Wrong sign in Conformal Casimir The quadratic conformal Casimir in $d$-dimensional Euclidean space is given by
\begin{equation}
C = \frac{1}{2}L_{\mu \nu}L^{\mu \nu} - D^2 -\frac{1}{2}\left(P^\mu K_\mu + K^\mu P_\mu \right)
\end{equation}
as given for example in the beginning of lecture 6 here http://pirsa.org/C14038. 
Since there is an isomorphism between the conformal group and $SO(d+1,1)$ it should be possible to get this result by simply expanding $\frac{1}{2} M^{ab}M_{ab}$ with the identifications (DiFrancesco Eq. (4.20))
\begin{equation}
\begin{split}
M_{-1,0} &= D \\
M_{-1,\mu}  &= \frac{1}{2} \left( P_\mu -K_\mu \right) \\
M_{0,\mu}\ &= \frac{1}{2} \left( P_\mu +K_\mu \right) \\
M_{\mu \nu}\ &= L_{\mu \nu} 
\end{split}
\end{equation}
and $\eta_{ab}= \mathrm{diag}(-1,1,...1)$. However absolutely every time I attempt to do this calculation I get
\begin{equation}
C = \frac{1}{2}L_{\mu \nu}L^{\mu \nu} - D^2 +\frac{1}{2}\left(P^\mu K_\mu + K^\mu P_\mu \right).
\end{equation}
There are many different sign conventions out there but I don't think that's the problem because my wrong Casimir really does not commute with the elements of the algebra. 
I know it's not the most exciting calculation to do but I would eternally grateful to whoever can point out where the flaw lies. 
 A: To do the computation, considering 
$$\frac{1}{2}M^{ab}M_{ab}=\frac{1}{2}(M^{\mu\nu}M_{\mu\nu}+M^{\mu0}M_{\mu0}+M^{\mu,-1}M_{\mu,-1}+M^{0\nu}M_{0\nu}+M^{0,-1}M_{0,-1}+M^{-1,\nu}M_{-1,\nu}+M^{-1,0}M_{-1,0})=\frac{1}{2}(L^{\mu\nu}L_{\mu\nu}-\frac{1}{2}(P+K)^2+\frac{1}{2}(P-K)^2-2D^2)= \frac{1}{2}L_{\mu \nu}L^{\mu \nu} - D^2 -\frac{1}{2}\left(P^\mu K_\mu + K^\mu P_\mu \right)
$$
where it's crucial that the $-1$ signature in the metric is in the $0$ th direction.
A: Let
$$
\frac{1}{2}M^{ab}M_{ab}=\frac{1}{2}L_{\mu\nu}L^{\mu\nu}-D^2-\frac{1}{2}\left(P\cdot K + K\cdot P\right)\ ,\quad(\star)
$$
so that I can make reference to it. Now do
\begin{align*}
M^{ab}M_{ab}&=2M^{-1,0}M_{-1,0} + M^{0,\mu}M_{0,\mu} + M^{-1,\mu}M_{-1,\mu} + M^{\mu\nu}M_{\mu\nu}\\
&=L_{\mu\nu}L^{\mu\nu}+\frac{\eta^{-1,-1}}{4}(P-K)\cdot(P-K)+\frac{\eta^{0,0}}{4}(P+K)\cdot(P+K)+2\eta^{0,0}\eta^{-1,-1}D^2
\end{align*}
Now, we don't know (as of yet) which of the components (0,0 or -1,-1) of the metric should be negative. But the product of them should be, such that the last term is $-2D^2$. We also know that $\eta^{0,0}+\eta^{-1,-1}=0$. Then
$$
M^{ab}M_{ab}=L_{\mu\nu}L^{\mu\nu}-2D^2+\frac{1}{2}\left(\eta^{0,0}-\eta^{-1,-1}\right)\left(P\cdot K+K\cdot P\right)\ .\quad (1)
$$
If you demand $\frac{(1)}{2}=(\star)$, then $\eta^{0,0}=-1$. Now let us check the commutator you mentioned. By the Lorentz algebra
$$
\left[M_{\mu\nu},M_{\rho\sigma}\right]=-i\left(\eta_{\nu\rho}M_{\mu\sigma}+\eta_{\mu\sigma}M_{\nu\rho}-\eta_{\mu\rho}M_{\nu\sigma}-\eta_{\nu\sigma}M_{\mu\rho}\right)\ ,
$$
we find
$$
\left[M_{0\nu},M_{0\sigma}\right]=i\left(\eta_{00}M_{\nu\sigma}\right)=-iM_{\nu\sigma}=-iL_{\nu\sigma}\ .
$$
Now we do
$$
\left[M_{0\nu},M_{0\sigma}\right]=\frac{1}{4}\left[P_\nu+K_{\nu},P_{\sigma}+K_{\sigma}\right]=\frac{1}{4}\left[2i\left(L_{\sigma\nu}-\eta_{\sigma\nu}D\right)+2i\left(\eta_{\nu\sigma}D-L_{\nu\sigma}\right)\right]=-iL_{\nu\sigma}\ .
$$
I didn't check the other relations yet but I think this is the right choice, i.e. $\eta^{-1-1}=1$ and $\eta^{00}=-1$.
Hope it helps!
A: I get 
$$
C_2=\frac{1}{2}L_{\mu \nu}L^{\mu \nu} + D^2 +\frac{1}{2}\left(P^\mu K_\mu + K^\mu P_\mu \right)
$$
In Euclidean signature 
