Enhanced D-Brane Gauge Symmetry and the string decoupling limit

Question

Consider a system of $N_C$ $Dp$-branes (in Type IIA or Type IIB string theory, depending on whether $p$ is even or odd).

In the Les Houches lectures on supersymmetric gauge theories by Berman and Rabinovici, on page 80 (section 7.3) the following paragraph appears:

The expectation values of the Higgs fields in the adjoint representation can be shown to have themselves a very transparent geometrical meaning:

$$\textbf{x}_i = \langle \textbf{X}_{ii} \rangle \qquad (i = 1, \ldots, N_C)$$

$\textbf{x}_i$ on the left hand side of the equation denotes the location of the $i^{th}$ brane, $\textbf{X}_{ii}$ on the right hand side of the equation denotes the component of the Higgs fields in the $i^{th}$ element of the Cartan subalgebra of $U(N_C)$. In this context, the mass formula

$$m_{ij} = \frac{1}{l_s^2}|\textbf{x}_i-\textbf{x}_j|$$

is just the usual mass obtained by the Higgs mechanism.

To keep fixed the mass of these ''W'' particles in the limit of the decoupling of the string states $(l_s \rightarrow 0)$, the separations between the branes should vanish themselves in that limit, that is, these separations should be sub-stringy. In that limit, one cannot resolve the $N_C$ different worldvolumes, so the theory is perceived as a $U(1)^{N_C}$ gauge theory on a single $(p+1)$-dimensional world volume.

My confusion stems from the text in boldface. If the $N_C$ branes coincide (which happens in the limit $|\textbf{x}_i-\textbf{x}_j| \rightarrow 0$) then wouldn't we expect a gauge symmetry of $U(N_C)$? Why is the symmetry in the sub-stringy limit equal to $U(1)^{N_C}$ then? I would have thought that the gauge symmetry would be $U(1)^{N_C}$ when the D-branes are far apart.

Answer 1

These branes are being smashed together so that once they are closer than the string length they become indistinguishable from a single brane. These branes have $U(N_c)$ gauge group, and in this space there is a vector. In a generic sense all Lie algebras are like the harmonic oscillator with $a$, $a^\dagger$ and $a^\dagger a$ in the structure of roots and weights. Each of these branes has a root vector pointing in some direction, and physically we can see this as a destructive interference process. However, the stabilizer of the $U(N_c)$, group elements $g$ such that for $x$ $gx = x$, and this happens with each brane.

The group $U(N)$ contains $G_s$ and the stabilizer is such that $G_s/\mathbb Z_2 \simeq U(1)$, and on the general gauge transformation $Du = 0$ as a split bundle that is $U(1)$ and reducible. For $U(N)$ this split bundle is $\oplus_{i=1}^{N_c}d_i$, where each $d_i$ is $u(1)$. I could go into considerably greater mathematical detail, but the up shot is that the stack of branes collapses into a single brane with a large twisted bundle consisting of many $U(1)$s, explicitly $U(1)^{N_c}$ from this bundle splitting. Going back back to the physics, the large destructive interference only leaves the stabilizer for each $N_c$ in the split bundle.