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1

The action is $$ S = \frac{1}{2\pi \alpha'} \int d^2 \sigma \sqrt{\gamma} \gamma^{ab} \partial_a X^\mu \partial_b X_\mu $$ The definition of the stress tensor from GR is $$ T_{ab} = \lambda\frac{4\pi}{\sqrt{\gamma}} \frac{ \delta S}{ \delta \gamma^{ab}} $$ Usually $\lambda = 1$, but different books use different conventions. I do not remember what convention ...


2

Note that it's the residue of the combination $j(z)\mathcal A(z_0,\bar{z_0})$ that we need: the poles come from the OPE where the operators come together. Now it could be that $j(z)\mathcal A(z_0,\bar{z_0})$ is more singular than $\frac{1}{z-z_0}$ at $z=z_0$. Only the simple poles contribute by the residue theorem. And if there is no simple pole (maybe ...


2

First of all, note that the radial operator ordering ${\cal R}$ is implicitly implied in many textbooks of CFT (e.g. Ref. 1). For instance, eq. (2.2.7) on p. 39 in Ref. 1 is discussing Wick's theorem between two operator ordering prescriptions. In this case between normal ordering $:~:$ and radial ordering ${\cal R}$. See also e.g. this Phys.SE post. The ...


0

The little group is the subgroup of the Lorentz group that leaves an arbitrary four-momentum vector invariant, i.e. for an element of the group $g$ and momentum $V$ we have $gV=V$. This group is in general different for massive and massless particles. If you now find that the little group of your holomorphic primaries corresponds to that of massive states, ...


2

A more physical attempt: In general relativity, the metric tensor represents local clock and ruler measurments. If I multiply the metric tensor by a scalar constant, it should be obvious that this is inequivalent (in general) to a set of coordinate transformations, but, at the same time, I'm affecting local clock and ruler measurments (the ratio of the ...


4

General relativity is only conformally invariant in two dimensions. This can be proven by making the transformation $g_{ab} \rightarrow \phi g_{ab}$, and seeing what transformation Einstein's equation${}^{1}$ makes. What you will find is that Einstein's equation will MOSTLY transform, but you will get terms proportional to $(d-2)(d-1)$ and derivatives of ...


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Newton's constant is dimensionful. Hence the theory is NOT conformally invariant. In 2 dimensions, newtons constant is dimensionless. But then the apparent conformal symmetry is actually only a REDUNDANCY in the description (sometimes called weyl symmetry).


1

The derivation of the Virasoro algebra is obviously a very important calculation in ST/CFT. For starters, OP (v1) does not mention explicitly the normal ordering ": :" in the definition of the Virasoro generators $$ L_n ~=~ \frac{1}{2}\sum_{k\in\mathbb{Z}} :\alpha^I_{n-k}\alpha^I_k: $$ Normal ordering is important for $L_0$. Here the index $I$ runs over ...



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