Let $E\to X$ be a holomorphic Hermitian vector bundle over a complex manifold. Let $\xi\in \Omega^1(X,\operatorname{End}(E))$ be an end-valued form. We define its adjoint $\xi^*$ by the identity
$$h(\xi v,w)=h(v,\xi^*w)\in \Omega^1(X).$$
for all $v,w\in E$. (The Hermitian product is linear in $\Omega(X))$
Now we have that the 2-form $\xi\wedge \xi^*$ is anti-self-adjoint. To see that let us consider two simple 1-forms $a=\alpha\otimes f, b=\beta\otimes g\in \Omega^1(X,\operatorname{End}(E))$, we have
\begin{align*}
(a\wedge b)^* &= (\alpha\wedge \beta\otimes f\circ g)^*\\
&= \overline{\alpha\wedge \beta}\otimes (f\circ g)^*\\
&= -\bar{\beta}\wedge \bar{\alpha}\otimes g^*\circ f^*\\
&= -b^*\wedge a^*.
\end{align*}
So in particular
$$(\xi\wedge \xi^*)^*=-\xi\wedge \xi^*.$$
So a priori we should have that
$$h(\xi\wedge \xi^* v,v)\in i\mathbb{R}$$
should be pure imaginary.
The problem is that again a priori we have
$$h(\xi\wedge \xi^* v,v)=h(\xi^*v,\xi^*v)=||\xi^*v||^2\in \mathbb{R}.$$
What did I get wrong?
Best Answer
You should be very careful about the last assertion $$ h(\xi\wedge\xi^* v,v) = h(\xi^* v, \xi^* v) = \|\xi^*v\|^2 \in \mathbb{R} $$ This should not be a real number! It should be a 2-form. Unless you assume that your base manifold is also a Riemannian one, you are not taking products of the form-part of End(E)-valued differential forms. Actually you should also be very careful about the definition of the extension of the Hermitian product to $E$-valued forms themselves.
$$h(\alpha\otimes v,\beta\otimes w) := \alpha\wedge \bar{\beta} h(v,w)$$
Note how we get now graded sesquilinearity for $E$-valued forms:
$$ h(\Phi,\Psi) = (-1)^{\deg \Phi \cdot \deg \Psi}\overline{h(\Psi,\Phi)} $$
Then, the graded adjoint rule should be (note that this is a definition, but the one that makes sense!) $$h(\xi \Phi,\Psi) = (-1)^{\deg \xi\cdot \deg\Phi}h(\Phi,\xi^*\Psi)$$ for $E$-valued forms $\Phi,\Psi$. With this in mind:
Claim: if $S$ is a anti-self-adjoint End(E)-valued p-form, $$ h(S v, v) \in i\Omega^P(X,\mathbb{R}) $$ for every $E$-section $v$.
Proof: $$h(S v, v) = (-1)^{0\cdot p}h(v,S^*v) = h(v,-S v) = {(-1)^{\deg v \cdot \deg S v}} \overline{h(-S v,v)}=-\overline{h(S v,v)}$$
Consistency check
If $S=\xi\wedge\xi^*$, which is as you proved anti-self-adjoint, then by graded-adjointness: $$ h(\xi\wedge\xi^* v, v) = (-1)^{\deg \xi\cdot \deg \xi^*}h(\xi^*v,\xi^*v) = -h(\xi^*v,\xi^*v) = -h(v,\xi\wedge\xi^* v) = -(-1)^2\overline{h(\xi\wedge\xi^* v,v)}=-\overline{h(\xi\wedge\xi^* v,v)} $$