symmetry - Conceptual question about field transformation

(c.f Conformal Field Theory by Di Francesco et al, p39) From another source, I understand the mathematical derivation that leads to eqn (2.126) in Di Francesco et al, however conceptually I do not understand why the equation is the way it is. The equation is

$$\Phi'(x) - \Phi(x) = -i\omega_a G_a \Phi(x)$$ G is defined to be the generator that transforms both the coordinates and the field as far as I understand (so it is the full generator of the symmetry transformation) and yet it seems there are no instances of the transformed coordinate system (the primed system) present on the LHS of the equation.

Another example is the equation $\Phi'(x') = D(g)\Phi(x)$, where the field transforms under a representation $D$ of some Lie group. $D$ is then decomposed infinitesimally as $1 + \omega \cdot S$ where $S$ is the spin matrix for the field $\Phi$. So, when $S$ acts on the field, should it not only transform the spin indices on the field? But it appears we are in the primed system of the coordinates on the LHS meaning that the coordinates have changed too?

Can someone provide some thoughts?

I was also wondering how Di Francesco obtained eqn (2.127). Here is what I was thinking: Expand $\Phi(x')$, keeping the 'shape' of the field (as Grenier puts it) the same, so $\Phi(x') \approx \Phi(x) + \omega_a \frac{\delta \Phi(x')}{\delta \omega_a}$. Now insert into (2.125) gives

$$\Phi'(x') = \Phi(x') - \omega_a \frac{\delta x^{\mu}}{\delta \omega_a}\frac{\partial \Phi(x')}{\delta x^{\mu}} + \omega_a \frac{\delta F}{\delta \omega_a}(x)$$ which is nearly the right result except Di Francesco has a prime on the x at the final term. My thinking was he defined an equivalent function F in the primed system, i.e $F(\Phi(x)) \equiv F'(\Phi(x'))$. Are these arguments correct?

Answer

Group actions in classical field theory.

Let a classical theory of fields $\Phi:M\to V$ be given, where $M$ is a ``base" manifold and $V$ is some target space, which, for the sake of simplicity and pedagogical clarity, we take to be a vector space. Let $\mathscr F$ denote the set of admissible field configurations, namely the set of admissible functions $\Phi:M\to V$ which we consider in the theory.

In a given field theory, we often consider an action of a group $G$ on the set $\mathscr F$ of field configurations which specifies how they ``transform" under the elements of the group. Let's call this group action $\rho_\mathscr F$, so $\rho_\mathscr F$ is a group homomorphism mapping elements $g\in G$ to bijections $\rho_\mathscr F(g):\mathscr F \to \mathscr F$. Another way of saying this is that

$$\begin{align} \rho_{\mathscr F}:G\to \mathrm{Sym}(\mathscr F) \end{align}$$ where $\mathrm{Sym}(\mathscr F)$ is simply the group of bijections mapping $\mathscr F$ to itself.

Now, it also often happens that we can write such a group action in terms of two other group actions. The first of these is an action of $G$ on $M$, namely an action of the group on the base manifold on which the field configurations are defined. The second is an action of $G$ on $V$, namely an action of the group on the target space of the field configurations. Let the first be denoted $\rho_M$ and the second $\rho_V$, so explicitly

$$\begin{align} \rho_M &: G \to \mathrm{Sym}(M)\\ \rho_V &: G \to \mathrm{Sym}{V}. \end{align}$$ In fact, usually the group action $\rho_\mathscr F$ is written in terms of $\rho_M$ and $\rho_V$ as follows: $$\begin{align} \rho_{\mathscr F}(g)(\Phi)(x) = \rho_V(g)(\Phi(\rho_M(g)^{-1}(x))), \tag{$\star$} \end{align}$$ or, writing this in a perhaps clearer way that avoids so many parentheses, $$\begin{align} \rho_\mathscr F(g)(\Phi) = \rho_V(g)\circ \Phi \circ \rho_M(g)^{-1}. \end{align}$$

Making contact with Di Francesco et. al.'s notation - part 1

To make contact with Di Francesco's notation, notice that if we use the notation $x'$ for a transformed base manifold point, $\mathcal F$ for the target space group action, and $\Phi'$ for the transformed field configuration, namely if we use the shorthands

$$\begin{align} x' = \rho_M(g)(x), \qquad \rho_V(g) = \mathcal F, \qquad \Phi' = \rho_\mathscr F(g)(\Phi), \end{align}$$ then ($\star$) can be written as follows: $$\begin{align} \Phi'(x') = \mathcal F(\Phi(x)). \end{align}$$ which is precisely equation 2.114 in Di Francesco

Lie group actions and infinitesimal generators.

Now, suppose that $G$ is a matrix Lie group with Lie algebra $\mathfrak g$. Let $\{X_a\}$ be a basis for $\mathfrak g$, then any element of $X\in\mathfrak g$ can be written as $\omega_aX_a$ (implied sum) for some numbers $\omega_a$. Moreover, $e^{-i\epsilon\omega_aX_a}$ is an element of the Lie group $G$ for each $\epsilon\in\mathbb R$. In particular, notice that we can expand the exponential about $\epsilon = 0$ to obtain

$$\begin{align} e^{-i\epsilon\omega_aX_a} = I_G - i\epsilon\omega_aX_a + O(\epsilon^2). \end{align}$$ where $I_G$ is the identity on $G$. So the $X_a$ are the infinitesimal generators of this exponential mapping. Notice that the generators are precisely the basis elements we chose for the Lie algebra. If we have chosen a different basis, the generators would have been different. But now, suppose we evaluate the various group actions $\rho_\mathscr F, \rho_M, \rho_V$ on an element of $G$ written in this way, then we will find that there exists functions $G_a^{(\mathscr F)}, G_a^{(M)}, G_a^{(V)}$ for which $$\begin{align} \rho_\mathscr F(e^{-i\epsilon\omega_aX_a}) &= I_\mathscr F -i\epsilon\omega_a G_a^{(\mathscr F)} + O(\epsilon^2) \\ \rho_M(e^{-i\epsilon\omega_aX_a}) &= I_M -i\epsilon\omega_a G_a^{( M)} + O(\epsilon^2) \\ \rho_V(e^{-i\epsilon\omega_aX_a}) &= I_V -i\epsilon\omega_a G_a^{(V )} + O(\epsilon^2), \end{align}$$ and these functions $G_a$ are, by definition, the infinitesimal generators of these three group actions. Notice, again, that the generators we get depend on the basis we chose for the Lie algebra $\mathfrak g$.

Making contact with Di Francesco et. al.'s notation - part 2

Di Francesco simply uses the following notations:

$$\begin{align} G_a^{(\mathscr F)} = G_a, \qquad G_a^{(M)}(x) = i \frac{\delta x}{\delta \omega_a}, \qquad G_a^{(V)}(\Phi(x)) = i\frac{\delta \mathcal F}{\delta \omega_a}(x). \end{align}$$ Don't you wish the authors would explain this stuff better? ;)

Example. Lorentz vector field.

As an example, consider a theory of fields containing a Lorentz vector field $A$. Then the base manifold $M$ would be Minkowski space, the group $G$ would be the Lorentz group,

$$\begin{align} M = \mathbb R^{3,1}, \qquad G = \mathrm{SO}(3,1), \end{align}$$ and the relevant group actions would be defined as follows for each $\Lambda \in \mathrm{SO}(3,1)$: $$\begin{align} \rho_M(\Lambda)(x) &= \Lambda x \\ \rho_V(\Lambda)(A(x)) &= \Lambda A(x) \\ \rho_\mathscr F(\Lambda)(A)(x) &= \Lambda A(\Lambda^{-1}x). \end{align}$$