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If $M$ is a Riemannian manifold and $E \to M$ a Riemannian bundle, then the Riemannian metric $g$ can be viewed as a section of the bundle $\bigotimes^2 T^*M$, which means that it is a $(0,2)$-tensor field.

If we instead consider a complex vector bundle $E$ over $M$ with a hermitian metric $h$, this wikipedia article claims that $h$ is a section of the bundle $(E\otimes \bar{E})^*$. Can someone here elaborate on why should $h$ be a section of this proposed bundle?

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It's not just any section. To quote from the Wikipedia article you linked:

A Hermitian metric on a complex vector bundle $E$ over a smooth manifold $M$ is a smoothly varying positive-definite Hermitian form on each fiber. Such a metric can be viewed as a smooth global section $h$ of the vector bundle $(E \otimes \bar{E})^*$ such that for every point $p$ in $M$, $$ h_p(\eta, \bar{\zeta}) = \overline{h_p(\zeta, \bar{\eta})} $$ for all $\zeta, \eta$ in the fibre $E_p$ and $$ h_p(\zeta, \bar{\zeta}) > 0 $$ for all nonzero $\zeta$ in $E_p$.

A section of $(E \otimes \bar{E})^*$ alone is a sesquilinear form on $E$, where "sesquilinear form" in the linear algebra setting is what we call a linear map $V \otimes \bar{V} \to \mathbb{C}$ where $V$ is a complex vector space (equivalently, a bilinear map $V \times \bar{V} \to \mathbb{C}$). "Section" here encodes "smoothly varying," as with Riemannian metrics. The first condition in the quote above is then exactly the condition we pose on a sesquilinear form to make it into Hermitian, and the second encodes positive-definiteness.

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