Timeline for Spaces over which every vector bundle is a summand of the trivial bundle
Current License: CC BY-SA 3.0
16 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 2, 2013 at 14:16 | vote | accept | Ali Taghavi | ||
| Nov 30, 2013 at 16:40 | comment | added | Ali Taghavi | @Qiaochu If f:X ---->Y is a homotopy equivalent then f*(pull back) gives a natural bijection between vect_{n}(X) and Vect_{n}(Y), the isomorphism class of n dimensional bundles on X and Y, respectively | |
| Nov 30, 2013 at 12:55 | history | edited | Igor Belegradek | CC BY-SA 3.0 |
Made the title more descriptive.
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| Nov 29, 2013 at 22:31 | comment | added | Qiaochu Yuan | @Ali: it's not clear to me that this condition is homotopy invariant. | |
| Nov 29, 2013 at 22:08 | answer | added | Igor Belegradek | timeline score: 13 | |
| Nov 29, 2013 at 20:47 | comment | added | Ali Taghavi | @Qiaochu a restriction is existence of a vector bundle which is not a summand of a trivial bundle Ex;the canonical line bundle on RP^\infty, see vector bundles and k theory by Allen Hatcher | |
| Nov 29, 2013 at 2:02 | comment | added | Denis Nardin | I don't know the answer to the question, but any differentiable manifold is homeomorphic to a quotient of the closure of some star-shaped set of the $n$-space (this can be easily proved using Riemannian geometry), so it would not be unconceivable that every manifold had the homotopy type of a compact space. | |
| Nov 28, 2013 at 23:17 | comment | added | Qiaochu Yuan | @Ricardo: oh, I see. If $X$ is such a space and $f : X \to D$ a homotopy equivalence to a discrete space then in particular $f$ is a continuous map inducing an isomorphism on $\pi_0$, so the inverse image of each point in $D$ is open and these must be the path components of $X$. Thanks! (If we weaken to weak homotopy equivalence then this argument fails, though, and in this case maybe everything has the weak homotopy type of a compact Hausdorff space? It's just that I'm hesitant to talk about the homotopy type of something that isn't homotopy equivalent to a CW complex.) | |
| Nov 28, 2013 at 23:05 | comment | added | Ricardo Andrade | @Qiaochu Yuan: A space homotopy equivalent to a discrete space (or a CW-complex, or any space with open path components) has open path components. Thus any space homotopy equivalent to an infinite discrete space would have an infinite open cover by pairwise disjoint, non-empty subsets. Such a cover admits no finite subcover. | |
| Nov 28, 2013 at 22:58 | comment | added | Qiaochu Yuan | @Eric: how do you show that an infinite discrete space isn't homotopy equivalent to a compact Hausdorff space? I'm worried because e.g. the Cantor set and the one-point compactification of $\mathbb{Z}$ are both compact Hausdorff spaces, so any argument based only on $\pi_0$ seems to fail. | |
| Nov 28, 2013 at 22:04 | comment | added | Eric Wofsey | An infinite discrete space is a counterexample, but finding a connected counterexample seems harder: I don't know how to show a connected space does not have the homotopy type of any compact Hausdorff space. | |
| Nov 28, 2013 at 20:53 | comment | added | Qiaochu Yuan | I don't know any restrictions on the homotopy type of a compact Hausdorff space. Can anyone name one? | |
| Nov 28, 2013 at 20:34 | comment | added | Igor Belegradek | If $X$ is a smooth manifold, then every vector bundle over $X$ is a summand of a trivial bundle, i.e. embed the total space of the vector bundle in some $\mathbb R^N$, and restrict the normal bundle to the zero section. Is every manifold homotopy to a compact Hausdorff space? This seems unlikely. | |
| Nov 28, 2013 at 19:03 | history | edited | David White | CC BY-SA 3.0 |
edited body
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| Nov 28, 2013 at 14:47 | history | asked | Ali Taghavi | CC BY-SA 3.0 |