Scalars, vectors and tensors from metric-affine gravity

Abstract

The metric-affine gravity provides a useful framework for analyzing gravitational dynamics since it treats metric tensor and affine connection as fundamentally independent variables. In this work, we show that, a metric-affine gravity theory composed of the invariants formed from non-metricity, torsion and curvature tensors can be decomposed into a theory of scalar, vector and tensor fields. These fields are natural candidates for the ones needed by various cosmological and other phenomena. Indeed, we show that the model accommodates TeVeS gravity (relativistic modified gravity theory), vector inflation, and aether-like models. Detailed analyses of these and other phenomena can lead to a standard metric-affine gravity model encoding scalars, vectors and tensors.

Appendices

Appendix A: Contraction tensors

Contraction of tensors becomes a tedious operation as their rank becomes larger and larger. Already at the rank-3 level, there arise various possibilities in contracting the indices. Indeed, if one defines

$$\begin{aligned} \mathbb A \bullet \mathbb B \equiv \mathbb A ^{\lambda }_{\alpha \beta } \Xi ^{\alpha \beta \mu \nu }_{\lambda \rho } \mathbb B ^{\rho }_{\mu \nu } \end{aligned}$$

(57)

the contraction tensor $\Xi ^{\alpha \beta \mu \nu }_{\lambda \rho }$ is found to have the most general form

$$\begin{aligned} \Xi ^{\alpha \beta \mu \nu }_{\lambda \rho }&= g_{\lambda \rho }\left(g^{\alpha \beta } g^{\mu \nu } \oplus g^{\alpha \mu } g^{\beta \nu } \oplus g^{\alpha \nu } g^{\beta \mu }\right)\nonumber \\&\oplus \, \delta _{\lambda }^{\alpha } \left( g^{\beta \nu } \delta _{\rho }^{\mu } \oplus g^{\beta \mu } \delta _{\rho }^{\nu } \oplus g^{\mu \nu } \delta _{\rho }^{\beta }\right)\nonumber \\&\oplus \, \delta _{\lambda }^{\beta } \left( g^{\alpha \nu } \delta _{\rho }^{\mu } \oplus g^{\alpha \mu } \delta _{\rho }^{\nu } \oplus g^{\mu \nu } \delta _{\rho }^{\alpha }\right)\nonumber \\&\oplus \, \delta _{\lambda }^{\mu } \left( g^{\beta \nu } \delta _{\rho }^{\alpha } \oplus g^{\alpha \beta } \delta _{\rho }^{\nu } \oplus g^{\alpha \nu } \delta _{\rho }^{\beta } \right)\nonumber \\&\oplus \, \delta _{\lambda }^{\nu } \left( g^{\beta \mu } \delta _{\rho }^{\alpha } \oplus g^{\alpha \beta } \delta _{\rho }^{\mu } \oplus g^{\alpha \mu } \delta _{\rho }^{\beta } \right) \end{aligned}$$

(58)

where $\oplus$ implies $+$ or $-$ depending on whether symmetric or antisymmetric combinations of the indices are involved. Clearly, $\oplus$ also contains the appropriate symmetry factors.

As an example, let us take $\mathbb B ^{\rho }_{\mu \nu } = \mathbb S ^{\rho }_{\mu \nu }$ which is antisymmetric in $(\mu ,\nu )$. In this case, when contracting $\mathbb S ^{\rho }_{\mu \nu }$ with $\Xi ^{\alpha \beta \mu \nu }_{\lambda \rho }$ only the anti symmetric part of $\Xi ^{\alpha \beta \mu \nu }_{\lambda \rho }$ in $(\mu ,\nu )$ matters. In other words, when $\mathbb B ^{\rho }_{\mu \nu } = \mathbb S ^{\rho }_{\mu \nu }$ we consider only

$$\begin{aligned} \Xi ^{\alpha \beta [\mu \nu ]}_{\lambda \rho }&= \frac{1}{2} \Bigg [ g_{\lambda \rho }\left(g^{\alpha \mu } g^{\beta \nu } - g^{\alpha \nu } g^{\beta \mu }\right)\nonumber \\&\oplus \, \delta _{\lambda }^{\alpha } \left( g^{\beta \nu } \delta _{\rho }^{\mu } - g^{\beta \mu } \delta _{\rho }^{\nu } \right)\nonumber \\&\oplus \, \delta _{\lambda }^{\beta } \left( g^{\alpha \nu } \delta _{\rho }^{\mu } - g^{\alpha \mu } \delta _{\rho }^{\nu } \right) \nonumber \\&\oplus \, \Big [\delta _{\lambda }^{\mu } \left( g^{\beta \nu } \delta _{\rho }^{\alpha } \oplus g^{\alpha \beta } \delta _{\rho }^{\nu } \oplus g^{\alpha \nu } \delta _{\rho }^{\beta } \right)\nonumber \\&- \delta _{\lambda }^{\nu } \left( g^{\beta \mu } \delta _{\rho }^{\alpha } \oplus g^{\alpha \beta } \delta _{\rho }^{\mu } \oplus g^{\alpha \mu } \delta _{\rho }^{\beta } \right)\Big ]\Bigg ] \end{aligned}$$

(59)

which is anti-symmetric in $(\mu , \nu )$.

If $\mathbb A ^{\lambda }_{\alpha \beta }$ in (57) is antisymmetric in $(\alpha ,\beta )$ then we consider antisymmetric part of (59).

$$\begin{aligned} \Xi ^{[\alpha \beta ][\mu \nu ]}_{\lambda \rho }&= \frac{1}{4} \Big [\delta _{\lambda }^{\alpha } \left( g^{\beta \nu } \delta _{\rho }^{\mu } - g^{\beta \mu } \delta _{\rho }^{\nu } \right) - \delta _{\lambda }^{\beta } \left( g^{\alpha \nu } \delta _{\rho }^{\mu } - g^{\alpha \mu } \delta _{\rho }^{\nu } \right) \Big ]\nonumber \\&+ \frac{1}{4}\Big [\delta _{\lambda }^{\mu } \left( g^{\beta \nu } \delta _{\rho }^{\alpha } - g^{\alpha \nu } \delta _{\rho }^{\beta } \right) - \delta _{\lambda }^{\nu } \left( g^{\beta \mu } \delta _{\rho }^{\alpha } - g^{\alpha \mu } \delta _{\rho }^{\beta } \right)\Big ]\nonumber \\&+ \frac{1}{2} g_{\lambda \rho }\left(g^{\alpha \mu } g^{\beta \nu } - g^{\alpha \nu } g^{\beta \mu }\right) \end{aligned}$$

(60)

For instance, $\mathbb S \bullet \mathbb S$ will be computed by using this contraction tensor.

However, if $\mathbb A ^{\lambda }_{\alpha \beta }$ in (57) is symmetric in $(\alpha ,\beta )$ then we have to consider symmetric part of (59).

$$\begin{aligned} \Xi ^{(\alpha \beta )[\mu \nu ]}_{\lambda \rho }&= \frac{1}{4} \Big [\delta _{\lambda }^{\alpha } \left( g^{\beta \nu } \delta _{\rho }^{\mu } - g^{\beta \mu } \delta _{\rho }^{\nu } \right) + \delta _{\lambda }^{\beta } \left( g^{\alpha \nu } \delta _{\rho }^{\mu } - g^{\alpha \mu } \delta _{\rho }^{\nu } \right) \Big ]\nonumber \\&+ \frac{1}{4}\Big [\delta _{\lambda }^{\mu } \left( g^{\beta \nu } \delta _{\rho }^{\alpha } + g^{\alpha \nu } \delta _{\rho }^{\beta } \right) - \delta _{\lambda }^{\nu } \left( g^{\beta \mu } \delta _{\rho }^{\alpha } + g^{\alpha \mu } \delta _{\rho }^{\beta } \right)\Big ]\nonumber \\&+ \frac{1}{2} g^{\alpha \beta } \left( \delta _{\lambda }^{\mu } \delta _{\rho }^{\nu } - \delta _{\lambda }^{\nu } \delta _{\rho }^{\mu }\right) \end{aligned}$$

(61)

For instance, $\mathbb Q \bullet \mathbb S$ should be computed by using this contraction tensor. In computing

$\mathbb Q \bullet \mathbb Q$ we should symmetrize in both $(\mu ,\nu )$ and $(\alpha ,\beta )$. Then contraction tensor of $\mathbb Q \bullet \mathbb Q$ is given

$$\begin{aligned} \Xi ^{(\alpha \beta )(\mu \nu )}_{\lambda \rho }&= g^{\alpha \beta }g^{\mu \nu }g_{\lambda \rho }+\frac{1}{2}\Big [g_{\lambda \rho }\left(g^{\alpha \mu } g^{\beta \nu }+g^{\alpha \nu }g^{\beta \mu }\right)\nonumber \\&+g^{\mu \nu }\left(\delta ^{\alpha }_{\lambda } \delta ^{\beta }_{\rho }+\delta ^{\beta }_{\lambda } \delta ^{\alpha } _{\rho }\right)+ g^{\alpha \beta }\left(\delta ^{\mu }_{\lambda } \delta ^{\nu }_{\rho }+\delta ^{\nu }_{\lambda }\delta ^{\mu }_{\rho }\right)\Big ]\nonumber \\&+ \frac{1}{4}\Big [\delta ^{\alpha }_{\lambda }\left(g^{\beta \nu } \delta ^{\mu }_{\rho }+g^{\beta \mu }\delta ^{\nu }_{\rho }\right) +\delta ^{\beta }_{\lambda }\left(g^{\alpha \nu }\delta ^{\mu }_{\rho } +g^{\alpha \mu }\delta ^{\nu }_{\rho }\right)\Big ]\nonumber \\&+\frac{1}{4}\Big [\delta ^{\mu }_{\lambda } \left(g^{\beta \nu }\delta ^{\alpha }_{\rho }+ g^{\alpha \nu }\delta ^{\beta }_{\rho }\right)+\delta ^{\nu }_{\lambda } \left(g^{\beta \mu }\delta ^{\alpha }_{\rho }+g^{\alpha \mu }\delta ^{\beta }_{\rho }\right)\Big ] \end{aligned}$$

(62)

In addition to these, one can compute contraction of divergence of tensors as

$$\begin{aligned} \nabla ^{{(\!\!)}}A\bullet \nabla ^{{(\!\!)}}B= \nabla ^{{(\!\!)}}_{\lambda }A^{\lambda }_{\alpha \beta }\Xi ^ {\alpha \beta \mu \nu }\nabla ^{{(\!\!)}}_{\rho }B^{\rho }_{\mu \nu } \end{aligned}$$

(63)

$\Xi ^{\alpha \beta \mu \nu }$ is contraction tensor and defined in general form as

$$\begin{aligned} \Xi ^{\alpha \beta \mu \nu }= g^{\alpha \beta }g^{\mu \nu }\oplus g^{\alpha \mu }g^{\beta \nu }\oplus g^{\alpha \nu }g^{\beta \mu } \end{aligned}$$

(64)

If A is symmetric in $(\alpha ,\beta )$ and B is symmetric in $(\mu ,\nu )$ contraction tensor takes the form

$$\begin{aligned} \Xi ^{(\alpha \beta )(\mu \nu )}= g^{\alpha \beta }g^{\mu \nu }+\frac{1}{2}\left(g^{\alpha \mu }g^{\beta \nu }+g^{\alpha \nu }g^{\beta \mu }\right) \end{aligned}$$

(65)

this contraction tensor can be used to compute $\nabla \mathbb Q \bullet \nabla \mathbb Q$ because $\mathbb Q$ is symmetric in $(\alpha \beta )$. To compute $\nabla \mathbb S \bullet \nabla \mathbb S$, one needs contraction tensor which is antisymmetric both couple of indices.So,

$$\begin{aligned} \Xi ^{[\alpha \beta ][\mu \nu ]}= \frac{1}{2}\left( g^{\alpha \mu }g^{\beta \nu }-g^{\alpha \nu }g^{\beta \mu }\right) \end{aligned}$$

(66)

If one writes contraction tensor of $\nabla \mathbb Q \bullet \nabla \mathbb S$, finds it as

$$\begin{aligned} \Xi ^{(\alpha \beta )[\mu \nu ]}=0 \end{aligned}$$

(67)

Appendix B: Positive-definite mass matrix

In the text, we mentioned that for a stable theory, each of the three eigenvalues must individually be positive. This leads to non-trivial constraints on the coefficients in (42). In this appendix we shall discuss certain related details. The eigenvalues of (41) follow from the cubic algebraic equation

$$\begin{aligned} -\!\lambda ^{3}+b\lambda ^{2}+c\lambda +d=0 \end{aligned}$$

(68)

where

$$\begin{aligned} b&= a_{VV}+a_{UU}+a_{WW}\nonumber \\ c&= -a_{VV}a_{UU}-a_{VV}a_{WW}-a_{UU}a_{WW}+a^{2}_{UW} +a^{2}_{VU}+a^{2}_{VW}\nonumber \\ d&= a_{VV}a_{UU}a_{WW}+2a_{VU}a_{UW}a_{VW}-a^{2}_{UW} a_{VV}-a_{VW}^{2}a_{UU}-a^{2}_{VU}a_{WW}.\qquad \end{aligned}$$

(69)

The roots of (68) must each be non-negative for guaranteeing absence of instabilities. The analytic expressions for roots are well-known. However, the constraint equations they lead to are too complicated to achieve specific statements about the elements of the mass matrix (41). Nevertheless, in a given specific problem, one can determine the allowed ranges for $a_{VV}, \ldots , a_{UW}$ at least numerically,

As an algebraically simpler case to exemplify, one can focus on the special case of vanishing discriminant, that is, one considers

$$\begin{aligned} \Delta =18 abcd - 4 b^{3}d + b^{2}c^{2}- 4 ac^{3}-27 a^{2} d^{2} \end{aligned}$$

(70)

so that only two independent eigenvalues are left. Indeed, one has

$$\begin{aligned} \lambda _{1}=-\frac{b}{3a}-\frac{2}{3a} \root 3 \of {\frac{1}{2}[2b^{3}-9abc+27a^{2}d]} \end{aligned}$$

(71)

and

$$\begin{aligned} \lambda _{2}=-\frac{b}{3a}+\frac{1}{3a}\root 3 \of {\frac{1}{2}[2b^{3}-9abc+27a^{2}d]}\,. \end{aligned}$$

(72)

For positive-definite mass matrix, $\lambda _{1}$ and $\lambda _{2}$ must each be positive:

$$\begin{aligned} \lambda _{1}> 0 \Rightarrow -\frac{b}{2} < \root 3 \of {\frac{1}{2}[2b^{3}-9abc+27a^{2}d]} \end{aligned}$$

(73)

and

$$\begin{aligned} \lambda _{2}> 0 \Rightarrow b > \root 3 \of {\frac{1}{2}[2b^{3}-9abc+27a^{2}d]}\,. \end{aligned}$$

(74)

These two constraints lead one at once to the bound

$$\begin{aligned} -\frac{b}{2} < \root 3 \of {\frac{1}{2}[2b^{3}-9abc+27a^{2}d]} < b. \end{aligned}$$

(75)

Similar bounds can be derived for general as well as special cases [8]. In general, constraints on various coefficients become more suggestive in some physically relevant special cases. We here thus exemplify two such cases: symmetric and antisymmetric connections.

• 1. Symmetric Connection : ${(\!\!)}^{\lambda }_{\alpha \beta }={(\!\!)}^{\lambda }_{ \beta \alpha }$ As we have mentioned in the text, in this case, torsion tensor identically vanishes ($\mathbb S ^{\lambda }_{\alpha \beta }=0$), and consequently $\mathtt V _{\alpha }=\mathtt U _{\alpha }$. The theory then reduces to a two-vector theory of $\mathtt V$ and $\mathtt W$. From Eq. (40) the mass-squared matrix of vectors is found to be

  $$\begin{aligned} \frac{1}{2} M_{Pl}^2 \left(\begin{array}{cc} a^{\prime }_{VV}+a^{\prime }_{UU}+2 a^{\prime }_{VU} \quad a^{\prime }_{VW}+a^{\prime }_{UW}\\ a^{\prime }_{VW}+a^{\prime }_{UW} \quad a^{\prime }_{WW} \end{array} \right) \end{aligned}$$

  (76)

  where various coefficients are given by

  $$\begin{aligned} a^{\prime }_{VV}&= \frac{1}{18} + \frac{44}{9} c_{Q} ,\nonumber \\ a^{\prime }_{UU}&= a_{WW} \,,\nonumber \\ a^{\prime }_{WW}&= \frac{1}{18} + \frac{14}{9} c_{Q},\nonumber \\ a^{\prime }_{VU}&= -\frac{1}{9} + \frac{20}{9} c_{Q} ,\nonumber \\ a^{\prime }_{VW}&= -\frac{1}{9} + \frac{20}{9} c_{Q} ,\nonumber \\ a^{\prime }_{UW}&= \frac{7}{18} + \frac{14}{9} c_{Q} . \end{aligned}$$

  (77)

  which follow from (40) for vanishing torsion. Clearly, $c_Q$ is the only variable. The eigenvalues of (76) follow from the quadratic algebraic equation;

  $$\begin{aligned} \lambda ^2+b^{\prime }\lambda +c^{\prime }=0 \end{aligned}$$

  (78)

  where

  $$\begin{aligned} b^{\prime }&= -a_{VV}^{\prime }-a_{UU}^{\prime }-2a_{UU}^{\prime }-a_{WW}^{\prime }\nonumber \\ c^{\prime }&= (a_{VV}^{\prime }+a_{UU}^{\prime }+2a_{UU}^{\prime })a_{WW}^{\prime }- (a_{VW}^{\prime }+a_{UW}^{\prime })^2\,. \end{aligned}$$

  (79)

  From the Eq. (78), one directly determines the discriminant

  $$\begin{aligned} \Delta = \frac{11680}{81}c_{Q}^2+\frac{872}{162}c_{Q}+\frac{109}{324} \end{aligned}$$

  (80)

  and eigenvalues

  $$\begin{aligned} \lambda _{1,2}=\frac{a_{VV}^{\prime }+a_{UU}^{\prime }+2a_{UU}^{\prime }+a_{WW}^{\prime }\pm \sqrt{\Delta }}{2}= \frac{-\frac{1}{18}+\frac{112}{9}c_{Q}\pm \sqrt{\Delta }}{2} \end{aligned}$$

  (81)

  For a physically sensible theory, the eigenvalues must all be positive. By considering the constraint of positive discriminant and roots, one finds two appropriate intervals

  $$\begin{aligned} c_{Q} < - 0.046 \quad c_{Q}> 0.68. \end{aligned}$$

  (82)

  This shows that except for the small interval containing origin, all values of $c_{Q}$ lead to a stable massive two-vector theory.

• 2. Anti-symmetric tensorial connection: $V_{\alpha }=-U_{\alpha }$ and $W_{\alpha }=0$ In this case we end up with a single-vector theory with mass-squared $\frac{1}{2}M_{Pl}^2\bar{a}_{VV}$ where $\bar{a}_{VV}=1/3+8c_{S}+2c_{Q}+8c_{QS}$. This coefficient must be positive and hence

  $$\begin{aligned} 4c_{S}+c_{Q}+4c_{QS}> -\frac{1}{6} \end{aligned}$$

  (83)

  A much more special arises when non-metricity vanishes. In this special case, the coefficients $c_{Q}$ and $c_{QS}$ both vanish, an one finds

  $$\begin{aligned} c_{S}>-\frac{1}{24} \end{aligned}$$

  (84)

  as a bound on $c_S$.