# 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?

• I think you are right, generators are defined by Di Francesco as : $\Phi'(x) - \Phi(x) = -i\omega_a G_a \Phi(x)$. So I delete my previous answer. However, $2.134$ is a direct application of $2.128$ Jul 7, 2014 at 7:56
• Yes, indeed, although what you said about rearranging eqn (2.128) into $$\frac{\delta \mathcal F}{\delta \omega_a} = \frac{\delta x^{\mu}}{\delta \omega_a} \partial_{\mu}\Phi - iG_a\Phi$$ so that the ${\it total}$ change of $\mathcal F$ due to $\omega$ is a piece transforming the coordinates and a piece transforming the fields makes sense. What do you think?
– CAF
Jul 7, 2014 at 14:45
• The answer by @joshphysics is clearer. Its (*) equation (and more clearly the following equation), with its final definitions of the $G_a$ implies $2.128$, that is $G_a^{(\mathscr F)} = G_a^{(V)} - G_a^{(M)}$ Jul 8, 2014 at 8:30
• When you said that $\Phi(x') \approx \Phi(x) + \omega_a \frac{\delta \Phi(x)}{\delta \omega_a}$, this makes more sense but then how does it lead to the correct equation (2.127)?
– CAF
Jul 8, 2014 at 8:34
• OK I get the point. The difference between $x$ and $x'$ is just infinitesimal ($\omega_a$ parameters). So at the first order in the ($\omega_a$ parameters), it makes no difference considering $\omega_a \frac{\delta \mathcal F}{\delta \omega_a}(x')$ and $\omega_a \frac{\delta \mathcal F}{\delta \omega_a}(x)$. The difference between the two is in the second order in the infinitesimal parameter $\omega_a$, so is "neglectible" Jul 8, 2014 at 8:43

## 1 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}

• Many thanks joshphysics! I have a few short questions: -Does $M$ here, the 'base' manifold, correspond to for example, some coordinate space or frame of reference embedded in a Minkowski space? -Does the fact that $V$ being a vector space imply that the field under consideration is a vector field? So, e.g V correspond to the space in which all admissible spin vectors of the field live?
– CAF
Jul 9, 2014 at 10:31
• Is the following accurate? In a Lorentz transformation, we have that $\rho_{M}(g)(x) \rightarrow \Lambda$, where $\Lambda$ is an element of the Poincare group. In this case, we also have that $\rho_{\mathcal F}(g)(\Phi) = \Lambda$. But what would $\rho_{V}(g) \equiv \mathcal F$ represent?
– CAF
Jul 9, 2014 at 10:34
• @CAF The base manifold would be, for example, Minkowski space for field theories like QED. The fact that $V$ is a vector space allows for all sorts of exotic things, like for the field to be a tensor field (tensors are themselves elements of a vector space). The field could also be a spinor field etc. I added an example at the end to illuminate the notation. Jul 9, 2014 at 20:53
• I see, thanks. If $x' = \Lambda x$, then $\Lambda^{-1} x' = x$. The second to last equation above in Di Francesco's notation is $\mathcal F (\Phi(x)) = \Lambda \Phi(x),$ (Di Francesco actually uses $L_{\Lambda}$.) Similarly, (and I am not sure about this) is the last equation you wrote equivalent to $\Phi'(x') = \Lambda \Phi (x)$ ?(hmm, which is the same as the one before)
– CAF
Jul 9, 2014 at 21:26
• @CAF Yes, the last equation I wrote is equivalent to $\Phi'(x') = \Lambda \Phi(x)$ which is not the same as the second to last equation. The second to last equation is only the target space transformation, while the last equation is the entire transformation of the field including both the target space transformation, and the base manifold transformation. Jul 9, 2014 at 22:58