Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In P. Di Francesco, P. Mathieu, D. Snchal they fix the generators of the conformal group acting on a scalar field by somewhat arbitrarily defining $$\Phi'(x)=\Phi(x)-i\omega_a G_a\Phi(x)$$ and by arbitrary I mean the sign. The "full" transformation would then be given by the exponentiation $$\Phi'(x)=e^{-i\omega_a G_a}\Phi(x)$$ the generators are given in eq. 4.18 and the commutation relations are in eq. 4.19. But by looking at the action of the group in the coordinates, I think we could get the same representation 4.18 by requiring $$x'^\mu=e^{ i\omega_aT_a}x^\mu=\alpha^\mu+\omega^\mu_{~~\nu} x^\nu + \lambda x^\mu +2(\beta \cdot x)x^\mu-\beta^\mu x^2$$ but the problem is that, in order to reproduce 4.18 the sign of the exponential is not the same as the one for $G_a$. I would have expected that the field transform in a similar way: $$\Phi'(x)=e^{i\omega_a G_a} \Phi(x)$$ In fact it seems that there is not a "standard" definition. For example in this article by Mack and Salam, as well as in this book by Fradkin and Palchik, the generators as well as the commutation relations disagree among each other.

Anyway I would like to know how the generators should be defined so that everything is consistent from the Lie theory/representation theory point of view.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Consider a conformal theory of fields $\Phi$ defined on Minkowski spacetime $M=\mathbb R^{3,1}$ and with target space $V$. Let $\rho_M$ denote a representation of the conformal group $G$ on $M$ and let $\rho_V$ denote a representation of $G$ on $V$. Then we can define an action $\rho_F$ of $G$ on fields $\Phi$ as follows: $$ (\rho_F(g)\Phi)(\rho_M(g)x) = \rho_V(g)\Phi(x) $$ This is a more notationally descriptive version of di-Francesco eq. 2.114; $$ \Phi'(x') = \mathcal F(\Phi(x)) $$ In other words, the action of $G$ on fields has two parts, a spacetime part, and a target space (aka internal) part. Now, suppose that we find that there is another representation $\rho_D$ of $G$ acting on fields (by way of differential operators for example) for which $$ \Phi(\rho_M(g)x) = \rho_D(g)\Phi(x) $$ Then notice that the first equation above can be written as follows: $$ (\rho_F(g)\Phi)(x) = \rho_D(g^{-1})\rho_V(g)\Phi(x) $$ How does this manifest itself in terms of signs and generators? Well, suppose that for any group representation $\rho$ of $G$, we have a corresponding, linear representation $R$ of its Lie algebra $\mathfrak g$ such that if we write an element $g\in G$ as the exponential of a Lie algebra element $X$; $g = e^X$, then $$ \rho(g) = e^{R(X)} $$ Then we can write the transformation of the fields as $$ (\rho_F(g)\Phi)(x) = e^{-R_D(X)}e^{R_V(X)}\Phi(x) $$ For any $X\in\mathfrak g$. If these Lie algebra representations commute (as they would if, for example, $R_D$ is a representation via differential operators and $R_V$ is some spacetime-independent representation on the target space), then we can write $$ (\rho_F(g)\Phi)(x) = e^{R_V(X)-R_D(X)}\Phi(x) $$ Now, suppose that we choose a basis $X_a$ for the Lie algebra $\mathfrak g$ such that every element $X\in \mathfrak g$ can be written as $$ X = i\omega_aX_a $$ for some real numbers $\omega_a$. Then using linearity of the representations $R_D, R_V$ we have $$ (\rho_F(g)\Phi)(x) = e^{-i\omega_a(R_D(X_a) - R_V(X_a))}\Phi(x) $$ So if we use di-Francesco's notation $$ (\rho_F(g)\Phi)(x) = e^{-i\omega_aG_a}\Phi(x) $$ then we have $$ G_a = R_D(X_a) - R_V(X_a) $$ To see that this leads to consistent signs etc., let's look at an example:

Example. Translations

Consider a field whose target space transforms trivially under translations; $$ (\rho_F(g)\Phi)(x+a) = \Phi(x) $$ Then if $ia^\mu P_\mu$ is the Lie algebra element that corresponds to translations $x\to x+a$, and if we define $$ R_V(P_\mu) = 0, \qquad R_D(P_\mu) = -i\partial_\mu $$ Then we have $$ G_\mu = -i\partial_\mu $$ so that $$ e^{-ia^\mu G_\mu}\Phi(x) = e^{-a^\mu\partial_\mu}\Phi(x) = \Phi(x-a) = (\rho_F(g)\Phi)(x) $$ as desired.

share|improve this answer
    
Thanks a lot for your answer. Now there's still something that is bugging me. In your first eq. you write the field on the LHS with a prime. In Mack and Salam paper they use a similar relation to see that $\rho_V(g,x=0)$ is a rep. of the stability subgroup of $x=0$. Then I guess that in order to be completely consistent with the notation we can replace $\Phi'$ by $(R(g)\Phi)$ right? –  Barefeg Aug 13 '13 at 1:25
    
@Barefeg Sure thing. I changed the notation a bit in line with your observation. Basically, you're right, if I want to be more descriptive, then I note that I can write $\Phi'$ as some group action $\rho_F$ of $G$ on the fields; $\Phi' = \rho_F(g)\Phi$. –  joshphysics Aug 13 '13 at 1:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.