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 $\epsilon$, $g^{\mu \nu} = g^{\mu \nu}_K - \epsilon h^{\mu \nu}$. – G. Smith Sep 3 at 16:32
• Can this be shown by inverting the covariant perturbed metric? – user1887919 Sep 3 at 16:34
• It can be shown by computing $g_{\mu\lambda}g^{\lambda\nu}$. The $O(\epsilon)$ terms cancel, leaving $\delta_\mu^\nu+O(\epsilon^2)$. – G. Smith Sep 3 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. – Avantgarde Sep 4 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
$$(g + h)^{-1} = g^{-1} - g^{-1} (I + hg^{-1})^{-1} h g^{-1}$$
$$(g + h)^{-1} = g^{-1} - g^{-1} h g^{-1} = g^{\mu\nu} - h^{\mu\nu}$$