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Central charge in string theory

  1. Dec 27, 2015 #1
    1. The problem statement, all variables and given/known data
    I'd like to better understand why the central charge is assumed to be equal to the number of dimensions in the spacetime background.

    2. Relevant equations
    By definition of Virasoro algebra and its operators:
    [Lm,Ln]=(m−n)Lm+n+c/12(m3−m)δm+n,0

    3. The attempt at a solution
    The commutator relation among the modes satisfying the string worldsheet-equation gives rise to the Virasoro relation with c=D. But I'm missing the details of exactly how it happens.
    As another attempt to answer my question, I think that maybe c=D can follow from the defintion of the Vertex Operator Algebra where D is the rank of space of states (but again I'm unclear if this second approach is equivalent to the first one ... and if/how the VOA is related with the first quantization of the bosonic string theory...)
     
    Last edited: Dec 27, 2015
  2. jcsd
  3. Dec 28, 2015 #2
    The closest reference I've found to answer my question is "Characters of Modules of Irrational Vertex Algebras" of Antun Milas, pag 9 to 11.
    In particular pag 11 says "we can transport the structure of a Virasoro algebra module to VL with the grading given by the action of L(0)" and "We keep the same conformal vector so the central charge of VL is rank(L)".
    Maybe the last sentence answers my question but is too synthetic for me and I'm unable to expand and elaborate it more.
    Also from the same source, in a later, more advanced example of a "root lattice of ADE type" (pag 23), they choose a conformal vector that is different from the "standard (quadratic) Virasoro generator" and they note that - only without the linear term - the central charge is rank(L).
    So, in conclusion, I think that the assumption (made in string theory) of a central charge equal to the spacetime dimensions deserves some closer scrutiny and challenge.
     
  4. Dec 28, 2015 #3
    I think that this is the answer. http://relativity.livingreviews.org/open?pubNo=lrr-2012-11&page=articlesu17.html [Broken] Central charge is normalized by a factor 12 to be equal to the charge J in the Hamiltonian. Thus since each boson is interpreted as a flat spacetime dimension ( https://en.m.wikipedia.org/wiki/Non-critical_string_theory#The_critical_dimension_and_central_charge ) , the concept is basically that the charge J is equal to the number of free bosons. Is my answer correct?
     
    Last edited by a moderator: May 7, 2017
  5. Jan 2, 2016 #4
    Unfortunately the best answer I've found till now is a very trivial and shallow one from Lubos Motl:

    Basically
    1) If your CFT is made of "d" bosons - describing
    spacetime dimensions - then its central charge is "c=d", more or less by
    definition. The central charge of the sum of two CFTs is the sum of the
    central charges.
    and
    2) the normalization is chosen in such a way that a single boson
    has c=1

    In fact the only relevant math point made by Lubos is here

    Note that the commutators like [L_2,L_{-2}] are always a sum over "mu", and therefore they're proportional to the number of bosons (dimensions) - therefore "c" is proportional as well.
     
    Last edited: Jan 2, 2016
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