I don't understand the geometry of what happens when you give a manifold a metric, in particular how the group structure reduces to the orthogonal group.

I understand that the group [itex]G[/itex]structure is a requirement for constructing a Principal Bundle because we want to act on the fibres by representatives of some group [itex]G[/itex]. And it is this continuous, free and transitive right action that causes the fibre to be homeomorphic to the group.

However, giving a manifold a metric is equivalent to giving the tangent bundle a map (or a tensor) from [itex]TM\times TM \rightarrow \mathbb{R}[/itex]. The tangent bundle is associated with it's frame bundle (which is where the basis comes from) which is also a torsor, so I understand how the fibre bundle has an intimate relationship with a group. But now I get stuck, because my book says that now the Frames (i.e. fibres of the Frame Bundle) are required to be orthonormal due to the presence of a metric. Why? What has caused the group to reduce from [itex]GL(n,\mathbb{R})[/itex] to [itex]\mathcal{O}[/itex]?

Essentially, because a metric gives you a notion of angles and lengths.

Given a (symmetric, positive-definite) metric tensor ##G##, you can always define an orthonormal frame ##E## via ##E^\top E = G##. The existence of this frame locally reduces the structure group to ##SO(n)##. Stitching various patches together, we can conclude that the global structure group is reduced to ##O(n)##.

So are you saying that the metric tensor [itex]G[/itex] defined between tangent spaces [itex]T_x M[/itex] makes them inner product spaces (which gives you angles and lengths), and then this somehow picks out an orthonormal frame fibre [itex]E[/itex]? Then, since we have the metric to preserve the inner product (and so we can stitch properly), the group action must also act on the frame bundle in a similar preservative way, and the only subgroup of the [itex]GL(n,\mathbb{R})[/itex] which accomplishes this is the orthogonal group? So without any loss of information, we can just restrict our attention to the orthogonal group structure?

What if I asked the inverse question? Let [itex]E[/itex] be a tangent bundle on a manifold and associate it with a frame bundle [itex]F[/itex] with group structure [itex]GL(n,\mathbb{R})[/itex]. Suppose I then reduce the group structure to the orthogonal group [itex]\mathcal{O}_n[/itex], does this explicitly impose a metric on the manifold?

Since ##O(n)## is defined as the group that preserves a positive-definite quadratic form, then yes. The quadratic form can be called the metric tensor.

Suppose one has a Riemannian metric on an oriented 2 dimensional surface. At each point the vectors of length 1 form a circle and SO(2) acts on them by rotation in the positively oriented direction. The circles together with these rotations are a principal SO(2) bundle over the surface.

Here is how to determine the rotations on a unit tangent vector, ##v.## Let ##w ## be the unit vector that is orthogonal to ##v ## such that the frame ##(v,w)## is positively oriented. ##w## is uniquely determined by the metric together with the orientation. You need both.

Then the action is ##e^iθ .v = cos(θ)v + sin(θ)w ##

And the action on ##w## is ##-sin(θ)v + cos(θ)w##

So this action is also an action on the oriented orthonormal frame bundle. In the case of surfaces there is really no difference.

Now start with a linear action of SO(2) on the tangent bundle of the surface and suppose there is a principal SO(2) subbundle. One gets a Riemannian metric by letting the vectors in the fiber circles be the unit vectors and by rotating each vector by 90 degrees to find the vector orthogonal to it in the positively oriented direction.

Note that this simultaneously definines both a metric and an orientation on the surface.