# Why are the eigenvalues of a linearized RG transformation real?

The RG transformation $R_\ell$ maps a set of coupling constants $[K]$ of a model Hamiltonian to a new set of coupling constants $[K']=R_\ell[K]$ of a coarse-grained model where the length scale is modified by a factor of $\ell>1$.

In the vicinity of a fixed point $[K^*]$, i.e. at $K_n=K_n^*+\delta K_n$, one linearizes the RG transformation, $$K_n' = K_n^*+\delta K_n' = K_n^*+\sum_m\underbrace{\left.\frac{\partial K_n'}{\partial K_m'}\right|_{[K]=[K^*]}}_{M_{nm}^{(\ell)}}\delta K_m+\mathcal{O}(\delta K^2),$$ and proceeds by relating the (right) eigenvalues of the matrix $M^{(\ell)}$ to the critical exponents.

My question is the following: the matrix $M^{(\ell)}$ is real but might not be symmetric. One thus has to distinguish left and right eigenvectors & the eigenvalues are not guaranteed to be real if they even exist. However, in the literature it is said at this point that nonetheless $M^{(\ell)}$ is generically found to be diagonalizable with real eigenvalues. Is there some (mathematical) rationalization of this statement?

• I may suggest you to look at chapter 6 of Gallavotti, Benfatto books: Renormalization Group. Perhaps the framework there could be a little different from the one of your reference, but the eigenvalues of the (linear) renormalization group are explicitly shown for a supposedly large class of theories. Sep 28, 2014 at 16:45
• here's a link: inis.jinr.ru/sl/vol2/Physics/… Sep 28, 2014 at 16:46
• If the eigenvectors need to be physically real, then the eigenvalues are necessarily real too. For instance, the scaling quantities $u'_i$ are obtained by multiplying eigenvectors $\phi^i_a$ by $(\delta K'_n)_a$. So, if the scaling quantities need to be real, then the eigenvectors have to be real too, no ? Sep 30, 2014 at 9:21

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 accounts for the flow away from the fixed point and their imaginary part makes them oscillate as the cut-off scale is lowered.

Something similar happens in the RG flow of three body bosonic systems: the Efimov effect. Bound states of three (bosonic) particles have an infinite ladder of bound states with binding energies that grow exponentially with a fixed ratio,

$$E_n = \text{const.} \times E_{n-1} \to E_n = \text{const.}^n$$

This is a result of a limit cycle of the RG flow. The system is not at a fixed point (and therefore not invariant under scale transformations) but the parameters cycle through a closed set of vales as the cut-off scale is changed. We find a system that is invariant under RG transformations with a finite scale only. If you flow just enough to go around the cycle once, you find the same coupling constants. The period of the cycle measured in cut-off scale is related to $\text{const.}$. We find a discreet scale invariance as a simple fractal.

Unfortunately I can't explain why $M^{l}$ can be diagonalised in the first place. I can only guess that there are situations where this is not the case and the RG flow must be re-interpreted. I would very much enjoy to read about an example of such a situation.

• Very nice. The question whether or not RG flows can exhibit limit cycles or attractors was another one of mine, actually. Coming from a statistical physics background I wasn't even aware that RG is being used in the context of quantum gravity, too. Oct 1, 2014 at 22:03
• I'm curious about the same thing, physics.stackexchange.com/questions/679946/… Did you manage to make much progress in learning anything about this? Dec 1, 2021 at 3:53