Timeline for Regular elements in the torus of a group of Lie type
Current License: CC BY-SA 3.0
9 events
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| Jun 12, 2016 at 14:15 | comment | added | LSpice | Oh, sorry! I read that sentence as "all non-degenerate maximal torus do not have rational regular points" (which is clearly false), but now I understand that you meant "some non-degenerate maximal tori do not have rational regular points" (which I agree is a very important point to realise). | |
| Jun 12, 2016 at 8:24 | comment | added | Jay Taylor | @LSpice I think this says what I want but maybe the wording is not very clear. I'm just trying to say that the implication "$T$ non-degenerate $\Rightarrow$ $T$ contains a regular semisimple element" does not hold in general, which was part of the question. Clearly there can be non-degenerate tori that contain regular semisimple elements and if $q$ is large enough then all will. The point here is just to say that non-degenerate tori can be much smaller than one might initially think. | |
| Jun 4, 2016 at 21:26 | comment | added | LSpice | "Once one has this statement one can see that a non-degenerate maximal torus can have no rational regular elements, as pointed out by Jim." Certainly there are some maximal tori in some groups with rational regular elements (in which case the torus is non-degenerate). I guess that this sentence is maybe the opposite of what you mean? | |
| Apr 21, 2016 at 9:46 | comment | added | Nick Gill | Great, thanks for this, very helpful. | |
| Apr 21, 2016 at 9:06 | comment | added | Jay Taylor | The special unitary groups $\mathrm{SU}_n(q)$, with $n \geqslant 3$, always have a non-degenerate maximally split maximal torus. This is easy to see from a matrix representation. This is because all bar one of the elements are contained in the field $\mathbb{F}_{q^2}$, so one has enough wiggle room. However, when $q=2$ then one can check that the maximally split torus in ${}^3\mathrm{D}_4(2)$, ${}^2\mathrm{D}_n(2)$ and ${}^2\mathrm{E}_6(2)$ are all degenerate, even though they're non-trivial. | |
| Apr 21, 2016 at 9:04 | comment | added | Jay Taylor | Yes, certainly! If $q=3$ and $F$ is split then for any root $\alpha$ if there exists a root $\beta$ with $\langle \alpha,\check{\beta}\rangle = -1$ then we have $\alpha(\check{\beta}(-1)) = (-1)^{\langle \alpha,\check{\beta}\rangle} = -1 \neq 1$ so the maximal torus is certainly non-degenerate. I didn't check the details but there should always be a root with this property. | |
| Apr 21, 2016 at 8:50 | comment | added | Nick Gill | Hi Jay, Thanks for you answer. I'll have to think some more to understand what you've written. I had a quick look at Carter's Proposition 3.6.1 (in Finite Groups of Lie type), and it seems to me that maximally split maximal tori may be non-degenerate for $q=3$ too (at least for some simple groups $G$ and some endomorphisms $F$). Does this seem reasonable? | |
| Apr 20, 2016 at 9:35 | history | edited | Jay Taylor | CC BY-SA 3.0 |
added 14 characters in body
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| Apr 20, 2016 at 9:25 | history | answered | Jay Taylor | CC BY-SA 3.0 |