This question is motivated by 2-dimensional CFTs where the Classical conformal group (defined by the Witt algebra) is modified to the Virasoro algebra in the quantum theory. In this question, it was pointed out to me that the extra term in the Virasoro algebra is due to the conformal anomaly (for example, when quantizing the theory on a curved surface, the Ricci scalar explicitly introduces a length scale). This led me to wonder about this:
Consider a field theory has a classical symmetry with algebra $G$. Is it true that if this symmetry is anomalous in the quantum theory the algebra $G$ is modified? If so, how?
For example, Yang-Mills theory with a massless Dirac fermion, has a classical symmetry group $G_{gauge} \times G_{axial}$, where $G$ is the gauge group. In the quantum theory, $G_{axial}$ is anomalous. Does this somehow also imply that the algebra of $G_{axial}$ is modified somehow?
Answer
I'm pretty sure the answer is "no." There's a conformal anomaly in any even number of dimensions. For instance, in 4 dimensions it's the statement that $T^\mu_\mu$ is nonzero for a CFT on a general curved background, equal to coefficients $a$ and $c$ times the Euler density and Weyl$^2$ terms. But there's no central extension of the conformal algebra the way there is in two dimensions.
(You'll notice in the 2D case that the SL(2) part of the algebra is unaffected by the central extension, so in some sense it's only the conformal generators that don't generalize to higher dimensions that have a modified algebra.)
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