Timeline for Viewing exceptional Lie algebras via the classical ones
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 22, 2022 at 20:04 | comment | added | Callum | As an example, the 56-dimensional representation of $\mathfrak{e}_7$ in my notation above is just $\bigwedge^2 V \oplus \bigwedge^2 V^*$ and the symplectic form on it is just $0$ on each of these summands and contraction between them which I find quite an easy way to view it. The invariant quartic form is a bit trickier but the reference I added to my question by Adams described it nicely in those terms so I thought I'd add it in in case it helped someone else as well. | |
| Sep 22, 2022 at 19:55 | comment | added | Callum | @LSpice yes exactly, so that I could view the exceptional Lie algebras as a classical Lie algebra together with one of its representations and thus help me understand them a bit better. (Obviously we are then still left with a challenge to understand the bracket on $I$, but it helped me get a better picture of what was going on) | |
| Sep 22, 2022 at 19:27 | comment | added | LSpice | Ah, I see. So you were looking not just for a direct-sum decomposition as vector spaces, nor for a direct-sum decomposition as Lie algebras (where the summands would commute), but a decomposition $L = L' + I$ with $I$ a subspace of $L$ normalised by $L'$. | |
| Sep 22, 2022 at 18:56 | comment | added | Callum | @LSpice I edited the question to include a reference I read more recently in case anyone was interested so that must have bumped it to the front somehow. Thanks for your comments nonetheless. I think the only thing I was missing when I wrote that comment was that if the first set of roots form a root subsystem then that subalgebra preserves the other span of root spaces which follows from basic properties of root spaces. | |
| Sep 22, 2022 at 17:17 | comment | added | LSpice | @Callum, re, a reductive Lie algebra always decomposes as the sum of a maximal Cartan subalgebra and the root spaces for that maximal torus, and those can be grouped as you please; for example, the maximal subalgebra can be grouped with enough root spaces to make up the Lie algebra of a maximal-rank subgroup, and the remaining ones added on separately. (Sorry—I saw this on the front page and thought it was a new post; just noticed now it's from 2021.) | |
| Jun 21, 2021 at 19:00 | comment | added | Callum | Thanks so much. Just to check I understand this. We can construct subalgebras of maximal rank like these by removing a node from the extended Dynkin diagram. Indeed this identifies some root subsystem of the original system. Is it then the case that the Lie algebra decomposes as the sum of this subalgebra and the span of the remaining root spaces on which the subalgebra acts? | |
| Jun 17, 2021 at 19:34 | vote | accept | Callum | ||
| Jun 16, 2021 at 17:52 | history | answered | Robert Bryant | CC BY-SA 4.0 |