Timeline for "Totally real" linear transformations
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 25 at 16:34 | vote | accept | CommunityBot | ||
| Jan 27 at 2:39 | |||||
| Jan 25 at 13:14 | comment | added | Tom Goodwillie | But my argument by characteristic classes does not prove it in the negative determinant case. | |
| Jan 25 at 13:03 | comment | added | user515519 | @TomGoodwillie actually the determinant of $A$ and $-A$ is the same because the action is on $\mathbb{C}^n=\mathbb{R}^{2n}$. But you are right in any case the answer to the question is no. | |
| Jan 25 at 11:39 | comment | added | Tom Goodwillie | When $n$ is odd then it doesn't matter if $A$ has odd determinant, becase $-A$ then has positive determinant and acts the same on the Grassmannian. And when $n$ is odd then the (mod $2$) self-intersection number is indeed non-zero; you can see this by calculating the mod $2$ Euler class (= Stiefel-Whitney class $w_{2n-2}$) of the normal bundle of the complex projective space in the Grassmannian. | |
| Jan 24 at 2:17 | comment | added | user515519 | @PaulCusson Yes. I mean by the action of $A$ on the Grassmannian of planes. | |
| Jan 24 at 1:43 | comment | added | Paul Cusson | Perhaps I'm misreading but could you clarify what is meant by "and $A$ being totally real means that $\mathbb{C}P (n-1)$ intersects its image trivially."? By image do you mean $A$ applied to the elements of $\mathbb{C}P (n-1)$ inside $\mathbb{R}^{2n}$? | |
| Jan 24 at 0:53 | history | edited | user515519 | CC BY-SA 4.0 |
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| Jan 23 at 2:09 | answer | added | KhashF | timeline score: 3 | |
| Jan 23 at 1:51 | history | edited | user515519 | CC BY-SA 4.0 |
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| Jan 22 at 23:56 | history | asked | user515519 | CC BY-SA 4.0 |