Let $G$ be the gauge group whose Yang-Mill's theory one is looking at and $A$ be its connection and $C$ be a loop in the space-time and $R$ be a finite-dimensional representation of the gauge group $G$. Then the classical Wilson loop is defined as, $W_R(C)(A) = \mathrm{Tr}_R[\mathrm{Hol}(A,C)]$, the trace in the representation $R$ of the gauge field $A$ around the curve $C$.

  • I want to know why the above can be written as, $W_R(C) = e^{i\int _D F}$ where $F$ is the curvature of the connection when $G$ is Abelian and $C$ I guess is the boundary of a contractible disk $D$.

{..the above claim reminds me of heuristic calculations (far from a proof!) that I know of where one shows that in the limit if infinitesimal loops, the eventual deviation of a vector on being parallel transported along it by the Riemann-Christoffel connection is proportional to the product of the corresponding Riemann curvature tensor and the area of the loop..}

  • In the above proof kindly indicate as to what is the subtlety with $G$ being non-Abelian? Isn't there a natural notion of "ordering" in some sense along the loop given by a parametrization or the trajectory of a particle?

  • In relation to discussions on confinement, what is the motivation for also looking at the cases where $R$ is a representation not of $G$ but of its simply connected cover? I mean - how does the definition for $W_R(C)$ even make sense if $R$ is not a representation of $G$?

I don't understand what is meant by statements like (from a lecture by Witten), "..if $R$ is a representation of $G$, then there are physical processes contributing to $\langle W_R(C) \rangle$ in which large portions of the Wilson line have zero-charge i.e carry trivial representations of $G$, because some particles in the theory have annihilated the charges on the Wilson line.."

I would have thought that its only a representation of $G$ that can be fixed and I don't see this possible imagery of seeing a representation attached to every point on the loop.

  • What is the subtlety about Wilson loops for those representations of $G$ which can come from a representation of its universal cover? If someone can precisely write down the criteria for when this will happen and then what happens...

2 Answers 2

  • Let me assume the bundle is trivial. Then $A$ is a $\mathfrak{g}$-valued 1-form on the base $M$. Let $\gamma(t):[0,1]\rightarrow M$ be a parametrization of $C$. A horizontal lift $\tilde{\gamma}:[0,1]\rightarrow P\cong G\times M$ is just a pair $\tilde{\gamma}(t)=(\gamma(t),g(t))$, where $g(t)\in G$. The parallel transport equation then says $g^*\theta = \gamma^* A$, where $\theta$ is the Maurer-Cartan form on $G$. Explicitly, for matrix groups the equation reads $$g(t)^{-1}dg(t) = A(\gamma(t)).$$ The solution to this equation is by definition the path-ordered exponential $$g(t)=\mathcal{P}\exp\left(\int_0^1 \gamma^*A\right)g(0)=\mathcal{P}\exp\left(\oint_C A\right)g(0).$$ This works for general groups as well if you use the exponential map.

For an abelian group $G$ one has $F=dA$, and so you can use Stokes theorem to write the exponent as $\int_D F$. Note, that any connected compact abelian group is isomorphic to $U(1)^n$. The only difference for non-abelian groups $G$ is that you cannot reduce the integral of the connection to an integral of the curvature.

  • To evaluate a Wilson loop, you need $R$ to be a representation of $\mathfrak{g}$, it does not have to exponentiate to $G$. It works as follows: you first use your representation to make $A$ a matrix-valued 1-form and then use the ordinary matrix path-ordered exponential. If the representation exponentiates, it coincides with the usual definition, where you first use the exponential map and then take the trace.

  • I understand that Witten's remark as follows. Since your action is $\frac{1}{e^2}\int F\wedge *F$, to use perturbation series you rescale $A\rightarrow e A$. That means you have to expand the exponential for the Wilson loop in Taylor series. If your connection is not coupled to anything, all your diagrams are photons emitted and absorbed by the Wilson loop.

For example, at order $e^2$ you would have a process like $\langle Tr(AA)\rangle$, which you can think of as a Wilson loop together with a photon propagator attached, where the photon carries the representation indices away. So, half of the Wilson loop "carries" a representation (it corresponds to the trace), while the other half has the zero representation (since you are multiplying $A$'s). I hope it's clear without a picture.

  • $\begingroup$ Thanks for your reply. The last part of of your reply in the language of Feynman diagrams is indeed not clear without a picture. Can you give a reference for that - where this Taylor series evaluation of Wilson loops is done. I have only seen something like this as here - on page 25 of this - webusers.physics.illinois.edu/~efradkin/phys582/… Here it would seem that that calculating <Tr(AA)> is "equivalent" to calculating the Wilson loop. Can you send me some references about the kind of calculations that you are referring to. $\endgroup$
    – Student
    Commented Jan 31, 2012 at 3:26
  • $\begingroup$ About the first part of your answer, can you explain as to why you did the horizontal lift? (..that looks like kind of choosing a particular gauge along the curve..) You and cdcustom (below) both seem to be "equating" my $Hol(A,C)$ to your $g(t)$. Why is that so? What is confusing is that my definition (from the lecture of Witten) $R$ has to be a representation of $G$ but in your thinking you are taking the character in some representation of $Lie(G)$ -- and that is quite different! $\endgroup$
    – Student
    Commented Jan 31, 2012 at 3:31
  • $\begingroup$ Much of the comments by Witten seem to originate from the subtlety that if the Wilson loop is taken in characters of a representation of $G$ whether or not it will lift to a representation of the universal cover of $G$. It would be great to know of references from you regarding these. $\endgroup$
    – Student
    Commented Jan 31, 2012 at 3:32
  • $\begingroup$ Holonomy of a loop by definition is how the horizontal lift of the loop differs at the beginning and at the end, i.e. $Hol=g(1)g(0)^{-1}$. Regarding Wilson loops: just write down the Feynman diagram expansion you get after Taylor expanding the exponential, which gives $\sum e^n Tr(A^n)/n!$ with appropriate orderings. By the way, this gives another way to see why you need $R$ to only be a rep of $\mathfrak{g}$. Unfortunately, I don't know a reference for that, perhaps someone more knowledgeable in Wilson loops can give you some. $\endgroup$
    – Pavel Safronov
    Commented Feb 1, 2012 at 2:29
  • $\begingroup$ Thanks for the insights. I am not very convinced about this representation part of the argument. Yes, it is true that to define the connection one necessarily needs a representation of the Lie algebra but because the holonomy is an exponential of that the final character has to be taken in the induced representation of the Lie group (and not the Lie algebra). But there is something more to it as is clear from Witten's lectures - since he emphasizes this point about whether R is just a representation of $G$ or also has a lift to its cover. $\endgroup$
    – Student
    Commented Feb 1, 2012 at 19:00

Possible answer to your first bullet: Stokes theorem. If $G=U(1)$, then $F=dA$ (not $dA+[A,A]$) and so $\int_{\partial D}A=\int_{D}dA=\int_D F.$ Further, if memory serves, $U(1)$ is the only Abelian Lie algebra (thinking of the Cartan classification) so $G$ is Abelian means $G=U(1)$ anyway.

EDIT: Coordinate expression for the Holonomy. $\mathrm{Hol}(A,C)$ is defined to be $\mathcal{P}\exp(\int_{C}A)$. The connection is usually written as a 1-form $A_{\mu}=A_{\mu}^a\tau^a$ that takes values in $\mathrm{Lie}(G)$ with generators $\tau^a$. Then the holonomy is

\begin{equation} \mathrm{Hol}(A,C)=\mathcal{P}\exp(-ig \int_C A_\mu dx^\mu). \end{equation}

Here $g$ is the coupling constant and the $-i$ is to make it look like QFT rather then LQG :-) To get the gauge-invariant Wilson loop, expand the exponential and trace over the group indices. Check out Peskin and Schroeder (or any other QFT book) for more details.

  • $\begingroup$ Dear cduston could you correct the spelling of Stokes? $\endgroup$
    – Giuseppe
    Commented Jan 23, 2012 at 10:22
  • $\begingroup$ @cdcustom I guess my question was not clear. I can see the calculation that you gave. What I was asking is probably the first part of it as to how does one reduce the idea of $Tr_R [Hol(A,C)] = Tr[ e^{ \int _{\partial D} A}]$ (probably) especially with the interpretation of $A$ as a $Lie(G)$ valued connection. I want to know how to write down a coordinate expression for $Tr_R [Hol(A,C)]$. $\endgroup$
    – Student
    Commented Jan 23, 2012 at 23:43
  • $\begingroup$ added what I think you mean by coordinate expression...let me know if it is not. $\endgroup$
    – levitopher
    Commented Jan 24, 2012 at 17:10
  • 1
    $\begingroup$ Pretty sure there should be a path ordering? $\endgroup$
    – genneth
    Commented Jan 25, 2012 at 23:56

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