Timeline for A torus bundle whose vertical tangent bundle is indecomposable
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2023 at 6:21 | vote | accept | Anon | ||
| Jun 26, 2023 at 12:06 | answer | added | Neil Strickland | timeline score: 2 | |
| Jun 21, 2023 at 19:21 | comment | added | HJRW | Then I think either of the latter two examples I suggested should work, shouldn't they? I'm not sure about the cohomological characterisation, but surely it should imply that the monodromy is decomposable, since the exponential map will identify the vertical tangent bundle with the universal cover of the torus? | |
| Jun 21, 2023 at 17:11 | comment | added | Anon | @HJRW by indecomposable, I mean that a real vector bundle $E \to X$ does not split into a direct sum $E_1 +E_2$ of real vector bundles on $X$. An example of an indecomposable bundle on $\mathbb P^1$ is $\mathcal O(1)$ viewed as a real bundle of rank two. Infact for a rank two bundle the obstruction to decomposition lies in $O(2)/(O(1) \times O(1)) \simeq S^1$ and this is equivalent to a second cohomology class vanishing. | |
| Jun 21, 2023 at 15:42 | comment | added | HJRW | Another possibility is to use the monodromy matrix ((1,1),(0,1)). This, of course, is reducible but not decomposable. | |
| Jun 21, 2023 at 12:34 | comment | added | HJRW | Thanks. I guess I don’t know exactly what you mean by “decomposable”. One could instead choose a base with fundamental group the closed surface of genus 2, and get it to act on the torus by a pair of matrices that cannot be simultaneously diagonalised. Would that give you what you want? | |
| Jun 21, 2023 at 8:34 | comment | added | Anon | @HJRW The Anosov matrix has irrational eigenvectors, and these give a trivialization of the tangent space. In your example, the eigenvectors are $(1,(-1 \pm \sqrt{5})/2)$ | |
| Jun 21, 2023 at 7:58 | comment | added | HJRW | I suspect any 2-torus bundle over the circle with monodromy an Anosov matrix (eg ((2,1),(1,1))) will give you what you want. | |
| Jun 21, 2023 at 6:30 | history | edited | gmvh | CC BY-SA 4.0 |
"decompasable" should be "decomposable", I believe
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| Jun 21, 2023 at 5:28 | comment | added | Anon | Michael Albanese, I corrected the question based on your example. What I am really looking for is that the directions in the torus mix. | |
| Jun 21, 2023 at 5:27 | history | edited | Anon | CC BY-SA 4.0 |
added 56 characters in body; edited title
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| Jun 20, 2023 at 16:13 | comment | added | Michael Albanese | If $K$ denotes the Klein bottle, then $S^1\to K \xrightarrow{p} S^1$ and $\ker dp$ is non-trivial. So $T^2 \to S^1\times K \xrightarrow{\pi} S^1$ where $\pi(x, y) = p(y)$ and $\ker d\pi \cong \varepsilon^1\oplus\ker dp$ which is again non-trivial. | |
| Jun 20, 2023 at 13:33 | comment | added | Anon | R. van Dobben de Bruyn, The $P$ was a typo, thanks for noticing. | |
| Jun 20, 2023 at 13:32 | history | edited | Anon | CC BY-SA 4.0 |
edited body
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| Jun 20, 2023 at 13:00 | comment | added | R. van Dobben de Bruyn | What is $P$ here? | |
| Jun 20, 2023 at 10:42 | comment | added | Denis T | I can suggest looking at first homotopy group of a base. Take any nontrivial 2-dimensional local system $L$ on $B$. Bundle with the fiber $(L \otimes \Bbb R)/L$ is of the form you want. | |
| Jun 20, 2023 at 9:50 | history | asked | Anon | CC BY-SA 4.0 |