4
$\begingroup$

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.

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

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.

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.