Quotient of trivial bundles Suppose you are on a manifold. Suppose you have a trivial bundle and a trivial subbundle of it. If you divide this trivial bundle with its trivial subbundle, do you get a trivial bundle as a quotient? The answer is generally no. What are the conditions on the manifold (s.c. etc.) and bundles (codim e.g.) to get a trivial quotient?
 A: Suppose you have a trivial bundle of rank $k$ inside a trivial bundle of rank $n$. Then you get an obvious map to the Grassmannian of $k$-planes in $\mathbb{R}^n$. The homotopy class of this map is an invariant. Equip both bundles with inner products. If the quotient bundle is trivial, then we can take a trivial orthonormal moving frame for the $k$-bundle and an orthonormal moving frame for the quotient bundle, giving a moving frame for the $n$-bundle. Compare this to the trivialization of the $n$-bundle; they are equal up to some element of $O(n)$. So this lifts the map to the Grassmannian into a map to $O(n)$. Conversely, the map to the Grassmannian lifts to a map to $O(n)$ just when the quotient bundle is trivializable. I believe that the homotopy class of the map to the Grassmannian is in the image of the homotopy class of a map to $O(n)$ just when the quotient bundle is trivializable. You can say something about that homotopy type (at least stably) in the form of characteristic classes, i.e. by pulling back the cohomology of the Grassmannian by the map. (Thanks to Ricardo for the correction.)
A: If the dimension of the base $X$ of the bundles is less than the difference $n-k$ of the fiber dimensions then the quotient bundle is trivial. To see this it is enough to consider the case $k=1$ and then induct. An embedding of a trivial rank one bundle in a trivial rank $n$ bundle amounts to a map $X\to S^{n-1}$. If the dimension of $X$ is less than $n-1$ then such a map is nullhomotopic.
