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One important point to add to the obvious answer "yes, that is right" is that you should be careful about your signature. Specifically, Euclidean conformal group $SO(1,d+1)$ describes symmetries of Euclidean correlation functions and has nothing to do with unitary operators on the Hilbert space of the CFT. On the other hand, the Lorentz group $SO(1,... 2 To see why it's true, consider the set of all lightlike rays through the origin in Minkowski spacetime. This set has the topology of a two-dimensional sphere$S^2$, with each point on the sphere corresponding to one lightlike ray through the origin in Minkowski spacetime. What do Lorentz transformations in 4d Minkowski spacetime do to the points on that ... 2 You cannot conformally map the Lorentzian cylinder$\mathbb{R} \times S^{d-1}$to the Lorentzian plane$\mathbb{R^{1,d-1}}$even if you allow the domain and range to be missing finitely many points, e.g. by mapping to$\mathbb{R^{1,d-1}} \backslash \{0\}\$. By conformal mapping, I mean a diffeomorphism for which the metric pulls back to a conformally ...

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This answer ended up being much longer than I expected so tl/dr: The "invariances" Witten is talking about are redundancies in our description of the theory. So if we want to integrate over the physical space, we must integrate "up to diff and Weyl". If we did integrate over everything, we basically end up integrating over the same theory ...

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@AccidentalFourierTransform drew your attention to the Dirac operator, which, of course, can go Euclidean and extend to all dimensions. It has little to do with conformal FT, if that's where you want to go, because the conformal group is finite in all dimensions different than d=2. So disaggregate the conformal gig. First appreciate that Leon's construction ...

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