# Raising and lowering the indices of a perturbed metric

I am looking at a metric which is defined as (Eq 2.4 Glampedakis & Babak)

$$g_{\mu \nu} = g_{\mu \nu}^K + \epsilon h_{\mu \nu}$$

where $$g_{\mu \nu}^K$$ is the original unperturbed metric (Kerr) and $$h_{\mu \nu}$$ some perturbation.

Now, I know that the indices of the perturbation are raised/lowered using the unperturbed metric i.e.

$$h_{\alpha \beta} = g_{\alpha \gamma}^K g_{\beta \delta}^K h^{\gamma \delta}$$

(see e.g. application in Eq 2. of Narzilloev et al. 2019)

My question is how to get the contravariant form of $$g^{\mu \nu}$$?

Option 1 is that simply,

$$g^{\mu \nu} = g^{\mu \nu}_K + \epsilon h^{\mu \nu}$$

Option 2 considers $$g^{\mu \nu}$$ as an independent matrix and we invert it the usual way.

However these two options seem to not be equivalent. For example consider the $$g^{03}$$ term.

Option 1 tells us that $$g^{03} = g^{0 3}_K + \epsilon h^{0 3}$$ but $$h^{0 3} = 0$$ and so $$g^{03} = g^{0 3}_K$$.

But Option 2 tells us that

$$g^{03} = - \frac{g_{03}}{ \tilde{g}} = \frac{-1}{\tilde{g}} (g_{03}^{K} + \epsilon h_{03}) \ne g^{0 3}_K$$

where $$\tilde{g} = g_{00} g_{33} - g_{03}^2$$ and we have exploited the symmetries of the matrix (e.g. Eq 19.13 of these notes)

Can anyone provide some guidance on where I am going wrong? Thanks.

• To first order in $\epsilon$, $g^{\mu \nu} = g^{\mu \nu}_K - \epsilon h^{\mu \nu}$. Sep 3, 2019 at 16:32
• Can this be shown by inverting the covariant perturbed metric? Sep 3, 2019 at 16:34
• It can be shown by computing $g_{\mu\lambda}g^{\lambda\nu}$. The $O(\epsilon)$ terms cancel, leaving $\delta_\mu^\nu+O(\epsilon^2)$. Sep 3, 2019 at 16:40
• I've answered a similar question here - physics.stackexchange.com/a/330277/133418. Also, note that using the unperturbed metric to raise/lower indices is just because we're using perturbation theory and we neglect higher order terms. In principle, and in full generality, one should use the full metric. Sep 4, 2019 at 14:53

There is no finite expression for the inverse metric of $$g$$. This is due to the binomial inverse theorem. What we want is to find the inverse of the matrix $$g + h$$. From this theorem, this is
$$(g + h)^{-1} = g^{-1} - g^{-1} (I + hg^{-1})^{-1} h g^{-1}$$
$$(g + h)^{-1} = g^{-1} - g^{-1} h g^{-1} = g^{\mu\nu} - h^{\mu\nu}$$