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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 $$\langle \mathcal{O}_1(\lambda x_1)\ldots\mathcal{O}_n(\lambda ... 1 I think the answer to the question is basically that not only the flow itself cannot be reversed, but more generally, and maybe more clearly, there is no flow which could take you in the reverse way, no matter what is the suggested path. Since there is a decreasing function characterizing any flow, then any RG flow violating this decreasing fashion is ... 1 Even if a theory is naively (classically) scale invariant (eg: the scalar theory with \lambda \phi^4 interaction therm), quantum mechanically, the 4-point scattering amplitude depends on the energy of the scattering particles (as can be shown by a one-loop computation. Tree level computations are the classical approximation). Suppose the scattering ... 1 I don't completely understand the question. How a scalar transforms is completely dictated by conformal symmetry. The transformation law is$$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)$$where a^\mu are the infinitesimal parameters of the ... 1 That g^{-1}\mathrm{d} g is Liealgebra-valued for a Lie group-valued function g has nothing to do with the particular model or with physics, it is true for all matrix groups. Write g(x) = \exp(k(x)), where k(x) is now Lie algebra-valued and \exp is the usual power series in the case of a matrix group. Then \partial_\mu g = \partial_\mu k\exp(k) by ... 1 A CFT is still a QFT, and the way to put it at finite temperature is standard for any quantum system - you take your Hamiltonian H and compute Z=\mathrm{tr}\,e^{-\beta H}, where the trace is over the Hilbert space of states living on \mathbb{R}^{d-1} if your CFT is in d dimensions. The thermal correlators are computed in a similar way, ... 1 To see why the descendants are primary, you can use$$ \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....