Let $\pi \colon E \to M$ be a smooth vector bundle. A Riemannian metric on $E$ can be regarded as a global section of the vector bundle $(E\otimes E)^{\ast}$, or more specifically, the subbundle $(S^2E)^{\ast} \subset (E\otimes E)^{\ast}$. However, not every global section $s$ corresponds to a Riemannian metric, as there is no guarantee that $s(x) : S^2E_x \to \mathbb{R}$, when viewed as a symmetric map $E_x \times E_x \to \mathbb{R}$, is positive-definite. So my question is whether we can form a bundle such that positive-definiteness is automatic. More succinctly:
Is there a vector bundle $F$ associated to $E$ such that the global sections of $F$ are precisely the Riemannian metrics on $E$?
If so, can we characterise all Hermitian metrics on a complex vector bundle in a similar way?
Best Answer
An orthogonal or hermitian structure on $E$ is a section of fibre bundle (which is not a vector bundle). I will deal with the complex case.
The Lie algebra $\mathfrak{gl}(n)$ decomposes into $$ \mathfrak{gl}(n)\simeq \mathfrak{u}(n)\oplus \textrm{Herm}_n, $$ where $\textrm{Herm}_n$ is the vector space of Hermitian $n\times n$ matrices. The exponential map sends it to the positive-definite hermitian matrices: $$ \exp: \textrm{Herm}_n\to \textrm{Herm}_n^+, $$ which form a convex domain ("cone") . More importantly, this cone is actually $$ \textrm{Herm}_n^+ = GL(n,\mathbb{C})/U(n). $$ To see this, consider the action of $GL(n,\mathbb{C})$ on the hermitian matrices by $(T,h)\mapsto \overline{T}^t h T$. Now, do all of this "fibrewise": if $P$ is the frame bundle of $E$, an hermitian metric is a section of the associated bundle with fibre $GL(n,\mathbb{C})/U(n)$.
More coneptually, a choice of reduction of the structure group of a principal $G$-bundle $P$ to a subgroup $K$ is eqivalent to a choice of section of the associated $G/K$ bundle. The hermitian metric is a reduction of the structure group from $GL(n,\mathbb{C})$ to $U(n)$.
Aside:
Since this question may be a related to your other question
Hermitian Christoffel Symbols
let me say that if you have a complex manifold $X=(M,I)$ with Riemannian metric $g$ on $M$ (compatible with $I$) you can extend $g$ to the complexified tangent bundle $T_{X,\mathbb{C}}$ either as a $\mathbb{C}$-bilinear pairing (as in Kobayashi & Nomizu), or as a sesquilinear pairing $g_{\mathbb{C}}$: see section 1.2 of Huybrechts, "Complex geometry". Now, $T_{X,\mathbb{C}}\simeq T^{1,0}\oplus T^{0,1}$. With the former choice $T^{p,q}$ are isotropic sub-bundles, and only the off-diagonal pairing is non-trivial. With the latter choice (Huybrechts, Griffiths & Harris), the two subbundles are orthogonal and $\left. g_{\mathbb{C}}\right|_{T^{1,0}}$ is $\frac{1}{2}h$, $h= g-i\omega$ (and the conjugate of that on $T^{0,1}$). This turns $E = T^{1,0}$ into an hermitian vector bundle.
In indices, in the first case you have $h_{ab}=0= h_{\overline{a}\overline{b}}$, $h_{a\overline{b}}\neq 0$, while in the second the other way around.