# A small issue in renormalisation group formalism

In the general RG formalism, suppose $\vec{\mu}$ represents a vector in parameter space and $\vec{\mu}^*$ is the fixed point under the transformation $R$. Then for $\vec{\mu}=\vec{\mu}^*+\delta \vec{\mu}$, we have $$R(\vec{\mu})=R(\vec{\mu}^*+\delta \vec{\mu})=\vec{\mu}^*+R'(\vec{\mu}^*)\delta \vec{\mu}$$ after which the small deviation $\delta \vec{\mu}$ is expanded as a linear combination of the eigenvectors of the matrix $R'(\vec{\mu}^*)$, denoted as $\{\vec{v}_i\}$ with $\{\lambda_i\}$ as the corresponding eigenvalues.

It seems that people normally just assume that the eigenvectors of $R'(\vec{\mu}^*)$ form a complete basis, which is obviously not true in general. Now let's assume that $\{\vec{v}_i\}$ does not form a complete basis, then we would have $$\delta \vec{\mu}=\sum_i c_i\vec{v}_i+\Delta \vec{\mu}$$ where $\Delta \vec{\mu}=\delta \vec{\mu}-\sum_i c_i\vec{v}_i$ and $\Delta \vec{\mu} \perp \vec{v}_i$ for all $i$. This simply means that $\Delta \vec{\mu}$ is the projection of $\delta \vec{\mu}$ onto the normal of the hyperplane spanned by $\{\vec{v}_i\}$. In this case, the linear transformation from the linearisation obviously keeps $\Delta \vec{\mu}$ invariant. Thus, if $\lambda_i <0$ for all $i$, we will get $\vec{\mu}^* +\Delta \vec{\mu}$ in the end, which effectively seems to be a new shifted fixed point. Indeed, we have $$R(\vec{\mu}^*+\Delta \vec{\mu})=\vec{\mu}^*+R'(\vec{\mu}^*)\Delta \vec{\mu}=\vec{\mu}^*+\Delta \vec{\mu}$$ Since $|\Delta \vec{\mu}|<<1$, it seems that the two fixed points are very close to each other and in fact they can be made arbitrarily close if the the $\delta \vec{\mu}$ is carefully chosen. This seems to be a quite strange result. Furthermore, along the way we only kept the linear terms of the expansions, whereas I suspect that higher order terms may play a role.

The overall picture seems rather blurred to me. Really appreciate if someone can help clarify.

EDIT: Thanks to @Brightsun, $\Delta \vec{\mu}$ is not necessarily invariant under the action of $R'(\vec{\mu}^*)$ (example given in the comment)

• Why is $\Delta \vec{\mu}$ invariant under $R'$? I can think of linear maps that do not leave the vectors orthogonal to their eigenspaces invariant. – Brightsun Oct 29 '15 at 17:45
• @Brightsun could you show me the example? – M. Zeng Oct 30 '15 at 9:24
• Consider the 2x2 lower triangular matrix $A$ with non-zero elements equal to 1: its only eigenspace is made by multiples of $(0,1)$, but $A$ acts nontrivially upon, say, $(1,0)$. – Brightsun Oct 31 '15 at 10:14
• thanks for pointing this out. will edit the question accordingly. – M. Zeng Nov 1 '15 at 2:22
• You have contradicted yourself: if $\Delta\mu$ is not an eigenvector, then $R'(\mu)\Delta\mu\neq\Delta\mu$. Perhaps you want $\Delta\mu$ to stay within a subspace on which $R'(\mu)$ isn't diagonalizable? – TLDR Nov 1 '15 at 2:46

A typical renormalization group flow can be thought of as a smooth vector field $\vec V(\mu)$ defined on parameter space. Starting with parameters $\vec\mu(\ell)$ at scale $\ell$, you obtain parameters at scale $\ell'$ by solving the differential equation $\frac{d\vec\mu}{d\ell}=\vec V(\vec\mu(\ell))$. The function $R$ referred to above can be thought of as the process of solving this equation over some change in length scale. Fixed points of RG flow (scale-invariant field theories) then correspond to points $\vec \mu^*$ where $\vec V(\vec \mu^*)=\vec 0$. Assuming $\vec V(\vec \mu)$ is smooth, we can expand near $\vec\mu^*$ (this information we can usually get from field theoretic methods like the $\epsilon$-expansion in the neighborhood of simple `Gaussian' fixed points).
The situation you describe above (where $R(\mu^*+\Delta\mu)=\mu^*+\Delta\mu$ for a continuum of values of $\Delta\mu$) corresponds to the case when the zeros of $V(\mu)$ are not always isolated points, but curves or even hypersurfaces in parameter space.
A trivial example where this happens is if some operator $K$ that is conventionally assigned scaling dimension 0 is rewritten as $K=K_1+K_2$ in an enlarged parameter space. Sometimes there is a physical motivation to enlarge parameter space this way, like when there is an anisotropic crystal with a direction-dependent speed of sound. For example, if anisotropy breaks an $SO(2)$ symmetry to a $\mathbb{Z}_2$ symmetry, the system will behave like an Ising model for a large part of parameter space, behaving like an $SO(2)$ model only for highly ungeneric parameter values.
For intuition about the eigenspaces of $R'(\mu^*)$, it may help to look up graphs of various real RG flows (like in superconductors, or liquid crystals) and see how the flows behave near fixed points.