Timeline for Satake Diagram determines Involution
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| Jul 17, 2017 at 0:29 | comment | added | Mikhail Borovoi | @TorstenSchoeneberg: This depends on the base field. Over a general field one starts from the split form, but over $\Bbb R$ one starts from the compact form. See my texts (with co-authors) on Galois cohomology over $\Bbb R$: this paper and this preprint. | |
| Jul 16, 2017 at 1:56 | comment | added | Torsten Schoeneberg | The preprint linked by Mikhail Borovoi (thanks!) is quite enlightening. Also look at example 4.16 in this earlier version. So my earlier comment is to be taken with a grain of salt, your $\theta$ and $\tau$ live in a different $H^1$ which however, as explained in that preprint, is isomorphic to the one I wrote. (And I finally understand that in the Cartan-involution description, people take the compact form as the basepoint of their $H^1$, whereas coming from a Galois viewpoint, one takes the split form.) | |
| Jul 15, 2017 at 22:13 | comment | added | Torsten Schoeneberg | But obviously you will have to use some "classifying property" of Satake diagrams because that is the only extra information linking $\theta$ and $\tau$ in the assertion. | |
| Jul 15, 2017 at 22:06 | comment | added | Torsten Schoeneberg | That the Satake diagram determines the iso class of a form of a semisimple Lie algebra (or, by Galois cohomology, the class of your $\theta$ in $H^1(Gal(\mathbb{C}|\mathbb{R}), Aut (\mathfrak{g})$) is a quite involved theorem. (It goes back to Borel-Tits for reductive groups and Satake for real Lie groups: for the Lie algebra setting my thesis on this will be published soon.) Maybe if one goes through the constructions in this case, there is an easier route. How much of Cartan theory is one allowed to use? E.g. can we assume the fact that there is a unique compact form up to isomorphism? | |
| Jul 15, 2017 at 21:27 | comment | added | Mikhail Borovoi | Table 9 of Onishchik and Vinberg 1990 lists Satake diagrams of real forms of simple complex Lie algebras. On page 274 of this book it is written: This table quite easily implies Theorem 3. Two semisimple Lie algebras over $\Bbb R$ are isomorphic if and only it so are (in the natural sense) their Satake diagrams. | |
| Jul 15, 2017 at 21:02 | comment | added | Mikhail Borovoi | Concerning relations between $\Bbb C$-linear and $\Bbb C$-antilinear involutions, see this preprint | |
| Jul 15, 2017 at 20:59 | comment | added | Mikhail Borovoi | Note that conjugacy classes of involutions of complex simple Lie algebras are classified by Kac diagrams. See Table 7 in the book: Onishchik and Vinberg, Lie Groups and Algebraic Groups, Springer 1990. For the corresponding theory see Onishchik and Vinberg (Eds.), Lie Groups and Lie Algebras III, EMS vol. 41, Springer 1994. | |
| Jul 15, 2017 at 19:02 | comment | added | freeRmodule | That's interesting. Is it easy to see why $\mathbb{C}$-antilinear involutions are determined up to isomorphism by their satake diagrams? Because that would be my next question. | |
| Jul 15, 2017 at 18:59 | comment | added | Torsten Schoeneberg | It seems to me that, if $\sigma$ is the complex conjugation, the elements fixed by $\theta \circ \sigma$ resp. $\tau \circ \sigma$ are real forms of $\mathfrak{g}$ which are uniquely determined up to isomorphism by the resp. Satake diagram. Since these are the same, there is a real isomorphism between them, and I guess this lifts to the one your are looking for, which conjugates $\theta$ and $\tau$. I'll think through it more later. | |
| Jul 15, 2017 at 18:20 | comment | added | freeRmodule | In fact, as I mentioned in my question, there are involutions that fix a Cartan subalgebra pointwise but are not the identity. | |
| Jul 15, 2017 at 18:19 | comment | added | freeRmodule | That's not quite what I want. I want $\tau=\sigma\circ\theta\circ\sigma^{-1}$ for some automorphism (not necessarily inner) $\sigma\in Aut(\mathfrak{g})$. i.e. I want $\tau$ and $\theta$ to be conjugate in the group $Aut(\mathfrak{g})$. And it's not true that conjugacy classes of automorphisms (or involutions) are determined by the induced diagram automorphism. | |
| Jul 15, 2017 at 17:30 | comment | added | Mikhail Borovoi | An idea: Consider a "Borel pair" $\frak h\subset\frak b\subset\frak g$. There exists an inner automorphism $g\in {\rm Inn}(\frak g)$ such that $g\circ\theta$ takes $(\frak h,\frak b)$ to itself and thus defines an automorphism $\theta_*$ of the Dynkin diagram $D(\frak g,\frak h,\frak b)$, which does not depend on the choice of $g$. I suspect that you can compute $\theta_*$ from the Satake diagram of $\theta$. On the other hand, if $\tau_*=\theta_*$, then $\tau=g\circ\theta$ for some inner automorphism $g$ of $\frak g$. Is this what you need? | |
| Jul 15, 2017 at 17:00 | comment | added | freeRmodule | Hope that helps to clarify. | |
| Jul 15, 2017 at 16:59 | history | edited | freeRmodule | CC BY-SA 3.0 |
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| Jul 15, 2017 at 6:00 | comment | added | Torsten Schoeneberg | Could you explain or give a reference for what you mean by an involution of the complex algebra, and its Satake diagram? | |
| Jul 14, 2017 at 19:03 | history | asked | freeRmodule | CC BY-SA 3.0 |