Timeline for Triviality of direct multiples of flat complex vector bundles
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 25, 2014 at 20:15 | comment | added | Paolo Antonini | Thanks. Indeed is sufficient to take $n$ to be greater tian the stable ranking of the base | |
| Mar 9, 2012 at 19:00 | comment | added | Johannes Ebert | two bundle monomorphisms $R^i\to R^{r+i}$, one with $V$ as complement, the other one with trivial complement. Taking any homotopy between them and put it into general position. Hence the two bundle mono's are isotopic; hence their complements, i.e. $V$ and $R^r$ are isomorphic. When done carefully, this argument gives the same bound on the dimensions as the obstruction theory argument. | |
| Mar 9, 2012 at 18:58 | comment | added | Johannes Ebert | That $BU(n) \to BU$ is highly connected follows from the long exact homotopy sequence of $S^{2n-1} \to BU(n-1) \to BU(n)$. The stable cancellation follows by obstruction theory. Alternatively, if you are talking about smooth vector bundles on smooth manifolds, you can use general position arguments, like the following. If $M^m$ is a manifold, $V$ and $W$ vector bundles on $M$ of rank $r$ and $s$. If $s$ is sufficiently large w.r.t. $m$ and $r$, then any bundle map $f:V \to W$ in general position is injective (use transversality). Now if $V^r \oplus R^i \cong R^{r+i}$, you have ... | |
| Mar 9, 2012 at 16:51 | comment | added | diverietti | Is there any down-to-earth proof of this fact? It is so innocent-sounding... | |
| Feb 23, 2012 at 9:17 | comment | added | Paolo Antonini | Sure the stable rank Theorem. Thank you. | |
| Feb 23, 2012 at 9:02 | history | answered | Johannes Ebert | CC BY-SA 3.0 |