A Concentration of Measure Phenomenon in Lattice Yang-Mills

Abstract

We demonstrate that the pushforward of the product of Haar measures by the lattice Yang-Mills action concentrates as a Gaussian. It is also sketched how, using this fact, one can recover the strong-coupling expansion.

Figures (4)

• Figure 1: Summary of a graphical notation used below. Note the order of the product in the sum in the rightmost expression. It is the one consistent with treating $U$ and $U'$ as being parallel transports (acting conventionally to the right).
• Figure 2: A pictorial representation of (\ref{['eqnarray:integral']}) in the cases $n=2$ and $n=4$. Graphically, integration over a bundle of edges amounts to cutting them into half-edges, and then linking or coupling the half edges in pairs on one side of the cut with each other, with matching coupling of the half-edges on the other side of the cut as well. Incidentally, note that in the first case, the dominant asymptotic happens to be the exact answer. There are no further corrections.
• Figure 3: A pictorial representation of the argument given in the main text. We perform the integral over the group elements living on the edges. Only one of these integrals is explicitly written out with the rest being represented by the $\dots$ under the integral sign. As represented in Figure \ref{['fig2']}, after performing the integral over the group, the edges carrying that group element and its inverse are cut into half-edges with these halves linked up in pairs in identical ways on the two sides of the cut. In the figure, we focus on two edges represented by solid lines with arrow whose half-edges will be coupled. We focus on only one coupled pair, denoted by a solid line. The dashed one stands for the other one. The dotted lines stand for the other copies of the same edge carrying $U$ and $U^\dagger$ which, after integration, are of course coupled among themselves as well, which is indicated by the break in the lines. The $+ \, \dots$ stands for all the other terms in the sum in (\ref{['eqnarray:integral']}) which correspond to a different coupling of the half-edges. Also, note that since our concern here is the factors of $N$ obtained by following the coupling of edges, we haven't indicated either the factors of $\frac{1}{N}$ coming from the definition of $t(U)$, or those on the right hand side of (\ref{['eqnarray:integral']}). Now, following the two singled out coupled half-edges, we will have one of three cases in the lower half of the figure. In case (a), they meet at the corner with two coupled half-edges forming a loop. Note that this loop is constructed out of four half-edges. Also note that the two plaquettes from which these four half-edges came must actually be the same plaquette, as two of their edges coincide. In cases (b) and (c), the original two half-edges are not coupled back immediately forming a loop, but instead, are coupled to other half edges. The difference between these two cases is that in (b), the two plaquettes from which these half-edges come coincide as in (a), while in (c) they don't.
• Figure 4: This is a graphical representation of the only way two plaquettes can be coupled together in order to solely produce instances of case (a) from Figure \ref{['fig3']}. To see this, suppose the half-edges are coupled to produce case (a) for e.g., the bottom left corner. It follows that the other two half-edges coming from the bottom edge must be coupled together as well. Following them around the bottom right corner, we conclude that the corresponding two half-edges coming from the right edge must also be coupled together, for otherwise, we would have either case (b) or (c) at the bottom right corner. Proceeding like this around the plaquette, we conclude that the two half-edges coming from these two plaquettes must only couple among themselves, giving the picture on the right.

Theorems & Definitions (2)

• Theorem
• proof