## August 30, 2004

### r-flatness = 2-associativity

#### Posted by Urs Schreiber

I am further discussing loop space-, surface- and and 2-group-holonomy with those interested. While trying to convince the 2-group-theorists that 2-associativity admits more solutions than currently considered in the literature, Orlando Alvarez keeps coming up with ever more such solutions to loop space r-flatness(which ensures that loop space connections assign well-defined surface holonomy)! But what is the systematic way to find these solutions? Well, the answer should be given by 2-Lie algebra theory…

It seems to me that hence some effort demonstrating the equivalence of loop space r-flatness with 2-group 2-associativity is in order.

And Ithink that I have now figured out how to do the higher order expansion of (differential) 2-group 2-associativity, demonstrating that it is indeed (essentially) the same as r-flatness in loop space.

One key observation is that for the purpose of defining surface holonomies the 2-associativity law (‘exchange law’)

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

(where a simple dot indicates the horizontal and an open dot the vertical product)

needs only be required to hold in the limit where the surface elements ${f}_{2}$ and ${f}_{2}^{\prime }$ become infinitesimal, since that’s sufficient to consistently integrate their contributions up to a macroscopic surface holonomy.

(Note that this makes good sense also in the light of the fact that the right factor in the horizontal product is parallel transported to the left factor along a single edge, something which manifestly should not in general be done to an extended object.)

For a surface element ${f}_{1}$ and another one ${f}_{2}^{\prime }$ to the right of it the 2-associativity condition says that the commutator

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

must vanish (when all elements take values in the same group and $t$ is trivial).

Because ${f}_{2}^{\prime }$ should be infinitesimal we write (following Girelli & Pfeiffer)

(3)${f}_{2}\approx \mathrm{exp}\left(-{ϵ}^{2}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{iB}}_{\mu \nu }\left(\sigma \right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

and let ${f}_{1}$ be the horizontal product of many small surface elements as in this figure

What looks like TESLA superconducting cavities is a ‘horizontal’ chain of small surface elements (drawn, however, diagonally) making up the total surface strip ${f}_{1}$. ${x}^{\mu }$ and ${x}^{\nu }$ are the coordinates parallel to the sides of the little squares and $\sigma =\frac{1}{2}\left({x}^{\mu }+{x}^{\nu }\right)$ runs ‘horizontally’ along the diagonal.

The whole trick is now to get an expression for ${f}_{1}$ in the limit where it becomes the horizontal product of all these squares

(4)${f}_{1}\to {f}_{1}\cdot {f}_{1}^{\prime }\cdot {f}_{1}^{\prime \prime }\cdots$

and where the left factor ${g}_{2}^{-1}{f}_{1}{g}_{1}$ becomes

(5)${g}_{2}^{-1}{f}_{1}{g}_{1}\to \left({g}_{2}{g}_{2}^{\prime }{g}_{2}^{\prime \prime }\cdots {\right)}^{-1}\left({f}_{1}\cdot {f}_{1}^{\prime }\cdot {f}_{1}^{\prime \prime }\cdots \right)\left({g}_{1}{g}_{1}^{\prime }{g}_{1}^{\prime \prime }\cdots \right)\phantom{\rule{thinmathspace}{0ex}}.$

This problem however becomes very simple once it is realized that the terms obtained by truncating ‘$\cdots$’ are recursively related as

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

where $X$ is the same expression with at most single primes, and so on.

By performing the lattice computations similar to those done by Girelli&Pfeiffer, but taking one more order in $ϵ$ into account one finds that

(7)${g}_{1}^{-1}{f}_{1}{g}_{1}\approx \mathrm{exp}\left(-{ϵ}^{2}\phantom{\rule{thinmathspace}{0ex}}\left(1-ϵ\phantom{\rule{thinmathspace}{0ex}}i{A}_{\sigma }\right)i{B}_{\sigma \tau }\left(1+ϵ\phantom{\rule{thinmathspace}{0ex}}i{A}_{\sigma }\right)\right)$

and

(8)${g}_{2}^{-1}\mathrm{exp}\left(-{ϵ}^{2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{iY}\right){g}_{1}\approx \mathrm{exp}\left(-{ϵ}^{2}\phantom{\rule{thinmathspace}{0ex}}i\left(1-ϵ\phantom{\rule{thinmathspace}{0ex}}i{A}_{\sigma }\right)\left(Y+{F}_{\sigma \tau }\right)\left(1+ϵ\phantom{\rule{thinmathspace}{0ex}}i{A}_{\sigma }\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$

Plugging this into the above recursion formula one finds that ${g}_{2}^{-1}{f}_{1}{g}_{1}$ for ${f}_{1}$ a long thin ‘horizontal’ strip becomes

(9)$\cdots \approx \mathrm{exp}\left(-iϵ{\int }_{0}^{\sigma }d{\sigma }^{\prime }\phantom{\rule{thinmathspace}{0ex}}{W}_{A}^{-1}\left({\sigma }^{\prime },\sigma \right)\left({F}_{\sigma \tau }\left({\sigma }^{\prime }\right)+{B}_{\sigma \tau }\left({\sigma }^{\prime }\right)\right){W}_{A}\left({\sigma }^{\prime },\sigma \right)\right)\phantom{\rule{thinmathspace}{0ex}},$

where $W$ is the holonomy of $A$ along the strip.

Inserting this result into the 2-associativity condition finally yields

(10)$\left[{\int }_{0}^{\sigma }d{\sigma }^{\prime }\phantom{\rule{thinmathspace}{0ex}}{W}_{A}^{-1}\left({\sigma }^{\prime },\sigma \right)\left({F}_{\sigma \tau }\left({\sigma }^{\prime }\right)+{B}_{\sigma \tau }\left({\sigma }^{\prime }\right)\right){W}_{A}\left({\sigma }^{\prime },\sigma \right)\phantom{\rule{thinmathspace}{0ex}},{B}_{\sigma ,\tau }\left(\sigma \right)\right]=0\phantom{\rule{thinmathspace}{0ex}}.$

This, I claim, is the ‘exchange law’ in 2-group theory in terms of the target space fields $A$ and $B$ which ensures that 2-group holonomy of continuous surfaces is well-defined.

It is essentially the same as the r-flatness condition found by Orland Alvarez. The latter involves another integral with $\sigma$ and ${\sigma }^{\prime }$ exchanged.

But for all solutions to r-flatness found so far the integrand actually vanishes for all ${\sigma }^{\prime }$ seperately. All these solutions hence also solve the above 2-associativity condition.

The interesting question arises whether the above is actually equivalent to the vanishing of the integrand all by itself.

Posted at August 30, 2004 7:02 PM UTC

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### Re: r-flatness = 2-associativity

This is what you wrote in the pdf file of your notes, right? I am reading that at the moment, and will get back to you with some comments.

Posted by: Amitabha on September 1, 2004 10:11 AM | Permalink | Reply to this

### Re: r-flatness = 2-associativity

Yes, this is what I sketched as an idea for a proof in the first version of my notes. But now I think I was able to do give a valid demonstration, taking care of the details. Please have a look at the most recent version of my notes, which can still be found here.

The changes with respect to the first sketchy draft include

- lots of figures to illustrate the algebra

- an argument saying that the computation of surface holonomy using 2-groups only requires that 2-associativty (the ‘exchange law’) holds for infinitesimally small surface elements, but not necessarily for larger surface elements (because the infinitesimal condition suffices to gives a well defined way to integrate up the infinitesimal surface contributions)

- the above mentioned demonstration that this infinitesimal 2-associativity is indeed the same as the (r-)flatness condition known from the loop space formalism

- a tidied-up version of the demonstration that loop space holonomy computes the same surface holonomy as 2-group holonomy

- a discussion/review of the case when $G\ne H$ and the homomorphism $t$ of the crossed module is not invertible (but see below)

- I also made the notation in the loop space section coherent and in agreement with the conventions that I used in hep-th/0407122

What is still missing (but which I will include in the course of today and the next days) is

- a discussion of the fact that the r-flatness condition as well as the differential 2-associatitivity very probably hold in the form

(1)$\left[{F}_{\sigma \tau }^{{W}_{A}}\left({\sigma }^{\prime }\right)+{B}_{\sigma \tau }^{{W}_{A}}\left({\sigma }^{\prime }\right)\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{B}_{\sigma \tau }^{{W}_{A}}\left(\sigma \right)\right]=0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\forall \sigma ,{\sigma }^{\prime },\Gamma$

without the integrals over $\sigma$ and/or ${\sigma }^{\prime }$. This should follow from the fact that the condition indeed has to hold on every single loop $\Gamma$, which should imply that it cannot depend on any given loop and hence must hold locally in $\sigma$ and ${\sigma }^{\prime }$.

For the 2-group case this is very easy to show, but for the loop space case I am still discussing with Orlando Alvarez ways to make this precise (if correct). I am pretty convinced that it must be true and will try to work out the proof.

- Then I encountered an unexpected issue when writing up some comments on the case $G\ne H$ and $t$ not invertible. Maybe I am confused, but it seems to me that when $t$ is not invertible even setting $t\left(f\right)={g}_{2}{g}_{1}^{-1}$ (as done by Baez, Girelli and Pfeiffer) does not guarantee that 2-associativity holds. I find instead that the necessary and sufficient condition for 2-associativity to hold is

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

(Here $g⊲f$ is Girelli&Pfeiffer’s notation for what John Baez writes as $\alpha \left(g\right)\left(f\right)$.)

Only after both sides of this equation are hit with $t$ does one obtain the commutator condition

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

which is solved (among other solutions) by the famous $t\left({f}_{1}\right)={g}_{2}{g}_{1}^{-1}$. But when $t$ is not invertible the above implication arrow $⇒$ cannot be reversed and hence $t\left({f}_{1}\right)={g}_{2}{g}_{1}^{-1}$ is not sufficient to solve the 2-associativity condition $\left(\star \right)$.

I’d be very grateful for all comment and/or cooperation.

Posted by: Urs Schreiber on September 1, 2004 10:47 AM | Permalink | Reply to this

### Re: r-flatness = 2-associativity

I wrote:

Maybe I am confused somewhere.

Yes, I was. I forgot to insert the relation between ${g}_{2}$ and ${g}_{1}$. If we use ${g}_{2}=t\left(h\right){g}_{1}$ in $\left(\star \right)$ it works out as expected.

Posted by: Urs Schreiber on September 1, 2004 2:35 PM | Permalink | PGP Sig | Reply to this

### Re: r-flatness = 2-associativity

Urs! :)

It’s been a long time :)

I have been trying to follow the general line of work you’ve been pursuing lately, but to be honest, it is completely beyond me :)

That never stopped me from asking wild questions though :)

I was just skimming through your latest notes (p11.pdf) and I got to the point where you are discussing holonomies on a 2-complex (the paper says 1-complex by the way).

One thing that immediately sprung to mind was the idea of Poincare duality. With standard lattice gauge theory, you have a complex and you

parallel transport from one 0-cell to the next 0-cell along a 1-cell.

However, if you think of this as happening on an $n$-complex and you “wave the wand of Poincare duality” over it, the statement becomes

parallel transport from one dual n-cell to the next dual n-cell along a dual (n-1)-cell.

Sticking in $n=2$, we get something that sounds like what you are talking about, i.e.

parallel transport from one dual 2-cell to the next dual 2-cell along a dual 1-cell.

I don’t know if this makes sense or is useful in any way, but it is my first reaction.

Cheers!
Eric

PS: Have you considered $n$-holonomies on diamonds? It seems like that would be a natural thing to do.

Posted by: Eric on September 5, 2004 3:06 PM | Permalink | Reply to this

### Re: r-flatness = 2-associativity

Hi Urs,

With a little more thought, I’m coming back to this:

Have you considered $n$-holonomies on diamonds?

By the very construction of an $n$-diamond complex, it seems to me that defining $n$-holonomies becomes completely trivial.

There is only one way a 1-diamond can be parallel transported across the face of a 2-diamond. In fact, there is only one way a $p$-diamond can be transported across a $\left(p+1\right)$-diamond. Since differential geometry on $n$-diamonds has a continuum limit, then that would seem to suggest that $n$-holonomies are naturally treated on $n$-diamonds in a way that has a direct analogy to the continuum theory.

In other words, I don’t understand why things need to seem so complicated :)

Eric

Posted by: Eric on September 5, 2004 3:31 PM | Permalink | Reply to this

### 2-flatness = 2-bianci?

Hi Urs,

I was just playing around with diamonds and $n$-holonomies and maybe found something interesting.

The way my write-up of our notes (that never got finished) was going was making a close analogy between $n$-diamonds and $n$-categories, i.e. an edge ${e}_{{i}_{0}{i}_{1}}$ may be thought of as an “arrow”

(1)${e}_{{i}_{0}{i}_{1}}:{e}_{{i}_{0}}\to {e}_{{i}_{1}}$

from the node ${e}_{{i}_{0}}$ to the node ${e}_{{i}_{1}}$. Likewise, a $p$-path ${e}_{{i}_{0}\dots {i}_{p}}$ may be thought of an a $p$-dimensional arrow

(2)${e}_{{i}_{0}\dots {i}_{p}}:{e}_{{i}_{0}\dots {i}_{p-1}}\to {e}_{{i}_{1}\dots {i}_{p}}.$

In this way, an $n$-diamond is like an $n$-category that has forgotten everything about associativity and only remembers what arrows go where.

As in our notes (math-ph/0407005), the group-valued 1-form

(3)${H}^{\left(1\right)}=\sum _{{i}_{0}{i}_{1}}{H}_{{i}_{0}{i}_{1}}{e}_{{i}_{0}{i}_{0}}$

assigns the holonomy ${H}_{{i}_{0}{i}_{1}}$ to the edge ${e}_{{i}_{0}{i}_{1}}$.

In the same way, we can define a group-valued 2-form

(4)${H}^{\left(2\right)}=\sum _{{i}_{0}{i}_{1}{i}_{2}}{H}_{{i}_{0}{i}_{1}{i}_{2}}{e}_{{i}_{0}{i}_{1}{i}_{2}}$

that assigns the 2-holonomy ${H}_{i0{i}_{1}{i}_{2}}$ to the 2-arrow

(5)${e}_{{i}_{0}{i}_{1}{i}_{2}}:{e}_{{i}_{0}{i}_{1}}\to {e}_{{i}_{1}{i}_{2}}.$

I’m sure you are aware of that neat way to view the Bianci identity as a parallel transport around a 3-cube in such a way that every edge gets traversed twice in opposite directions so the holonomies cancel.

A similar tracing of 2-paths on the 3-cube gives a consistency requirement for ${H}^{\left(2\right)}$ that is obviously satisfied if

(6)${H}_{{i}_{0}{i}_{1}{i}_{2}}={H}_{{i}_{0}{i}_{1}}{H}_{{i}_{1}{i}_{2}}.$

This must have something to do with 2-associativity or something you are talking about :)

Note that

(7)${H}_{{i}_{0}{i}_{1}{i}_{2}}={H}_{{i}_{0}{i}_{1}}{H}_{{i}_{1}{i}_{2}}.$

is equivalent to requiring that

(8)${H}^{\left(2\right)}={\left[{H}^{\left(1\right)}\right]}^{2}=F.$

I hope that some day you will really learn to appreciate our paper :)

Eric

Posted by: Eric on September 5, 2004 5:18 PM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Hi Eric -

nice to see you at the Coffee Table again!

Thanks for finding that typo in my notes. It is surprising how much simple nonsense one can overlook.

I’m sure you are aware of that neat way to view the Bianci identity as a parallel transport around a 3-cube in such a way that every edge gets traversed twice in opposite directions so the holonomies cancel.

Sorry, not right off the top of my head. I would need to write that out to see it.

A similar tracing of 2-paths on the 3-cube gives a consistency requirement for H ( 2) that is obviously satisfied if ${H}_{{i}_{0}{i}_{1}{i}_{2}}={H}_{{i}_{0}{i}_{1}}{H}_{{i}_{1}{i}_{2}}.$

[…]

is equivalent to requiring ${H}^{\left(2\right)}=...=F$.

This sounds very interesting indeed, but right now I don’t see how exctly you obtain this result. Could you spell out the details?

Posted by: Urs Schreiber on September 6, 2004 10:04 AM | Permalink | PGP Sig | Reply to this

### Re: 2-flatness = 2-bianci?

Good morning!

Sorry, not right off the top of my head. I would need to write that out to see it.

You should :) The first time I saw it was in Baez & Munian, but it might also be in Gambini & Pullin.

This sounds very interesting indeed, but right now I don’t see how exctly you obtain this result. Could you spell out the details?

I can try :)

A 2-diamond may be thought of as being composed of two 2-paths. If we label the 4 nodes of a 2-diamond counterclockwise by $i\to j\to k\to l$, then we can write the 2-diamond as

(1)${e}_{\mathrm{ijk}}=-{e}_{\mathrm{ilk}}.$

A 3-diamond is composed of 6 2-diamonds. Maybe you could draw (a 3d perspective view of) a 3-cube and label the edges accordingly (maybe from bottom left to top right) to indicate the “time flow”.

Your drawing actually defines a cycle across the surface of the 3-diamond. To see this, consider a 2-path

(2)${e}_{\mathrm{ijk}}$

and draw a line

(3)${e}_{\mathrm{ij}}\to {e}_{\mathrm{jk}}$

from the midpoint of ${e}_{\mathrm{ij}}$ to the midpoint of ${e}_{\mathrm{jk}}$. The midpoint of ${e}_{\mathrm{jk}}$ connects to an adjacent 2-path, so draw a line to the midpoint of the corresponding opposite edge.

If you continue in this manner, you will find that you eventually end up at the same point you started with. The cycle will consist of 12 segments.

It is important to get that drawing right, so I hope I explained it clearly enough.

Ok. I had mercy and just mailed you an .eps figure I threw together :)

Now, with the figure in mind, we think of

(4)${H}_{\mathrm{ijk}}$

as a 2-holonomy transporting something from the edge ${e}_{\mathrm{ij}}$ to the edge ${e}_{\mathrm{jk}}$. Each segment in the cycle in the figure corresponds to a 2-holonomy. If you transport something around the entire cycle, you will get a total 2-holonomy. If the result after transporting around the cycle is not the identity, you will get something that could probably be referred to as 3-curvature :)

If the 2-holonomy satsfies

(5)${H}_{\mathrm{ijk}}={H}_{\mathrm{ij}}{H}_{\mathrm{jk}},$

then the 3-curvature vanishes due to the Bianci identity. However, there could be other 2-holonomies ${H}^{\left(2\right)}$ with vanishing 3-curvatures that were not obtained by some 1-holonomy ${H}^{\left(1\right)}$.

I could be (and probably am) wrong, but it seems like saying the the 2-holonomy is obtained from a 1-holonomy, i.e.

(6)${H}^{\left(2\right)}={\left[{H}^{\left(1\right)}\right]}^{2}=F$

is like the special case you are talking about where

(7)$F=-B$

if we associate the 2-holonomy

(8)${H}^{\left(2\right)}=-B$

or something like that :)

It almost seems like non-abelian cohomology or something :)

If ${d}_{A}B=0$, is $B={d}_{A}\Psi$ for some $\Psi$?

Or something :)

Gotta run!
Eric

Posted by: Eric on September 6, 2004 4:18 PM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Hi Eric -

concerning your computation of 2-holonomy: Did you consider the problem that there are different orders in which you could decide to multiply the surface holonomies of the six faces? This is what these constraints usually come from.

On the other hand, vanishing 3-curvature ensures a certain 2-path independence in the usual way, so if that’s ensured the result is is always the trivial group element.

I’d have to think about how to directly translate from what you have written to what I am currently concerned with. Looks like you found a very fast way to get the \$B + F = 0\$ condition!

Posted by: Urs Schreiber on September 6, 2004 6:23 PM | Permalink | PGP Sig | Reply to this

### Re: 2-flatness = 2-bianci?

Hi Urs :)

Today was a holiday for us and I just spent 6 hours doing yard work. It was the first weekend that my wrist felt strong enough to do anything so I had to make up for 6 weeks of neglect :)

Did you consider the problem that there are different orders in which you could decide to multiply the surface holonomies of the six faces? This is what these constraints usually come from.

I don’t understand the question :) On a 3-diamond there is absolutely no ambiguity in the 2-holonomy calculation. In that figure I sent, if you begin at the midpoint of any edge on the 3-diamond, you have a choice between two possible directions to begin your journey. However, the two choices give the same constraint because they amount to the same cycle traversed in opposite directions. The “horizontal” versus “vertical” stuff does not appear because you can only parallel transport an edge quantity forward (or backward) in time, not sideways. In other words, you have to follow the flow dictated by the $n$-diamond.

Note: In that figure I sent, time flows from the bottom left of the 3-diamond to the top right.

I doubt this is satisfactory to you so it is probably good if we try to hash this out better (if you think it is worth it).

Hmm… a random thought (I need to run soon)…

If you think of the two processes

(1)$\mathrm{horizontal}\to \mathrm{vertical}$

versus

(2)$\mathrm{vertical}\to \mathrm{horizontal}$

as enclosing some 3d surface, then you might get some interpretation similar to that of the 3-diamond. The two processes correspond to the two choices on the 3-diamond. Neat! :)

Discrete geometry to the rescue! ;)

On the other hand, vanishing 3-curvature ensures a certain 2-path independence in the usual way, so if that’s ensured the result is is always the trivial group element.

The more I think about it, the more neat this is. The conditions you have seem to relate to the vanishing of the 3-curvature. The $H\to V$ versus $V\to H$ thing amounts to the different ways you can wrap around the 3-diamond.

Cheers!
Eric

Posted by: Eric on September 6, 2004 11:12 PM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Hi Eric -

I feel a little unsure about what you are saying. Here is why:

Consider a cube made up from six faces, each of which is labeled with a group element ${h}_{i}\in H$.

So we have a surface holonomy for each face and want to multiply these somehow to get a surface holonomy for the entire cube surface.

The problem is of course that there is no unique order in which the ${h}_{i}$ are to be multiplied.

You seem to be arguing that since the concept of an $n$-diamond implies a preferred direction in space (or on spacetime if you wish) it also induces a preferred order in which these group elements ${h}_{i}$ are to be multiplied.

I have to admit that I still don’t understand how you imagine that the single preferred direction on the $n$-diamond completely fixes one of the 720 ways to multiply the six group elements.

Even if you argue that one of these ways is special (but you have to consider rotations around the preferred direction of the diamond, too) and that this is the one you declare to be used for computing surface holonomy of cubes, this does not really solve the problem.

For one, it seems to me that when you start composing cubes to larger and larger objects you have to keep specifying preferred orders of multiplication for all kinds of configurations. Seems like it needs an infinite number of rules, one for each shape constructed from plaquettes of the diamond.

Even if you specify all these rules, the result is not quite what one would like to have. You end up with a gauge theory that assigns different surface holonomy to configurations that can be turned into each other by rotations of the diamond (around the preferred axis, if you want to preserve that, but still).

Is this what you have in mind?

Posted by: Urs Schreiber on September 8, 2004 5:08 PM | Permalink | PGP Sig | Reply to this

### Re: 2-flatness = 2-bianci?

Hi Urs,

One of the advantages I have with not understanding anything is that I don’t see why things are difficult :)

You and I spent several months working together to show that discrete differential geometry just simply does not work on any arbitrary graphs. You need to have certain structure.

Now you are using arbitrary graphs to show that 2-holonomies don’t work (or aren’t interesting or whatever). Is it possible that you are simply violating the rules of discrete geometry?

I know you have a million things going on, but when you get a chance I would suggest that you think about how everything you are doing would work on a diamond complex.

On a diamond, there just simply is not a “spatial” 3-cube at a particular instant in time. Instead you have a 3-diamond where each edge is light-like and the faces all extend into both space and time. Just like lattice Yang-Mills worked out so beautifully.

What I am suggesting is that transporting an edge quantity across a face to another edge, i.e. a 2-holonomy, requires an increment in time (and the edges must share a common node).

Therefore, the $V\to H$ and $H\to V$ things trace out different surfaces in space-time.

You are much more capable than I am to do the exercise, but I guarantee you will not regret it if you try to merge ideas from our discrete paper with this 2-holonomy stuff.

Was that figure I sent with the 2-holonomies enclosing the 3-diamond giving the 3-curvature not clear? That figure really holds the key I think (but what do I know? :)). I should try to explain it better if my point wasn’t clear. Did you see how I got $F=-B$?

Best wishes,
Eric

Posted by: Eric on September 8, 2004 6:16 PM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Ok. This is my third time retyping this. Maybe it is a sign I should just give up :)

I have to admit that I still don’t understand how you imagine that the single preferred direction on the $n$-diamond completely fixes one of the 720 ways to multiply the six group elements.

[snip]

Is this what you have in mind?

No, not at all. I am not explaining myself clearly. Sorry about that.

Let’s think about what a 1-holonomy ${H}_{\mathrm{ij}}$ does on a diamond complex. This parallel transports a node quantity from node ${e}_{i}$ to node ${e}_{j}$ along the edge ${e}_{\mathrm{ij}}$. In the process, one tick of the clock goes by (since ${e}_{\mathrm{ij}}$ is light like).

With that in mind, let’s think about what a 2-holonomy ${H}_{\mathrm{ijk}}$ does on a diamond complex. This parallel transports an edge quantity from the (light-like) edge ${e}_{\mathrm{ij}}$ to the (light-like) edge ${e}_{\mathrm{jk}}$ along the surface ${e}_{\mathrm{ijk}}$. In the process, one tick of the clock goes by.

Am I confused already or does that make a little sense? :)

If it makes a little sense, then look at that 3-diamond figure I sent you. That cycle is the unique cycle on the 3-diamond you get by parallel transporting an edge quantity over the surface of the 3-diamond. There is no other cycle of 2-holonomies over the surface of a single 3-diamond. This does not, however, mean that there is only one way to glue 2-holonomies together. What it means is that if you choose any two edge in the 3-diamond, you have exactly two paths (a path can be thought of as a choice of “gluing”) over the surface of the 3-diamond to to get from one of the chosen edges to the other because the two edges split the cycle into two paths. This imposes consistency conditions on the two paths (gluings) because for vanishing 3-curvature, the two paths combined should give the trivial 2-holonomy, i.e. the identity.

Now, if you choose two “causally connected” edges in a diamond complex that are not part of the same 3-diamond, then there may be MANY paths connecting the two edges. This corresponds to MANY ways to glue the 2-holomies along the path together. All of these paths may have consistency conditions among them, but the basic idea is already there for the single 3-diamond.

Finally (I’m running out of time again), if you set

(1)${H}_{\mathrm{ijk}}={H}_{\mathrm{ij}}{H}_{\mathrm{jk}},$

then follow that cycle around the surface in the figure while reinterpretting it in terms of transport along the edges, you will see that each edge gets traversed exactly twice in opposite directions so that each edge holonomy gets cancelled and you are left with the identity signifying vanishing 3-curvature. This is nothing but the Bianci idea! :)

Therefore, if your 2-holonomy is obtained from the 1-holonomy, then you have vanishing 3-curvature by the Bianci identity. It is really pretty cute.

I hope that I am making sense, but I need to run for now.

Good night! :)
Eric

Posted by: Eric on September 9, 2004 5:20 AM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Therefore, if your 2-holonomy is obtained from the 1-holonomy, then you have vanishing 3-curvature by the Bianci identity. It is really pretty cute.

Maybe I am making some silly mistake, but I just failed to verify this claim. Maybe I should be prepared to eat my words in shame :)

Posted by: Eric on September 9, 2004 5:48 AM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Hi Eric -

concerning the Bianchi identity, maybe you are looking for the following:

(1)${F}_{A}=\mathrm{dA}+A\wedge A$
(2)$⇒{d}_{A}{F}_{A}=\left[\left(d+A\right),{F}_{A}\right]=d\left(A\wedge A\right)+A\wedge \mathrm{dA}-\left(\mathrm{dA}\right)\wedge A=d\left(A\wedge A\right)-d\left(A\wedge A\right)=0\phantom{\rule{thinmathspace}{0ex}}.$

Concerning the 2-holonomy ${H}_{\mathrm{ijk}}$ I still don’t seem to understand what you have in mind. Looking at the figure that you kindly provided does not currently help me following your words.

For communication purposes, let me put coordinates on that 3-diamond, along the lightlike directions. So there is a cartesian coordinate system with coordinates ${x}^{1},{x}^{2},{x}^{3}$ such that the past tip of the diamond is at $\left(0,0,0\right)$. At this point three edges are emanating $\left(0,0,0\right)\to \left(1,0,0\right)$, $\left(0,0,0\right)\to \left(0,1,0\right)$ and $\left(0,0,0\right)\to \left(0,0,1\right)$. From each of the target vertices there are two edges $\left(1,0,0\right)\to \left(1,1,0\right)$, $\left(1,0,0\right)\to \left(1,0,1\right)$ and $\left(0,1,0\right)\to \left(1,1,0\right)$, $\left(0,1,0\right)\to \left(0,1,1\right)$ and $\left(0,0,1\right)\to \left(0,1,1\right)$, $\left(0,0,1\right)\to \left(1,0,1\right)$. Finally there are three edges to the topmost tip $\left(1,1,0\right)\to \left(1,1,1\right)$, $\left(1,0,1\right)\to \left(1,1,1\right)$ and $\left(0,1,1\right)\to \left(1,1,1\right)$.

Do we agree so far?

Now I think you are saying that for instance the edge $\left(0,0,0\right)\to \left(1,0,0\right)$ can be parallel transported to the edge $\left(0,1,0\right)\to \left(1,1,0\right)$ by ${H}_{\left(0,0,0\right)\left(1,0,0\right)\left(1,1,0\right)}$. This I understand and agree with.

Now how do you want to proceed moving the edge $\left(0,1,0\right)\to \left(1,1,0\right)$ over the surface? What is the unique path-between-edges that you have in mind?

Sorry for still not understanding what you mean. But I’ll understand eventually!

Posted by: Urs Schreiber on September 9, 2004 10:09 AM | Permalink | PGP Sig | Reply to this

### Re: 2-flatness = 2-bianci?

Finally there are three edges to the topmost tip $\left(1,1,0\right)\to \left(1,1,1\right)$, $\left(1,0,1\right)\to \left(1,1,1\right)$ and $\left(0,1,1\right)\to \left(1,1,1\right)$.

Do we agree so far?

Yep :)

Now I think you are saying that for instance the edge $\left(0,0,0\right)\to \left(1,0,0\right)$ can be parallel transported to the edge $\left(0,1,0\right)\to \left(1,1,0\right)$ by ${H}_{\left(0,0,0\right)\left(1,0,0\right)\left(1,1,0\right)}$. This I understand and agree with.

I am probably wrong, but this is not what I am saying. I am saying that ${H}_{\left(0,0,0\right)\left(1,0,0\right)\left(1,1,0\right)}$ is the 2-holonomy that parallel transports the edge $\left(0,0,0\right)\to \left(1,0,0\right)$ to the edge $\left(1,0,0\right)\to \left(1,1,0\right)$.

Here is the 2-holonomy cycle in all its glory:

(1)$\left(0,0,0\right)\to \left(1,0,0\right)⇒\left(1,0,0\right)\to \left(1,1,0\right)⇒\left(1,1,0\right)\to \left(1,1,1\right)⇒\left(0,1,0\right)\to \left(1,1,0\right)⇒\left(0,0,0\right)\to \left(0,1,0\right)⇒\left(0,1,0\right)\to \left(0,1,1\right)⇒\left(0,1,1\right)\to \left(1,1,1\right)⇒\left(0,0,1\right)\to \left(0,1,1\right)⇒\left(0,0,0\right)\to \left(0,0,1\right)⇒\left(0,0,1\right)\to \left(1,0,1\right)⇒\left(1,0,1\right)\to \left(1,1,1\right)⇒\left(1,0,0\right)\to \left(1,0,1\right)⇒\left(0,0,0\right)\to \left(1,0,0\right)$

Unless I’ve made a typo, this cycle corresponds to the figure I sent you and each edge appears only once (I repeated the first edge at the end so you can see the path close on itself). Note that in the figure, I just drew segments connecting the midpoints of each edge.

What is the unique path-between-edges that you have in mind?

Now, if you want to parallel transport an edge to some other edge in the 3-diamond, you locate the two edges in the cycle. You then have a choice as to which direction you want to go to reach the second edge from the first. The two choices have a consistency condition due to the vanishing 3-curvature. The cycle is unique, but you have a choice between two paths depending on which partition of the cycle you choose that connects the two edges.

Sorry about the confusion :) Hopefully this helps clarify things. Once things are clarified, we might find that it is useless, but let’s determine that when we get there :)

Eric

Posted by: Eric on September 9, 2004 7:22 PM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Hi Eric -

sorry, sorry, I just realized that I was looking at the wrong figure all along!

Now I finally know which edge-to-edge transition you are talking about!

The cycle is unique

Is it? If I just rotate the cylce (while leaving the ${H}_{\mathrm{ijl}}$ labels alone) it becomes a different cycle, I’d say.

But maybe I am, now that I finally have become aware of one misunderstanding, already subject to the next one.

Do I understand correctly that given that path-between-edges you are now thinking of the surface holonomy associated to the path-between-edges as the product

(1)${H}_{\left(0,0,0\right)\left(1,0,0\right)\left(1,1,0\right)}{H}_{\left(1,1,0\right)\left(1,1,1\right)\left(0,1,0\right)}{H}_{\left(0,1,0\right)\left(0,0,0\right)\left(0,1,1\right)}\cdots$

?

Posted by: Urs Schreiber on September 9, 2004 8:39 PM | Permalink | PGP Sig | Reply to this

### Re: 2-flatness = 2-bianci?

The cycle is unique

Is it? If I just rotate the cylce (while leaving the ${H}_{\mathrm{ijl}}$ labels alone) it becomes a different cycle, I’d say.

Well, this is where causality comes in. Consider the 2-diamond ${e}_{\mathrm{ijk}}=-{e}_{\mathrm{ilk}}$. You can only transport

(1)${e}_{\mathrm{ij}}\to {e}_{\mathrm{jk}}$

or

(2)${e}_{\mathrm{il}}\to {e}_{\mathrm{lk}}.$

Transporting ${e}_{\mathrm{ij}}\to {e}_{\mathrm{il}}$ is not allowed because it does not involve a click of the clock. It is a “sideways” move :) This means that you can only rotate the 3-diamond about the “time axis”, i.e. the major diagonal across the diamond. This axis has a three-fold symmetry and the cycle is unchanged by such a rotation.

The causality issue is similar to the way lattice Yang-Mills worked out in our paper. We did not parallel transport things at a fixed point in time. The process of parallel transportation itself required the evolution in time.

Do I understand correctly that given that path-between-edges you are now thinking of the surface holonomy associated to the path-between-edges as the product

(3)${H}_{\left(0,0,0\right)\left(1,0,0\right)\left(1,1,0\right)}{H}_{\left(1,1,0\right)\left(1,1,1\right)\left(0,1,0\right)}{H}_{\left(0,1,0\right)\left(0,0,0\right)\left(0,1,1\right)}\dots$

?

Something like this, yes. I haven’t got the details worked out though :)

Eric

PS: I think the condition for vanishing 3-curvature is

(4)$\left[{H}^{\left(1\right)},{H}^{\left(2\right)}\right]=0,$

which corresponds to ${d}_{A}B=0$. This is why setting

(5)${H}^{\left(2\right)}={\left[{H}^{\left(1\right)}\right]}^{2}$

corresponds to the Bianchi identity.

Posted by: Eric on September 9, 2004 9:57 PM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Hi Eric -

Something like this, yes. I haven’t got the details worked out though :)

I thought that you were saying you have a presription to associate a 2-holonomy to the cube and that this 2-holonomy must vanish?

(1)$\left[{H}^{\left(1\right)},{H}^{\left(2\right)}\right]\approx {d}_{A}B$

gives the 3-curvature and that at least one case where it vanishes is ${H}^{\left(2\right)}=\left({H}^{\left(1\right)}{\right)}^{2}\approx F$ and that this has a very nice geometric interpretation generalizing the one for the 2-curvature which we discuss in out notes.

Posted by: Urs Schreiber on September 10, 2004 10:12 AM | Permalink | PGP Sig | Reply to this

### Re: 2-flatness = 2-bianci?

Good morning! :)

I thought that you were saying you have a presription to associate a 2-holonomy to the cube and that this 2-holonomy must vanish?

What I have, as usual, is just a crazy idea :)

I started thinking about 2-holonomies on a 3-diamond and tried to make analogies to 1-holonomies on a 2-diamond.

The 2-curvature is obtained by parallel transporting a node quanity around the boundary of a 2-diamond.

An analogous statement bumbed up a degree is:

The 3-curvature is obtained by parallel transporting an edge quantity around the boundary of a 3-diamond.

I’m not sure if that last statement makes sense or not, but that figure illustrates a (seemingly unique) way to parallel transport an edge quantity over the surface of a 3-diamond in such a way that the surface swept out by the edge completely covers the boundary and is not self intersecting.

But when I look at that picture, I don’t see anyway to reconcile it with $G={d}_{A}B$. However, I do see some way to reconcile the picture with the quantity

(1)$\left[{H}^{\left(2\right)},{H}^{\left(2\right)}\right]=0.$

So what would this mean? This almost looks like a 4-curvature :)

(2)${d}_{B}B=0.$

Is there such a thing?

Whatever the case may be, the point I’ve been trying to make all along is that the 2-holonomy stuff you’ve been trying to work out via 2-group stuff, would probably end up being quite simple if you considered it as just an application of discrete differential geometry.

Best wishes,
Eric

PS: In the process of writing this, I seem to have worked out the details. The reason I was hesitant about your question was that I wasn’t sure if I needed any “+” symbols thrown in the expression you wrote, i.e.

(3)${H}_{\left(0,0,0\right)\left(1,0,0\right)\left(1,1,0\right)}{H}_{\left(1,1,0\right)\left(1,1,1\right)\left(0,1,0\right)}{H}_{\left(0,1,0\right)\left(0,0,0\right)\left(0,1,1\right)}\dots$

Now I see that you will only have products of two at a time (because it takes two ticks of the clock to reach the end of the 3-diamond just like LYM in our notes) with pluses and minuses in between, where the correct expression is actually

(4)${F}^{\left(4\right)}={\left[{H}^{\left(2\right)}\right]}^{2}=0.$
Posted by: Eric on September 10, 2004 1:35 PM | Permalink | Reply to this

### Re: 2-flatness = 2-bianci?

Now I see that you will only have products of two at a time (because it takes two ticks of the clock to reach the end of the 3-diamond just like LYM in our notes) with pluses and minuses in between, where the correct expression is actually

(1)${F}^{\left(4\right)}={\left[{H}^{\left(2\right)}\right]}^{2}=0.$

I’m stumbling along, but am making progress I suppose.

The “correct” expression above is, in fact, not correct. Surprise surprise :)

If it were correct, we’d have terms like

(2)${H}_{\left(0,0,0\right)\left(1,0,0\right)\left(1,1,0\right)}{H}_{\left(1,1,0\right)\left(2,1,0\right)\left(2,2,0\right)},$

where the last node of the first term corresponds to the first node of the last term.

Instead, from the figure, we would expect to see things like

(3)${H}_{\left(0,0,0\right)\left(1,0,0\right)\left(1,1,0\right)}{H}_{\left(1,0,0\right)\left(1,1,0\right)\left(1,1,1\right)},$

where the last two nodes of the first term correspond to the first two nodes of the last term.

Why is it so difficult to convert a pretty picture into meaningful algebraic symbols?! :)

Eric

Posted by: Eric on September 10, 2004 8:43 PM | Permalink | Reply to this

### Re: 2-flatness = 2-bianchi?

Hi Eric -

Seems to me that what you have in mind is precisely encoded in

(1)${d}_{A}B\approx \left[{H}^{\left(1\right)},{H}^{\left(2\right)}\right]$

which we know to be correct. This equation tells you how that 2-path along the cube has to look like.

First of all, it really involves a 3-form instead of the 4-form that you arrived at and which has the wrong degree, I think.

Next… But I don’t need to spell that all out, because there is a beautiful lattice derivation of all this in equations (3.26)-(3.28) of

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

This derivation actually also illuminates how correctly interpreting this lattice result really requires all the 2-group ideas. You can’t do away with them because things should be made as simple as possible, but not simpler - as somebody once noted.

Posted by: Urs Schreiber on September 12, 2004 10:10 PM | Permalink | PGP Sig | Reply to this

### Re: 2-flatness = 2-bianchi?

Hello :)

Next… But I don’t need to spell that all out, because there is a beautiful lattice derivation of all this in equations (3.26)-(3.28) of

I actually tried to read this paper way back when it first came out. I’ll consider it homework to figure out what is going on :)

This derivation actually also illuminates how correctly interpreting this lattice result really requires all the 2-group ideas. You can’t do away with them because things should be made as simple as possible, but not simpler - as somebody once noted.

Oops! I didn’t mean to suggest that all the 2-group stuff could be done away with. I only meant to suggest that it might work out more naturally in the framework of discrete differential geometry, i.e. I think that some things that might seem mysterious actually become obvious when you think in terms of forms on a diamond complex. It’s just a hunch.

Eric

PS: Of course I have a hidden agenda. I want discrete differential geometry to work its way into the paper you are writing with Baez ;)

Posted by: Eric on September 13, 2004 4:14 AM | Permalink | Reply to this

### Re: r-flatness = 2-associativity

Hi Eric -

I just came across a text where indeed Poincaré duality of n-category diagrams is considered: See p. 14 of

Low dimensional topology and higher-order catergories (by a surprisingly large amount of authors which I won’t list here…)

Posted by: Urs Schreiber on September 6, 2004 5:53 PM | Permalink | PGP Sig | Reply to this

### Girelli & Pfeiffer

Good morning,

I spent a little time looking at the paper by Girelli & Pfeiffer last night and had some questions. I know you are busy, so feel free to ignore the questions temporarily or indefinitely :)

First, looking at Equation (2.10), it seems like this already requires the 3-curvature to vanish because the 2-connection is obtained from a 1-connection. I’m probably wrong about that, but I don’t understand the motivation for that equation.

Second, $g▹h$ seems to be like some kind of parallel transport. I don’t yet understand exactly how it works. It looks like somehow ${g}_{\mathrm{ij}}▹{h}_{\mathrm{jkl}}$ pulls ${h}_{\mathrm{jkl}}$ back from node $j$ to node $i$.

Well, technically those are more like observations than questions, but if you get the urge to set me straight, I wouldn’t resist :)

Gotta run for now,
Eric

Posted by: Eric on September 13, 2004 3:08 PM | Permalink | Reply to this

### Re: Girelli & Pfeiffer

Hi Eric -

your observations are perfectly correct and highlight two very important points.

Indeed, that equation (2.10) is the macroscopic version of the differential $\mathrm{dt}\left(B\right)+F=0$ constraint.

Unfortunately, the reason for writing down this pivotal equation is not really explained in that paper by Girelli&Pfeiffer, and it took me a while, and some additional information from John Baez, to fully figure it out:

There is a theorem telling you that from every crossed module you get a 2-group and from every 2-group you get a crossed module and, most importantly, that these processes are mutually inverse up to isomorphism.

(As you will have seen, a crossed module is just two groups $G$ and $H$ and a way for $G$ to act on $H$ by automorphisms).

This means that given any 2-group whatsoever you can think of it as coming from a crossed module in a standard way. This standard way involves imposing equation (2.10).

So that’s where (2.10) comes from. But there is a subtlety (really two of them):

There is (at least one) non-standard way to get a 2-group from a crossed module. This non-standard way is more directly related to the loop space formalism. It is very similar to the standard way, but slightly different. In the non-standard way the equation (2.10) does not appear, but instead the weaker r-flatness condition appears.

Still, since it is a theorem that every 2-group comes from a crossed module in the standard way, it must follow that any 2-group coming from a crossed module in a non-standard way and not satisfying (2.10) must be isomorphic to a 2-group coming from another crossed module in the standard way.

Indeed, this can be checked to be true. In section 3.2.4 (‘Examples’) of my notes I demonstrate in detail how the isomorphisms look like which relate the non-standard 2-groups that follow from known solutions of the r-flatness condition to the standard 2-groups.

But there is a further subtlety, and this is a more important one:

In order to compute surface holonomy we do not really need a 2-group connection. What we really need is a 2-functor from the strict 2-groupoid of bigons to what I call a ‘very weak 2-group’.

The reason is simple: The crucial ‘exchange law’ in 2-group theory which says that horizontal and vertical composition of surface elements must commute really only needs to hold for adjacent surface elements. But if we demand the exchange law to hold in the 2-group target of the 2-connection functor it must then also hold non-locally, which is too strict a condition.

You can find more details on what I am talking about here in the section 3.2 (‘Highbrow apprpoach’) of my notes. I believe that I have an example of an interesting weak 2-connection which is not a strict 2-connection and hence in particular avoids the restrictive equation (2.10).

Next, $g⊲h$ is indeed precisely the parallel transport of with respect to $g$. This is a crucial insight for relating 2-group formalism to loop space formalism and I work out the details of that in section 3.1 (‘Lowbrow approach’) and section 4 (‘Translating between 2-groups and loop space’).

Posted by: Urs Schreiber on September 13, 2004 8:28 PM | Permalink | PGP Sig | Reply to this

### Discrete p-Connections

Hi Urs,

In the treatment of lattice Yang-Mills in our notes, it is beginning to occur to me that maybe we made quite a few assumptions. I’d like to revisit those for a second.

There we had $G$-valued cochains, where $G$ is some group algebra. For example, a $G$-valued 0-cochain is written

(1)$\varphi =\sum _{i}{\varphi }_{i}{e}^{i}$

with ${\varphi }_{i}\in G$. We assumed that given a second $G$-valued 0-cochain $\psi$ we would have

(2)$\varphi \psi =\sum _{i}{\varphi }_{i}{\psi }_{i}{e}^{i}.$

Now, I claim that this is only true if we have a vanishing 0-connection giving rise to a trivial 0-holonomy

(3)${H}^{\left(0\right)}=\sum _{i}{H}_{i}{e}^{i}.$

For a trivial 0-holonomy, ${H}_{i}=1$ for all $i$. If the 0-connection does not vanish, then I claim we have nontrivial commutation relations between elements of $G$ and the bases ${e}^{i}$. In particular, we will have

(4)${e}^{i}g={H}_{i}^{-1}g{H}_{i}{e}^{i},$

i.e. $g$ gets conjugated by the 0-holonomy upon commuting to the other side. In this case, we would have

(5)$\varphi \psi =\sum _{i}{\varphi }_{i}{H}_{i}^{-1}{\psi }_{i}{H}_{i}{e}^{i}.$

In particular

(6)${H}^{\left(0\right)}\varphi =\varphi {H}^{\left(0\right)}.$

The generalization is pretty obvious. If we have a nonvanishing 1-connection, then we’d have

(7)${e}^{\mathrm{ij}}g={H}_{\mathrm{ij}}^{-1}g{H}_{\mathrm{ij}}{e}^{\mathrm{ij}}.$

If we have a nonvanishing $p$-connection, we’d have

(8)${e}^{{i}_{0}\dots {i}_{p}}g={H}_{{i}_{0}\dots {i}_{p}}^{-1}g{H}_{{i}_{0}\dots {i}_{p}}{e}^{{i}_{0}\dots {i}_{p}}.$

It’s late and I gotta run and I admit that this is a half-baked idea, but I think something like this will turn out to be true if you want to study higher holonomies on a diamond complex.

Good night!
Eric

Posted by: Eric on September 15, 2004 5:23 AM | Permalink | Reply to this

### Re: Discrete p-Connections

True, that’s a good point. I’ll have to think about it.

Posted by: Urs Schreiber on September 15, 2004 8:46 AM | Permalink | PGP Sig | Reply to this

### Re: Discrete p-Connections

A quick note before I head to work…

We can define a $\left(p+1\right)$-curvature by

(1)${F}^{\left(p+1\right)}=\left[{H}^{\left(1\right)},{H}^{\left(p\right)}\right].$

If I define the lattice 1-Yang-Mills action by

(2)$〈{F}^{\left(1\right)}\mid {F}^{\left(1\right)}〉,$

I get

(3)$〈{F}^{\left(1\right)}\mid {F}^{\left(1\right)}〉=2N\left\{1-\frac{1}{N}\Re \left[tr\left(\sum _{\mathrm{ij}}{H}_{i}^{-1}{H}_{\mathrm{ij}}^{-1}{H}_{j}{H}_{\mathrm{ij}}\right)\right]\right\}.$

This has a similar spirit as the Wilson action and obviously vanishes for trivial 0-holonomies.

Maybe there is something interesting here :)

Gotta run!
Eric

Posted by: Eric on September 15, 2004 1:10 PM | Permalink | Reply to this

### Re: Discrete p-Connections

But what is ${H}_{\mathrm{ij}}^{\left(0\right)}$?? :-)

Posted by: Urs Schreiber on September 15, 2004 8:35 PM | Permalink | PGP Sig | Reply to this

### Re: Discrete p-Connections

What?!?! :)

There is no ${H}_{\mathrm{ij}}^{\left(0\right)}$!! :)

That is expression is you get when you expand

(1)$〈{F}^{\left(1\right)}\mid {F}^{\left(1\right)}〉,$

where

(2)${F}^{\left(1\right)}=\left[{H}^{\left(1\right)},{H}^{\left(0\right)}\right].$

Boy. I was so excited to see a response from you and that is what I get. hehe :)

Go to sleep! :)
Eric

Posted by: Eric on September 15, 2004 9:01 PM | Permalink | Reply to this

### Re: Discrete p-Connections

Ok. Since this is more fun than work anyway, let me spell it out.

(1)${H}^{\left(0\right)}=\sum _{i}{H}_{i}{e}^{i}$
(2)${H}^{\left(1\right)}=\sum _{\mathrm{ij}}{H}_{\mathrm{ij}}{e}^{\mathrm{ij}}$
(3)${e}^{\mathrm{ij}}{H}_{j}={H}_{\mathrm{ij}}^{-1}{H}_{j}{H}_{\mathrm{ij}}{e}^{\mathrm{ij}}$
(4)${e}^{i}{H}_{\mathrm{ij}}={H}_{i}^{-1}{H}_{\mathrm{ij}}{H}_{i}{e}^{i}$
(5)${H}^{\left(1\right)}{H}^{\left(0\right)}=\sum _{\mathrm{ij}}\left({H}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)\left({H}_{j}{e}^{j}\right)=\sum _{\mathrm{ij}}{H}_{\mathrm{ij}}{H}_{\mathrm{ij}}^{-1}{H}_{j}{H}_{\mathrm{ij}}{e}^{\mathrm{ij}}=\sum _{\mathrm{ij}}{H}_{j}{H}_{\mathrm{ij}}{e}^{\mathrm{ij}}.$
(6)${H}^{\left(0\right)}{H}^{\left(1\right)}=\sum _{\mathrm{ij}}\left({H}_{i}{e}^{i}\right)\left({H}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)=\sum _{\mathrm{ij}}{H}_{i}{H}_{i}^{-1}{H}_{\mathrm{ij}}{H}_{i}{e}^{\mathrm{ij}}=\sum _{\mathrm{ij}}{H}_{\mathrm{ij}}{H}_{i}{e}^{\mathrm{ij}}.$
(7)${F}^{\left(1\right)}=\left[{H}^{\left(1\right)},{H}^{\left(0\right)}\right]=\sum _{\mathrm{ij}}\left({H}_{j}{H}_{\mathrm{ij}}-{H}_{\mathrm{ij}}{H}_{i}\right){e}^{\mathrm{ij}}.$
(8)$〈{F}^{\left(1\right)}\mid {F}^{\left(1\right)}〉=tr\left[\sum _{\mathrm{ij}}{\left({H}_{j}{H}_{\mathrm{ij}}-{H}_{\mathrm{ij}}{H}_{i}\right)}^{†}\left({H}_{j}{H}_{\mathrm{ij}}-{H}_{\mathrm{ij}}{H}_{i}\right)\right]$

Expand this out and you’ll get the expression I wrote before :)

Eric

Posted by: Eric on September 15, 2004 9:20 PM | Permalink | Reply to this

### Re: Discrete p-Connections

Looking at the expressions

(1)${e}^{\mathrm{ij}}{H}_{j}={H}_{\mathrm{ij}}^{-1}{H}_{j}{H}_{\mathrm{ij}}{e}^{\mathrm{ij}}$

and

(2)${e}^{i}{H}_{\mathrm{ij}}={H}_{i}^{-1}{H}_{\mathrm{ij}}{H}_{i}{e}^{i}$

it seems like maybe we want to think of the elements

(3)${H}_{i}{e}^{i}$

and

(4)${H}_{\mathrm{ij}}{e}^{\mathrm{ij}}$

as your basis elements. If we do this, something cute happens.

(5)${F}^{\left(1\right)}=\sum _{\mathrm{ij}}\left({H}_{j}{H}_{\mathrm{ij}}-{H}_{\mathrm{ij}}{H}_{i}\right){e}^{\mathrm{ij}}=\sum _{\mathrm{ij}}\left({H}_{j}-{H}_{\mathrm{ij}}{H}_{i}{H}_{\mathrm{ij}}^{-1}\right){H}_{\mathrm{ij}}{e}^{\mathrm{ij}}.$

This expression carries the essence of parallel transport. The term

(6)${H}_{\mathrm{ij}}{H}_{i}{H}_{\mathrm{ij}}^{-1}$

says that if you have an element ${g}_{j}\in G$ at node I and operate on it with the above, then the result is a parallel transport of ${g}_{j}$ from $j$ to $i$ via ${H}_{\mathrm{ij}}^{-1}$, then multiply by ${H}_{i}$ and finally parallel tranport the result back from $i$ to $j$ via ${H}_{\mathrm{ij}}$. In other words, you can only compare elements in the same “tangent space.”

Neat, eh? :)

Eric

Posted by: Eric on September 15, 2004 9:32 PM | Permalink | Reply to this

### Re: Discrete p-Connections

I guess a better way to say it is that

(1)${H}_{\mathrm{ij}}{H}_{i}{H}_{\mathrm{ij}}^{-1}$

is the parallel transport of ${H}_{i}$ from node $i$ to node $j$. Duh! Sorry :)

Eric

Posted by: Eric on September 15, 2004 9:40 PM | Permalink | Reply to this

### Re: Discrete p-Connections

As you could probably have guessed, 0-holonomies and 1-holonomies worked out beautifully, but I do not yet see a way to handle 2-holonomies :)

There are some strange associativity issues creeping up for some reason. I wonder what those could be? :)

Eric

Posted by: Eric on September 16, 2004 3:23 AM | Permalink | Reply to this

### Re: Discrete p-Connections

There are some strange associativity issues creeping up for some reason.

Hmm…

There are already some associativity issues appearing for 0-holonomies.

(1)$\left[\left({a}_{i}{e}^{i}\right)\left({b}_{i}{e}^{i}\right)\right]\left({c}_{i}{e}^{i}\right)=\left({a}_{i}{H}_{i}^{-1}{b}_{i}{H}_{i}{e}^{i}\right)\left({c}_{i}{e}^{i}\right)={a}_{i}{H}_{i}^{-1}{b}_{i}{c}_{i}{H}_{i}{e}^{i}.$
(2)$\left({a}_{i}{e}^{i}\right)\left[\left({b}_{i}{e}^{i}\right)\left({c}_{i}{e}^{i}\right)\right]=\left({a}_{i}{e}^{i}\right)\left({b}_{i}{H}_{i}^{-1}{c}_{i}{H}_{i}{e}^{i}\right)={a}_{i}{H}_{i}^{-1}{b}_{i}{H}_{i}^{-1}{c}_{i}{H}_{i}{H}_{i}{e}^{i}.$

Associativity requires that

(3)${c}_{i}={H}_{i}^{-1}{c}_{i}{H}_{i}$

or

(4)${H}_{i}{c}_{i}={c}_{i}{H}_{i}.$

In other words, associativity requires that the 0-holonomy commute with all ${c}_{i}\in G$, i.e the 0-holonomy ${H}^{\left(0\right)}$ must be trivial.

This might not be too surprising I guess. I almost expected something like this. A nontrivial 0-holonomy doesn’t really make much sense :)

Vanishing 0-holonomy also fixes the associativity with 1-holonomies

(5)$\left[\left({a}_{i}{e}^{i}\right)\left({b}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)\right]\left({c}_{j}{e}^{j}\right)=\left({a}_{i}{H}_{i}^{-1}{b}_{\mathrm{ij}}{H}_{i}{e}^{\mathrm{ij}}\right)\left({c}_{j}{e}^{j}\right)={a}_{i}{H}_{i}^{-1}{b}_{\mathrm{ij}}{H}_{i}{H}_{\mathrm{ij}}^{-1}{c}_{j}{H}_{\mathrm{ij}}{e}^{\mathrm{ij}}.$
(6)$\left({a}_{i}{e}^{i}\right)\left[\left({b}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)\left({c}_{j}{e}^{j}\right)\right]=\left({a}_{i}{e}^{i}\right)\left({b}_{\mathrm{ij}}{H}_{\mathrm{ij}}^{-1}{c}_{j}{H}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)={a}_{i}{H}_{i}^{-1}{b}_{\mathrm{ij}}{H}_{\mathrm{ij}}^{-1}{c}_{j}{H}_{\mathrm{ij}}{H}_{i}{e}^{\mathrm{ij}}.$

Associativity requires

(7)${H}_{i}{H}_{\mathrm{ij}}^{-1}{c}_{j}{H}_{\mathrm{ij}}={H}_{\mathrm{ij}}^{-1}{c}_{j}{H}_{\mathrm{ij}}{H}_{i}.$

However, since the 0-holonomy must be trivial, this reduces to

(8)${H}_{\mathrm{ij}}^{-1}{c}_{j}{H}_{\mathrm{ij}}={H}_{\mathrm{ij}}^{-1}{c}_{j}{H}_{\mathrm{ij}}.$

Voila :) We can have nontrivial 1-holonomies (thank goodness) :)

I haven’t worked it out yet (it’s not obvious to me how to proceed), but I’m willing to bet that enforcing associativity for

(9)$\left({a}_{i}{e}^{i}\right)\left({b}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)\left({c}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)$

and

(10)$\left({a}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)\left({b}_{\mathrm{jk}}{e}^{\mathrm{jk}}\right)\left({c}_{\mathrm{kl}}{e}^{\mathrm{kl}}\right)$

gives something equivalent to r-flatness.

Eric

Posted by: Eric on September 17, 2004 4:56 AM | Permalink | Reply to this

### Re: Discrete p-Connections

Typo

(1)$\left({a}_{i}{e}^{i}\right)\left({b}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)\left({c}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)$

should be

(2)$\left({a}_{i}{e}^{i}\right)\left({b}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)\left({c}_{\mathrm{jk}}{e}^{\mathrm{jk}}\right)$
Posted by: Eric on September 17, 2004 5:15 AM | Permalink | Reply to this

### Re: Discrete p-Connections

Hi Eric -

sorry for being so slow with replying.

This parallel tranport implicit when in going from right multiplication to left multiplication with a basis element looks like a very good idea to me. Indeed, this is what all the 2-group business is dealing with, too, though not all people working on 2-groups will admit it immediately. ;-)

In 2-group theory the edge and surface holonomy is in a sense which I could make precise always located at a point. Comparing and/or composing 2-holonomies always involves parallel transporting the group elements sitting at these points to a common position, and only there they live in the same space.

This is exactly what you are doing here, too. And I agree that this should have been included already in our notes, probably. I haven’t followed all your equations in full detail (but I see my previous mistake now) but it seems like even with this new parallel transport introduced you get the ordinary expression for the gauge curvature by evaliuating ${F}^{\left(2\right)}$, right?

When the 1-form case works out nicely one should try to see how this generalizes to 2-forms. Of course you are doing that right now!

The associativity condition that you have considered should somehow extend to a notion of 2-associativity. For this to work one will have to consider product other than the ordinary products of the paths/cochains, because this ordinary product increases the degree of the object: What one would need is a way to compose two 2-cochains such that one gets a total 2-cochain being the concatenation of both.

Do you have an idea for that?

Posted by: Urs Schreiber on September 17, 2004 2:22 PM | Permalink | Reply to this

### Re: Discrete p-Connections

Hi Urs,

it seems like even with this new parallel transport introduced you get the ordinary expression for the gauge curvature by evaliuating ${F}^{\left(2\right)}$, right?

Naively, it seems so, but I am not sure. The thing I am struggling with at the moment is what to do with terms like

(1)${e}^{\mathrm{ij}}{H}_{\mathrm{jk}}{e}^{\mathrm{jk}}.$

I know I want to move ${H}_{\mathrm{jk}}$ over to the left of ${e}^{\mathrm{ij}}$, but I’m not sure if this should involve a 1-holonomy, a 2-holonomy, or both. If I was forced to guess, I’d say it would involve a 2-holonomy, but that would mean the transporting something to the left over a basis depends on what is to the right, if you know what I mean.

Let me write stuff down and see if it inspires any ideas…

(2)${e}^{\mathrm{ij}}{H}_{\mathrm{jk}}{e}^{\mathrm{jk}}={H}_{\mathrm{ijk}}^{-1}{H}_{\mathrm{jk}}{H}_{\mathrm{ijk}}{e}^{\mathrm{ijk}}$

Therefore,

(3)$\left[{H}^{\left(1\right)}{\right]}^{2}=\sum _{\mathrm{ijk}}\left({H}_{\mathrm{ij}}{e}^{\mathrm{ij}}\right)\left({H}_{\mathrm{jk}}{e}^{\mathrm{jk}}\right)=\sum _{\mathrm{ijk}}{H}_{\mathrm{ij}}{H}_{\mathrm{ijk}}^{-1}{H}_{\mathrm{jk}}{H}_{\mathrm{ijk}}{e}^{\mathrm{ijk}}.$

Let me introduce some notation. Define

(4)${H}_{j\to i}={H}_{\mathrm{ij}}^{-1}{H}_{j}{H}_{\mathrm{ij}}.$

Similarly, define

(5)${H}_{\mathrm{jk}\to \mathrm{ij}}={H}_{\mathrm{ijk}}^{-1}{H}_{\mathrm{jk}}{H}_{\mathrm{ijk}}.$

Ok. I’m stuck. Now what? :)

Eric

Posted by: Eric on September 17, 2004 9:51 PM | Permalink | Reply to this

### Re: Discrete p-Connections

Hi Eric -

I am not sure how to proceed, but I can provide the followig observation from 2-group reasoning: There you need a 2-connection only to trasport an edge vertically, which means that the endpoints of the edge are kept fix while the rest of the edge may move. In such a process the surface traced out by moving the edge is the one from which the group element for the 2-parallel transport is obtained. No such complication arises when moving an edge ‘horizontally’ in such a way that it does not sweep out any surface at all.

Posted by: Urs Schreiber on September 19, 2004 5:11 PM | Permalink | PGP Sig | Reply to this

### Re: Discrete p-Connections

Hi Urs,

I don’t really know how to proceed either, but it might be a good idea to try to make closer analogies with 2-group stuff as you seem to be suggesting.

I have been trying to work with a single diamond complex D. On a single diamond, the edges you would transport would be the actual edges of a graph, e.g. ${e}^{\mathrm{ij}}$. However, the 2-group stuff seems to be more closely related to loop space, so many I should be dealing with “polygon space” as we were talking about before. In that case, the edges that would get transported would not be things like ${e}^{\mathrm{ij}}$, but would rather be things like ${e}^{i}\otimes {e}^{j}$ on a product diamonds. In other words, ${e}^{i}\otimes {e}^{j}$ is a 0-cochain on $D×D$.

In this case, transporting an edge amounts to just transporting the endpoints and we know how to do that. This would also seem to give something like a “vertical” transport.

Just a thought…

Eric

Posted by: Eric on September 19, 2004 5:33 PM | Permalink | Reply to this

### Re: Discrete p-Connections

Hi Urs,

I don’t know if I am moving forward or backward, but here are some more thoughts :)

If ${e}^{i}\otimes {e}^{j}$ is a 0-cochain in polygon space, then what is $d\left({e}^{i}\otimes {e}^{j}\right)$? I’m thinking that it should be

(1)$d\left({e}^{i}\otimes {e}^{j}\right)=\left({\mathrm{de}}^{i}\right)\otimes \left({\mathrm{de}}^{j}\right).$

Then it would follow that an “edge” in polygon space from the “node” ${e}^{i}\otimes {e}^{k}$ to ${e}^{j}\otimes {e}^{l}$ would be written

(2)${e}^{\mathrm{ij}}\otimes {e}^{\mathrm{kl}}.$

This makes a certain amount of sense.

Then the 2-holonomy is given by

(3)${H}^{\left(2\right)}={H}^{\left(1\right)}\otimes {H}^{\left(1\right)}.$

I’m not really sure where I am going with this, but I thought it wouldn’t hurt to throw some ideas out there and see if anything makes sense (as usual) :)

Cheers,
Eric

Posted by: Eric on September 19, 2004 7:56 PM | Permalink | Reply to this

### Knock knock?

Urs!

Where are you? You’ve disappeared :)

How is the progress with your collaboration with Baez going? Are you becoming an internet hermit like he did? :)

Best wishes,
Eric

Posted by: Eric on October 1, 2004 5:19 AM | Permalink | Reply to this

### Re: Knock knock?

Hi Eric -

I wouldn’t say that I have completely disappeared. You find my name here and there in sci.physics.strings, sci.physics.research - and recently also on sci.math.reserach :-)

But I haven’t posted to the SCT so much lately. There are several reasons for that. I am glad that Robert McNees posted a new entry and I’d enjoy hearing back from him about news from that conference on black holes.

Next week I’ll be on a small meeting of our local research groups and right after that our semester starts again, which means I’ll be busy busy busy.

What is the status of our project to write an exposition on loop space (or path space) differential geometry?

John Baez pointed out to me that in Gambini & Pullin’s book the multi-integrals on loops on whose relevance we had a long discussion are also discussed (in chapter 2). I once started reading that book but dropped it as soon as coming on p.3 across the statement that it only deals with unparameterized loops. But maybe it is a good starting point. I recall that you are quite fond of Gambini&Pullin, so maybe we could restart our project based on the material of that book. It seems that they don’t discuss exterior differentials and differential forms and tensor calculus on loop space, so that would be something for us to do.

I have now ordered the book with Chen’s collected papers. That might help us get some of the math largely rigorous. If you are interested, that is…

Posted by: Urs Schreiber on October 1, 2004 11:08 AM | Permalink | PGP Sig | Reply to this

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