When going through the calculation of the conformal algebra, I find that
$$[J_{ab},J_{cd}]=i(g_{ad}J_{bc}+g_{bc}J_{ad}-g_{ac}J_{bd}-g_{bd}J_{ac})$$
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 $SO(d+1,n+1)$?