Differences and Similarities between the Lorentz and Conformal Groups 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 $SO(d+1,n+1)$?
 A: 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.
