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$T$ is a second-level descendant of the identity operator. It is therefore annihilated by $L_n$ with $n>2$. Also, identity doesn't have any first-level descendants, so T is also annihilated by $L_1$ (which is to say that $T$ is a quasi-primary). I am not sure whether the formula for $l_n$ you write is correct when acting on $T$, since it is not a ...

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I do not agree that $X$ is a primary field. Primary field is defined by its transformation properties under the conformal group (see e.g. yellow book). In particular, under scaling transformation, a correlation function involving primary operators, transforms as $$\langle \mathcal{O}_1(\lambda x_1)\ldots\mathcal{O}_n(\lambda ... 0 Note that under an infinitesimal change in the metric of the form g \to g + \delta g the action changes to$$ \delta S = \int T^{ab} \delta g_{ab} $$Now, under Weyl transformations we have$$ g_{ab} \to e^{2\omega} g_{ab} \qquad \implies \qquad \delta g_{ab} = 2 \omega g_{ab} $$For Weyl transformations \omega is completely arbitrary. If we consider a ... 1 To see why the descendants are primary, you can use$$ \partial\left(T(z)X(w,\overline{w})\right) = T(z)\partial X(w,\overline{w}) = \frac{\partial^2 X}{z-w} + \frac{\partial X}{(z-w)^2} $$And see that it is a primary field of weight h = 1, \overline{h} = 0, and similarly for the other field.... 1 A CFT is still a QFT, and the way to put it at finite temperature is standard for any quantum system - you take your Hamiltonian H and compute Z=\mathrm{tr}\,e^{-\beta H}, where the trace is over the Hilbert space of states living on \mathbb{R}^{d-1} if your CFT is in d dimensions. The thermal correlators are computed in a similar way, ... 1 That g^{-1}\mathrm{d} g is Liealgebra-valued for a Lie group-valued function g has nothing to do with the particular model or with physics, it is true for all matrix groups. Write g(x) = \exp(k(x)), where k(x) is now Lie algebra-valued and \exp is the usual power series in the case of a matrix group. Then \partial_\mu g = \partial_\mu k\exp(k) by ... 1 I don't completely understand the question. How a scalar transforms is completely dictated by conformal symmetry. The transformation law is$$K_\mu \phi(x) = \big(\Delta x_\mu + x_\mu \, x_\nu \partial_\nu -x^2 \partial_\mu \big) \phi(x)$$or if you wish$$\delta_K \phi(x) = a^\mu K_\mu \phi(x)$$where a^\mu are the infinitesimal parameters of the ... 1 Even if a theory is naively (classically) scale invariant (eg: the scalar theory with \lambda \phi^4 interaction therm), quantum mechanically, the 4-point scattering amplitude depends on the energy of the scattering particles (as can be shown by a one-loop computation. Tree level computations are the classical approximation). Suppose the scattering ... 0 The usual definition of normal ordered product is:$$:X^\mu(z,\bar z)X^\nu(w,\bar w): = X^\mu(z,\bar z) X^\nu(w, \bar w) - \langle X^\mu(z,\bar z) X^\nu(w, \bar w) \rangle $$As you said, this is the regular part of the OPE, since only the divergent part of two operators gives non vanishing contribution to the correlator. Of course$$\langle ...

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I think the answer to the question is basically that not only the flow itself cannot be reversed, but more generally, and maybe more clearly, there is no flow which could take you in the reverse way, no matter what is the suggested path. Since there is a decreasing function characterizing any flow, then any RG flow violating this decreasing fashion is ...

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Free theories can be built out of non-interacting scalars, fermions and vectors, and therefore have a Lagrangian description. There may be exceptions for higher-spin fields or exotic SUSY multiplets etc. but those are not so interesting for your question. Next, a weakly coupled fixed point normally means starting with a free theory (which always has a ...

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