My question is from P.98 of the book by Di Francesco on Conformal Field theory. He gives the six non-vanishing commutation relations between the elements $P_{\mu}, D, L_{\mu \nu}$ and $K_{\mu}$ comprising the four dimensional space of generators of the conformal group. These are then rewritten by defining a set of four more generators: $$J_{\mu \nu} = L_{\mu \nu}\,\,\,,\,\,\,J_{-1, \mu} = \frac{1}{2}(P_{\mu} - K_{\mu})\,\,\,,\,\,\,J_{-1,0} = D\,\,\,,\,\,\,J_{0,\mu} = \frac{1}{2}(P_{\mu} + K_{\mu}),$$ and I believe the motivation for doing so is so that the six commutation relations mentioned above can be eloquently recast in a single one line commutation relation $$[J_{ab}, J_{cd}] = i(\eta_{ad}J_{bc} + \eta_{bc}J_{ad} - \eta_{ac}J_{bd} - \eta_{bd}J_{ac})$$ It also says that $a,b \in \left\{-1,0,1,\dots,d\right\}$ and that the new generators obey the $SO(d+1,1)$ commutation relations (which I think is the one-line equation above).
My question is: What do the indices $a,b$ represent and what does the notation $SO(d+1,1)$ mean? I think there are $d$ spatial dimensions but I can't see what the significance of the $-1$ and $0$ elements are.
Many thanks.
Answer
The indices $a$ and $b$ are chosen in such a way that inserting different combinations of values from $-1$ to $d$ gives just the original six commutation relations for the generators of conformal symmetry transformations. As you have correctly suggested, the point of this is to rewrite the six relations in compact form.
The latter shows that the conformal group is actually given by $\text{SO}(d+1,1)$, which is the group of special orthogonal transformations in $d+1$ spacelike and $1$ timelike dimensions, where $d$ is the number of spatial space-time dimensions. $0$ represents the time-component, while $-1$ is spacelike and appears because we have dilatations and special conformal transformations. Note that that $-1$ is not a spatial dimension with respect to spacetime but rather an additional dimension with respect to the group action.
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