Inverse of a metric tensor on a Hermitian manifold

Question

Let $(M, g)$ be a Hermitian manifold. We have a metric tensor $g^{i \bar j} dz_i \otimes d\bar{z_j}$, where $(g_{i \bar j})$ is a hermitian positive definite matrix. Now we naturally get the inverse of the metric $(g^{i \bar j})$. I have been told being inverse to each other would imply: $g^{p \bar k} g_{q \bar k} = \delta_{pq}$ which makes no sense to me. I think matrix multiplication should give us $g^{p \bar k} g_{k \bar q} = \delta_{pq}$.

Answer 1

Metric tensors are usually assumed to be symmetrical, i.e. $g_{\mu\nu} = g_{\nu\mu}$, so \begin{equation} g_{\mu\nu}g^{\nu\epsilon} = g_{\nu\mu}g^{\nu\epsilon} = \delta_\mu^{\ \ \epsilon} \end{equation}

The symmetry is due to the fact that the metric is used to compute the line elements $ds^2$ and the follwing holds \begin{equation} ds^2 = g_{\mu\nu}dx^\mu dx^\nu = g_{\nu\mu} dx^\nu dx^\mu \end{equation} (I used the Einstein convention on repeated indices).

Answer 2

The inverse property implies $$\sum_k(g^{-1})^{pk} g_{kq}+\sum_{\bar{k}}(g^{-1})^{p\bar{k}} g_{\bar{k}q} ~=~ \delta^p_q.$$ It is standard convention to not write the power "$-1$" explicitly for the inverse metric. Next use symmetry $g_{\bar{k}q}=g_{q\bar{k}}$ and that for a Hermitian metric $g_{kq}=0$ to obtain the sought-for relation.