Conformal linear gravity in de Sitter space II

Abstract

From the group theoretical point of view, it is proved that the theory of linear conformal gravity should be written in terms of a tensor field of rank-3 and mixed symmetry (Binegar et al. in Phys. Rev. D 27:2249, 1983). We obtained such a field equation in de Sitter space (Takook et al. in J. Math. Phys. 51:032503, 2010). In this paper, a proper solution to this equation is obtained as a product of a generalized polarization tensor and a massless scalar field and then the conformally invariant two-point function is calculated. This two-point function is de Sitter invariant and free of any pathological large-distance behavior.

Appendices

Appendix A: Some mathematical preliminaries

In this appendix we first review the Krein space briefly and then collect some useful relations.

Hilbert space is built by a set of modes with positive norms:

$${\mathcal{H}}= \biggl\{ \sum_{k\geq0}\alpha_k \phi_k;\sum_{k\geq0}|\alpha_k|^2< \infty \biggr\}, \quad \mbox{with}\ (\phi_1,\phi_2)>0. $$

Krein space is defined as a direct sum of an Hilbert space and an anti-Hilbert space (negative inner product space):

$${\mathcal{K}}= {\mathcal{H}}\oplus\bar{\mathcal{H}}, $$

where $\bar{\mathcal{H}}$ stands for the anti-Hilbert space. Note that due to the indefinite inner product space, some states are allowed to have negative norm. These modes are only used as a mathematical tool in renormalization procedure and are ruled out by imposing some conditions. In fact as discussed in [26], in Krein space setup, minimally coupled scalar field is defined on non-Hilbertian Fock space. This is followed by the fact that the one-particle sector is itself not a Hilbert space since the total space (Krein space) is equipped with an indefinite inner product. The physical space is the quotient space: Krein space/negative-norm space. This is a Hilbert space carrying the UIR of the de Sitter group. It is shown that quantization in Krein space either removes some infinities (for example the vacuum energy vanishes without any need of reordering the terms), or at least regularizes the theory (for more details see [58] and references therein).

In what follows, some useful relations that are used in this paper, are listed: $\bar{\partial}_{\alpha}$ is the tangential (or transverse) derivative on dS space, defined by

$$\bar{\partial}_{\alpha}=\theta_{\alpha\beta}\partial^{\beta }= \partial_{\alpha}+x_{\alpha}x\cdot\partial,\quad \mbox{with}\ x\cdot \bar{\partial}=0, $$

and also one can define

$$ \begin{aligned} &M_{\alpha\beta}\equiv-i(x_{\alpha} \partial_{\beta}-x_{\beta }\partial_{\alpha})= -i(x_{\alpha}\bar{\partial}_{\beta}-x_{\beta}\bar{\partial}_{\alpha }), \\ &S_{\alpha\beta} \mathcal{K}_{\gamma\delta}\equiv-i(\eta_{\alpha\gamma } \mathcal{K}_{\beta \delta}-\eta_{\beta\gamma} \mathcal{K}_{\alpha\delta} + \eta_{\alpha\delta} \mathcal{K}_{\beta\gamma}-\eta_{\beta\delta} \mathcal{K}_{\alpha\gamma}). \end{aligned} $$

(A.1)

Operator $Q^{(1)}_{2}$ commutes with the action of the group generators, thus, it is constant in each UIR. The eigenvalues of $Q^{(1)}_{2}$ can be used to classify the UIR’s i.e.,

$$ \bigl(Q^{(1)}_2-\bigl\langle Q^{(1)}_2 \bigr\rangle\bigr){\mathcal{K}}(x)=0. $$

(A.2)

Following Dixmier [59], one can get a classification scheme using a pair (p,q) of parameters involved in the following possible spectral values of the Casimir operators:

$$ \begin{aligned} &Q^{(1)}_{p}= \bigl(-p(p+1)-(q+1) (q-2) \bigr)I_d , \\ &Q^{(2)}_{p}= \bigl(-p(p+1)q(q-1) \bigr)I_d. \end{aligned} $$

(A.3)

Three types of UIR are distinguished for SO(1,4) according to the range of values of the parameters q and p [59, 60], namely: principal, complementary and discrete series. The flat limit indicates that for the principal and complementary series the value of p bears the meaning of spin. For example in discrete series p=q=2 have a Minkowskian interpretation as a massless spin-2 particle.

The action of the Casimir operators Q ₁ and Q ₂ can be brought in the more explicit form

$$ \begin{aligned}[b] Q_1 K_{\alpha} &=(Q_0-2)K_{\alpha}+2x_{\alpha}\partial\cdot K -2 \partial_{\alpha} x\cdot K \\ &=(Q_0-2)K_{\alpha}+2x_{\alpha}\bar{\partial}\cdot K -2 \bar{\partial}_{\alpha} x\cdot K \\ &\quad +2x_\alpha(x \cdot K), \end{aligned} $$

(A.4)

$$ \begin{aligned}[b] Q_2 {\mathcal{K}}_{\alpha\beta}&=(Q_0-6){ \mathcal{K}}_{\alpha \beta}+2{\mathcal{S}}x_{\alpha}\partial\cdot{ \mathcal{K}}_{\beta} \\ &\quad -2{\mathcal{S}}\partial_{\alpha}x\cdot{ \mathcal{K}}_{\beta}+2 \eta_{\alpha\beta}{ \mathcal{K}}'. \end{aligned} $$

(A.5)

The two-point function (4.8) with the choice of x′ reads

$${\mathcal{W}}_{\alpha\beta\alpha'\beta'}\bigl(x,x'\bigr)=\Delta'_{\alpha\beta \alpha'\beta'} \bigl(x,\partial,x',\partial'\bigr){\mathcal{W}} \bigl(x,x'\bigr), $$

where

$$ \begin{aligned}[b] &\Delta'_{\alpha\beta \alpha'\beta'}\bigl(x, \partial,x',\partial'\bigr) \\ &\quad =-\frac{2}{3}{\mathcal{S}}\theta' \theta \cdot \bigl( \theta' \cdot \theta -D'_{1}\bigl[x' \cdot \theta+\theta \cdot \bar{\partial}'\bigr] \bigr) \\ &\qquad +{\mathcal{S}} {\mathcal{S}}'\theta \cdot \theta' \bigl( \theta' \cdot \theta -D'_{1}\bigl[x' \cdot \theta+\theta \cdot \bar{\partial}'\bigr] \bigr) \\ &\qquad +\frac{1}{3} D'_2{\mathcal{S}} \bigl(4 \bigl(x' \cdot {\theta}\bigr) + \bigl({\theta} \cdot {\bar{\partial}}' \bigr) - x'({\theta}) \bigr) \\ &\qquad \times \bigl(\theta' \cdot \theta -D'_{1} \bigl[x' \cdot \theta+\theta \cdot \bar{\partial}'\bigr] \bigr), \end{aligned} $$

(A.6)

note that the primed operators act only on the primed coordinates.

To obtain the two-point function, the following identities become important:

(A.7)

(A.8)

(A.9)

Appendix B: Mathematical relations underlying Eq. (3.8)

Substituting K in Eq. (3.3-II), yields

(B.1)

then the general solution for ϕ ₃ can be written as

$$ \phi_3= c_1(x \cdot Z_2) \phi_2 + c_2(Z_2 \cdot \bar{\partial}) \phi_2, $$

(B.2)

where c ₁ and c ₂ are two constants. In order to find c ₁ and c ₂, we do the following steps:

Step(I):

the divergenceless condition $(Q_{0}\phi_{3}=4x \cdot Z_{2}\phi_{2}+Z_{2} \cdot \bar{\partial}\phi_{2})$, together with (B.1), results in

(B.3)

on the other hand from Eq. (3.7) we have

(B.4)

from (B.3) and (B.4), one obtains

$$ \begin{aligned} &Q_0(Q_0-2)^2(Z_2 \cdot \bar{\partial})\phi_2=0, \\ &Q_0(Q_0-2)^2(Q_0+4) (x \cdot Z_2)\phi_2=0. \end{aligned} $$

(B.5)

Substituting (B.5) in (B.2) results in

$$ Q_0(Q_0-2)^2(Q_0+4) \phi_3=0. $$

(B.6)

After doing some straightforward algebra, one obtains the reduced forms of (B.5) and (B.6) as follows:

(B.7)

or

$$ (Q_0-2)^2(Q_0+4)\phi_3=0. $$

(B.8)

Step(II):

Using the divergenceless condition and (B.1), one gets

(B.9)

Combining (B.9) and (B.4) results in

(B.10)

From (B.3), (B.10), and (B.7), we get

(B.11)

and using the divergenceless condition and (B.7), the above equation can be written as

(B.12)

From Eqs. (B.2) and (B.12), we obtain c ₁−2c ₂=1, and substituting (B.2) into (B.1) together with (B.7) results in c ₁=c ₂=−1.