\alpha \left(g\right)\left(h\right)={\mathrm{ghg}}^{-1}$

2) Take $G=\mathrm{Aut}\left(H\right)$ be the group of all automorphisms of $H$.

Let $t:H\to G$ be the map sending any element $h$ to the corresponding ‘inner automorphism’ of $H$, that is, the automorphism given by conjugating by $h$:

(2)$t\left(h\right)\left({h}^{\prime }\right)={\mathrm{hh}}^{\prime }{h}^{-1}$

Let $\alpha$ be the obvious action of $G=\mathrm{Aut}\left(H\right)$ as automorphisms of $H$.

Now I am a little confused. I think I do understand these two examples. But aren’t both just two special cases of the general notion of 2-group as discussed in your paper?

- From a pure mathematical viewpoint, construction 2) is actually a lot simpler and more important than construction 1), even though it doesn’t look simpler in the lowbrow way I just described it.

I think I roughly do appreciate the relation to 2-categories that you explain somewhere:

We can think of a group $H$ as a category with a single object where group elements are morphism with source and target all the same single object.

The invertible functors between such group categories are the group automorphisms. These can be taken to be the objects of a category called $\mathrm{Aut}\left(H\right)$.

The morphisms of $\mathrm{Aut}\left(H\right)$ must be morphisms between these functors. Let $g$ be such an automorphism of $H$. Then any element $h$ of $H$ maps $g$ to another automorphism ${g}^{\prime }$ by

(3)${g}^{\prime }\left(x\right):=h\left(g\left({h}^{-1}x\right)\right).$

So the elements of $H$ correspond to morphisms between the $g$.

Since $\mathrm{Aut}\left(H\right)$ of which $g$ are the elements is itself a group and since it morphisms $H$ are a group, too, we are dealing with a 2-category being a group, i.e. a 2-group.

I hope that’s roughly correct. (But it doesn’t affect anything of what I say below.)

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 $t\left(B\right)$s. But I will make that more precise and explicit in the future.

Okay… it may be no big deal for your applications, but it can be…

Ok. If it doesn’t disturb you too much I would be very happy if we could for the moment restrict the discussion to the case where t is the identity and try to clarify the question I am thinking about in this special case first. If and when we think we clarified this special case I believe it will be easy to hit everything in sight with a nontrivial t and see what we get in this more general case.

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 $B+F\ne 0$.

Of course it’s possible to have $G=H$ but $t:H\to G$ not equal to the identity! That would be one possible explanation of their work, since this can give $B+\mathrm{dt}\left(F\right)=0$ but $B+F\ne 0$.

Hm. Over at the SCT Amitabha Lahiri made a possibly similar suggestion. But I am not sure yet if this really helps. See below for more on that.

I don’t know what the ‘2-associativity’ condition is.

I mean the condition that it does not matter whether we first perform the horizontal and then the vertical products, or the other way round. The equation in definition 1 of your paper. Ah, let me see, you call it the ‘exchange law’. Sorry, somehow I thought it should be called 2-associativity.

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

I should note that it’s not ‘2-group theory’ which makes this claim, but a paper by Girelli and Pfeiffer.

Oh, good that you are saying this. This is getting to the heart of the matter. From how Girelli&Pfeiffer quote your paper as a proof for their claim I took it that they are referring to your ‘proposition 5’ as implying this. There you show that in order to get from a crossed module with elements $h$ and $g$ to a 2-group one has to identify 2-morphisms with pairs $\left(h,g\right)$ and set

(4)$\mathrm{source}\left(h,g\right)=g$

and

(5)$\mathrm{target}\left(h,g\right)=\mathrm{hg}$

(as I said, I’ll drop the $t$ that should be appearing here for the time being, if you allow).

But this means that if $\left(h,g\right)$ is the morphism ${g}_{1}\genfrac{}{}{0}{}{h}{\to }{g}_{2}$

we have ${g}_{2}=h{g}_{1}$ and hence

(6)$h=g2g{1}^{-1}.$

The differential version of this, i.e. the second order term in epsilon when you write

(7)$h=1+{ϵ}^{2}B$
(8)$g=1+ϵA$

is

(9)$B+{F}_{A}=0.$

When I read your and Pfeiffer’s papers I derived the same in a similar but somewhat more direct way:

When the 2-associativity law (in definition 1 on p. 8 of your paper) is explicitly written in terms of the definition of the vertical and the horizontal product it is equivalent to

(10)$\left(\star \right){f}_{1}{g}_{1}{f}_{2}^{\prime }{g}_{1}^{-1}={g}_{2}{f}_{2}^{\prime }{g}_{2}^{-1}{f}_{1}\phantom{\rule{thinmathspace}{0ex}}.$

(For more details on what I mean here please see my recent reply to Amitabha at the SCT.

My very point which I tried to make in the comment starting this discussion is that we should try to find all solutions to this equation in order to find all conditions under which we get a consistent surface holonomy from egde and surface group labels.

One can easily see that _one_ solution to $\left(\star \right)$ is

(11)${f}_{1}={g}_{2}{g}_{1}^{-1},$

which is precisely the solution implying $B+F=0$ discussed above. So that’s how this condition comes up.

My point was that there are in fact other solutions of this equation. For instance let the edge labels be arbitrary and restrict the surface labels to take values in an abelian ideal of the group. Then this equation is satisfied, too. This corresponds to one of the classes of surface holonomies that Alverz et al. discuss in terms of loop space formalism.

Moreover, I pointed out that when we go from the condition $\left(\star \right)$ to its infinitesimal/differential version by again setting $g=1+ϵA$ etc., we obtain an equation which admits still one more solution. Namely flat edge holonomies together with covariantly constant surface holonomies. Because it is really this infinitesimal version which is needed to compute surface holonomies in the continuum, this does give a consistent surface holonomy, and indeed it corresponds to the other class of loop space connections discussed by Alvarez et al.

I guess I’ll just have to read their stuff. Of course I wouldn’t mind a wee summary in plain English….

There are a lot of ideas in that paper, but for our discussion really only a couple of them are relevant. I have reviewed and briefly discussed them in the second half of section 3.4 of hep-th/0407122which again is based on this SCT entry:

The idea is to write down a connection $𝒜$ on loop space of the form

(12)$𝒜={\int }_{0}^{2\pi }d\sigma \phantom{\rule{thinmathspace}{0ex}}W\left(\sigma \right){B}_{\mu \nu }\left(\sigma \right){W}^{-1}\left(\sigma \right){X}^{\prime \mu }\left(\sigma \right){\mathrm{dX}}^{\nu }\left(\sigma \right)$

where the integral is over the given loop, $W$ is the holonomy of $A$ along that loop and ${X}^{\prime }$ is the $\sigma$-derivative of the loop $X:\left(0,2\pi \right)\to M$.

This connection gives us a consistent surface holonomy of surfaces which are images of curves in based loop space in particular if it is flat (since then it is independent of the parameterization of the surface in terms of sigma and in terms of the foliation by loops. (Flatness is not necessary, but sufficient for this to be true.)

Alvarez et al. thought they need to set ${F}_{A}=0$. When one does that it is pretty easy to see that there are two cases in which $𝒜$ is flat, namely

1) When $B$ is in an abelian ideal of the common algebra in which $A$ and $B$ take values.

2) When $B$ is covariantly constant with respect to $A$.

These two conditions dont have $B+F=0$, obviously (in general).

I have disucssed above and in more detail at the SCT how both these conditions are mapped to additional solutions of the ‘exchange law’ that differ from

(13)$h={g}_{2}{g}_{1}^{-1}.$

It turned out that Alvarez et al. were overly restrictive by assuming they need ${F}_{A}=0$, but this is how they found these 2 classes of solutions. The first even gives consistent surface holonomy for ${F}_{A}\ne 0$.

But, while still unaware of the work by Alvarez et al., I derived in hep-th/0407122 that there are more loop space connections of the above form which are flat. Namely those obtained by setting $B+{F}_{A}=0$. These are moreover precisely those that preserve the above form of the connection under local loop space gauge transformations. This is one sign of many why $B+{F}_{A}=0$ is the most ‘natural’ class of surface holonomies.

If you instead make a loop space gauge transformation to any of the loop space connections by Alvarez et al. the result is something which can no longer be written in the above form, but which involves in general several 1-forms together with the 2-form. So in a sense the surface holonomies by Alvarez et al live in a framework larger that the standard 2-group theory.

But all this I have already said at the SCT in reply to Amitabha Lahiri, where you can find it, so I’ll stop here. I am in the proces of typing some LaTeX notes about some of the ideas that we are here discussing in ASCII. That might help the discussion.

Meanwhile you can find pretty-printed versions of the formulas that I am considering at the SCT. With MSIE it takes just the free download of the MathPlayer plugin, with Mozilla just the free download of a certain font to read these equations. More details can be found here.

Thanks a lot for your time and comments.

Posted by: Urs Schreiber on August 25, 2004 4:45 PM | Permalink | PGP Sig | Reply to this

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

That’s a lot of comments! :-) I can’t get around to comment on all the issues you raised – it’s almost time for me to go home now … I’ll just clarify a couple of things related to what I said in my earlier comment.

One is that an Abelian ideal is not necessarily the center. It’s an invariant subalgebra which is Abelian, but it need not commute with the rest of the algebra. See eq.3.30 on p.17 of the Alvarez paper.
This subtlety may not be important, but I would like to keep it in mind, because I don’t quite see if you can still define a surface holonomy if A is not flat in this more general case.

In the special case where the Abelian ideal also commutes with the rest of the algebra, you can have a non-flat A which lives in the subalgebra, but in order to have a consistent surface holonomy you should have F vanishing on the rest of the algebra, I think.

This is because there is no canonical surface ordering prescription, so if you have a non-zero non-Abelian F (and Abelian everything else), you will get group elements for every little surface element when you break up the big surface, and there is no standard way of ordering those little pieces.

This is basically my point. Whichever way you want to construct a surface holonomy, it has to be independent of ordering. The condition B+F = 0 ensures that when both B and F are in non-Abelian algebras. Yes, you should write t(B), which ensures that the things being added are in the same space. But if you have an alternative prescription, the `total’ surface holonomy for each infinitesimal surface must still be either Abelian or identity.

If you want to avoid _that_, you need an alternative prescription for ordering surfaces … actually even before that I would like to know the meaning of surface holonomy, and what sort of objects it acts on. Like path holonomy mixes states of particles, which are vectors in a rep. of that group. I suppose a non-Abelian string state ought to be a rep. of a 2-group?

Posted by: Amitabha on August 26, 2004 12:54 PM | Permalink | Reply to this

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

Yes, sorry, I realized that I said ‘abelian ideal’ when I was really thinking of the center. You are of course right, that when we don’t restrict to this very special case it is hard to see if we can lift the ${F}_{A}=0$ condition. Still, the faces taking values in the center of the group and the edges being arbitrary is one solution of the exchange law that is not of the form considered in the literature. I am still in the process how the same is true for the more general conditions discussed by Alvarez et al.

Whichever way you want to construct a surface holonomy, it has to be independent of ordering.

Yes, that’s precisely what I am talking about. This independence of ordering is ensured when the exchange law

(1)$\left({f}_{1}\circ {f}_{1}^{\prime }\right)\cdot \left({f}_{2}\circ {f}_{2}^{\prime }\right)=\left({f}_{1}\cdot {f}_{2}\right)\circ \left({f}_{1}^{\prime }\cdot {f}_{2}^{\prime }\right)$

is satisfied, where the $f$ are surface labels as on p.8 of hep-th/0206130 a simple do is the horizontal product and an open do the vertical product.

When you insert the definition of horizontal and vertical multiplication into this equation it becomes equivalent to

(2)${f}_{1}{g}_{1}{f}_{2}{g}_{1}^{-1}={g}_{2}{f}_{2}{g}_{2}^{-1}{f}_{1}$

where $g$ are the corresponding edge labels.

So we can label our edges and faces whichever way we want, so long as our assignment does satisfy this condition which ensures that when we compute the holonomy of the total surface the result is independent of the order in which we did this.

The choice used in the 2-group literature is to set

(3)${f}_{1}={g}_{2}{g}_{1}^{-1}\phantom{\rule{thinmathspace}{0ex}}.$

This is one solution of the above condition and hence does give us a well-defined notion of surface holonomy.

But there are other solutions of this equation. One simple solution is to have the $g$s arebitrary and $f$ in the center of the group. The resulting surface holonomy is rather restricted, but it is an honest solution of this condition which is not of the form $B+F=0$.

I believe that all the cases disucssed by Alvarez et al can be shown to correspond to additional, previously unrecognized, solutions of this equation.

But if you have an alternative prescription

See, I don’t have an alternative prescription. Using the standard horizontal and vertical product I am saying that there are more conditions under which they are ‘2-associative’ and give consistent surface holonomy. This must be true in one way or another, due to the result by Alvarez et al.

I suppose a non-Abelian string state ought to be a rep. of a 2-group?

I am not sure about that. The boundary states that I have considered, which involve the non-abelian 2-form, are elements of the fundamental rep of the group. This might be related to the fact that the non-abelian 2-form in string theory perhaps (nobody seems to know for sure) membranes end on NS5 branes. In such a case the boundaries of the membranes should carry a fundamental group index each, like the ends of an open string do in one dimension less. Since the boundaries of the memnranes are the open/closed strings that couple to the non-abelian 2-form, it seems reasonable that the boundary state associated with them also carries that one index.

So I think this is indicative of how the big picture will look like. But to date nobody seems to know any details.

Posted by: Urs Schreiber on August 26, 2004 2:44 PM | Permalink | PGP Sig | Reply to this

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

You want arbitrary $g$ on the edges (or rather arbitrary $A$) and some $f$ on the face which is valued in the center. What do you now calculate as the surface holonomy?

Let’s make this simple – take $f$ to be the identity. Then you have only edge contributions. In other words you have the (untraced) holonomy of $A$ along the edge. You can define that to be the surface holonomy, but that doesn’t contain any information about the surface.

That is, you can define the surface holonomy of any surface with a boundary to be the (untraced) holonomy of some gauge field $A$ along the boundary, multiplied by some $U\left(1\right)$ or even ${U\left(1\right)}^{k}$ factor for the face, which could even hold interesting physics, but it’s not really non-Abelian surface holonomy.

And I think the purpose of defining a surface holonomy is relevant in figuring out what we are looking for – if in the end you are going to find only a group valued object, the 2-group must be reduced to a group. This is if the non-Abelian string is a vector in the rep. in the same sense as fields. (Which cannot be true.) If the string carries a rep., it must carry a vector at each point along the string, and these vectors must be related (by parallel transport along the string), and the surface holonomy must compare the vectors in one configuration with another.

Anyway, that’s it for this week. Back on Monday. Thanks for the interesting discussion.

Posted by: Amitabha on August 27, 2004 12:45 PM | Permalink | Reply to this

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

What do you now calculate as the surface holonomy?

I just follow the standard rules of horizontal and vertical multiplication.

The vertical product of face labels is

(1)${f}_{1}\circ {f}_{1}^{\prime }={f}_{1}^{\prime }{f}_{1}$

and the horizontal product is

(2)${f}_{1}\cdot {f}_{2}={f}_{1}{g}_{1}{f}_{2}{g}_{1}^{-1}\phantom{\rule{thinmathspace}{0ex}},$

where all products by juxtaposition are the ordinary group product.

For the order of these two products to be irrelevant the exchange law has to hold which is equivalent to

(3)$\left[{g}_{2}^{-1}\phantom{\rule{thinmathspace}{0ex}}{f}_{1}\phantom{\rule{thinmathspace}{0ex}}{g}_{1}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{f}_{2}\right]=0\phantom{\rule{thinmathspace}{0ex}}.$

The solution of this consistency condition considered in the 2-group literature is

(4)${g}_{2}^{-1}{f}_{1}{g}_{1}=1\phantom{\rule{thinmathspace}{0ex}}.$

But obviously that’s not the most general solution. I am trying to identify the two classes of solutions given by Alvarez et al. in this framework, but this I have so far only managed to do perturbatively in the lattice constant. But one further solution which can be written down without much ado is

(5)$g\mathrm{arbitrary}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\wedge \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}f\mathrm{in}\mathrm{the}\mathrm{center}\phantom{\rule{thinmathspace}{0ex}}.$

This solves the above condition and it is non-abelian, because $g$ is.

Still, one sees that this is a very simple class of solutions only, because $f$ and $g$ ‘decouple’ and one is left with something abelian only.

Namely for $f$ in the center of the group the horizontal product becomes

(6)${f}_{1}\cdot {f}_{2}={f}_{1}{g}_{1}{f}_{2}{g}_{1}^{-1}={f}_{1}{g}_{1}{g}_{1}^{-1}{f}_{2}={f}_{1}{f}_{2}\phantom{\rule{thinmathspace}{0ex}}$

i.e. just the ordinary group product again. So the surface holonomy obtained this way is just that obtained by forgetting about the edge labels alltogether and using the abelian way to compute surface holonomy.

So this is not a very interesting solution of the exchange law consistency condition, but it is one solution not of the form ${g}_{2}^{-1}{f}_{1}{g}_{1}=1$. That’s my point.

And indeed my point is that there should be even more, less trivial solutions, namely those corresponding to the two classes of consistency conditions found by Alvarez et al. in loop space formalism. But I am still not completely done with showing that. It must be true, however, we know that from loop space formalism.

Posted by: Urs Schreiber on August 27, 2004 1:08 PM | Permalink | PGP Sig | Reply to this

### Lattice-expansion of 2-associativity

I think I am beginning to see how the exchange law or 2-associativity condition (as I like to call it) of 2-group theory is equivalent to the condition of flatness of the corresponding connection on loop space. I indeed begin to believe both these conditions express precisely the same fact.

By 2-associativity I mean the well-known condition

(1)$\left({f}_{1}\circ {f}_{1}^{\prime }\right)\cdot \left({f}_{2}\circ {f}_{2}^{\prime }\right)=\left({f}_{1}\cdot {f}_{2}\right)\circ \left({f}_{1}^{\prime }\cdot {f}_{2}^{\prime }\right)$

that the order of horizontal products $a\cdot b$ and vertical products $a\circ b$ is irrelevant. Using $G=H$ for notational simplicity and $t=\mathrm{id}$ the above is first of all equivalent to (as I have discussed before)

(2)$\left[{g}_{2}^{-1}{f}_{1}{g}_{1},{f}_{2}^{\prime }\right]=0$

where all products are group products. Now this condition should be expanded in the small lattice spacing. For higher order this is straightforward but quite tedious. That’s why I would like to share the lowest order result here and discuss how it correctly gives the first terms of the corresponding condtition of loop space flatness.

I’ll follow the notation of section 3.3 of

F. Girelli & H. Pfeiffer: Higher gauge theory - differential versus integral formulation (2004)

and in particular their figure 1. In the little square depicted there you have to identify the surface label with ${f}_{1}$ from above, the lower right edge with ${g}_{1}$ and the upper left edge with ${g}_{2}$. At the upper right corner you have to imagine one further square sitting whose surface is labeled by ${f}_{2}$.

Now make an expansion of the above consistency condition

(3)$\left[{g}_{2}^{-1}{f}_{1}{g}_{1},{f}_{2}^{\prime }\right]=0$

following the general method of section 3.3 of Girelli&Pfeiffer’s paper. One finds

(4)${g}_{2}^{-1}{f}_{1}{g}_{1}\approx \mathrm{exp}\left({a}^{2}\left({F}_{\sigma \tau }+{B}_{\sigma \tau }\right)+{a}^{3}\frac{1}{2}\left[{F}_{\sigma \tau }+{B}_{\sigma \tau },{A}_{\sigma }+{A}_{\tau }\right]\right)$

and of course

(5)${f}_{2}^{\prime }\approx \mathrm{exp}\left({a}^{2}{B}_{\sigma \tau }+{a}^{3}\left({\partial }_{\sigma }+{\partial }_{\tau }\right){B}_{\sigma \tau }\right)\phantom{\rule{thinmathspace}{0ex}}.$

Using this one can convince oneself that the first few orders of the lattice spacing $a$ in the expansion of

(6)$\left[{g}_{2}^{-1}{f}_{1}{g}_{1},{f}_{2}^{\prime }\right]=0$

coincide with the terms of a respective expansion in the flatness condition on the loop space connection

(7)$𝒜={\int }_{0}^{2\pi }d\sigma \phantom{\rule{thinmathspace}{0ex}}{W}_{A}\left(\sigma \right){B}_{\mu \nu }\left(\sigma \right){W}_{A}^{-1}\left(\sigma \right){X}^{\prime \mu }\left(\sigma \right){\mathrm{dX}}^{\nu }\left(\sigma \right)\phantom{\rule{thinmathspace}{0ex}}.$

Namely, as has been noted by Orlando Alvarez a while ago in a private memo, this flatness condition is equivalent to

(8)$0=-{\int }_{0}^{2\pi }d{\sigma }^{\prime }\phantom{\rule{thinmathspace}{0ex}}\theta \left(\sigma -{\sigma }^{\prime }\right)\left[{F}_{{\sigma }_{\tau }}^{W}\left({\sigma }^{\prime }\right)+{B}_{{\sigma }_{\tau }}^{W}\left({\sigma }^{\prime }\right),{B}_{\sigma \tau }^{W}\left(\sigma \right)\right]+\theta \left({\sigma }^{\prime }-\sigma \right)\left[{B}_{\sigma \tau }^{W}\left({\sigma }^{\prime }\right),{F}_{{\sigma }_{\tau }}^{W}\left(\sigma \right)+{B}_{{\sigma }_{\tau }}^{W}\left(\sigma \right)\right]$

(where ${B}^{W}={\mathrm{WBW}}^{-1}$ etc.)

But getting the higher order terms seems like a lot of work.

Does anyone see how to speed up this calculation?

Posted by: Urs Schreiber on August 26, 2004 5:56 PM | Permalink | PGP Sig | Reply to this

### Re: Lattice-expansion of 2-associativity

Does anyone see how to speed up this calculation?

Yes, there is a much simpler way. Just let ${f}_{1}$ be a very thin but long piece of surface and then use the expansion for the continuous horizontal product mentioned here… This seems to prove the claimed equivalence between the exchange law and loop space flatness.

But I’ll better stop commenting on my own questions and write up some readable notes instead…

Posted by: Urs Schreiber on August 26, 2004 7:14 PM | Permalink | PGP Sig | Reply to this

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