Timeline for Quotient of trivial bundles
Current License: CC BY-SA 3.0
8 events
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| Mar 13, 2013 at 22:04 | comment | added | Ricardo Andrade | 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 sub-bundle. In this case, the relevant fibration sequence is $O(n)\to V_k(\mathbb{R}^n)\to BO(n-k)\to BO(n)$. | |
| Mar 13, 2013 at 22:03 | comment | added | Ricardo Andrade | @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 $O(n)\to Gr_k(\mathbb{R}^n)\to BO(k)\times BO(n-k)\to BO(n)$. | |
| Mar 13, 2013 at 14:38 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Mar 13, 2013 at 12:17 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Mar 13, 2013 at 12:03 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Mar 13, 2013 at 5:36 | comment | added | Ricardo Andrade | 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. | |
| Mar 13, 2013 at 4:09 | comment | added | kalafat | 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. | |
| Mar 12, 2013 at 12:33 | history | answered | Ben McKay | CC BY-SA 3.0 |