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In the textbook of TASI 2009, section "Introduction to extra dimension" i can find the answer as follows. They state that $5D$ or higher dimensional gauge coupling has a negative mass dimension, so the 5d or higher dimensional gauge theory is non-renormalizable.


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If you are thinking to apply renormalization to gravity, I would suggest to look at the explanation from "String Theory" of Kevin Wray: the problem is that we need more and more parameters to absorb the infinities that occur in the theory. String theory solve this particular problem because a string has finite extent lp, the divergent integral is cutoff at ...


3

Firstly, renormalisation isn't really inherently to do with the UV divergences you get: it's more to do with the idea that interactions with other fields change a particle's energy from its inherent 'bare' mass, so the measured mass isn't the same as the value that appears in the quadratic part of the action. You play a renormalisation game in perturbation ...


2

What I think one needs to internalize conceptually is that the program of renormalization is always favourable (and almost always required) in physical theories, be they fundamental or effective phenomenological ones (including condensed matter field theories), be there infinities or not. I think the last point is by far the most important. Yes, ...


9

You seem to be confusing regularization with renormalization. Regularization is the process of removing (or, more properly, parameterizing) infinities in loop integrals. Often in elementary texts a "cutoff" representing an energy scale above which the theory is assumed to be invalid is discussed, and counterterms are added to the Lagrangian in order to make ...


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Comments to the question (v1): Ref.1 is considering $\varphi^3$ theory $$\tag{1} {\cal L}(J)~=~\frac{1}{2}Z_{\varphi}\partial^{\mu}\varphi\partial_{\mu}\varphi - \frac{1}{2}Z_{m}m^2\varphi^2 - \frac{1}{6}Z_{g}g\varphi^3+(Y+J)\varphi.$$ To read the Feynman diagrams in Ref. 1, note that the source $J(x)$ is drawn as a black bullet $\bullet$, and the ...


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The fact that the boundary theory is conformal means that renormalization does not induce running of the coupling. However, there are divergences which have to be regularized and renormalized. The regularization requires the introduction of an arbitrary scale, which is not Weyl invariant and leads to a conformal anomaly (in even dimensions). ...



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