Identifying the relevant directions in the Ising model renormalization

Question

I'm reading the chapter about the renormalization group in Yeoman's book "Statistical mechanics of phase transitions" and I'm puzzled about how the author relates the scaling of the RG with the critical exponents. We have some RG map on the Hamiltonian $H\rightarrow R(H)$. We suppose that we are close to the fixed point $H^* $, so

$$H'=R(H^*+\delta H)=H^* + A(H^*)\delta H$$

where $A$ is a matrix and $\delta H$ is seen as a vector with the coupling constants as components. This matrix can be diagonalized and we can write

$$ A(H^*)\delta H= A(H^*)\sum_k\mu_k \Phi_k=\sum_k\lambda_k\mu_k \Phi_k\tag {$\star$}$$

where $\Phi_k$ are functions of the lattice and $\lambda_k$ are the eigenvalues of $A$. It's easy to argue that they must have the form

$$ \lambda_k=b^{y_k}$$

where $b$ is the scaling factor of the map. No problem until here. If $y_k>0$ we call it relevant, otherwise irrelevant.

Then the author says that for the Ising model the relevant couplings are the temperature and the magnetic field, and here I'm already confused. First, because $A$ depends on what renormalization scheme we choose, so how can we predict what the eigenvectors are without saying more about $R$? Second, the Ising Hamiltonian (if we absorb temperature in it) is

$$ H_I=-\beta J \sum_{\langle i,j\rangle}s_is_j-\beta h\sum_i s_i$$

I don't see how $\beta$ and $h$ could ever appear as in $(\star)$ as two linearly superposed terms. We would need something like

$$ H_I=H^*+\beta\Phi_\beta+h\Phi_h$$

but I don't see how this can be true because $\beta h$ appears in the Hamiltonian, it seems to me that $\beta h$ must be treated as a single coupling, and you can't take $\beta$ and $h$ separately. In short, my first question is

How can one treat temperature as a coupling constant if it appears in all the coupling constants?

There were originally two parts of this question, following advice of Adam I posted the second part in a separate question

Answer 1

While saying that the temperature and the magnetic field are relevant parameters is standard, it is somewhat sloppy, and confusing at first.

What is really meant is that both the temperature (more precisely $\beta J$ for the Ising model) and the magnetic field (or $\beta h$ for the Ising model) have non-trivial projections onto the two distinct relevant directions of the fixed point. Calling $t$ the coefficient of the operator $\Phi_t$ associated with the exponent $\nu$, and $H$ the coefficient of the symmetry breaking operator $\Phi_H$ associated to the exponent $\delta$, what is meant is that $t$ depends on $\beta J$ (but not $h$), and $H$ depends on $\beta J$ and $\beta h$ (since $H$ does not vanish at the critical temperature if $h$ is finite, its temperature dependence will give rise to corrections to scaling).

But in fact, all the coefficients $\mu_k$ do depend on $\beta J$ (and $\beta h$ if they correspond to symmetry breaking operators), but since most are irrelevant, we don't really care (unless we want to describe correction to scaling, see also below).

Furthermore, if the model is more complicated (with for instance more interactions $J'$, $J''$ between spins), then generically any of the couplings will be relevant (that is, one can induce a transition by changing $J'$, keeping $J$ constant), though the phenomenology might be complicated.

Finally, why do we usually assume that $t$ is linear in the temperature? For this, we use our physical intuition. First, the easiest thing we can change experimentally is most of the time the temperature, so it is the relevant physical quantity. Since changing the temperature does induce a transition (otherwise, we would use another physical quantity), the relevant parameter $t$ must depend on temperature. Furthermore, since to get close to the fixed point, only a finite number of RG transformation is necessary, and since each transformation is analytic in all quantities, $t$ is an analytic function of the temperature. By definition, it vanishes at the critical temperature. By Taylor expanding around $0$, it will thus be a linear function of temperature close enough to the critical temperature. In principle, the linear coefficient could vanish, but this is not the case generically (it can happen if there are some additional constraints that are robust, due to some additional symmetry, for instance). The fact that $t$ is in fact a complicated function of the temperature will induce correction to scaling is we are not close enough to the transition.