I Conformal CR's From Group?

1. Jun 29, 2017

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 page fourteen to sixteen.

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

2. Jul 1, 2017

samalkhaiat

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

3. Jul 11, 2017

formodular

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.

4. Jul 11, 2017

samalkhaiat

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 $\mbox{CFT}_{n}$.

5. Jul 12, 2017

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.