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?

$\begingroup$ At least in the algebraic setting, I feel like the quotient is always a trivial bundle because a map between two trivial bundles is just a vector of scalars... $\endgroup$ – IMeasy Mar 12 '13 at 8:13

$\begingroup$ If you equip the bundle with the constant metric, then the quotient can be identified with the orthogonal complement and thus is trivial... $\endgroup$ – Alex_K Mar 12 '13 at 8:58

22$\begingroup$ @Alex_K: The normal bundle of $S^2$ in $\mathbb{R}^3$ is trivial. So the tangent bundle of $S^2$ is the quotient of a trivial bundle (the tangent bundle of $\mathbb{R}^3$ restricted to $S^2$) by a trivial 1dimensional subbundle (the normal bundle of $S^2$). Yet $S^2$ is not parallelizable. $\endgroup$ – Ricardo Andrade Mar 12 '13 at 9:15

$\begingroup$ Oh yes...thank you for pointing out my mistake...every stably trivial bundle is a counter example to what I said... $\endgroup$ – Alex_K Mar 12 '13 at 12:08

$\begingroup$ Taking that one step further, you can always stabilize by adding trivial summands: $TM\oplus\epsilon^m= \epsilon^N$ and then do your division. $\endgroup$ – Chris Gerig Mar 12 '13 at 19:38
If the dimension of the base $X$ of the bundles is less than the difference $nk$ 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 $X\to S^{n1}$. If the dimension of $X$ is less than $n1$ then such a map is nullhomotopic.


$\begingroup$ No. We are given a trivial rank one subbundle. $\endgroup$ – Tom Goodwillie Mar 13 '13 at 16:27


$\begingroup$ A small comment since the question is about manifolds. When $X$ is a smooth manifold of dimension $d$ whose interior has no compact component, you can get away with only $d\leq nk$. This follows from Tom's answer because $X$ is then homotopy equivalent to a CWcomplex of dimension $d1$: $X$ admits a handlebody decomposition with no handles of dimension $d$. $\endgroup$ – Ricardo Andrade Mar 13 '13 at 21:42

$\begingroup$ Yes. In fact, if $X$ has the homotopy type of a complex of dimension less than $nk$, or if it has no cohomology in degrees less than that (even with twisted coefficients), that's enough. $\endgroup$ – Tom Goodwillie Mar 13 '13 at 22:27
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.)

$\begingroup$ Good start. But you should choose a reference point and the map into $G_k\mathbb R^n$ has to be independent of the path. For that the manifold you are on has to be s.c. otherwise this is a map with twisted(local) coefficients or say a section of the locally constant sheaf of $G_k\mathbb R^n$ on the manifold. $\endgroup$ – kalafat Mar 13 '13 at 4:09

$\begingroup$ I just want to make a small clarification. If the map into the Grassmannian (described in Ben's answer) is null homotopic, then the quotient bundle is indeed trivializable. The converse is not true in general, however. $\endgroup$ – Ricardo Andrade Mar 13 '13 at 5:36

$\begingroup$ @Ben: You are welcome. I agree with the current statement of your answer. An alternative homotopy theoretical method to reach the same conclusion consists of analysing the long exact sequence for $\pi_\ast\operatorname{Map}(X,)$ of the fibration sequence $O(n)\to Gr_k(\mathbb{R}^n)\to BO(k)\times BO(nk)\to BO(n)$. $\endgroup$ – Ricardo Andrade Mar 13 '13 at 22:03

1$\begingroup$ By the way, Ben's answer would also hold (pretty much word for word) if one were to replace the map to the Grassmannian with its lift to the Stiefel manifold $V_k(\mathbb{R}^n)$ of $k$frames in $\mathbb{R}^n$. This latter map is simply remembering the trivialization of the subbundle. In this case, the relevant fibration sequence is $O(n)\to V_k(\mathbb{R}^n)\to BO(nk)\to BO(n)$. $\endgroup$ – Ricardo Andrade Mar 13 '13 at 22:04