Timeline for Question about an example in symplectic geometry
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2020 at 21:17 | comment | added | Maria | This is good to know ! Thanks a lot ! | |
| Nov 13, 2020 at 19:24 | history | edited | LSpice | CC BY-SA 4.0 |
Characteristic 0
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| Nov 13, 2020 at 18:52 | comment | added | LSpice | (I don't know why I went with such a complicated example, which, moreover, works on the wrong side of the duality; it can happen in $\mathfrak{su}(3)^*$, over $\mathbb F_3$, as evidenced by $\varpi_\alpha + \varpi_\beta$, where $\alpha$ and $\beta$ are simple roots swapped by the Galois involution, which is regular but fixed by the long element $s_\alpha s_\beta$.) | |
| Nov 13, 2020 at 16:48 | comment | added | LSpice | It is possible for an element to be regular, in the sense that no root vanishes on it, but not strongly regular, in the sense that it can still be fixed by some Weyl-group element. I think, though I'm not positive, that, for Lie-algebra elements this does not happen in characteristic $0$, but it can happen: for example, in adjoint $\mathsf B_2$ over $\mathbb F_4$, with $\alpha$ long simple and $\beta$ short simple, $\varpi_\alpha^\vee(1) + \varpi_\beta^\vee(\theta)$ (with $\theta \notin \mathbb F_2$) is regular but fixed by $s_\beta$. | |
| Nov 13, 2020 at 13:39 | comment | added | Maria | Thank you for your answer @LSpice! But could you explain please what do you mean by " since regular elements in this case are strongly regular" ? | |
| Nov 13, 2020 at 13:37 | vote | accept | Maria | ||
| Nov 12, 2020 at 21:48 | history | answered | LSpice | CC BY-SA 4.0 |