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.

  • $\begingroup$ Presumably the $-1$ in the signature is in the zeroth direction not in the -1th direction. $\endgroup$ – user110373 Nov 22 '17 at 21:21

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.

  • $\begingroup$ Did you mean to say in the "-1th direction"? Contradicting your previous comment? Because if so, I can't follow your computation. If you are right about your comment (and it really is the 0th direction), then the calculation comes out right but I believe that a) that disagrees with the books definition and b) that it breaks the commutation relations. I could be wrong though. $\endgroup$ – AlexM Nov 22 '17 at 21:43
  • $\begingroup$ Sorry that was a typo. We had this as homework before where $M_{-1,\mu}=\frac{1}{2}(K_{\mu}+P_{\mu})$, with $-1$ signature at $-1$th direction. In order to use the definition written in your problem, you need the $-1$ to be at the zeroth direction. $\endgroup$ – user110373 Nov 22 '17 at 22:03
  • $\begingroup$ If you redefine the metric the way you did, it seems like the commutator $[M_{0,\mu},M_{0,\nu}]=-i\eta_{00}L_{\mu \nu}$ gives the wrong sign compared to $\frac{1}{4}[P_\mu+K_\mu,P_\nu+K_\nu]=-iL_{\mu \nu}$. It really seems like the Casimir with the "wrong" plus sign is the correct version. Also after redoing some calculations it seems that only my "wrong" Casimir commutes with $P^\mu$ $\endgroup$ – AlexM Nov 23 '17 at 15:28


$$ \frac{1}{2}M^{ab}M_{ab}=\frac{1}{2}L_{\mu\nu}L^{\mu\nu}-D^2-\frac{1}{2}\left(P\cdot K + K\cdot P\right)\ ,\quad(\star) $$

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!

  • $\begingroup$ Thanks a lot for the explicitness. Your check of the commutators depends crucially on the sign convention in the Lorentz algebra which is however opposite to most references I have seen, including DiFrancesco. Using his definitions, it really seems like the Casimir has a plus sign. $\endgroup$ – AlexM Nov 24 '17 at 10:09

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


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