Timeline for Is the space of real conics with a singular point an orientable manifold?
Current License: CC BY-SA 3.0
21 events
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| Dec 18, 2014 at 7:39 | history | edited | Ritwik | CC BY-SA 3.0 |
Pointed out the mistake regarding the calculation of the first stiefel whitney class
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| Dec 18, 2014 at 7:37 | vote | accept | Ritwik | ||
| Dec 17, 2014 at 15:55 | history | edited | Ritwik | CC BY-SA 3.0 |
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| Dec 17, 2014 at 12:11 | history | edited | Ritwik | CC BY-SA 3.0 |
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| Dec 17, 2014 at 11:51 | answer | added | Robert Bryant | timeline score: 10 | |
| Dec 17, 2014 at 11:48 | comment | added | Alex Degtyarev | OK, manifold or not, the loop I suggested in my comment is disorienting: it disorients the Moebius band part $(S^1\times S^1\smallsetminus\text{diagonal})/\sim$ of the fiber, which is a manifold, while being constant in the $\Bbb R\rm p^2$ direction). So, you must have miscalculated either the normal bundle or its $w_1$. | |
| Dec 17, 2014 at 11:00 | history | edited | Ritwik | CC BY-SA 3.0 |
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| Dec 17, 2014 at 10:58 | comment | added | Ritwik | I have given a slightly more detailed explanation as to why the mat $\psi$ has $(0,0,0)$ as a regular value (in the usual calculus sense). The map $\psi$ is probably not a submersion......but I don't see how that contradicts the previous assertion. | |
| Dec 17, 2014 at 10:55 | history | edited | Ritwik | CC BY-SA 3.0 |
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| Dec 17, 2014 at 10:18 | comment | added | Ritwik | @ Ketil: I agree with what you said. That's indeed $\mathbb{RP}^1$. Therefore? | |
| Dec 17, 2014 at 9:20 | comment | added | Ketil Tveiten | Look at e.g. the subspace of double lines through a given point. That's $\mathbb{RP}^1$ right there. | |
| Dec 17, 2014 at 8:35 | comment | added | Alex Degtyarev | Anyway, if the fiber of a fibration is nonorientable, so is the total space, just because the normal bundle of a fiber is trivial. You don't need a spectral sequence for that; any disorienting cycle in any fiber is disorienting for the total space. Thus, you can construct such cycles very explicitly: I guess it would suffice to just interchange two lines intersecting at a point (e.g., $xy=0$ homotoped to itself by a rotation through $\pi/2$). | |
| Dec 17, 2014 at 8:06 | comment | added | Alex Degtyarev | Exactly my point. Just try to differentiate carefully. | |
| Dec 17, 2014 at 8:05 | comment | added | Ritwik | @Alex: Take partial derivatives with respect to $\rho_{00}$, $\rho_{10}$ and $\rho_{01}$. Then plug in x=0 and y=0. | |
| Dec 17, 2014 at 8:00 | comment | added | Alex Degtyarev | Sorry, but I fail to see how it is transverse. It is not even a submersion! | |
| Dec 17, 2014 at 7:48 | history | edited | Ritwik | CC BY-SA 3.0 |
Gave a justification as to why X is a manifold.
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| Dec 17, 2014 at 7:36 | comment | added | Ritwik | @Alex: I can give a a proof that X is indeed a manifold (even around double lines). The evaluation map and the vertical derivative are sections of the two bundles I have written. The section is transverse to the zero set (in fact the linearization restricted to the tangent space of RP^5 is surjective). I will try to write a proper justification of this soon. | |
| Dec 17, 2014 at 6:41 | comment | added | Alex Degtyarev | I think that your "manifold" has singularities (two smooth components intersecting at conics degenerating to a double line), so the normal bundle computation is not quite convincing. | |
| Dec 17, 2014 at 6:01 | comment | added | Ritwik | @Tehrani: I confess I myself do not understand the details of the calculation; but I have been told by algebraic topologists that a very standard calculation using spectral sequence gives you that H^4(E, Z) = 0, if E is an RP^2 fiber bundle over RP^2 (something about looking at the E2 page). | |
| Dec 17, 2014 at 2:44 | history | edited | Peter Crooks | CC BY-SA 3.0 |
edited title
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| Dec 17, 2014 at 2:41 | history | asked | Ritwik | CC BY-SA 3.0 |