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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. So the homotopy type 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.)

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.)

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. So the homotopy type 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 Richard for the correction.)

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. So the homotopy type 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.)

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 $\mathbb{R}^n$. This map is homotopic to a constant map just when the quotient bundle is trivializable. So the homotopy type of that map is your invariant. 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.

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. So the homotopy type 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 Richard for the correction.)