I am having a slight problem regarding the specific relationship between adjoint and dual maps in the context of complex vector spaces endowed with Hermitian metrics. I know that a Hermitian metric ($\mathbb{C}$-linear in the 1st argument) gives an isomorphism
$$\overline{V}_i\simeq_{h^*} V^*$$
where $\overline{V}$ is the same real vector space as $V$ but endowed with the opposite complex structure, and $h^*$ is the deal metric in $V^*$. This isomorphism sends an element $v\in \overline{V}$ to $\langle \cdot, v\rangle_h \in V^*$.
Statement of the problem: let $\phi: V_2 \rightarrow V_1$ be a complex-linear map between complex vector spaces endowed with Hermitian metrics $h_2,h_1$ respectively. Consider the adjoint map
$$\phi^* : V_1 \rightarrow V_2$$
satisfying $\langle \phi^* v_1,v_2\rangle_{h_2} = \langle v_1, \phi v_2\rangle_{h_1}$ for all $v_i \in V_i$. Consider also the dual map
$$\tilde{\phi}:V_1^* \rightarrow V_2^*$$
which acts by $(\tilde{\phi}\theta_2)(v_1) := \theta_2(\phi v_1)$ for all $\theta_2\in V_2^*,v_1\in V_1$. Here the dual spaces are endowed with the natural complex structure they inherit as dual spaces ($\tilde{I}\theta := \theta\circ I$). My intuition tells me that the relationship between $\tilde{\phi}$ and $\phi^*$ is given by a commutative diagram like
$\require{AMScd}$
\begin{CD}
V_1^* @>{\tilde{\phi}}>> V_2^*\\
@VVV @VVV\\
\overline{V}_1 @>{c_2\circ \phi^* \circ c_1}>> \overline{V}_2
\end{CD}
where $c_i$ is the conjugation anti-isomorphism between $V_i$ and $\overline{V}_i$, and the vertical arrows denote the isomorphism induced by the metrics.
I am having trouble trying to prove this and, since I am following solely my intuition, I am not sure this relationship $\tilde{\phi}-\phi^*$ is given by that diagram, so any insights would be very much appreciated. As a side note, I am interested in this situation not for vector spaces rather as Hermitian vector bundles, but since this is a purely linear problem I figured I could prove this fiberwise.
Best Answer
I think it's less confusing if we just ignore the complex conjugate vector spaces, and consider the Hermitian forms on each vector space as giving antilinear maps $h_i \colon V_i \to V_i^*$ defined by $h_i(v_i) = (w \mapsto \langle w, v_i\rangle_i)$. By an antilinear map I mean an $\mathbb{R}$-linear map obeying $f(\lambda v) = \overline{\lambda} f(v)$. The composition of an antilinear map with a linear map is again antilinear; the composition of two antilinear maps is $\mathbb{C}$-linear.
If $\phi \colon V_2 \to V_1$ is a linear map, then its dual $\tilde{\phi} \colon V_1^* \to V_2^*$ is the precomposition $\tilde{\phi} f := f \circ \phi$. Now we just have to check whether $h_2^{-1} \circ \tilde{\phi} \circ h_1$ is the adjoint map: since we have composed two antilinear maps and one antilinear map, this composition is linear. We have $$ \begin{aligned} \langle v_2, h_2^{-1} \tilde{\phi} h_1 v_1 \rangle_2 &= \langle v_2, h_2^{-1}\tilde{\phi}(v \mapsto \langle v, v_1\rangle_1) \rangle_2 \\ &= \langle v_2, h_2^{-1}(v \mapsto \langle \phi v, v_1\rangle_1) \rangle_2 \\ &= \langle v_2, h_2^{-1}(v \mapsto \langle v, \phi^* v_1\rangle_2) \rangle_2 \\ &= \langle v_2, \phi^* v_1 \rangle_2, \end{aligned}$$ and since this equality holds for all $v_i \in V_i$ we have that $h_2^{-1} \circ \tilde{\phi} \circ h_1 = \phi^*$.