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Oct 27, 2021 at 13:42 comment added Carlo Beenakker yes, the map changes the action $G$ (it is a many-to-one map), but it does not change the partition function $Z$; in the simplest case (dimensions greater than 4) all nonlinear terms in $\phi$ flow to zero except the quadratic term, and then you have a simple Gaussian integral; more generally you hope/expect that the nonlinear terms will become small enough to be able to treat them perturbatively.
Oct 27, 2021 at 13:04 comment added JustWannaKnow Also, what is the role of the estimates I mentioned when one does perturbative RG? Are this attempts to prove that higher order powers of $\phi$ are small enough so they do not contribute, as you mentioned?
Oct 27, 2021 at 13:03 comment added JustWannaKnow Carlo, thanks for your answer. The lecture notes will certainly be helpful. But could you clarify a little more your answer? I mean, the renormalization group $\mathscr{R}$ I mentioned in my post changes the effective action, so your answer seem to indicate that one is actually trying to use this map $\mathscr{R}$ so the flow $G \to G' \to \cdots$ end up having a nicer/simpler form. Is this correct? If so, the idea is to get (hopefully) get something one can actually integrate? Or a fixed point, as I mentioned?
Oct 26, 2021 at 20:27 history answered Carlo Beenakker CC BY-SA 4.0