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.


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.

  • $\begingroup$ Thanks for the detailed reply Lawrence but isn't it the common lore that $N$ coincident branes lead to a $U(N)$ symmetry? I guess what I should then ask is: what is the difference between coincident branes and these destructively colliding branes? And how is it clear from the context of the text that they are referring not to a coincident stack of branes but to a destructive interference of branes? $\endgroup$ – leastaction May 23 '16 at 22:32
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    $\begingroup$ This is the case, $N$ branes form a "moose," a term that I have not heard much recently, and they then form a chain. I said above, which is a bit confusing that each brane has $U(N)$, when I meant all of the branes have $U(N)$. I'd say that the statement is they coincide, which to my mind is a picture of an N-stacked sandwich being compressed by an elephant. I might be wrong here, but this looks like a split bundle situation when the branes are all in this somewhat singular or smashed configuration. $\endgroup$ – Lawrence B. Crowell May 23 '16 at 23:05
  • $\begingroup$ So to summarize the discussion, there are two possibilities: colliding branes (split bundle, with symmetry $U(1)^N$, with the interpretation that there's just one brane), and coincident branes (with symmetry $U(N)$, which I guess you call a moose). Both cases correspond to $x_i - x_j \rightarrow 0$, so how does one distinguish between them? For instance, if I look at the standard description in a textbook like Becker, Becker and Schwarz, I am only told that when $N$ Dp-branes coincide, the symmetry group is $U(N)$. $\endgroup$ – leastaction May 24 '16 at 10:57
  • $\begingroup$ This is similar in some ways to singular points on a moduli space for $SU(2)$ instantons in 4-dim. The smashing of these branes together is a sort of singularity, or thought another way a situation where the $U(N)$ vector orientation on them "destructively interfers." From a mathematical perspective this looks a lot like a singular condition. Once the distance between the branes is less than the string length there is no manner in which information can be determined. The Planck length is the min-length one can isolate a qubit. Maybe that is another way to see it. I could be wrong of course. $\endgroup$ – Lawrence B. Crowell May 24 '16 at 19:43

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