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RR charges in string theory

  1. Oct 3, 2008 #1
    Hi there, I'm searching for an argument according to which fundamental strings cannot carry RR charge. I'm aware of the couplings between, say, the Maxwell field and the point particle, the string and the B_{\mu\nu}, and the D-branes and the various RR potentials.

    One should form Lorentz scalars, right? But why can't a string couple to a C_{\mu\nu} RR potential? Thank you in advance!
  2. jcsd
  3. Oct 3, 2008 #2


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    As I recall (I don't have a string theory textbook in front of me and am far from a specialist), the R-R states are some functions involving the R-R fields exclusively (for both type IIA and type IIB) so the vertex operators only involve R-R fields and not R-R potentials. Therefore only the field strengths couples to the string.
  4. Oct 3, 2008 #3
    Thank you for your answer. Although what you say is true, I'm still somewhat confused about the way the R-R charges couple with D-branes. They couple through their potentials C_(p), not through their field strengths..
  5. Oct 3, 2008 #4


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    Actually, a string can couple to C_{\mu\nu}. However, this is not the fundamental string, but rather a D-string, known also as D1-brane.

    So, the right question is: Why a FUNDAMENTAL string cannot couple to C_{\mu\nu}? To be honest, I don't know. In fact, I am not even sure that it is true. There could be some dualities between fundamental strings and D-strings ...
  6. Oct 3, 2008 #5
    Yes, I was aware of that, I meant the fundamental string...but wow...there really doesn't exist an explicit argument according to which the fundamental string cannot carry R-R charges? I've already searched the literature, but still can't find one..
  7. Oct 4, 2008 #6
    Dear PhysiSmo,

    to determine the interactions of fundamental string "fields" (their terms in the effective action), one has to calculate the scattering amplitude of the corresponding excitations (string states). That's computed from the correlation function of vertex operators corresponding to the fundamental string states (evaluated on Riemann surface worldsheets). But here it's important how exactly it is computed.

    As Haelfix correctly says, the term in the effective action that involves RR fields is written by replacing the RR vertex operators by the field strengths (spacetime derivatives of the RR potentials) rather than the RR potentials themselves.

    Why? Because we know exactly what these vertex operators are, of course. In the RNS formalism, they must include both left-moving and right-moving spin fields, Q* and Q, those that create the branch cuts for fermions psi. A Dirac-like matrix Gamma_{abc...} must be inserted in between these two Q's, to get a scalar. The a,b,c... indices must be contracted with a p-form of coefficients. It turns out that p is even or odd in type IIA and type IIB, respectively. So the tensor to contract must be the field strength that has the same property.

    This argument is just about one sign, so let me say it on an example. In type IIB, there is a RR 0-form, the axion (scalar). But the left-moving and right-moving spin fields on the worldsheet transform as spacetime spinors of the same chirality (the same GSO projection on both sides, as far as the spacetime chirality goes), so by taking a tensor product, the lowest-spin term in the decomposition is not a scalar but a vector. You get a scalar from 16* times 16 in the Euclidean signature or 16 times 16' in the Minkowski one but 16 times 16 gives you no scalar invariant, for example.

    So all terms in the effective action that you calculate from the perturbative amplitudes of string states only depend on the field strengths which are clearly the correct RR-related fields with the right number of indices. You may imagine that 1/2 of the derivative comes from each spin field. Consequently, there is no explicit potential "A" in the action, without derivatives: the action doesn't change if you shift "A" goes to "A plus constant". But the change is what would define the charge (compare with A_m j^m couplings in electromagnetism), so the elementary string states carry no RR charge. The charges are zero.

    The D-branes do carry them. They are not represented by local vertex operators described above. Instead, there are extra boundaries of the worldsheet. They add the explicit A_abc j^abc... couplings. See e.g. Polchinski's 1995 famous paper about it.

    The absence of RR charges of fundamental strings doesn't contradict S-duality. This duality only appears in type IIB, not type IIA. In that case, it means that fundamental strings don't carry charges under the RR axion, RR two-form etc. But the RR axion is mapped under S-duality to (parts of) the dilaton while the RR two-form is mapped to the normal NS-NS (two-form) B-field. So the S-dual statement of "fundamental strings carry no RR two-form charge" is that "D1-branes carry no B-field charge" which is correct, too. The discussion for the 0-forms would be a bit different because the dilaton-axion complex field behaves kind of nonlinearly and one would have to be careful what the absence of coupling exactly means and how the S-duality acts on them.

    Best wishes
  8. Oct 4, 2008 #7
    Hi Lubos,

    thank you very much for the clarifications, you've been really helpful. Still, though, I don't understand clearly the statement above. In classical EM, gauge transformations leave the action

    [tex]q\int A_{\mu}(x)dx^{\mu}[/tex]

    invariant, so I can't fully realize the ''definition of charge'' you provided above. Moreover, could you pinpoint me a reference (or some) to study about the vertex operators and the interactions? Taking a look at the equations is always some help.

    Thank you again for your response!
  9. Oct 6, 2008 #8
    Dear PhysiSmo, thanks for your answer and interest. Indeed, the whole action has to be gauge-invariant. But in general, this statement only applies to gauge transformations that become trivial at infinity.

    The transformation that uniformly shifts the value of A_m in electromagnetism - the simplest example - clearly does change the action. Just check it. If "A" is shifted by a constant vector everywhere in space, q times the integral of A_m dx^m changes by q times a divergent integral (length of the infinite worldline projected to the constant vector, so to say). For a more decent gauge transformation, q times the integral changes by q times a constant (the shift of a Wilson line). Only if the product is a multiple of 2.pi, the gauge transformation is unphysical - because 2.pi shifts of actions don't affect the path integral. That's why Wilson lines take values on a "circle", the are periodic variables, and why the "lambda" generating the gauge transformations must be single-valued function living at a circle, too.

    I feel that these comments are independent of what I wrote above. I just wrote that the terms in the action of perturbative string theory that you may deduce by computing the scattering amplitudes only depend on "F", not on "A", for the RR fields. I didn't say that all gauge-invariant terms are allowed. Indeed, they are not. You are right that the whole action, even with some terms that have an explicit potential like "A", can be (and must be) gauge invariant. That's why D-branes can carry RR charges, for example. But this fact doesn't change anything about the proposition that perturbative string states don't carry RR charges - the latter is justified by something else than gauge invariance that you eagerly (and incorrectly) try to squeeze instead of the correct argument that you apparently don't want to hear. ;-) The correct argument, once again, is that there is a full procedure to reconstruct the action from the amplitudes and in this procedure, the vertex operator for the RR field is not replaced by "A" in the action but by "F" in the action. The hypothesis that they should be replaced by "A" can be ruled out by seeing that the required "A" has a wrong number of indices (even vs odd, or vice versa).

    The famous Polchinski paper about the RR charges of D-branes - that also writes some vertex operators for strings and says most of the things I wrote, I believe, is

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