Timeline for How does all of the bundles over a certain manifold characterize the homotopy class of the base manifold?
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20 events
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| Apr 22, 2011 at 11:06 | vote | accept | Honglu | ||
| Apr 22, 2011 at 11:06 | vote | accept | Honglu | ||
| Apr 22, 2011 at 11:06 | |||||
| Apr 21, 2011 at 2:27 | answer | added | Sean Tilson | timeline score: 2 | |
| Apr 21, 2011 at 1:55 | history | edited | Honglu | CC BY-SA 3.0 |
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| Apr 21, 2011 at 1:15 | history | edited | Honglu |
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| Apr 20, 2011 at 13:48 | answer | added | Oscar Randal-Williams | timeline score: 10 | |
| Apr 20, 2011 at 13:21 | answer | added | Mark Grant | timeline score: 3 | |
| Apr 20, 2011 at 11:22 | history | edited | Honglu | CC BY-SA 3.0 |
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| Apr 20, 2011 at 11:13 | history | edited | Honglu | CC BY-SA 3.0 |
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| Apr 20, 2011 at 5:48 | answer | added | algori | timeline score: 5 | |
| Apr 20, 2011 at 5:29 | comment | added | Somnath Basu | @ lethe - That is part of Steenrod's thesis. There's a quick proof (for closed manifolds) via differential cohomology machinery - $M$ is orientable means $w_1(TM)=0$. Since $M$ is odd dimensional, $\chi(M)=0$ which implies $w_3(TM)=0$. It is known that the total Stiefel-Whitney class $w=1+w_1+w_2+w_3$ is the Steenrod square of $v=1+v_1+v_2+v_3$, called the Wu class. Using this one can show that $w_2(TM)=0$. Since $w_2$ is the obstruction to the existence of a $2$-frame (which gives $2$ lin. ind. nowhere zero vector fields) we can use the orientation frame to get $3$ such vector fields. | |
| Apr 20, 2011 at 5:21 | comment | added | algori | lethe -- what kind of bundles are you considering? Real vector bundles? Complex vector bundles? Some other kind of bundles? And what are $M$ and $N$? | |
| Apr 20, 2011 at 4:53 | comment | added | Honglu | To budney, I mean the map induces the 1-1 correspondence between all bundles, not the induced map between a particular bundle and its pullback. Btw, why do all orientable 3 mfd have trivial tangent bundle? | |
| Apr 20, 2011 at 4:47 | comment | added | Andy Putman | @lethe : The set of isomorphism classes of bundles of a rank $k$ on $X$ is equal to $H^1(X;F)$, where $F$ is the sheaf of continuous functions with values in $GL_k(\mathbb{R})$. | |
| Apr 20, 2011 at 4:45 | comment | added | Dan Ramras | Ryan, that sounds different from what lethe was asking. His condition is stronger, for example, than requiring that $f$ induce an isomorphism on K-theory. | |
| Apr 20, 2011 at 4:44 | history | edited | Honglu | CC BY-SA 3.0 |
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| Apr 20, 2011 at 4:39 | comment | added | Honglu | Ahh, indeed, my question was vague. Actually I'm merely curious about how to derive some information about bundles. Let me edit it. | |
| Apr 20, 2011 at 4:36 | comment | added | Ryan Budney | There are many counter-examples. All orientable 3-manifolds have trivial tangent bundles, so all maps between them induce isomorphisms of their tangent bundles, yet most such maps are not homotopy-equivalences. | |
| Apr 20, 2011 at 4:34 | comment | added | Mariano Suárez-Álvarez | What do you mean in the last question by "can we determine?"? In what sense? We can considering the set of all bundles, for example... is that "determining" them? | |
| Apr 20, 2011 at 4:30 | history | asked | Honglu | CC BY-SA 3.0 |