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I'm not being precise, but morally: Imagine you were integrating out all modes above a frequency $b\Lambda$. Consider $\omega < b\Lambda < 3 \omega$. A mode $\phi$ with frequency $\omega$ when cubed, will have some part of it as mode of frequency $3 \omega$, since: $\sin (3x) = 3 \sin (x) - 4 \sin^3 (x)$. (Easier to see that ${(e^{i \omega t})}^3 = ... 2 Let us start with a simple analogy that helps grasping the concept. Say, we need to calculate a electric field of a charged cylinder. What is the first thing you do? You choose your$z$-axis along the axis of the cylinder. Why won't you choose your axes differently, like pointing whatever they want? Well, you could, but that will make the problem ... 7 The advantage of unitary gauge is that it completely removes unphysical fields, while adding additional degrees of freedom to the gauge bosons, which consequently become massive. This gauge works well for tree-level calculations, but complications arise when considering loops: The propagators of gauge fields and ghosts (which are needed to impose the ... 0 For what it's worth, it is claimed that the critical exponents differ above and below the critical point for some exactly solvable 2-dimensional model: http://www.ujp.bitp.kiev.ua/files/journals/49/11/491114p.pdf (Ukr. J. Phys., v.49, #11, p.1122 (2004)). 2 Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly ... 4 This is a good question. I can only answer half of it: The eigenvalues of$M^{l}\$ are some times complex. Take a look at the RG flow of Einstein gravity close to its UV fixed point. The two relevant couplings (newton's constant and cosmological constant) spiral around the fixed point as they move away from it. The real part of the critical exponents ...