# Casimir operators of de Sitter space

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De-Sitter space can be thought of as a 4 dimensional hyperboloid embedded in 5D Minkowski space. Hence, the symmetry group of dS is $SO(1,4)$ whose generators are,

$J_{AB}=i\left(X_A\frac{\partial}{\partial X_B}-X_B\frac{\partial}{\partial X_A}\right)$

where $A,B=0,1,2,3,4$. They follow the standard $SO(1,4)$ commutation relations,

$\left[J_{AB},J_{CD}\right]=i\left(\eta_{BC}J_{AD}+\eta_{AD}J_{BC}-\eta_{AC}J_{BD}-\eta_{BD}J_{AC}\right)$

where $\eta_{AB}=(-1,1,1,1,1)$. The coordinates on the dS hyperboloid (whose equation is, $\eta_{AB}X^AX^B=\frac{1}{H^2}$)  are,

$X^0=-\frac{1}{2H}\left(H\eta-\frac{1}{H\eta}\right)+\frac{x^2}{2\eta}$

$X^i=\frac{x^i}{H\eta}$

$X^4=-\frac{1}{2H}\left(H\eta+\frac{1}{H\eta}\right)+\frac{x^2}{2\eta}$

Here $i=1,2,3$.  $(\eta,\vec{x})$ are parameters on the hyperboloid. The 5D Minkowski metric restricted to the hyperboloid, in terms of these coordinates, becomes,

$ds^2=\frac{-d\eta^2+d\vec{x}^2}{H^2\eta^2}$.

Split the $SO(1,4)$ generators as,

$L_{ij}=J_{ij}$

$D=J_{04}$

$P_i=J_{0i}+J_{4i}$

$K_i=J_{4i}-J_{0i}$

It can be easily checked that in terms of the $\eta,\vec{x}$ coordinates,

$L_{ij}=i\left(x_i\partial_j-x_j\partial_i\right)$

$D=-i\left(\eta\partial_{\eta}+x^i\partial_i\right)$

$P_i=\frac{-i}{H}\partial_i$

$K_i=i\left[-2Hx^i\left(\eta\partial_{\eta}+x^i\partial_i\right)+H\left(-\eta^2+x^2\right)\partial_i\right]$

These generators follow the standard conformal algebra (check the one given in the big yellow book). The point $\eta,\vec{x}=0$ is left invariant by $D,L_{ij},K_i$. Hence, if $\phi(\eta,\vec{x})$ is a classical field,

$L_{ij}\phi(0)=S_{ij}\phi(0)$

$D\phi(0)=-i\Delta\phi(0)$

$K_i\phi(0)=0$

Now, $SO(1,4)$ has two Casimir operators,

$C_1=-\frac{1}{2}J^{AB}J_{AB}$

$C_2=-W^AW_A$

where $W^A=\tfrac18 \epsilon^{ABCDE}J_{BC}J_{DE}$. In terms of the conformal genertors,

$C_1=D^2-\tfrac12L_{ij}L^{ij}-\tfrac12\{P_i,K_i\}$

by making $C_1$ act on $\phi(0)$, the eigenvalues turn out to be,

$C_1=-s(s+1)-\Delta(\Delta-3)$

In terms of the conformal generators, the components of $W_A$ are,

$W^0=\tfrac12\epsilon_{ijk}J^{ij}J_{k4}=\tfrac12L_k\left(P_k+K_k\right)$

$W^i=\tfrac12\epsilon_{ijk}J_{jk}J_{04}-\epsilon_{ijk}J_{j4}J_{0k}=-L^iD+\tfrac14\epsilon_{ijk}\{K_j,P_k\}$

$W^4=-\tfrac12\epsilon_{ijk}J^{ij}J_{0k}=\tfrac12L_k\left(P_k-K_k\right)$

where $L_k=-\tfrac12\epsilon_{kij}L_{ij}$ and it has the following commutation rules,

$\left[L_i,L_j\right]=i\epsilon_{ijk}L_k$

$\left[L_i,D\right]=0$

$\left[L_i,P_j\right]=i\epsilon_{ijk}P_k$

$\left[L_i,K_j\right]=i\epsilon_{ijk}K_k$

Now, $W^2=-W_0^2+W_i^2+W_4^2=(W_4+W_0)(W_4-W_0)-[W_0,W_4]+W_i^2$, so when it acts on a classical field, the first term does not contribute since it has a $K_i$ on the right which annihilates the field. Also, it is easy to see that $L_k$ commutes with $W_0$ and $W_4$. Therefore,

$[W_0,W_4]=-L_k[J_{k4},L_iJ_{0i}]=-iL_kW_k$.

In the formula for $W_k$, commuting the $P_i$ past the $K_i$ gives a factor of $iL_k$ so that

$W_k\phi(0)=-L_k(D-i)\phi(0)$.

Thus, we get,

$C_2\phi(0)=s(s+1)(\Delta+1)(\Delta+2)\phi(0)$

which is incorrect according to this reference (their $q$ is my $\Delta-1$). The first Casimir is correct. The correct answer for $C_2$ should be $-s(s+1)(\Delta-1)(\Delta-2)$. Can somebody point out where I made a mistake?

asked Apr 11, 2020 in Q&A

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