## April 20, 2005

#### Posted by urs

Recently I mentioned here an idea by E. Akhmedov to construct a nonabelian surface holonomy using 2D TFTs. Today he has a new preprint on that issue.

It is well known that for every associative semi-simple algebra with structure constants ${C}^{i}{}_{\mathrm{jk}}$ there is a 2-dimensional topological field theory whose partition sum is computed by triangulating the surface, assigning one ${C}^{i}{}_{\mathrm{jk}}$ to each surface element and contracting all indices using an orientation on the edges of the triangulation.

It seemed to me that E. Akhmedov was proposing to generalize this by allowing to replace the ${C}^{i}{}_{\mathrm{jk}}$ with ${C}^{i}{}_{\mathrm{jk}}+{B}^{i}{}_{\mathrm{jk}}\left(x\right)$, where the additional tensor $B$ depends on the triangle $x$ that it is assigned to. This will of course no longer give a topological field theory but if this could be given a well defined continuum limit it might have the interesting interpretation as a way to compute some notion of ‘nonabelian’ surface holonomy.

I don’t see yet, though, that this idea has been shown to have a well-defined implementation, some aspects of which I discussed on sci.physics.strings.

After seeing his hep-th/0503234 I had emailed E. Akhmedov asking some technical questions and mentioning that there has been previous work on this question.

Unfortunately I never received a reply. But today a new paper appears on the arXiv:

E. Akhmedov, V. Dolotin & A. Morozov
Comment on the Surface Exponential for Tensor Fields
hep-th/0504160

In this paper some aspect of the previous proposal is being examined more closely.

This paper now does cite previous work on related issues. In the introduction it mentions that

the problem is known under many names, from topological models [2,3] to Connes-Kreimer theory [4]-[6] and that of the 2-categories [7]

This is not quite correct. 2-categories would be needed for volume-holonomy, since using them you can construct 3-groups. What is true is that general 2-groups, which are 1-categories have been considered as a tool for investigating surface holonomy.

The reference number [7] in the above preprint is our From Loop Groups to 2-Groups. This, incidentally, is not concerned with the question that E. Akhmedov is addressing in his papers.

(It rather studies possibly interesting structure 2-groups for 2-bundles. In these 2-bundles we can study nonabelian surface holonomy, but this is discussed in hep-th/0412325).

After having had a closer look at E. Akhmedov’s proposal I have become a little sceptical that the technical details are being appropriately addressed. If anyone thinks I am wrong about this I would kindly ask him or her to help me for instance clarify the following question:

As long as we work with $x$-independent quantities ${C}^{i}{}_{\mathrm{jk}}$ it is a theorem that for the number we compute by assigning them to a triangulation of a surface to be well defined, the $C$ have to be the structure constants of a semisimple algebra.

Now if we let these quantities become position dependent by setting $C\to C+B\left(x\right)$ there must be some condition on the $B$ to ensure that the ‘partition sum’ we compute now is well defined and in particular has a well defined continuum limit.

The requirements may now be a little different, since we don’t want the surface field theory given by $C+B$ to be topological anymore. But there must still be some requirement.

What is this sufficient condition on $B$ and how can it be solved?

Posted at April 20, 2005 5:43 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/558

Dear Urs,
I have sent to you already
4 messages including two in reply to your
first message! I have checked with our system administrator the situation with
my messages to you and he says that
for some reason your computer rejects
my e-mails!?

First of all, i do not see any problems
for the situation when B-field depends on
x: this is explained in my first paper.
But i will be happy to hear more details
from you.

Second, we have cited that paper which you
mention rather than another because
it is this paper which we have read.
I will read soon another one as well.

Third, every thing we do can be
generalized to four-index tensors and
orderings over 3-volumes and so on:
this is mentioned in the concluding
section of my first paper. I do not
see any conceptual difficulties, but
ofcause one have to find an algorithm
for looking for the analogs of the matrix
I in those situations.

Fourth, concerning the situation
with ambigous two-forms. I can say
the following (at least as far as i understand your question): Consider a section of a
standard vector bundle. It is a “function” defined at
each point of the base of the bundle. I put function
into brakets because it is not globally defined! To deal
with such objects one uses the standard gauge connection,
which tells us how the “function” changes as we take one or
another path on the base. Similarly if a one-form gauge connection is
not globally defined (the only explicite case i know is
the Wu-Yang monopole) i suggest to deal with the connection
on the bundle whose base is the space of loops, i.e.
with the non-Abelian two-tensor field: standard one-form
gauge conncetion is “attached” to paths rather than points
(like the section of a vector bundle). Then the non-Abelian
two-tensor field defines how to “parallel” transport obgects
attached to paths! Then, extrapolating this idea, i suggest
to consider a three-tensor gauge field (with four “color” indices)
to define a parallel transport on the space of closed two-dimensional surfaces, i.e. describes the situation with non-globally defined
B-field attached to two-dimensional surfaces. And so on.

I hope my picture is clear and obvious.
Hopefully it will be useful for you.

Regards,
Emil.

Posted by: Emil Akhmedov on April 22, 2005 8:15 AM | Permalink | Reply to this

### Position-dependant coloring

Hi,

thanks for your reply. I am not sure what caused the problem with your mail, usually all mail gets through to me. The only thing I know is that, as an anti-spam measure, our server checks if the domain that incoming mail comes from is responsive.

I did, however, today get a message from Y. Lyublev, with a forwarded message from you where you ask if I did receive your last two messages.

But anyway, we can talk here in the comment section.

So:

The issue concerning the position-dependant $B$-field that I mentioned is for instance the following:

Consider a surface being a single square with four edges. The two simplest triangulations of this square are obtained by inserting an edge on either of the two diagonals. This gives rise to four different triangles. Call them ${x}_{L},{x}_{R}$ for one of the two triangulations and ${x}_{U},{x}_{D}$ for the other.

In figure 7 of hep-th/9212154 ${x}_{L}$ is the triangle $\left(\mathrm{ijp}\right)$, ${x}_{R}$ is the triangle $\left(\mathrm{pkl}\right)$, ${x}_{U}$ is the triangle $\left(\mathrm{ipl}\right)$ and ${x}_{D}$ is the triangle $\left(\mathrm{pjk}\right)$.

Switching from the triangulation $\left({x}_{L},{x}_{R}\right)$ to $\left({x}_{U},{x}_{D}\right)$ is the fusion move. I guess we want to have invariance of our ‘2-holonomy’ to-be under this move?

Now, if I understood correctly, you are essentially proposing to assign triangle-dependant tensors

(1)${H}_{\mathrm{ij}}{}^{k}\left(x\right)={C}_{\mathrm{ij}}{}^{k}+{B}_{\mathrm{ij}}{}^{k}\left(x\right)$

to each triangle $x$.

Here $C$ are the structure constants of a semisiple associative algebra. For $B=0$ this ensures invariance under the fusion move since in that case

(2)${H}_{\mathrm{ij}}{}^{p}\left({x}_{L}\right){H}_{\mathrm{pk}}{}^{l}\left({x}_{R}\right)={C}_{\mathrm{ij}}{}^{p}{C}_{\mathrm{pk}}{}^{l}={C}_{\mathrm{jk}}{}^{p}{C}_{\mathrm{ip}}{}^{l}={H}_{\mathrm{jk}}{}^{p}\left({x}_{D}\right){H}_{\mathrm{ip}}{}^{l}\left({x}_{U}\right)$

due to the associativity of the algebra. That’s the usual part.

But now if $B$ is nonvanishing we get a more intricate condition. Now from one triangulation we get

(3)${H}_{\mathrm{ij}}{}^{p}\left({x}_{L}\right){H}_{\mathrm{pk}}{}^{l}\left({x}_{R}\right)=\left({C}_{\mathrm{ij}}{}^{p}+{B}_{\mathrm{ij}}{}^{p}\left({x}_{L}\right)\right)\left({C}_{\mathrm{pk}}{}^{l}+{B}_{\mathrm{pk}}{}^{l}\left({x}_{R}\right)\right)={C}_{\mathrm{ij}}{}^{p}{C}_{\mathrm{pk}}{}^{l}+{C}_{\mathrm{ij}}{}^{p}{B}_{\mathrm{pk}}{}^{l}\left({x}_{R}\right)+{B}_{\mathrm{ij}}{}^{p}\left({x}_{L}\right){C}_{\mathrm{pk}}{}^{l}+{B}_{\mathrm{ij}}{}^{p}\left({x}_{L}\right){B}_{\mathrm{pk}}{}^{l}\left({x}_{R}\right)$

and from the other triangulation we get

(4)${H}_{\mathrm{jk}}{}^{p}\left({x}_{D}\right){H}_{\mathrm{ip}}{}^{l}\left({x}_{U}\right)=\left({C}_{\mathrm{jk}}{}^{p}+{B}_{\mathrm{jk}}{}^{p}\left({x}_{D}\right)\right)\left({C}_{\mathrm{ip}}{}^{l}+{B}_{\mathrm{ip}}{}^{l}\left({x}_{U}\right)\right)={C}_{\mathrm{jk}}{}^{p}{C}_{\mathrm{ip}}{}^{l}+{C}_{\mathrm{jk}}{}^{p}{B}_{\mathrm{ip}}{}^{l}\left({x}_{U}\right)+{B}_{\mathrm{jk}}{}^{p}\left({x}_{D}\right){C}_{\mathrm{ip}}{}^{l}+{B}_{\mathrm{jk}}{}^{p}\left({x}_{D}\right){B}_{\mathrm{ip}}{}^{l}\left({x}_{U}\right)\phantom{\rule{thinmathspace}{0ex}}.$

Equating these gives the following condition on $B$:

(5)${C}_{\mathrm{ij}}{}^{p}{B}_{\mathrm{pk}}{}^{l}\left({x}_{R}\right)+{B}_{\mathrm{ij}}{}^{p}\left({x}_{L}\right){C}_{\mathrm{pk}}{}^{l}+{B}_{\mathrm{ij}}{}^{p}\left({x}_{L}\right){B}_{\mathrm{pk}}{}^{l}\left({x}_{R}\right)={C}_{\mathrm{jk}}{}^{p}{B}_{\mathrm{ip}}{}^{l}\left({x}_{U}\right)+{B}_{\mathrm{jk}}{}^{p}\left({x}_{D}\right){C}_{\mathrm{ip}}{}^{l}+{B}_{\mathrm{jk}}{}^{p}\left({x}_{D}\right){B}_{\mathrm{ip}}{}^{l}\left({x}_{U}\right)\phantom{\rule{thinmathspace}{0ex}}.$

It are conditions like these that I believe have to be taken care of.

Of course we really just want this condition to hold in the continuum limit. So we could introduce coordinates $v$ and $w$ on our little square of side-length $4ϵ$, say, with the origin $v=0=w$ in the center and such that

(6)$B\left({x}_{R}\right)=B\left(0\right)+ϵ\left({\partial }_{v}B\right)\left(0\right)+O\left({ϵ}^{2}\right)$
(7)$B\left({x}_{L}\right)=B\left(0\right)-ϵ\left({\partial }_{v}B\right)\left(0\right)+O\left({ϵ}^{2}\right)$
(8)$B\left({x}_{U}\right)=B\left(0\right)+ϵ\left({\partial }_{w}B\right)\left(0\right)+O\left({ϵ}^{2}\right)$
(9)$B\left({x}_{D}\right)=B\left(0\right)-ϵ\left({\partial }_{w}B\right)\left(0\right)+O\left({ϵ}^{2}\right)\phantom{\rule{thinmathspace}{0ex}},$

where $0$ denotes the origin $v=0=w$.

Then we could insert this into the above condition and solve that condition to order ${ϵ}^{2}$, only. This turns the consistency condition on $B$ into a nonlinear differential equation which depends on the choice of structure constants $C$.

I think one would have to find solutions to that nonlinear differential equations.

The bubble move will probably have to be treated completely analogously. It will yield another condition in terms of a differential equation on $B$, I think.

When the $B$ field is not globally defined, one would need to worry about transitions and cocycle conditions and stuff like that. If there really is a way to get a surface holonomy out of this then the relevant cocycles almost certainly must be those known from nonabelian gerbes or 2-bundles, probably in their weak form. But maybe we can discuss this after we have discussed the above issue.

Posted by: Urs Schreiber on April 22, 2005 12:49 PM | Permalink | Reply to this

### Re: Position-dependant coloring

Hi, Urs!

I understand your point and my paper
resolves it: it was exactly the reason to write
it.

Let us first discuss the situation with constant
B-field. Ofcause I+B does not obey fusion rules
unless B is specially chosen. Hence the product
of (I+B)’s over a graph DOES depend
on this graph: depends on a concrete triangulation
of the surface in question even if B is the same
at each triangle!

But the
pont is that the exponent as defined in my paper
does NOT depend on the choise of the
sequence of graphs (triangulating
a genus g surface), which is picked up to take
M->\infty limit (M is the number
of triangles of the triangulation)!
The result of the limit is:

E(B, I,\kappa,g)_{j_1,…j_m} =
\sum_n 1/n! (B^(3))^n I^g_(j_1,…j_n| 3n)

where I^g_(j_1,…,j_m| 3n)
is the matrix with m + 3n indexes
which is obtained as the product of I_(3) over ANY
triangulation of the genus g surface with n+1 holes
and 3 external legs at each of the n holes and m
legs at one extra hole.

Now to construct area ordering we take
E(B(x), I,\kappa,0)_{j_1,j_2,j_3} and put it at each
triangle (x) of a triangulated surface (approximation
of the smooth surface over which we would
like to make the ordering) inside a
target space (Do not confuse this surface
with the ones which are used to define the exponent
itself!). Then we take the continuum limit from
the triangulation surfaces to the smooth one.
The result of such definition is that
the area ordering DOES not depend on the way
the limit is taken!!!

Rufly speaking this happens due to the
fact that all indexes of
different B-s (taken at different triangles)
are contructed via I’s,
but not directly to each other: i.e. in Teylor
expanding the area ordering
we do not encounter such situations as
B_{ijk}(x) B^k_{lm}(x’) rather we have something
like B_{ijk}(x) I^k_{lm} B^l_{nf} and etc.!

I hope it is clear what i mean!

Regards,
Emil.

Posted by: Emil on April 23, 2005 12:27 PM | Permalink | Reply to this

### Re: Position-dependent coloring

I understand that and why the contributions vanish where one $B$ is contracted with another, as in ${B}_{\mathrm{ijk}}{B}^{k}{}_{\mathrm{lm}}$.

But how do you know that the reamining terms give a triangulation-independent value?

It seems to me that you are saying that in the equation that I wrote down above the terms linear in $B$, namely

(1)${C}_{\mathrm{ij}}{}^{p}{B}_{\mathrm{pk}}{}^{l}\left({x}_{R}\right)+{B}_{\mathrm{ij}}{}^{p}\left({x}_{L}\right){C}_{\mathrm{pk}}{}^{l}-{C}_{\mathrm{jk}}{}^{p}{B}_{\mathrm{ip}}{}^{l}\left({x}_{U}\right)-{B}_{\mathrm{jk}}{}^{p}\left({x}_{D}\right){C}_{\mathrm{ip}}{}^{l}$

automatically cancel. I don’t see why that should be true. Do you discuss that in your paper?

Posted by: Urs on April 25, 2005 4:38 PM | Permalink | Reply to this

### Re: Position-dependent coloring

Dear Urs,
due to mine definition of the exponent
i do not encounter such a situation (as
the one you are ralking about) in the
definition of the area-ordering.
Rather one encounters the situation as follows:

I_{ijk|m_1m_2m_3} B^{m_1m_2m_3}(x) \times
I^k_{lm|k_1k_2k_3} B^{k_1k_2k_3}(y) =
I_{ijlm|m_1m_2m_3|k_1k_2k_3} \times
B^{m_1m_2m_3}(x) B^{k_1k_2k_3}(y),
where the matrixes I_{ijk|m_1m_2m_3}, etc.
are defined in my first paper.

Regards,
Emil.

Posted by: Emil on April 26, 2005 10:56 AM | Permalink | Reply to this

### Re: Position-dependent coloring

I had another look at your paper. So if I understand correctly, the situation I was talking about does not arise in your construction because the graph that I was talking about is not one obtained by starting with a closed graph and removing non-adjacent triangles. I missed this part on first reading.

OK, so then I am back to my first question, which maybe you answered in one of those mails that didn’t reach me: Your construction seems to amount to saying that you pick a 2-form $B$ which takes values in vertices of a topological 2D field theory and define the surface holomy ${U}_{B}\left(\Sigma \right)$ to be

(1)${U}_{B}\left(\Sigma \right)=〈\mathrm{exp}\left({\int }_{\Sigma }B\right){〉}_{g}$

where $〈\cdot {〉}_{g}$ is the TFT correlator for a closed surface $\Sigma$ of genus $g$.

Posted by: Urs on April 26, 2005 9:13 PM | Permalink | Reply to this

### Re: Position-dependent coloring

Dear Urs,
what i was declearing in my paper is that
I^g_{i_1…i_n} are correlators in TFT.
B-field itslef has nothing to do with TFT:
rather it is a perturbation over the TFT!

Concerning the formula for U_B(\Sigma)
which you are writing: I would like to
have such a formula. However, i do not know
how to get it!!! This is the problem i
am working on now in a sense.

Emil.

Posted by: Emil on April 28, 2005 11:33 AM | Permalink | Reply to this

### correlators

what i was declearing in my paper is that ${I}_{{i}_{1};\cdots ;{i}_{n}}^{g}$ are correlators in TFT. $B$-field itslef has nothing to do with TFT: rather it is a perturbation over the TFT!

Yes, I understand that. My question further above resulted from a misreading of your discussion in section 3 of hep-th/0503234.

Concerning the formula for ${U}_{B}\left(\Sigma \right)$ which you are writing: I would like to have such a formula. However, i do not know how to get it!!! This is the problem i am working on now in a sense.

I am not sure what you mean by ‘how to get it’. You can just take this as the definition of your surface holonomy!

And in fact it seems to me that this is what you are doing. Isn’t your equation (16) in hep-th/0503234 precisely saying that ${U}_{B}\left(\Sigma \right)=〈\mathrm{exp}\left({\int }_{\Sigma }B\right){〉}_{\Sigma }$ for the case of a discrete TFT where B carries these three indices and where $〈\cdots {〉}_{g}={I}_{\cdots }^{g}$?

To me it seems that you don’t really need the discussion of the exponent in section 3 to get this formula. You can just declare this to be your definition.

And if somebody hands you a continuum version of a 2D TFT then by just using the definition ${U}_{B}\left(\Sigma \right)=〈\mathrm{exp}\left({\int }_{\Sigma }B\right){〉}_{\Sigma }$ this idea of surface holonomy also generalizes to the continuum case, directly.

Don’t you think so?

P.S. You can insert formulas in your comment by simply typing ordinary LaTeX code in your text inside of dollar signs, as usual, and by choosing in the ‘Text Filter’ menu right above the edit pane the filter ‘itex to MathML with parbreaks’.

If you want to see more details on this have a look at this.

Posted by: Urs on April 28, 2005 12:11 PM | Permalink | Reply to this

### Open TFT

One more thing:

We get a 2-holonomy for a strict abelian 2-group from your proposal as follows:

A bigon in spacetime $M$ is a thin homotopy equivalence class of smooth maps

(1)$\Sigma :\left[0,1{\right]}^{2}\to M$

such that $\Sigma \left(0,t\right)=s\in M$ and $\Sigma \left(0,t\right)=t\in M$.

The image has disc topology and hence should be associated to an operator $V\to ℂ$ by the TFT, where $V$ is some vector space. This it does by setting the value of the surface holonomy of that bigon to be

(2)$V\ni v\stackrel{U\left(\Sigma \right)}{\to }{〈\mathrm{exp}\left({\int }_{\Sigma }B\right)〉}_{\Sigma }\in ℂ$

as I said before.

Now, by defining horizontal and vertical composition of these things by gluing sqares $\left[0,1{\right]}^{2}$ horizontally and vertically and reparameterizing them appropriately, one easily sees that this gives a strict abelian 2-group.

One way to make this a little bit more non-abelian would be to go to boundary TFT. For instance as described in Moore’s lecture.

For such a boundary TFT we could presumeably introduce a boundary-vertex-valued 1-form $A$ and generalize the above by

(3)$V\ni v\stackrel{U\left(\Sigma \right)}{\to }{〈P\mathrm{exp}\left({\int }_{\partial \Sigma }A\right)\mathrm{exp}\left({\int }_{\Sigma }B\right)〉}_{\Sigma }\in ℂ\phantom{\rule{thinmathspace}{0ex}},$

where $P$ indicates path ordering, as usual.

I’d say this now gives a nonabelian strict 2-group, if done right, where all the nonabelian nature comes from the $A$.

Posted by: Urs on April 28, 2005 4:41 PM | Permalink | Reply to this

### Re: Open TFT

Hi, Urs! Sorry for such a long delay: i was bisy with some beurokasy. Now i am in Japane and have more time for communication.

Actually, in August i will be i MPI, Golm and, as far as i understand you are in Germany… Hence, we can talk personally because i find it difficult to communicate via e-mail…

Concerning the formular

${U}_{B}\left(\Sigma \right)=〈\mathrm{exp}\left({\int }_{\Sigma }B\right){〉}_{\Sigma }$

I do not think that it is correct… To see that just try to make it explicite: what do you mean by ${e}^{\int B}$? With what do you conract the indices of B-field? What do you mean by $〈...{〉}_{\Sigma }$? Even if you will write this explicitely try to Teylor expand the expression over the powers of B: you will see that it will contain wrong terms and the numerical coefficients will not be the same as in the case of ${E}^{B}$!!! Apart from that the toplogical field theory expression does not obey eq (13) in my joint paper with Dolotin and Morozov. As well i do not know how to make explicite in this context the equation (2) in my first paper…

Posted by: emil on May 16, 2005 3:35 AM | Permalink | Reply to this

### Notation

Actually, in August i will be i MPI, Golm and, as far as i understand you are in Germany… Hence, we can talk personally because i find it difficult to communicate via e-mail…

Maybe I could visit the MPI. I’ll be in Oberwolfach on a gerbe workshop from August 14 to August 20, though.

what do you mean by ${e}^{\int B}$? With what do you conract the indices of B-field? What do you mean by $〈\cdots {〉}_{\Sigma }$?

I simply mean the standard things:

$〈\cdots {〉}_{\Sigma }$ is the correlator of the TFT on the surface $\Sigma$. So if ${V}_{1},{V}_{2},\dots ,{V}_{N}$ are vertex operators of the TFT then

(1)${〈{V}_{1}{V}_{2}\dots {V}_{N}〉}_{\Sigma }$

is the $N$-point function of the TFT (maybe up to normalization), i.e. the correlator obtained by taking $\Sigma$, cutting $N$ holes in it and inserting the ${V}_{\cdot }$ in these holes. For the discretized TFT that we are talking about, the vertices ${V}_{\cdot }$ will be rank-3 tensors in the vector space of the underlying semisimple associative algebra (i.e. they will ‘carry three indices’) and the prescription for computing the correlator is to choose any triangluation of $\Sigma$ with $N$ triangles removed, to insert the ${V}_{n}^{\mathrm{ijk}}$ at these triangles and compute the partition function of the triangulation. So ${〈\cdots 〉}_{\Sigma }$ with $N$ insertions is precisely what you write as ${I}_{\dots }^{g}$ with lots of indices.

Now, I said that $B$ is a 2-form taking values in vertices, so for the TFTs that we are talking about $B$ is a 2-form on $\Sigma$ (or pulled back to it) taking values in rank-3 tensors in that vector space of our algebra.

Hence

(2)$\begin{array}{ccc}{〈\mathrm{exp}\left({\int }_{\Sigma }B\right)〉}_{\Sigma }& =& \sum _{N=0}^{\infty }\frac{1}{N!}{〈\left({\int }_{\Sigma }B{\right)}^{N}〉}_{\Sigma }\\ \phantom{\rule{thinmathspace}{0ex}}& =& \sum _{N=0}^{\infty }\frac{1}{N!}\left(\prod _{n=1}^{N}{\int }_{\Sigma }{\mathrm{dx}}_{n}^{{\mu }_{n}}{\mathrm{dx}}_{n}^{{\nu }_{n}}\right){〈{B}_{{\mu }_{1}{\nu }_{1}}\left({x}_{1}\right){B}_{{\mu }_{2}{\nu }_{2}}\left({x}_{2}\right)\cdots {B}_{{\mu }_{N}{\nu }_{N}}\left({x}_{N}\right)〉}_{\Sigma }\\ \phantom{\rule{thinmathspace}{0ex}}& =& \sum _{N=0}^{\infty }\frac{1}{N!}\left(\prod _{n=1}^{N}{\int }_{\Sigma }{\mathrm{dx}}_{n}^{{\mu }_{n}}{\mathrm{dx}}_{n}^{{\nu }_{n}}\right){I}_{{j}_{1}{k}_{1}{l}_{1}\mid \dots \mid {j}_{N}{k}_{N}{l}_{N}}{B}_{{\mu }_{1}{\nu }_{1}}^{{j}_{1}{k}_{1}{l}_{1}}\left({x}_{1}\right){B}_{{\mu }_{2}{\nu }_{2}}^{{j}_{2}{k}_{2}{l}_{2}}\left({x}_{2}\right)\cdots {B}_{{\mu }_{N}{\nu }_{N}}^{{j}_{N}{k}_{N}{l}_{N}}\left({x}_{N}\right)\end{array}\phantom{\rule{thinmathspace}{0ex}},$

where in the last line we have the expression that you give in equation (16) of hep-th/0503234.

As well i do not know how to make explicite in this context the equation (2) in my first paper…

That equation says that taking $〈\mathrm{exp}\left({\int }_{{\Sigma }_{i}}{t}_{i}B\right){〉}_{{\Sigma }_{i}}$ for ${\Sigma }_{1}$, ${\Sigma }_{2}$ and ${\Sigma }_{3}$ closed surfaces with a single boundary component (i.e. each with ‘1 triangle removed’) and then gluing these three surfaces along their boundaries to a single closed surface ${\Sigma }_{\mathrm{tot}}$ with one connected boundary component is the same as computing

(3)${〈\mathrm{exp}\left({\int }_{{\Sigma }_{\mathrm{tot}}}\left({t}_{1}+{t}_{2}+{t}_{3}\right)B\right)〉}_{{\Sigma }_{\mathrm{tot}}}\phantom{\rule{thinmathspace}{0ex}}.$

This does hold for CFT correlators because we are dealing with topological field theories and the position of the insertion does not matter for the correlator.

Posted by: Urs Schreiber on May 17, 2005 10:26 AM | Permalink | Reply to this

### Re: Notation

But why do you think that at the $N$-th order there will not appear the contact terms? Say at the second order:

${I}_{{j}_{1}\phantom{\rule{thinmathspace}{0ex}}{k}_{1}\phantom{\rule{thinmathspace}{0ex}}{l}_{1}\phantom{\rule{thinmathspace}{0ex}}{j}_{2}\phantom{\rule{thinmathspace}{0ex}}{k}_{2}\phantom{\rule{thinmathspace}{0ex}}{l}_{2}}\phantom{\rule{thinmathspace}{0ex}}{B}^{{j}_{1}\phantom{\rule{thinmathspace}{0ex}}{k}_{1}\phantom{\rule{thinmathspace}{0ex}}{l}_{1}}{B}^{{j}_{2}\phantom{\rule{thinmathspace}{0ex}}{k}_{2}\phantom{\rule{thinmathspace}{0ex}}{l}_{2}}$, which are absent in the exponent in question. But from the point of view of the averaging $〈...{〉}_{\Sigma }$ it is not obvious why such terms do vanish…

As well i would like to see more explicite construction: Say we can easily extract the correlation functions like ${I}_{{j}_{1}{k}_{1}{l}_{1}\mid ...\mid {j}_{\mathrm{nk}}{}_{\mathrm{nl}}{}_{n}}$ from the matrix integral:

$\int {\prod }_{i=1}^{N}d{\stackrel{̂}{\Phi }}_{i}\phantom{\rule{thinmathspace}{0ex}}\mathrm{exp}\left(-\mathrm{Tr}\left[{\kappa }^{\mathrm{ij}}{\stackrel{̂}{\Phi }}_{i}\phantom{\rule{thinmathspace}{0ex}}{\stackrel{̂}{\Phi }}_{j}+{I}^{\mathrm{ijk}}\phantom{\rule{thinmathspace}{0ex}}{\stackrel{̂}{\Phi }}_{i}\phantom{\rule{thinmathspace}{0ex}}{\stackrel{̂}{\Phi }}_{j}\phantom{\rule{thinmathspace}{0ex}}{\stackrel{̂}{\Phi }}_{k}\right]\right)$, where ${\stackrel{̂}{\Phi }}_{i}=\mid \mid {\Phi }_{i}^{\mathrm{ab}}\mid \mid$ and $a=1,...,K$. In fact, the necessary correlator can be extracted by extracting the corresponding topology Feynman graph contribution to the corresponding correlation function: the only difficulty i see is how to distinguish the contect terms from generic ones: say the term ${I}_{{j}_{1}{k}_{1}{l}_{1}\mid ...\mid {j}_{\mathrm{nk}}{}_{\mathrm{nl}}{}_{n}}$ from ${I}_{{j}_{1}{k}_{1}{l}_{1}...{j}_{\mathrm{nk}}{}_{\mathrm{nl}}{}_{n}}$

But even if this is done it is not clear for me how to write the exponent in question as such a matrix integral, i.e. to write the averaging $〈{e}^{B}〉$ as the matrix integral!!!

Posted by: emil on May 18, 2005 2:57 AM | Permalink | Reply to this

### Re: Notation

But why do you think that at the $N$-th order there will not appear the contact terms? Say at the second order:

(1)${I}_{{j}_{1}{k}_{1}{l}_{1}{j}_{2}{k}_{2}{l}_{2}}{B}^{{j}_{1}{k}_{1}{l}_{1}}{B}^{{j}_{2}{k}_{2}{l}_{2}}\phantom{\rule{thinmathspace}{0ex}},$

which are absent in the exponent in question. But from the point of view of the averaging $〈\cdots {〉}_{\Sigma }$ it is not obvious why such terms do vanish…

I think these terms do not appear by the very definition of the correlator. By definition $〈{V}_{1}\dots {V}_{N}{〉}_{\Sigma }$ is the $N$-point function of the TFT on $\Sigma$, which means that it is the result of taking $\Sigma$, poking $N$-holes in it and inserting the vertices ${V}_{n}$ at these holes. For the discrete TFTs that we are talking about this gives just your prescription for how to compute ${I}_{{j}_{1}{k}_{1}{l}_{1}\mid \dots \mid {j}_{N}{k}_{N}{l}_{N}}^{\Sigma }$.

So for instance in order to compute

(2)${〈B\left({x}_{1}\right)B\left({x}_{2}\right)〉}_{\Sigma }$

the definition of the correlator tells us to take a triangulation of $\Sigma$ with any two seperate triangles removed and insert $B\left({x}_{1}{\right)}^{{i}_{1}{j}_{1}{k}_{1}}$ at one triangle and $B\left({x}_{2}{\right)}^{{i}_{2}{j}_{2}{k}_{2}}$ at the other. That holds true even if ${x}_{1}$ happens to equal ${x}_{2}$. The correlator does not know or care about if we think of the vertices inserted in it as functions of some position variable or not.

In order to emphasize this point we could do the following:

Let’s write ${\delta }_{\left(a,b,c\right)}$ for the vertex operator which is a rank-3 tensor with the following components

(3)${\delta }_{\left(a,b,c\right)}^{\mathrm{ijk}}=\left\{\begin{array}{cc}1& \mathrm{for}\phantom{\rule{thinmathspace}{0ex}}i=a,\phantom{\rule{thinmathspace}{0ex}}j=b\phantom{\rule{thinmathspace}{0ex}},k=c\\ 0& \mathrm{otherwise}\end{array}$

Then we can rewrite equivalently

(4)$B\left(x\right)=B\left(x{\right)}^{\mathrm{ijk}}{\delta }_{\left(i,j,k\right)}$

(with the sum over $i,j,k$ implicit, as always) and hence

(5)${〈B\left({x}_{1}\right)B\left({x}_{2}\right)〉}_{\Sigma }=B\left({x}_{1}{\right)}^{{i}_{1}{j}_{1}{k}_{1}}B\left({x}_{2}{\right)}^{{i}_{2}{j}_{2}{k}_{2}}{〈{\delta }_{\left({i}_{1},{j}_{1},{k}_{1}\right)}{\delta }_{\left({i}_{2},{j}_{2},{k}_{2}\right)}〉}_{\Sigma }\phantom{\rule{thinmathspace}{0ex}}.$

This makes it more vivid how the correlator does not know or care about the ‘position’ of the insertions ${V}_{1}={\delta }_{\left({i}_{1},{j}_{1},{k}_{1}\right)}$ and ${V}_{2}={\delta }_{\left({i}_{2},{j}_{2},{k}_{2}\right)}$. All we have to do to evaluate the above expression is to evaluate the general 2-point function

(6)${I}_{{i}_{1}{j}_{1}{k}_{1}\mid {i}_{2}{j}_{2}{k}_{2}}^{\Sigma }={〈{\delta }_{\left({i}_{1},{j}_{1},{k}_{1}\right)}{\delta }_{\left({i}_{2},{j}_{2},{k}_{2}\right)}〉}_{\Sigma }$

and then insert the result in the above expression to get

(7)$\begin{array}{ccc}{〈B\left({x}_{1}\right)B\left({x}_{2}\right)〉}_{\Sigma }& =& B\left({x}_{1}{\right)}^{{i}_{1}{j}_{1}{k}_{1}}B\left({x}_{2}{\right)}^{{i}_{2}{j}_{2}{k}_{2}}{〈{\delta }_{\left({i}_{1},{j}_{1},{k}_{1}\right)}{\delta }_{\left({i}_{2},{j}_{2},{k}_{2}\right)}〉}_{\Sigma }\\ \phantom{\rule{thinmathspace}{0ex}}& =& B\left({x}_{1}{\right)}^{{i}_{1}{j}_{1}{k}_{1}}B\left({x}_{2}{\right)}^{{i}_{2}{j}_{2}{k}_{2}}{I}_{{i}_{1}{j}_{1}{k}_{1}\mid {i}_{2}{j}_{2}{k}_{2}}^{\Sigma }\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

You also wrote:

But even if this is done it is not clear for me how to write the exponent in question as such a matrix integral, i.e. to write the averaging $〈{e}^{\in B}〉$ as the matrix integral!!!

Hm, I am not familiar with the matrix-theory reformulation of 2-dimensional TFTs. Before I can say anything about this I’d need to know more details. For starters, how would you specify the topology of the surface $\Sigma$ in the corresponding matrix integral? Say you wanted to compute the partition function ${Z}_{g}$ of some closed orientable surface of genus $g$ in your TFT. How would you implement the information about the genus into the matrix integral?

Posted by: Urs Schreiber on May 18, 2005 9:05 AM | Permalink | Reply to this

### Re: Notation

Remeber that the integral goes over $K×K$ matrices. Then to consider genus $g$ surface i have to pick up the terms of order ${K}^{\chi g}$. But there is the difficulty (exactly the one which concerns me) how to make n separate holes with 3 wedges at each hole. Basically we can calculate correlator as follows:

$〈{\Phi }_{{j}_{1}}\phantom{\rule{thinmathspace}{0ex}}{\Phi }_{{k}_{1}}\phantom{\rule{thinmathspace}{0ex}}{\Phi }_{{l}_{1}}...{\Phi }_{{j}_{n}}\phantom{\rule{thinmathspace}{0ex}}{\Phi }_{{k}_{n}}\phantom{\rule{thinmathspace}{0ex}}{\Phi }_{{l}_{n}}{〉}_{\mathrm{Matrix}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{integral}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{genus}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g}$. This correlator will contain all kinds of contrubutions corresponding to seprate holes or joint ones!!! And as far as i know there is no any way how to separate this kind of contributions: The contributions do not come with ${K}^{\chi \left(g,n\right)}$!!!

And this is the main point why i say that i do not know any explicite way of writing the formula ${U}_{\Sigma }\left(B\right)=〈{e}^{B}{〉}_{\Sigma }$. Your formulae are rather abstract to me: you just define that there should be a TFT which gives this formula. It is okay with me and i beleave that there should be something like that. But i would like to see an explicite definition of this TFT: through matrix (Kontsevich or Gross-Migdal formulation) or path (Witten formulation) integral!

Posted by: emil on May 19, 2005 3:17 AM | Permalink | Reply to this

### Re: Notation

Your formulae are rather abstract to me: you just define that there should be a TFT which gives this formula.

But, no, I am using explicitly the TFTs that you were originally talking about, namely those whose partition function is obtained by triangulating the surface and inserting the structure constants of a semisimple algebra at every triangle, summing all the indices consistently.

The definition of correlator that I am talking about is precisely the one also used, to name an example, by the original authors

M. Fukuma, S. Hosono & H. Kawai, hep-th/9212156

(For instance the 2-point function is discussed on top of p. 8)

The only, inessential, difference is that these authors concentrate on vertices that contract one edge of the dual graph, while you proposed to use vertices that contract three edges of the original graph.

I am not sure why you now want to express the nice elementarily defined (just contract some tensors) partition function and correlator by means of matrix integrals or functional integrals. What would it buy you in as far as you are concerned with getting hold of a notion of surface holonomy?

Posted by: Urs Schreiber on May 19, 2005 1:40 PM | Permalink | Reply to this

### Re: Notation

Urs, i published another paper in hep-th
where i discuss in grater details some of the points we have discussing here.

Posted by: emil on June 13, 2005 3:55 AM | Permalink | Reply to this

### hep-th/0506032

I am not sure about what is new in hep-th/0506032. But I would be interested in seeing the announced details to equation (36).

By the way, I find some of your statements a little puzzling. Right at the beginning of the introduction for instance you write that

knowing the value of a section of a fiber bundle at a point $x$, we can find its value at any other point $y$

You must be thinking of parallel transport in a bundle with connection here.

Consider for instance of a trivial $U\left(1\right)$-bundle over the interval $\left(0,1\right)$. Sections are all unimodular complex functions on $\left(0,1\right)$. Certainly these are not specified by their value at a single point, whether there is a connection on the bundle or not.

Posted by: Urs Schreiber on June 14, 2005 11:18 AM | Permalink | Reply to this

### Global Issues, Continuum Limit and Weak 2-Group Formulation

Concerning the issue of how to figure out the correct global definition as well as the continuum limit of this idea, if possible, I would like to repeat how I speculate this construction might be expressible in terms of a weak 2-group, if it works. Once one has a local 2-holonomy in terms of a weak 2-group it is straightforward to get the global construction by means of weak 2-bundles.

I have mentioned the following before on sci.physics.strings:

Start with the reformulation of the TFT-like 2-holonomy at finite resolution in terms of that ‘category of triangles’ that I mentioned recently. There seems to be an obvious structure that this construction asymptotes to in the continuum limit:

Let $G$ be the space in which the internal indices take values. For instance for the TFT defined from discrete group algebras, $G$ could be the limiting Lie group. But it might be something else, too (and one should not confuse this group with any ‘gauge group’ here).

Pick any connected collection of triangles, i.e. a small surface element. It defines a tensor with $n$ incoming and $m$ outgoing indices. In the limit this gives a map from pairs $\left({\gamma }_{1},{\gamma }_{2}\right)$ of smooth paths in $G$ to the base field

(1)$T\left({\gamma }_{1},{\gamma }_{2}\right)$

such that composition is given by ‘continuous index contraction’

(2)$\left(T\circ T\prime \right)\left({\gamma }_{1},{\gamma }_{2}\right)=\int D\left[\gamma \right]T\left({\gamma }_{1},\gamma \right)T\left(\gamma ,{\gamma }_{2}\right)\phantom{\rule{thinmathspace}{0ex}}.$

(Here the functional integral is over a reparameterization gauge slice, the choice of which is free due to the properties mentioned above.)

These have to be invertible and we can impose a ‘star-condition’ if we like. But since these $T$ are nothing but integral kernels we see that the group they form is that of unitary operators on something like ${L}^{2}\left(\mathrm{PG}\right)$, where $\mathrm{PG}$ is the path space over $G$. But in fact without any loss of generality we can assume that all paths start and end at a given point in $G$ and have sitting instant at that point and are parameterized by a parameter in $\left[0,1\right]$, so we really get $\Omega G$, the based loop space over $G$ and unitary operators on ${L}^{2}\left(\Omega G\right)$.

In addition to this ‘vertical product’ coming from composition there is a ‘horizontal’ product on these guys (coming from literally horizontally composing triangles) given by

(3)$\left(T\cdot T\prime \right)\left({\gamma }_{1},{\gamma }_{2}\right)=T\left(L\left({\gamma }_{1}\right),L\left({\gamma }_{2}\right)\right)T\prime \left(R\left({\gamma }_{1}\right),R\left({\gamma }_{2}\right)\right)$

where $L\left(\gamma \right)$ is the path that traces out the first half of $\gamma$ at twice the speed, and $R\left(\gamma \right)$ similarly gives the right half and it is implicit that $T$ and $T\prime$ vanish when their arguments are not based loops with sitting instant at the base.

In fact, I believe that this defines on the $T$s the structure of a weak monoidal category with all morphisms invertible. By throwing in weak formal horizontal inverses (representing the ‘zig-zag symmetry’ of holonomy which is otherwise not captured) we should get a weak 2-group.

At this vague level this is kind of obvious. The hard part is to take care of subtleties with these functional integrals and things like that.

Posted by: Urs Schreiber on April 22, 2005 7:05 PM | Permalink | Reply to this

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