= Solution
Use the source's $+---$ <metric signature> and $x^0=\eta$, with <conformal time> <derivatives> denoted by primes. For this <vector cosmological perturbation>, the <metric tensor> and <inverse metric> through first order are
$$
g_{00}=a^2,\quad g_{0i}=0,\quad g_{ij}=-a^2(\delta_{ij}-J_{ij}),\qquad
g^{00}=a^{-2},\quad g^{0i}=0,\quad g^{ij}=-a^{-2}(\delta_{ij}+J_{ij}).
$$
Spatial indices on perturbations are raised with the background <Kronecker delta>. The <Christoffel symbol> formula is
$$
\Gamma^\alpha_{\beta\gamma}=\frac12g^{\alpha\sigma}(\partial_\beta g_{\sigma\gamma}+\partial_\gamma g_{\sigma\beta}-\partial_\sigma g_{\beta\gamma}).
$$
Because $a=a(\eta)$ only, it gives
$$
\boxed{\Gamma^0_{00}=\frac12a^{-2}(2aa')=\mathcal H,\qquad \Gamma^0_{0i}=\frac12a^{-2}\partial_i(a^2)=0.}
$$
For the other requested <Christoffel symbols>, only the time <derivative> of the spatial <metric tensor> contributes:
$$
\Gamma^0_{ij}=-\frac12a^{-2}\partial_\eta g_{ij}
=\frac12a^{-2}\left[2aa'(\delta_{ij}-J_{ij})-a^2J'_{ij}\right].
$$
Thus
$$
\boxed{\Gamma^0_{ij}=\mathcal H(\delta_{ij}-J_{ij})-\frac12J'_{ij}.}
$$
Finally contract the first upper and first lower spatial indices, summing over $i$:
$$
\gamma^i_{ij}=\frac12(\partial_iJ_{ij}+\partial_jJ_{ii}-\partial_iJ_{ij})=\frac12\partial_jJ_{ii}.
$$
Since $J_{ii}=2\partial_iF_i=0$ by the <transverse vector field> condition on $\mathbf F$,
$$
\boxed{\gamma^i_{ij}=0.}
$$
The useful <Christoffel symbol> traces are consequently $\Gamma^\mu_{\mu0}=4\mathcal H$ and $\Gamma^\mu_{\mu j}=0$ to first order: the first uses $J'_{ii}=0$ and the second uses the contraction just proved.
Back to article page
