Suppose $\pi:E\to M$ is a smooth $S^{k-1}$-bundle (a fiber bundle with fiber $S^{k-1}$), and suppose that the structure group of $E$ can be reduced to the orthogonal group $O(k)$. Then is it true that there is a vector bundle $E'\to M$ with a Riemannian metric such that the unit sphere bundle of $E'$ is $E$?
Since the transition functions of $E\to M$ have values in $O(k)\subset GL(k,\Bbb R)$, we can form a vector bundle $E'\to M$ with the same transition functions as $E$. But I don't see how to define a Riemannian metric on $E'$ to get $S(E')=E$.
Suppose that $E'$ is defined by the trivialization $(U_i)$ (we suppose the fibre isomorphic to the vector space $F$)and the transition functions are defined by $h(x,y)=(x,g_{ij}(x)(y)), g_{ij}(x)\in O(k)$ and $y$ is an element of $E'_x$ the fibre of $x$. Let $b$ be the Euclidean metric. We can define the metric $b_i$ on $U_i\times F$ by $b_i^x(u,v)=b(u,v)$ where $u,v\in E_x$ the fibre of $x$. This metric induces a metric on $E$.