I Can Conformal Transformations be Derived from Group Composition Rules?

[formodular]

Hi!

Is there a way to end up with the algebra

i) quickly

ii) starting from a group, as how one gets the CR's from the Lorentz group composition rules, as on http://www.krassnigg.org/web/physics/wp-content/uploads/hoqft12-skriptum.pdf.

The other relations are quite complicated and the composition rules well are not clear.

 

[samalkhaiat]

Hi!

Is there a way to end up with the algebra

i) quickly

ii) starting from a group, as how one gets the CR's from the Lorentz group composition rules, as on http://www.krassnigg.org/web/physics/wp-content/uploads/hoqft12-skriptum.pdf.

The other relations are quite complicated and the composition rules well are not clear.

The algebra of the conformal group \mbox{Con}(1,n-1) is isomorphic to that of the Lorentz group \mbox{SO}(2,n). The latter can be obtained from the infinitesimal form of the group multiplication law U(\Lambda) U(\bar{\Lambda}) = U(\Lambda \bar{\Lambda}), by setting U(\Lambda) = 1 + \frac{i}{2} \omega_{AB}M^{AB}, where \omega_{AB} + \omega_{BA} = 0, and A , B = -2, -1, 0, 1, \cdots , n-1. Doing the easy exercise gives you [i M^{AB } , M^{CD}] = \eta^{BC}M^{AD} - \eta^{AC}M^{BD} + \eta^{AD}M^{BC} - \eta^{BD} M^{AC} , where \eta^{AB} = (1 , -1, \eta^{\mu\nu}) with \eta^{\mu\nu} being the Lorentz metric on the n-dimensional Minkowski space-time \mu, \nu = 0, 1, \cdots n-1. Now, the algebra of \mbox{Con}(1,n-1) is obtained from the above by defining the following generators D = M^{-2 , -1} , \ \ \ J^{\mu\nu} = M^{\mu\nu} , \frac{1}{2} (P^{\mu} - K^{\mu}) = M^{-2 , \mu} , \frac{1}{2}(P^{\mu} + K^{\mu}) = M^{-1, \mu} .

You may want to look at

https://www.physicsforums.com/showthread.php?t=172461

 

[formodular]

The algebra of the conformal group \mbox{Con}(1,n-1) is isomorphic to that of the Lorentz group \mbox{SO}(2,n).

Yes!

This is ultimately the answer, and seeing it would give the commutation relations very quickly if you knew you could end up doing this. I must say understanding this is what I am having difficulty with and have spent a week trying to see this, and the link to DeSitter/Anti-DeSitter spaces, in a way that predicts everything in advance, and need to think some more before formulating a question on how to see this in advance, e.g. in an old Dirac paper he just goes to DeSitter space in seconds with no work.

 

[samalkhaiat]

Yes!
This is ultimately the answer

The Lie algebra isomorphism \mathfrak{so}(n,2) \cong \mathfrak{con}(n-1,1), which I sketched in #2, is the mathematical base for the correspondence between field theory in the bulk of the space \mbox{Ads}_{n+1} and a conformal field theory on its boundary \mbox{CFT}_{n}. This is because of the fact that the group SO(n,2) is the isometry group of \mbox{Ads}_{n+1}, and \mbox{Con}(n-1,1) is (obviously) the symmetry group of \mbox{CFT}_{n}.

 

[formodular]

Well...

I do not understand that yet, so from a more rudimentary place - if you view conformal transformations as transformations preserving the lightcone,
$$ds^2 = x^2 + y^2 + z^2 - t^2 = 0 \ \ \to \ \ ds'^2 = f(x)ds^2 = 0$$
then we can go from here to what we will call AdS space by treating this as embedded in
$$x^2 + y^2 + z^2 - t^2 = 0 \ \ \to \ \ x^2 + y^2 + z^2 - t^2 - w^2 = - R^2 $$

https://www.jstor.org/stable/1968649

where when $x,y,z,t$ are small in comparison with $R$ we have $w$ equal to $R$ to first order so that, as $R \to \infty$ we obtain our lightcone, and so it seems reasonable why conformal transformations arise in this AdS context,
$$x'^2 + y'^2 + z'^2 - t'^2 - w^2 = - R^2 \ \ \to \ \ f(x)(x^2 + y^2 + z^2 - t^2) - w^2 = - R^2$$
and why the Lie algebras might be linked.

Then there is another viewpoint

https://www.jstor.org/stable/1968455

where you view your $4$ dimensional surface as a surface in $5$ dimensional projective space which, in homogeneous coordinates, gives $6$ coordinates $x^{\mu}$ and you can go from $x_{\mu} x^{\mu} = 0$ to conformal transformations as in that paper.

So yes this is some rudimentary pieces of this puzzle so far, not enough yet.