So, I know that in orthogonal spaces (real vector spaces with a symmetric bilinear form) there is a canonical isomorphism bewtween $E$ and $E^*$ induced by the bilinear form $\langle\vec{v}|\vec{w}\rangle=v^Tgw$, which at the end of the day allows us to raise and lower indices:
$$\langle\vec{v}|\space\space\rangle=v_k\epsilon^k=v^ig_{ik}\epsilon^k \implies v_k=v^ig_{ik}$$
where $\{\epsilon^k\}$ represents the dual basis of our vector basis and repeated indexes are summed according to the Einstein summation convention.
So in a Hermitian space, where we have a Hermitian form instead of a symmetric bilinear form, and thus $\langle\vec{v}|\vec{w}\rangle=v^Hgw$ in matrix notation, does $$v_k=v^ig_{ik}\tag{1}$$ or does hold
$$v_k=\bar{v}^ig_{ik}? \tag{2}$$
Best Answer
Let $(e_i)_{i\in I}$ be a basis in $E$ and $(e^{\ast j})_{j\in I}$ be the dual basis for $E^{\ast}$ such that $e^{\ast j}(e_i)=\delta^j_i $.
Then we can expand vectors $v=\sum_{i\in I} v^i e_i\in E$ and covectors $f=\sum_{j\in I} f_j e^{\ast j}\in E^{\ast}$.
The isomorphism $\Phi:E\to E^{\ast}$ is given by via the sesquilinear form$^1$ $\langle \cdot, \cdot\rangle: E\times E \to \mathbb{C}$ as $$\Phi(v)~=~ \langle v, \cdot \rangle~=~\sum_{i,j\in I} \bar{v}^i g_{ij} e^{\ast j},$$ where $g_{ij}:=\langle e_i, e_j \rangle$.
Therefore OP's last suggestion (2) applies $$\Phi(v)_j~=~\sum_{i\in I} \bar{v}^i g_{ij}.$$
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$^1$In our convention the sesquilinear form is conjugated $\mathbb{C}$-linear in the first entry and $\mathbb{C}$-linear in the second entry. Be aware that the opposite convention also exists in the literature.