Intuition for gauge parallel transport (Wilson loops)

Question

I'm looking for a geometrical interpretation of the statement that "Wilson loop is a gauge parallel transport".

I have seen QFT notes describe U(x,y) as "transporting the gauge transformation", and some other sources referred to U as "parallel transport of identity wrt connection A".. The only other place i have encountered parallel transport thus far is GR, and there I had a clear geometrical picture of what parallel transport of tangent vectors is, while the QFT setting does not yield itself to this interpretation. Could somebody shed light on my confusion or point me in the right direction?

Answer 1

This is the definition of the gauge field. Suppose you have an SU(2) symmetry, for definiteness, consider isospin. So the notion of "proton" and "neutron" define two axes in isospin space, and you might want to say that it is arbitrary which two linear combinations of proton and neutron are the right basis vectors. So that someone defines one basis of "proton" and "neutron" at one point, and someone else defines a different basis at the same point, and you can't tell which one is right (pretend there is no charge on the proton, and the masses are exactly equal).

So you have the freedom to redefine the proton and neutron by a different SU(2) rotation at every point. This is the gauge freedom, you can multiply by a different group element G(x). Now to compare a proton at a point x with a proton at a point y, you have to transport the proton along a curve from x to y.

The gauge connection tells you what matrix you multiply by when you move in an infinitesimal direction $\delta x_\alpha$. The SU(2) matrix you rotate by is

$$ M^i_j = I + i A_{\alpha j}^i \delta x^\alpha$$

This is infinitesimally close to the identity, so the A part is in the Lie algebra of SU(2). The "i" is conventional in physics, to make the A matrix hermitian as opposed to anti-hermitian, as is the cleaner convention and the one used in mathematics. This means that A is a linear combination of Pauli matrices. This gives you a concrete representation of the gauge field (suppressing the i,j indices):

$$ A_\alpha = A_{\alpha k}\sigma^k $$

You assumed that the parallel transport is linear in the $\delta x$'s, this is so that the notion is compatible with the notion of spacetime as a differential manifold--- if you double the displacement you double the infinitesimal rotation angle. You assume it's infinitesimal by physical continuity.

From this, it is obvious that the parallel transport along a curve is the product of A's along each of the infinitesimal segments that make up the curve:

$$ \prod (I+ A_k dx^k) = \mathrm{Pexp}(i \int A dx )$$

Where the path-ordered exponential is defined as the limit of the product on the left. This is the nonabelian generalization of the phase acquired by a charge particle in an electromagnetic field along a path.

The gauge field is then a map between curves and SU(N) matrices with the property that if you join paths end-to-end, the matrices multiply. The matrix associated to an infinitesimal closed loop is called the curvature, and it is proportional to the element of area enclosed in the loop. This is identical to general relativity. The whole exercize is a generalization of the connection of general relativity to cases where the groups are not rotations. Specializing to the rotation case gives GR.

Answer 2

It is simple to describe mathematically.

First I will recall what the equation for the Aharonov-Bohm phase means, and then I will describe (without proof) its relationship to parallel transport for $G$-bundles, which I define.

The gauge potential $A$ is a connection on some principal $G$-bundle, where $G$ is the gauge group. Principal $G$-bundles over a manifold $M$ are classified by the second cohomology $H^2(M,G)$. For a curve $\gamma$ in $M$, we can pull back any principal bundle to $\gamma$, and since $\gamma$ is one dimensional, $H^2(\gamma,G)=0$, so the pullback of the principal bundle is trivial. Accordingly, the pullback of the connection $A$ is just a 1-form on $\gamma$ (once we pick a trivialization). This means the expression $exp(i \int_\gamma A)$ makes sense. To make it not depend on the choice of trivialization (by which a change in trivialization causes a conjugation of this quantity, which is an element of $G$) we need to take a class function $\chi :G\rightarrow \mathbb{C}$ (that is, one which only depends on the conjugacy classes in $G$) and consider $\chi (exp(i\int_\gamma A))$. This only depends on $A$, $\gamma$, and $\chi$. A class function $\chi$ is the same as a trace in a complex representation $R$ of $G$ (such objects are in bijection) and so one can rewrite this in the normal way:

$Tr_R exp(i\int_\gamma A)$.

This is the Aharonov-Bohm phase.

We can rephrase this in terms of parallel transport as follows. The fiber over a basepoint $x \in M$ is a set on which $G$ acts faithfully and transitively. That is, any particular choice of basepoint $\tilde x$ in the fiber can be sent to the identity of $G$ and so defines an isomorphism between the fiber and the regular representation of $G$ on itself. Once a choice of basepoint in the fiber is made, just as in parallel transport of vector bundles, any curve from $x$ in $M$ lifts uniquely to a curve in the total space of the bundle which begins at $\tilde x$. For a loop $\gamma$, this lifts to a curve beginning at $\tilde x$ and ending somewhere else in the fiber. In other words, the lift defines a map of $G$ representations $G\rightarrow G$. This is the analogue of parallel transport for principal $G$-bundles. With a choice of basepoint, we can identify this set of maps with $G$, so every loop gives us an element of $G$ depending on a trivialization of the bundle at $x$. Change of trivialization corresponds to conjugation in $G$, so we can pick a representation and again calculate the Aharonov-Bohm phase in this abstract picture.

The de Rham theorems (or maybe an easy extension thereof, just use some partitions of unity so it suffices to show for a contractible space) ensure that every such parallel transport map arises from a connection form $A$ as in the first paragraph. That is, the map $G \rightarrow G$ of $G$-representations is just $exp(i\int_\gamma A)$ (depending on a trivialization).

A good reference for this material can be found in Nakahara's book Geometry and Topology in Physics.