What does reduction of structure group of principal bundle say? $\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}$Let $G$ be a Lie group and $\pi:P\rightarrow M$ be a principal $G$ bundle.
The notion of reduction of structure group is standard but I will recall here in case some one needs it.
Let $f:P(M,G)\rightarrow P'(M',G')$ be a morphism of principal bundles such that $f:P\rightarrow P'$ is an imbedding and $f:G\rightarrow G'$ is a monomorphism. If $M=M'$ and the induced map $f:M\rightarrow M'$ is identity map, we call $P(M,G)$ to be reduced bundle for $P'(M,G').$
Given a principal bundle $P’(M’,G’)$ and a Lie subgroup $G$ of $G’,$ we say the structure group $G’$ is reduced to $G$ if there is a reduced bundle $P(M,G).$
Reduction of structure group says some thing interesting about manifolds involving it. For example,

*

*A manifold  admits an almost-complex structure if the frame bundle on the manifold, whose fibers are $\GL(2n,\mathbb{R})$, can be reduced to the group $\GL(n,\mathbb{C})\subset \GL(2n,\mathbb{R})$.

*A manifold is orientable if and only if its frame bundle can be reduced to the special orthogonal group, $\SO(n,\mathbb{R})\subset \GL(n,\mathbb{R})$.

I am interested in knowing similar results about reduction of structure group. Please add references (if possible, a sketch of the proof) for results you quote here. One result in one answer please.
 A: Let $G$ be a topological group and $M$ be a smooth manifold. Then, a reduction of the structure group of the frame bundle from $\mathrm{GL}_n(\mathbb R)$ to $\mathrm{GL}_n(\mathbb R)\times G$ is equivalent data to a principal $G$-bundle $Q\to M$.
Here's a proof sketch: if $H$ and $H'$ are groups, an $(H\times H')$-torsor is the same thing as a product of an $H$-torsor and an $H'$-torsor. So the fiber of the principal $(\mathrm{GL}_n(\mathbb R)\times G)$-bundle we obtained from the frame bundle at some $x\in M$ is a product of a $\mathrm{GL}_n(\mathbb R)$-torsor $P_x$ and a $G$-torsor $Q_x$, and $P_x$ is the $\mathrm{GL}_n(\mathbb R)$ of bases of $T_xM$. Since $Q_x$ varies smoothly, it defines a principal $G$-bundle $Q\to M$.
A: My answer is a very standard fact, but I am surprised nobody wrote it. Please, if I am wrong let me know.
Consider the frame bundle $P'=FM$ of a manifold $M$. In this case, $G'=GL(n)$, and consider $G=O(n)$. The orbit space $P'/G=FM/O(n)$ is also a bundle over $M$.
The fibre $(FM/O(n))_x$ consists of classes of bases of $T_x M$, where the equivalence of bases is given by the existence of an orthogonal transformation between them.
Now, if we have a section $\sigma$ of the bundle $P'/G$, we are assigning, in a smooth way, a class of bases of $T_x M$ to every $x\in M$. We have, then, a new principal bundle $P=\bigsqcup\limits_{x\in M} \{x\}\times \sigma(x)$ with group $O(n)$. It can be shown that is a reduction of the frame bundle.
On the other hand, every class $\sigma(x)$ determines a bilinear form in $T_x M$, indeed, a metric. So we have a Riemannian metric on $M$. On the contrary, a Riemannian metric let us specify a section of $FM/O(n)$, and therefore a principal bundle reduction of the frame bundle.
So the principal bundle reduction, in this case, is giving us the data provided by a Riemannian metric.
By the way, as we know, every manifold can be equipped with such a metric, and so we always have this kind of principal bundle reduction.
