The question Disappearance of moduli for condensate of open strings made me think.

Suppose we have a Dp-brane completely wrapped over a $T^d$ compactification with $p\leqslant d$. Look at an open string worldsheet with both ends on this wrapped brane. A Fourier decomposition can be made for the $X$ fields over this worldsheet. Look at the mode with no worldsheet Fock excitations for worldsheet momenta 2 or greater, a single excitation for n=1, and consider the values for $P_\mu$ for the worldsheet n=0 mode. Assume in addition the winding numbers for the d-p compactified dimensions normal to the wrapped brane are zero. For those $9-p$ dimensions normal to the Dp-brane, $P_\mu=0$. That leaves us with p spatial dimensions (let's give them the indices i,j,k,...) and 1 time dimension, index 0. The string modes we're interested in are massless. This means $P_0^2=P_i P_i$. Let $\epsilon^\mu$ be the orientation for the excited n=1 mode. The Lorentz invariant norm is given by $\langle \epsilon^\mu | \epsilon^\nu \rangle = \eta^{\mu\nu}$ ($-+...+$ signature). The condition for BRST-closure is $\epsilon^0 P_0 + \epsilon^i P_i =0$. The null BRST-exact state has $\epsilon^\mu$ parallel to $P^\mu$. Because of the compactification, $P_i = n_i/R_i$ where $R_i$ is the radius along the ith dimension and the $n_i$'s are integers. In other words, the spectrum for the $P_\mu$'s is discrete along all directions. If there is at least one nonzero $n_i$, the BRST-closed subspace has positive semidefinite norm, and after quotienting over the BRST-exact null state, we are left with a positive definite space.

What if all the $n_i$'s are zero? Then, $P_0=0, P_i=0$, all polarizations are BRST-closed, and there is no nontrivial BRST-exact subspace. So, the BRST cohomological space has an indefinite norm! Can someone please help me here?

Suppose we have two D0-branes. They are separated by a distance $L$. Consider an open string connecting both branes. Assume it has the same properties as given in the previous section. As long as $L$ is nonzero, the BRST cohomology has positive definite norm. What happens when $L=0$? An indefinite norm again!


1 Answer 1


There are no actual states with zero energy, $P_0=0$ (I don't even need to talk about $P_i$), unless they correspond to modes of massless states and only massless scalars – moduli – should be considered real. Note that relativity implies that $P_0=0$ states would have to have zero mass and zero spatial momenta in both compact and noncompact dimensions, too.

Your argument has nothing to do with the compactification of some dimensions or the dimensionality of D-branes. By T-duality and other reasons, the spectrum of states – as well as the number of states in the physical subspace – is clearly totally the same independently of $p$ and $d$ for a fixed choice of the allowed zero modes such as the total momenta and/or windings. At most, you introduced a regulator by your compactification so that you're sure that you get these zero-energy states.

Again, the zero-energy (and therefore zero momentum) states of scalars correspond to the ability to change the vev of the corresponding moduli in the whole spacetime. The zero-energy states of vector bosons are closed but they're exact, too. They're pure gauge. If you have $A_0=1$, for example, it may be written as $\partial_\mu\lambda$ where $\lambda$ is linear in time $x_0$. For this reason, these modes also decouple – all their S-matrix amplitudes are zero.

Similar zero-energy spin-two states would be unphysical states of the graviton, changing $g_{\nu\nu}$ to another value in the whole spacetime which is just diffeomorphism. Spin-1/2 particles don't produce any negative-norm polarizations. Spin-3/2 gravitino states do but they decouple just like in the case of vector bosons.

  • $\begingroup$ At the level of open string worldsheets, under worldsheet BRST cohomology, there is an indefinite norm! At the level of string field theory, that's an entirely different matter. $\endgroup$
    – Puzzled
    Nov 16, 2012 at 10:58
  • $\begingroup$ Unlike the closed string graviton scenario, the $P_\mu$ spectrum is discrete in my examples. If the spectrum is continuous, the indefinite norm cohomology can be interpreted as an IR regularization issue. However, with a discrete spectrum, it can't! $\endgroup$
    – Puzzled
    Nov 16, 2012 at 11:04
  • $\begingroup$ Maybe, as Hecles suggested, open string theory based upon worldsheets is incomplete. $\endgroup$
    – Puzzled
    Nov 16, 2012 at 11:16
  • $\begingroup$ Well, Wilson lines of completely wrapped D-branes aren't pure gauge, are they? Even though they come from a condensate of open strings with $P_0=0$. $\endgroup$
    – Puzzled
    Nov 16, 2012 at 14:07
  • $\begingroup$ Dear @Puzzled, concerning first comment, there are various simple unphysical states in the BRST formalism. Those with an extra $b$ excitation, a $c$ excitation, but those without extra $b,c$ excitations are the same states that exist in the "old covariant quantization" as well and their norm or their being pure gauge reproduces the rules from normal gauge theory without BRST. So be sure that if I can write the polarization vector as one coming from $A_\mu\sim\partial_\mu\lambda$, then it is pure gauge, i.e. BRST-exact, even in the BRST formalism. BRST-exact states are automatically null. $\endgroup$ Nov 16, 2012 at 16:28

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