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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, $\eta_{AB}X^AX^B=-X_0^2+X_i^2+X_4^2=\frac{1}{H^2}$) 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 field operator with primary states $|\phi⟩=\phi(0)|0⟩$,

$L_{ij}|\phi⟩=S_{ij}|\phi⟩$

$D|\phi⟩=-i\Delta|\phi⟩$

$K_i|\phi⟩=0$

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 $|\phi⟩$, the eigenvalues turn out to be,

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

Here $s$ is the spin and $\Delta$ is the scaling dimension of the field. 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 state, the first term does not contribute since it has a $K_i$ on the right which annihilates the state. 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

$W_k|\phi⟩=-L_k(D-i)|\phi⟩$.

Thus, we get,

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

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?

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    $\begingroup$ Note that the error in the second Casimir can be spotted simply from the fact that the eigenvalue should be invariant under the group generated by $(\Delta,s)\to (3-\Delta,s)$ and $(\Delta,s)\to (1-s,1-\Delta)$. So it not that there is a typo in the reference. $\endgroup$ – Peter Kravchuk Apr 18 '20 at 3:56
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    $\begingroup$ Other than that, it seems that you are using a solid logic, so your error is somewhere in the details. I don't think it is very constructive to ask people to do the dirty work of checking the algebra for you.:) $\endgroup$ – Peter Kravchuk Apr 18 '20 at 4:01
  • $\begingroup$ I know it seems like I'm asking people to check my calculation but it's perfectly fine if someone can come up with an alternate derivation of the eigenvalues of the second Casimir. I know of only one other way in which you can obtain this result, it's outlined in a paper by LH Thomas from 1941, but it's very lengthy. The one that I've given is much shorter but i was worried because it ran into this problem. $\endgroup$ – Sounak Sinha Apr 18 '20 at 4:28
  • $\begingroup$ So you're saying that if I make $(\Delta,s)\rightarrow(3-\Delta,s)$ and $(\Delta,s)\rightarrow(1-s,1-\Delta)$ in the second Casimir, the answers match? $\endgroup$ – Sounak Sinha Apr 18 '20 at 4:32
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    $\begingroup$ No, I am saying that Casimir eigenvalues are invariant under affine Weyl reflections, which in this case are generated by these two transformations. That is, if you make any of the two replacements in the correct eigenvalue, it shouldn't change. $\endgroup$ – Peter Kravchuk Apr 18 '20 at 22:49

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