When going through the calculation of the conformal algebra, I find that


where $J_{xy}$ are the generators of the conformal field theory, which consists of translations, dilations, rotations, and the special conformal group.

Many sources say that the above is equivalent to an $SO(d+1,n+1)$ special pseudo-orthogonal group, where $d$ is the number of space dimensions and $n$ is the number of time dimensions. However, I also know that the generators of the Lorentz group satisfies a similar algebra:

\begin{align} [\mathcal{M}^{\mu \nu},\,M^{\rho \sigma}]=i(g_{\nu \rho}\mathcal{M}^{\mu \sigma}-g_{\nu \sigma}\mathcal{M}^{\mu \rho}-g_{\mu \rho}\mathcal{M}^{\nu \sigma}+g_{\nu \rho}\mathcal{M}^{\mu \sigma}) \end{align}

which is $SO(3,1)$ My question is why do we say the Lorentz group is $SO(3,1)$ while the conformal group is $SO(d+1,n+1)$? Wouldn't the metric be the same in both--i.e., be a diagonal matrix with three +1's and one -1? If we consider a general metric in some arbitrary dimension, wouldn't the Lorentz group also be $SO(d+1,n+1)$?


1 Answer 1


You don't have all algebra to the conformal-group.

The conformal-group have rotations and boost $M_{\nu\mu}$, the Lorentz group ($SO(d,n)$) $$ [M_{\nu\mu},M_{\rho\gamma}]=4\eta_{[\mu[\rho}M_{\sigma]\mu]}. \text{ (This is yours equation for $J_{ab}\leftrightarrow \mathcal{M}_{\nu\mu}$)} $$ Also have translations $P^\mu$, the Poincaré Group $$ [P_\mu, M_{\nu\rho}]=2\eta_{\,\mu[ \nu}P_{\rho]}. $$ But, the conformal group also have, the special conformal transformations $K^\mu$ $$ [K_\mu, M_{\nu\rho}]=2\eta_{\,\mu[ \nu}K_{\rho]};\quad [P_\mu,K_\nu]=2(\eta_{\mu\nu}D+M_{\mu\nu}), $$ and the dilations $$ [D,P_\mu]=P_\mu;\quad [D,K_\mu]=-K_\mu. $$ And this is,of course, the algebra of $SO(D,2)$ whit $D=d+n$ and $n=1$ (I don't remember if this algebra is valid to $n>1$).

Let $\hat\mu=0,\cdots,D,D+1$, the generators can be codified in a compact way by $$ M^{\hat \mu\hat \nu}= \begin{pmatrix} M^{\mu\nu}&\frac{1}{2}(P^\mu-K^\mu)&\frac{1}{2}(P^\mu +K^\mu)\\ -\frac{1}{2}(P^\mu -K^\mu)& 0 &-D\\ -\frac{1}{2}(P^\mu +K^\mu)& D &0 \end{pmatrix}, $$ which the indices are raised by $\hat \eta=\text{diag}(-1,1,1,\cdots,1,-1)$.

Therefore the Lorentz Group only have rotations and boost, and it is a subgroup of the conformal-Group.


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