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  Is the stability matrix of a linearised RG flow always diagonalisable?

+ 6 like - 0 dislike

This is a follow up on "Why are the eigenvalues of a linearized RG transformation real?".

My question is simple: Is there some physical (or mathematical) reason for the stability matrix of Renormalisation Group flows close to fixed points to be diagnoalisable? What is it? If there isn't: Are there known counter examples? How do we deal with them?

This post imported from StackExchange Physics at 2015-08-04 09:52 (UTC), posted by SE-user Steven Mathey

asked Oct 2, 2014 in Theoretical Physics by Steven Mathey (350 points) [ revision history ]
edited Oct 16, 2017 by Dilaton

There is nothing that forces diagonalizability. I believe there are even counterexamples, where the Jordan structure (which replaces the diagonal structure in the nondiagonalizable case) produces logarithmic terms in the corresponding expansion. This should be discussed somewhere in Fisher's work around 1974.

Read "A renormalisation-group treatment of two-body scattering". After equation 9 : If, for a particular system, we find that the rescaled potential tends towards this fixed point as we lower the cut-off towards zero, then etc ... and the citations 11/13/18/19 with the conclusion.

1 Answer

+ 3 like - 0 dislike

I don't believe there is a mathematical reason, especially if there is latitude in reverse-engineering the field theory or stat mech system to evince such a behavior. Indeed, if Lorentz-nonivariant systems are examined, things like limit cycles , e.g. this one are not hard to concoct. As for physical reasons, they might well be easy to bypass/moot if one argued for them. I don't know of any systems, however, with this property, which might not say much.

As a mathematical wisecrack, I could manufacture a simple toy system with two couplings, x and y and logarithmic scale variable t : $$ \dot{x}=-x + ay, \qquad \dot{y}= -y , $$ with evident solutions stable around the fixed point (0,0), $$ y= e^{-t}, \qquad x= (c +at) e^{-t} . $$ The stability matrix of the ODE system is $$ \left( \begin{array}{cc} -1 & a \\ 0 & -1 \\ \end{array} \right) $$ which is not diagonalizable, with only one eigenvector, $$ \left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right) $$ of eigenvalue -1. This is not to say the system is not stable, however, if one could solve the ODE, somehow, as here.

This post imported from StackExchange Physics at 2017-10-16 12:31 (UTC), posted by SE-user Cosmas Zachos
answered May 14, 2016 by Cosmas Zachos (370 points) [ no revision ]

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