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Consider a Dp-brane. Compactify $d$ spatial dimensions over a torus $T^d$. Suppose $d\geqslant p$, and that the Dp-brane is completely wrapped around the compactified dimensions.

Look at the open string modes ending on this wrapped D-brane. There is a zero energy open string mode associated with each spatial dimension. That corresponds to the orientation of the worldsheet field excitation. If the direction is along an uncompactified spatial dimension, that corresponds to quanta of brane displacements along that direction. These cases needn't concern us. If the direction is normal to the brane but lies along a compactified dimension, that corresponds to quanta of brane displacements along that direction. If it's tangent to the brane, it corresponds to quanta of the Wilson line of the brane gauge field along that wrapped direction.

Here's the question. Suppose the wrapped D-brane has a total mass of $M$. Suppose there is a compactified spatial dimension of radius $R$ along which the brane isn't wrapped, i.e. $p<d$. The brane has Kaluza-Klein momenta along that direction of value $n/R$ where n is integral. The energy spectrum is given by $$\sqrt {M^2 +n^2/R^2} ~\approx~ M + \frac{n^2}{2MR^2} +\mathcal{O}(M^{-3}).$$ A condensate of zero energy open strings with orientation along that direction ought to give a continuous moduli? Why is there no moduli then, and why is the energy spectrum discretized? Or consider a direction in which the brane is wrapped. We ought to have a continuous modulus of Wilson line of the brane gauge field along that dimension? Once again, we have discretization. Why?

Anyway, how can a completely wrapped D-brane have KK momentum? In the string worldsheet picture, we have open strings ending at the D-brane background at a fixed position. Along those compactified dimensions along which the brane isn't wrapped, we have Dirichlet boundary conditions. Such open strings can only have winding numbers, but no KK momentum either. Even then, we're dealing with a condensate of open strings with zero winding number.

This discretization doesn't occur if the brane remains unwrapped under at least two uncompactified spatial dimensions because the brane now has infinite mass. If it remains unwrapped only along one uncompactified spatial dimension, there's the Mermin-Wagner theorem, meaning there’s no fixed brane position or Wilson line.

PS: Maybe this question can be rephrased in terms of BPS. We have a wrapped BPS brane. But nonperturbatively, somehow, the BPS state has to be delocalized along the compactified dimension or its Wilson line in a superposition over all possible values? Open strings with no energy are also BPS. So, we can have any condensate of them and still remain BPS? This clearly isn't the case with a lifted modulus, and a discretization of the energy spectrum.

PPS: How do you even express the KK modes of the D-brane, or the dual to its Wilson lines, in terms of a condensate of open strings? Suppose you have the lowest energy state with zero KK momentum. Then, disregarding transverse displacements along the uncompactified spatial dimensions, there is an energy gap to the next energy state. However, a condensate of open strings with internal mode excitations along the compactified dimensions would naively give no energy gap.

PPPS: Is the number of open string modes corresponding to these "disappearing moduli" even a well-defined operator? This is precisely because at the perturbative level, such open string modes have no energy. If it's not a well defined operator, just how do you even express this in terms of open string worldsheets?

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By T-dualities along the finite directions on which the D-brane is wrapped, your first problem is equivalent to the problem of a D0-brane moving on a circle.

The massless open-string scalar corresponding to the direction of the circle produces the field $X(\tau)$ living on the world line of the D0-brane. And indeed, by assumption, the global topology of the configuration space for this field $X(\tau)$ includes the periodic identification $X(\tau)\sim X(\tau)+2\pi R$. This is equivalent to the quantization for the complementary momentum, $P=N/R$.

But none of these two equivalent signs of the periodicity/quantization of the transverse location/momentum is visible at the level of the spectrum of the open strings. In the perturbative treatment of the D-brane dynamics, we effectively expand around a classical shape of the "heavy" D-brane, i.e. around a classical solution. Imagine $X(\tau)=0$. The open-string excitation corresponding to the transverse field is a quantum of a "wave" that only deviates "infinitesimally" from $X(\tau)=0$. The quantitative distance scale at which $X$ deviates is parameterically shorter than then string scale. So the open-strings know about all the potential of the D-brane to change its shape and vibrate but a finite number of open-string excitations still leaves you near $X(\tau)=0$ so it isn't capable of probing the global properties of the $X$ configuration space. To reach the point $X=2\pi R$, you would need an infinite number of open strings – the number would scale like $1/g$ or something like that – and the perturbative expansion would break down or would become insufficient at that point because it's an expansion around the point $g=0$ at which point the required number of open strings is strictly infinite.

So at the leading order, the open-string calculations will be totally insensitive to the global properties of the D-brane configuration space. This is also reflected by your $n^2/2MR^2$ kinetic term from the D-brane. Because $M\sim 1/g$, this kinetic energy scales like $g$ and vanishes in the $n=0$ limit. But yes, once you go to the subleading order in $g_s$ when you're calculating the energy or evolution, you will be able to see that the field on the D0-brane is periodic and the coefficient of this $1/2M$-like kinetic term is a quantized $n^2/R^2$.

Analogously, you may solve the other universal problem with the Wilson line which is just the T-dual of the D0-brane above performed along the $X$ axis. You get a D1-brane with a Wilson line that plays exactly the role of the periodic $X$ for the D0-brane. The dual quantity to this periodic Wilson line is the electric flux, $\int d \tilde X\,F_{01}$, along the D-string, which is quantized much like the momentum of the D0-brane – they're dual descriptions of each other. Again, the open-string excitations only produce "infinitesimal variations" (or Planckian, substringy variations) of the Wilson line that are too small to see the periodicity of the Wilson line or the quantization of the complementary electric flux.

It's helpful, at least for me, to look at this problem via Matrix theory which picks some low-energy limit of the D-brane dynamics. Indeed, for D0-branes on a circle, one may also describe the situation by D1-branes on the dual circle. The matrix model, matrix string theory, will automatically give you the right gauge theory with the right periodicity conditions of the Wilson line and the quantization for the electric flux. You may be asking how one derives that the gauge theory has the right identifications of the configuration space. One may be more or less rigorous but it's clear that it has to have this identification. This identification boils down to the fact that e.g. the D0-brane at places $X$ and $X+2\pi R$ is really the same state – it's the same state from the perturbative string theory viewpoint as well because the D0-brane locations define two totally identical (i.e. one) boundary conditions for the world sheet, the same boundary CFT. The configuration space for D-branes is the space of different boundary conditions and because $X$ and $X+2\pi R w$ give the same boundary conditions, they're identified.

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I know this is a nonperturbative effect. You are trying to reexpress it in terms of matrix theory, but how do you even deal with this question at the level of a condensate of open strings? Or is the description in terms of open strings incomplete, with matrix theory being the complete theory? –  Hecles Nov 6 '12 at 14:36
    
If path integrals over string worldsheets is fundamental, surely there ought to be a way to answer my question purely at the level of open string worldsheets, shouldn't there? A failure to do so is an indictment of the completeness of the string worldsheet description. –  Hecles Nov 6 '12 at 14:38
    
@Hecles: Yes, the perturbative string picture is incomplete. –  user1504 Nov 6 '12 at 14:57
    
Dear @Hecles, yes, I confirm user1504, perturbative string theory is an incomplete description of string theory. It only describes things that exist at $g=0$, and what they're doing at a general $g$ is expressed as a Taylor expansion - which isn't the most general function even among functions that are smooth near $g=0$. Things that are singular or nonperturbative near $g=0$ are invisible in perturbative string theory. Moduli spaces whose volumes go like $1/g$ look infinite in perturbative expansion, and so on... D-branes give you a way to "extend" perturbative expansions if done right. –  Luboš Motl Nov 16 '12 at 10:38
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