The idea of an orthogonal structure is a generalisation to arbitrary vector bundles of the distinguishing ingredient of a Riemannian manifold: the Riemannian metric. In short, an orthogonal structure on a vector bundle is a smooth choice of inner product on the fibres.
In finite dimensions, that is pretty much all that can be said about the basic definition. For infinite dimensional manifolds, however, there is considerably more to be said because the linear notion of an inner product is much, much richer.
Although the definition makes sense in the topological category, we shall confine ourselves to the smooth category for the time being.
Let $\pi \colon E \to M$ be a smooth vector bundle over a smooth manifold. An orthogonal structure on $E$ is a smooth choice of inner product on $E$. That is to say, an orthogonal structure $g$ on $E$ is a section of $E^* \otimes E^*$ which restricts to an inner product on each fibre.
As for Riemannian metrics, a standard partition of unity argument yields the following existence result.
Let $\pi \colon E \to M$ be a smooth vector bundle with finite dimensional fibres which admits a trivialisation over a numerable cover. Then $E$ admits an orthogonal structure.
Moreover, in finite dimensions all orthogonal structures are equivalent in that there is a bundle isomorphism between any two.
In finite dimensions, to put an orthogonal structure on a vector bundle is to give a reduction of its structure group from the general linear group to the corresponding orthogonal subgroup along the defining inclusion $O(n) \hookrightarrow GL(n)$. Such a reduction is also known as a choice of vielbein.
Thus the existence in general is equivalent to the fact that the inclusions $O(n) \to GL(n)$ and $U(n) \to GL(n)$ are homotopy equivalences, as for the inclusion of any maximal compact subgroup. A similar statement can be made about the fact that any two orthogonal structures are equivalent.
The situation in infinite dimensions is more complicated due to the fact that there are more situations to consider. In finite dimensions, any two orthogonal structures are equivalent, which stems from the fact that any two inner products define equivalent norms in finite dimensions. In infinite dimensions, it is no longer true even that all locally convex topological vector spaces admit inner products, and when one does then it is highly unlikely that it will admit only one (up to equivalence).
What gives Riemannian geometry much of its power is the induced isomorphism between the tangent and cotangent bundles. Thus a natural first question to ask of an orthogonal structure in infinite dimensions is whether or not this is still true.
Question: Does the orthogonal structure induce an isomorphism $E \cong E^*$?
Yes: It is strong.
No: It is weak.
The condition that an orthogonal structure be strong is very restrictive. The bundle must be modelled on Hilbert spaces (with the correct topology) and the inner product must be the right choice. Many orthogonal structures that occur “in the wild” do not fit those two conditions and are thus relegated to the “weak” bin. However, all is not lost and it is still possible to work with such orthogonal structures. Although a weak orthogonal structure means that the bundle is not modelled on Hilbert spaces, it does mean that there are Hilbert spaces nearby and it is possible to further refine the notion of a “weak orthogonal structure” by asking how “nice” are those associated Hilbert spaces.
The idea is quite simple. To put an orthogonal structure on a vector space, say $V$, is to give an injection $V \to H$ in to some Hilbert space. We can assume that $V$ has dense image, but this is not strictly necessary. Thus an inner product is a diagram:
and the classification of weak orthogonal structures is essentially a study of how well that diagram extends over the base space.
A choice of orthogonal structure on a manifold is also equivalently called a choice of vielbein. See there for more details.
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