So, when we have any symmetic bilinear form $g = g_{ij} \epsilon^i \otimes \epsilon^j$, we can write $(A^{-1})^\mu{}_i = A_{\space i}{}^\mu$. This is one of the most beautiful things that index notation brings.

The proof (which I will need to refer to) is quite simple:

$$A^i{}_\mu(A^{-1})^\xi{}_i=\delta_\mu{}^\xi=g_{\mu\nu}g^{\nu\xi}=A^i{}_\mu g_{ij} A^j{}_\nu g^{\nu\xi} = A^i{}_\mu A_i{}^\xi$$

Now, this assumes that $v_j = v^i g_{ij}$, which is true if we have a symmetric bilinear metric. But if we have a hermitian metric, this becomes $v_j = \bar{v}^i g_{ij}$...

My question is if the former proof is also true for hermitian metrics in spite of that. I tried repeating the same proof but ended up with

$$(A^{-1})^\xi{}_i = \bar{A}_i{}^\xi$$

but I'm not sure at all that's correct, because raising and lowering indices with a hermitian metric makes me really confused.


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