How does index notation work in Hermitian spaces?

Question

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 $$v_k=\bar{v}^ig_{ik}? \tag{2}$$

Answer 1

$(2)$ is correct as Index raising and lowering is based on a canonical map which is conjugate-linear.

Let $H$ denote the given Hermitian space with Hermitian form $\,\langle\cdot|\cdot\rangle\,$, assumed to be conjugate-linear in its first slot and linear in the second. Synonymously, many people would say that $H$ is a finite-dimensional complex Hilbert space.
Denote by $H^*$ the dual space. Each $v\in H$ defines a linear functional $\,{}_v\ell\in H^*$ by $${}_v\ell (x)\: =\:\langle\, v\mid x\,\rangle\quad\forall\,x\in H.$$ The correspondence $\,v\longmapsto{}_v\ell\,$ is, by the Riesz representation theorem, a conjugate-linear isometric isomorphism from $ H$ onto $H^*$. (This remains true in the $\infty$-dimensional case.)

Coordinates come into play after choosing a basis $\{\epsilon_j\}\subset H$. The latter uniquely determines the dual basis $\{\epsilon^k\}$ of $H^*$ satisfying $\epsilon^k(\epsilon_j)=\delta^k_j$. Let $g_{jk}:=\langle\epsilon_j|\epsilon_k\rangle$ which is a Hermitian matrix because $\,g_{jk} = \overline{g_{kj}}\,$.

The summation convention is adopted in the sequel. Apply $\,{}_v\ell\,$ to the basis vectors, first in the Riesz representation $${}_v\ell(\epsilon_k)\: =\: \langle\,v^{\,j}\epsilon_j\mid\epsilon_k\,\rangle \: =\: \overline v^{\,j}g_{jk}$$ then in writing it w.r.to the dual basis $${}_v\ell(\epsilon_k)\: =\: \left({}_v\ell_{\!j}\,\epsilon^{\,j}\right)(\epsilon_k)\: =\: {}_v\ell_k\,.$$ This yields $${}_v\ell_k\:=\:\overline v^{\,j}g_{jk}\:=\:\overline{g_{kj}v^{\,j}}$$ which is $(2)$, the desired Index lowering: The matrix $\,g\,$ maps the coordinate vector of $v\in H$ to the coordinate vector of ${}_v\ell\in H^*$, and the overline shows the overall conjugate-linear relationship.

Answer 2

1. 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 $.

2. Then we can expand vectors $v=\sum_{i\in I} v^i e_i\in E$ and covectors $f=\sum_{j\in I} f_j e^{\ast j}\in E^{\ast}$.

3. 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 $g_{ij}:=\langle e_i, e_j \rangle$.

4. 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.