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In his paper Constructive Renormalization Theory, V. Rivasseau describes the idea of Wilson's approach of solving path integrals step by step. In section 1.4, page 5, however, there is a statement which I do not follow. He discusses a $\phi^{4}$ theory and decomposes the covariance into a sum: \begin{eqnarray} C_{0}(p) = \int_{1}^{\infty}e^{-\alpha(p^{2}+m^{2})}d\alpha \quad \mbox{and} \quad C^{j}(p) = \int_{M^{-2j}}^{M^{-2(j-1)}}e^{-\alpha(p^{2}+m^{2})}d\alpha \tag{1}\label{1} \end{eqnarray} in such a way that: \begin{eqnarray} C_{\rho}(p) = \sum_{j=0}^{\rho}C^{j}(p) \tag{2}\label{2} \end{eqnarray} He proceeds to explain how the partition function can be obtained by performing interativelly $\rho+1$ (convolution) integral. At some point, he states the following:

We see that constructing the ultraviolet limit is the same as finding a bare action $S_{\rho}(\phi)$ such that the effective action, or renormalized action $S_{0}(\phi)$ converges as $\rho \to \infty$.

I might be missing something, but I don't fully understand this statement. First, it seems that, in the limit $\rho \to \infty$ one recovers the the regularized theory, because (\ref{1}) is a telescoping series with UV cutoff and as $j \to +\infty$ one ends up with regularized covariance \begin{eqnarray} C_{0}(p) = \frac{1}{p^{2}+m^{2}}e^{-(p^{2}+m^{2})}. \tag{3}\label{3} \end{eqnarray} In other words, I don't see how it is possible to take the UV limit using this decomposition into steps as a limiting case. Second, it is a little bit odd for me to say that one finds $S_{\rho}(\phi)$ so that $S_{0}(\rho)$ exists when $\rho \to \infty$. The flow is going backwards, from $\rho$ to zero, how can you then $\rho \to \infty$ afterwards? Is it just because $S_{0}$ should depend on $\rho$ somehow?

Add: A related question is the following. I think the main idea is to proceed with the integrating steps until one achieves a fixed point. What does one expect of this fixed point? I mean, is it dependent on the UV cutoff so one has to deal with removing this cutoff or does one expect that it is independent on this cutoff so the removal is trivial?

In short, I am really confused about how one plans to remove the cutoff. I am not convinced that this can be done by a limiting case of the RG iteraction as Rivasseau stated. However, if this limiting case is achieved and it does depend on the cutoff, how to remove it? I know these topics are really complicate in practice and there is no recipe, but I would be happy with a picture of what one is trying to do or expect to have. In other words, what is the philosophy?

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The telescopic argument you mentioned is incorrect. If you carefully look at Rivasseau's notations, you will see that the first term $C^{0}(p)$ in the sum $$ \sum_{j=0}^{\rho}C^{j}(p) $$ is defined as $$ \int_{1}^{\infty} e^{-\alpha(p^2+m^2)}d\alpha\ . $$ So the sum is just the integral over the range $[M^{-2\rho},\infty)$.

Note that kind of momentum slicing is also used in harmonic analysis, i.e., Littlewood-Paley decompositions, here, in the nonhomogeneous case where the large scales/low frequencies are combined into a single big chunk. The slices are indexed by $\mathbb{N}$ instead of $\mathbb{Z}$ used in the homogeneous case.

For the addendum question about philosophy, that requires a long answer which I already gave at

Formal mathematical definition of renormalization group flow

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  • $\begingroup$ Abdelmalek, thanks for your help. I now got it. However, isn't it more common to first regularize the covariance, say, putting $\hat{C}_{\Lambda_{0}}(p) = \frac{1}{p^{2}+m^{2}}e^{-\frac{(p^{2}+m^{2})}{\Lambda_{0}^{2}}}$ for some $\Lambda_{0} > 0$ and then decomposing it into scales, say, by setting $\Lambda(t) = \Lambda_{0}e^{-t}$ and summing it using a geometric progression? This seems rather different than Rivasseau's proposal, since in this case I am mentioning, the $\Lambda_{0}$ remains in the end and one ends up with the initial regularized theory instead of the continuum theory. $\endgroup$
    – JustWannaKnow
    Commented Nov 19, 2022 at 16:47
  • $\begingroup$ Putting another way, if the decomposition of Rivasseau promptly gives the continuum limit as $\rho \to \infty$, why do people still use a regularized theory with an UV (and possibly an IR) cutoff and use the same machinery to get a a $\Lambda_{0}$ dependent theory in the end? I think this is what I meant by my added question. $\endgroup$
    – JustWannaKnow
    Commented Nov 19, 2022 at 16:50
  • $\begingroup$ That's because of different points of view (dimensionful vs dimensionless) going around, and explaining the difference is one of the main points of my other MO answer I linked to. I think you should read it before coming back with new questions. The original intent in constructing the UV limit is to take $\Lambda_0$ to zero, as Rivasseau does by taking $\rho$ to infinity. However there is a rescaling trick which allows one to always revert to the situation where $\Lambda_0$ is fixed say to the value 1, i.e., alwasy working on the unit lattice. This what I explained in that other answer. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Nov 19, 2022 at 19:01

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