Appendix A
The straightforward algebra gives
$$\begin{aligned} {{\mathcal {L}}}&= \frac{1}{2}\sqrt{-g}(g^{ik}R_{ik} + {\tilde{g}}^{ik}{\tilde{R}}_{ik})\nonumber \\&= [{\bar{g}}^{(ik)}(R_{ik} + {\tilde{R}}_{ki}) + {\bar{g}}^{[ik]}(R_{ik} - {\tilde{R}}_{ki})]\nonumber \\&= {\bar{g}}^{(ik)}[R_{ik} - \Gamma ^{ \lambda }_{[i\xi ]}\Gamma ^{ \xi }_{[\lambda k]}]\nonumber \\&\quad + {\bar{g}}^{[ik]}(\Gamma ^\lambda _{[ik], \lambda } -\Gamma ^\lambda _{[i\lambda ], k} +\Gamma ^\xi _{ik}\Gamma ^\lambda _{\xi \lambda } - \Gamma ^\xi _{i\lambda }\Gamma ^\lambda _{\xi k}\nonumber \\&\quad - \Gamma ^\xi _{ki}\Gamma ^\lambda _{\lambda \xi } + \Gamma ^\xi _{\lambda i}\Gamma ^\lambda _{k\xi })\nonumber \\&= {\bar{g}}^{(ik)}[R_{ik} - \Gamma ^{ \lambda }_{[i\xi ]}\Gamma ^{ \xi }_{[\lambda k]}] + {\bar{g}}^{[ik]}[\Gamma ^\lambda _{ik; \lambda }]. \end{aligned}$$
(A1)
Following ref. [17] but using \(\Gamma _\mu = 0\), let
$$\begin{aligned} {{\mathcal {L}}} = H + \frac{\mathrm{d}X^\lambda }{\mathrm{d}x^\lambda } \end{aligned}$$
(A2)
with
$$\begin{aligned} X^\lambda&= {\bar{g}}^{(ik)}\Gamma _{(ik)}^\lambda - {\bar{g}}^{(i\lambda )}\Gamma ^k_{(ik)} + {\bar{g}}^{[ik]}\Gamma ^\lambda _{[ik]}, \end{aligned}$$
$$\begin{aligned} H&= - {\bar{g}}^{(ik)}_{ ,\lambda }\Gamma ^\lambda _{(ik)} + {\bar{g}}^{(i\lambda )}_{ ,\lambda }\Gamma ^k_{{(ik)}}\\&\quad + {\bar{g}}^{(ik)}(\Gamma ^\xi _{(ik)}\Gamma ^\lambda _{(\xi \lambda )} - \Gamma ^\xi _{(i\lambda )}\Gamma ^\lambda _{(\xi k)} \\&\quad - \Gamma ^\lambda _{[i\xi ]}\Gamma ^\xi _{[\lambda k]})- {\bar{g}}^{[ik]}_{ ,\lambda }\Gamma ^\lambda _{[ik]}\\&\quad + {\bar{g}}^{[ik]}[- \Gamma ^\lambda _{[i\xi ]}\Gamma ^\xi _{(\lambda k)} \\&\quad - \Gamma ^\lambda _{[\xi k]}\Gamma ^\xi _{(i\lambda )}+ \Gamma ^\xi _{[ik]}\Gamma ^\lambda _{(\xi \lambda )}]. \end{aligned}$$
Thus, H is free of the partial derivatives of \(\Gamma ^\lambda _{(ik)}\) and \(\Gamma ^\lambda _{[ik]}\), and the four-divergence term in the action integral is equal to a surface integral at infinity on which all arbitrary variations are taken to vanish.
The variations of H with respect to \(\Gamma ^\lambda _{(ik)}\) and \(\Gamma ^\lambda _{[ik]}\) give
$$\begin{aligned}&{\bar{g}}^{(ik)}_{ ,\lambda } + {\bar{g}}^{(i\alpha )}\Gamma ^k_{(\lambda \alpha )} + {\bar{g}}^{(\alpha k)}\Gamma ^i_{(\alpha \lambda )} - {\bar{g}}^{(ik)}\Gamma ^\alpha _{(\lambda \alpha )} \nonumber \\&\quad = -[{\bar{g}}^{[i\alpha ]}\Gamma ^{ k}_{[\lambda \alpha ]} + {\bar{g}}^{[\alpha k]}\Gamma ^{ i}_{[\alpha \lambda ]}], \end{aligned}$$
(A3)
$$\begin{aligned}&{\bar{g}}^{[ik]}_{ ,\lambda } + {\bar{g}}^{[i\alpha ]}\Gamma ^{k}_{(\lambda \alpha )} + {\bar{g}}^{[\alpha k]}\Gamma ^{i}_{(\alpha \lambda )} - {\bar{g}}^{[ik]}\Gamma ^\alpha _{(\lambda \alpha )} \nonumber \\&\quad = - [{\bar{g}}^{(i\alpha )}\Gamma ^{ k}_{[\lambda \alpha ]} + {\bar{g}}^{(\alpha k)}\Gamma ^{ i}_{[\alpha \lambda ]}]. \end{aligned}$$
(A4)
On adding (A3) and (A4), we get
$$\begin{aligned}&{\bar{g}}^{ik}_{ ,\lambda } + {\bar{g}}^{i\alpha }(\Gamma ^k_{(\lambda \alpha )} + \Gamma ^{k}_{[\lambda \alpha ]}) + {\bar{g}}^{\alpha k}(\Gamma ^i_{(\alpha \lambda )} + \Gamma ^{i}_{[\alpha \lambda ]})\nonumber \\&\quad - {\bar{g}}^{ik}\Gamma ^\alpha _{(\lambda \alpha )} = 0. \end{aligned}$$
(A5)
This can be written as
$$\begin{aligned}&{\bar{g}}^{ik}_{ ,\lambda } + {\bar{g}}^{i\alpha }\Gamma ^{\prime k}_{\lambda \alpha } + {\bar{g}}^{\alpha k}\Gamma ^{\prime i}_{\alpha \lambda } - {\bar{g}}^{ik}\Gamma ^\alpha _{(\lambda \alpha )} = 0,\nonumber \\&\Gamma ^{\prime k}_{\lambda \alpha } = \Gamma ^{k}_{(\lambda \alpha )} + \Gamma ^{k}_{[\lambda \alpha ]},\nonumber \\&\Gamma ^{\prime i}_{\alpha \lambda } = \Gamma ^{i}_{(\alpha \lambda )} + \Gamma ^{i}_{[\alpha \lambda ]}. \end{aligned}$$
(A6)
This is eq. (15).
By contracting (A6) once with respect to \((k, \lambda )\), then with respect to \((i, \lambda )\), and subtracting the equations term by term, one gets eq. (16).
Appendix B
In this appendix we shall use the Greek symbols \(\mu ,\nu \) instead of i, k. Define
$$\begin{aligned} s^{\mu \nu }&= \frac{1}{2}\sqrt{-g}\left( g^{\mu \nu } + g^{\nu \mu }\right) \equiv \frac{1}{2}\left( {\bar{g}}^{\mu \nu } + {\bar{g}}^{\nu \mu }\right) \nonumber \\&=\frac{1}{2}\sqrt{-g}g^{(\mu \nu )}, \end{aligned}$$
(B1)
$$\begin{aligned} a^{\mu \nu }&= \frac{1}{2}\sqrt{-g}\left( g^{\mu \nu } - g^{\nu \mu }\right) \equiv \frac{1}{2}\left( {\bar{g}}^{\mu \nu } - {\bar{g}}^{\nu \mu }\right) \nonumber \\&= \frac{1}{2} \sqrt{-g} g^{[\mu \nu ]}. \end{aligned}$$
(B2)
The equations of connection in the broken symmetric theory are obtained by writing
$$\begin{aligned} {{\mathcal {L}}} = H + \frac{\mathrm{d} X^\lambda }{\mathrm{d}x^\lambda } \end{aligned}$$
(B3)
with
$$\begin{aligned}&X^\lambda = s^{\mu \nu }\Gamma _{(\mu \nu )}^\lambda - s^{\mu \lambda }\Gamma ^\nu _{(\mu \nu )} + a^{\mu \nu }Q^\lambda _{\mu \nu } \nonumber \\&\qquad \quad + \frac{2}{3} a^{\mu \lambda }\Gamma _\mu + \Gamma ^\lambda ,\nonumber \\&H = - s^{\mu \nu }_{, \lambda }\Gamma ^\lambda _{(\mu \nu )} + s^{\mu \lambda }_{, \lambda }\Gamma ^\nu _{{(\mu \nu )}} \nonumber \\&\qquad \quad + s^{\mu \nu } (\Gamma ^\xi _{(\mu \nu )}\Gamma ^\lambda _{(\xi \lambda )} - \Gamma ^\xi _{(\mu \lambda )}\Gamma ^\lambda _{(\xi \nu )})\nonumber \\&\qquad \quad +s^{\mu \nu }(-Q^\lambda _{\mu \xi }Q^\xi _{\lambda \nu } + x\Gamma _\mu \Gamma _\nu ) \nonumber \\&\qquad \quad + s^{\mu \nu }\Gamma ^\lambda _{(\mu \nu )}\Gamma _\lambda - a^{\mu \nu }_{, \lambda }Q^\lambda _{\mu \nu } \nonumber \\&\qquad \quad + a^{\mu \nu }( - Q^\lambda _{\mu \xi }\Gamma ^\xi _{(\lambda \nu )} - Q^\lambda _{\xi \nu }\Gamma ^\xi _{(\mu \lambda )} + Q^\xi _{\mu \nu }\Gamma ^\lambda _{(\xi \lambda )} ) \nonumber \\&\qquad \quad - y a^{\mu \lambda }_{, \lambda }\Gamma _\mu ,\nonumber \\&Q^\lambda _{\mu \nu } = \Gamma ^\lambda _{[\mu \nu ]} + \frac{1}{3}\delta ^\lambda _\mu \Gamma _\nu - \frac{1}{3}\delta ^\lambda _\nu \Gamma _\mu . \end{aligned}$$
(B4)
Note that
$$\begin{aligned} \delta \int \frac{\mathrm{d}X^\lambda }{\mathrm{d}x^\lambda } \mathrm {d}^4 x = \delta \int _\sigma X_\lambda \,\mathrm{d}\sigma ^\lambda = 0. \end{aligned}$$
(B5)
Hence, only variations of H will contribute.
It is easy to see that variations of the function
$$\begin{aligned} H - 2k^\mu Q^\lambda _{\mu \lambda }, \end{aligned}$$
(B6)
where \(k^\mu \) is a four-vector Lagrange multiplier, with respect to \(\Gamma ^\lambda _{(\mu \nu )}, Q^\lambda _{\mu \nu }\) and \(\Gamma _\mu \) give, respectively, the three equations
$$\begin{aligned}&s^{\mu \nu }_{, \lambda } + s^{\mu \alpha }\Gamma ^\nu _{(\lambda \alpha )} + s^{\alpha \nu }\Gamma ^\mu _{(\alpha \lambda )} - s^{\mu \nu }\Gamma ^\alpha _{(\lambda \alpha )}\nonumber \\&\quad = -[a^{\mu \alpha }Q^\nu _{\lambda \alpha } + a^{\alpha \nu }Q^\mu _{\alpha \lambda }], \end{aligned}$$
(B7)
$$\begin{aligned}&a^{\mu \nu }_{, \lambda } + a^{\mu \alpha }\Gamma ^\nu _{(\lambda \alpha )} + a^{\alpha \nu }\Gamma ^\mu _{(\alpha \lambda )} - a^{\mu \nu }\Gamma ^\alpha _{(\lambda \alpha )} \nonumber \\&\quad - k^\mu \delta ^\nu _\lambda + k^\nu \delta ^\mu _\lambda = - [s^{\mu \alpha }Q^\nu _{\lambda \alpha } + s^{\alpha \nu }Q^\mu _{\alpha \lambda }] \end{aligned}$$
(B8)
and
$$\begin{aligned} ya^{\mu \nu }_{, \nu } + x s^{\mu \nu }\Gamma _\nu = 0. \end{aligned}$$
(B9)
It follows from the first two equations that
$$\begin{aligned}&s^{\mu \alpha }_{,\alpha } + s^{\alpha \beta } \Gamma ^\mu _{(\alpha \beta )} + a^{\alpha \beta }Q^\mu _{\alpha \beta } = 0, \end{aligned}$$
(B10)
$$\begin{aligned}&a^{\mu \nu }_{, \nu } = 3k^\mu . \end{aligned}$$
(B11)
Hence, it follows from eq. (B9) and the last equation that
$$\begin{aligned} k^\mu&= \theta s^{\mu \nu }\Gamma _\nu , \quad \theta = -\frac{x}{3y},\end{aligned}$$
(B12)
$$\begin{aligned} k^\mu _{,\mu }&= 0. \end{aligned}$$
(B13)
Equations (B11) and (B12) result in eq. (27) in §3.
Adding (B7) and (B8), we get
$$\begin{aligned}&{\bar{g}}^{\mu \nu }_{,\lambda } + {\bar{g}}^{\mu \alpha }(\Gamma ^\nu _{(\lambda \alpha )} + Q^\nu _{\lambda \alpha }) + {\bar{g}}^{\alpha \nu }(\Gamma ^\mu _{(\alpha \lambda )} + Q^\mu _{\alpha \lambda })\nonumber \\&\quad - {\bar{g}}^{\mu \nu }\Gamma ^\alpha _{(\lambda \alpha )}= k^\mu \delta ^\nu _\lambda - k^\nu \delta ^\mu _\lambda , \end{aligned}$$
(B14)
where \({\bar{g}}^{\mu \nu } = \sqrt{- g} g^{\mu \nu }\). Multiplying (B14) by \({\bar{g}}_{\mu \nu }\) and using the results
$$\begin{aligned} {\bar{g}}^{\mu \nu }{\bar{g}}_{\mu \lambda } = \delta ^\nu _\lambda , \quad {\bar{g}}^{\mu \nu }{\bar{g}}_{\lambda \nu } = \delta ^\mu _\lambda , \quad Q^\lambda _{\alpha \lambda } = 0, \end{aligned}$$
(B15)
we first observe that
$$\begin{aligned} \Gamma ^\alpha _{(\lambda \alpha )}= & {} \frac{\vert g\vert _{, \lambda }}{2 \sqrt{- g}} + \frac{1}{2}({\bar{g}}_{\lambda \beta } - {\bar{g}}_{\beta \lambda })k^\beta \nonumber \\\equiv & {} \frac{\vert g\vert _{, \lambda }}{2 \sqrt{- g}} + {\bar{g}}_{[\lambda \beta ]}k^\beta . \end{aligned}$$
(B16)
Hence, dividing (B14) by \(\sqrt{- g}\), and also using (B16) and the results
$$\begin{aligned} g^{\mu \alpha }g_{\beta \alpha }k^\beta = k^\mu \quad \mathrm{and}\quad g^{\alpha \nu }g_{\alpha \beta }k^\beta = k^\nu , \end{aligned}$$
(B17)
we get
$$\begin{aligned}&g^{\mu \nu }_{, \lambda } {+} g^{\mu \alpha }\left( \Gamma ^\nu _{(\lambda \alpha )} {+} Q^\nu _{\lambda \alpha } {+} \frac{1}{\sqrt{- g}}(g_{\lambda \beta }k^\beta \delta ^\nu _\alpha {-} g_{\beta \alpha }k^\beta \delta ^\nu _\lambda ) \right) \nonumber \\&\qquad + g^{\alpha \nu }\left( \Gamma ^\mu _{(\alpha \lambda )} {+} Q^\mu _{\alpha \lambda } {+} \frac{1}{\sqrt{- g}}(g_{\alpha \beta }k^\beta \delta ^\mu _\lambda {-} g_{\beta \lambda }k^\beta \delta ^\mu _\alpha )\right) \nonumber \\&\quad = 3g^{\mu \nu }\frac{g_{[\lambda \beta ]}k^\beta }{\sqrt{- g}}. \end{aligned}$$
(B18)
Now, define the new affine coefficients
$$\begin{aligned} \Gamma ^{\prime \prime \nu }_{\lambda \alpha }&= \left( \Gamma ^\nu _{(\lambda \alpha )} {+} Q^\nu _{\lambda \alpha } {+} \frac{1}{\sqrt{- g}}(g_{\lambda \beta }k^\beta \delta ^\nu _\alpha {-} g_{\beta \alpha }k^\beta \delta ^\nu _\lambda ) \right) , \end{aligned}$$
(B19)
$$\begin{aligned} \Gamma ^{\prime \prime \mu }_{\alpha \lambda }&= \left( \Gamma ^\mu _{(\alpha \lambda )} {+} Q^\mu _{\alpha \lambda } {+} \frac{1}{\sqrt{- g}}(g_{\alpha \beta }k^\beta \delta ^\mu _\lambda {-} g_{\beta \lambda }k^\beta \delta ^\mu _\alpha )\right) \end{aligned}$$
(B20)
and
$$\begin{aligned} \Phi _\lambda = \frac{1}{2}g_{[\lambda \beta ]}k^{\beta }, \end{aligned}$$
(B21)
which is eq. (25) in §3. Then, eq. (B18) can be written in the form
$$\begin{aligned} g^{\mu \nu }_{, \lambda } + g^{\mu \alpha }\Gamma ^{\prime \prime \nu }_{\lambda \alpha } + g^{\alpha \nu }\Gamma ^{\prime \prime \mu }_{\alpha \lambda } = 3g^{\mu \nu }\Phi _\lambda . \end{aligned}$$
(B22)
This is eq. (24) in §3.
