Timeline for cup-length of the first Chern class of complex grassmannian
Current License: CC BY-SA 3.0
15 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| May 11, 2015 at 6:20 | comment | added | Francois Ziegler | @RSQ: For the (indeed, true) relation $c_1 = [\omega]$, see e.g. S. S. Chern, Complex manifolds without potential theory, p. 82. | |
| May 11, 2015 at 4:35 | vote | accept | Shiquan Ren | ||
| May 8, 2015 at 18:51 | answer | added | Tom Goodwillie | timeline score: 7 | |
| May 8, 2015 at 16:53 | history | reopened |
Francois Ziegler Joonas Ilmavirta Yemon Choi Andrey Rekalo Johannes Hahn |
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| May 7, 2015 at 19:56 | review | Reopen votes | |||
| May 8, 2015 at 16:53 | |||||
| May 7, 2015 at 7:24 | history | closed |
abx Alex Degtyarev coudy Stefan Waldmann Dima Pasechnik |
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| May 6, 2015 at 8:09 | comment | added | Jason Starr | @RSQ. For the theorem abx states, you can look up "Hard Lefschetz Theorem" in Griffiths and Harris, among others. There are, of course, simpler proofs for the Grassmannian. I am sure that the description of the complex Grassmannian in Griffiths and Harris includes your result. In particular, $c_1^{2n-2}$ is computed as a Catalan number. | |
| May 6, 2015 at 5:49 | comment | added | Shiquan Ren | Dear Prof. abx, could you also give any references? Thanks! | |
| May 6, 2015 at 5:49 | comment | added | Shiquan Ren | Dear Prof. Francois Ziegler, could you give any references? Thanks! | |
| May 6, 2015 at 4:43 | review | Close votes | |||
| May 7, 2015 at 7:24 | |||||
| May 6, 2015 at 4:25 | comment | added | abx | Any Grassmannian is a Fano variety, that is, $c_1$ is an ample class. This implies that your "cup-length" is the dimension plus one, in your case $2n-1$. | |
| May 6, 2015 at 3:22 | comment | added | Francois Ziegler | $G_2(\mathbb C^{n+1})$ is a real symplectic manifold of complex dimension $d=2(n+1-2)$. Its $2$-form $\omega$ is a representative of $c_1$, isn't it? If I'm not mistaken in that, you get as an immediate consequence that $c_1^{2n-2} = [\omega^{\wedge d}]\ne0$, hence $l\geqslant 2n-1$. | |
| May 6, 2015 at 2:19 | history | edited | Shiquan Ren |
edited tags
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| May 6, 2015 at 2:11 | history | asked | Shiquan Ren | CC BY-SA 3.0 |