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By considering $\langle T^\alpha_\alpha\rangle$, the Weyl anomaly, we can show that the critical dimension, $D=26$ is the only possible choice of dimension for the bosonic string.

However, how do we know we can stop there? How do we know there doesn't exist a similar anomaly, that perhaps gives us a different critical dimension - in which case there is no choice of $D$ which is does not give some anomaly?

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  • $\begingroup$ A good question. The anomaly is a phenomena of nonconservation on the quantum level of a classically conserved current. It seems to me, that the conserved currents in the bosonic are given by the Virasoro generators. Are they the only conserved charges in the theory? $\endgroup$ Sep 1 '20 at 11:34
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    $\begingroup$ There is some issue with this question I think (the question itself I am in no way able to answer) that doesn't warrant reporting, but you should avoid asking the proof of a negative. Asking to be sure there isn't a contradiction to some theory, is equivalent (though less extreme perhaps) than say asking to prove god's nonexistence. Possibly this question is more nuanced than I am qualified to comment on though and this is entirely reasonable $\endgroup$
    – MC2k
    Sep 1 '20 at 23:06
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Anomalies arise as the quantum nonconservation of a classical symmetry. The answer to your question is simply that the only classical symmetries of the Polyakov action are the diffeomorphism and Weyl ones.

It may be the case that bosonic string theory could develop an anomaly after quantization, but that is actually not the case; your favourite way to quantize the classical Polyakov action (cannonical/light-cone/Gupta-Bleuler/BRST) is the proof that the latter is actually not the case. No anomalus quantum symmetries are developed after quantization.

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  • $\begingroup$ Let's take the BRST quantisation method. From what I can tell, what we are really trying to impose is $\langle T_\alpha^\alpha\rangle_g = \langle T_\alpha^\alpha\rangle_{g'}$, where $g$, $g'$ are two different fixed metrics. To do so, we need the central charge, $c=0$. Suppose we take a different operator $O$, and compare $\langle O\rangle_g = \langle O\rangle_{g'}$, conserved current or otherwise. Then how do we know this doesn't give a different 'central charge' equivalent, which gives $D\neq 26$ $\endgroup$
    – awsomeguy
    Sep 2 '20 at 9:07
  • $\begingroup$ The Weyl anomaly is explicitly computable, and we take it zero to guarantee Lorentz invariance. The second case you mention cannot happen. There are no global symmetries in string theory (see Motl's answer) nor local worldsheet symmetries because the worldsheet can always be taken locally flat, and no local invariants exist over Riemann surfaces. Such operators $O$ (charged with a possible anomalous symmetry) cannot exist in the worldsheet CFT. $\endgroup$ Sep 2 '20 at 15:14

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