Timeline for singularities of the dual variety of a surface
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2020 at 12:31 | vote | accept | IMeasy | ||
| Jan 13, 2020 at 12:30 | vote | accept | IMeasy | ||
| Jan 13, 2020 at 12:31 | |||||
| Jan 6, 2020 at 5:40 | answer | added | AG learner | timeline score: 5 | |
| Nov 8, 2013 at 7:25 | answer | added | IMeasy | timeline score: 3 | |
| Nov 8, 2013 at 5:50 | comment | added | naf | The dual variety need not be normal. | |
| Nov 7, 2013 at 20:13 | comment | added | IMeasy | @aginesky For a surface it is enough that it be smooth and non-linear. [P. Griffiths, J. Harris, Algebraic geometry and local differential geometry, Ann. Sci. Ec. Norm. Super., 12 (1979), 355–432)] | |
| Nov 7, 2013 at 19:16 | comment | added | meh | I don't have the reference handy, but isn't the dual variety a hypersurace iff there is no ruling (family of lines) on the variety ? | |
| Nov 7, 2013 at 11:53 | comment | added | IMeasy | The generic plane section of a normal variety is smooth by Seidenberg | |
| Nov 7, 2013 at 11:07 | comment | added | naf | From Bertini's theorem applied to the normalisation you can see that the curve is irreducible, but I don't see why it says anything about the singularities. | |
| Nov 7, 2013 at 8:17 | comment | added | IMeasy | But maybe we use different definitions of duality? | |
| Nov 7, 2013 at 8:17 | comment | added | IMeasy | @Allen: I am afraid that the dual of the Veronese surface is the (symmetric) determinantal cubic in $P^5$. Since basically you are considering $Sym^2V$ with $dim(V)=3$, there are no self dual orbits. The rank 1 (the surface) is the dual of the rank 2 (the cubic HS). Of course the generic tensor is rank 3. | |
| Nov 7, 2013 at 8:07 | comment | added | Allen Knutson | Oops, embarassingly bad example. How about the 2nd Veronese of $P^2$, in $P^5$? I think that's self-dual too (and hope I haven't made as stupid a mistake this time). | |
| Nov 7, 2013 at 7:56 | comment | added | IMeasy | @Jason: yes I am working over $\mathbb{C}$. It is the first time that I check Jouanolou's book: what is exactly the argument you are mentioning? | |
| Nov 7, 2013 at 7:52 | history | edited | IMeasy |
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| Nov 7, 2013 at 6:33 | comment | added | IMeasy | @Allen, via the Segre embedding $P^1 \times P^1$ is a quadric hypersurface in $P^3$. Am I wrong? | |
| Nov 7, 2013 at 4:54 | comment | added | Allen Knutson | In particular I think Jason is saying that the fact that $S^*$ is the dual of a surface is irrelevant. But I'm confused by your first claim, since e.g. the Segre embedding of $P^1 \times P^1$ is self-dual. | |
| Nov 6, 2013 at 23:08 | comment | added | Fernando Muro | at.algebraic-topology? | |
| Nov 6, 2013 at 22:47 | comment | added | Jason Starr | Are you in char 0? If so, you can use Bertini applied to the normalization of $S^*$, as described in Jouanolou's book. | |
| Nov 6, 2013 at 21:36 | history | asked | IMeasy | CC BY-SA 3.0 |