Timeline for Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
Current License: CC BY-SA 3.0
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| Mar 27, 2015 at 13:57 | comment | added | Peter Crooks | Thank you! This is exceedingly helpful. In light of the two given answers, I should probably require $\mathfrak{g}$ to be simple. | |
| Mar 26, 2015 at 20:19 | comment | added | Dave Witte Morris | What I mean is that the representations of $\phi_1$ and $\phi_2$ on $\mathfrak{g}$ both contain three 3-dimensional irreducibles, and the rest of each representation is a big trivial module. Specifically, each is the direct sum of three 3-dimensional irreducibles and ten 1-dimensional trivial modules. So the two representations are isomorphic. | |
| Mar 26, 2015 at 20:07 | comment | added | Vincent | I am probably misreading something but it sounds like you say that if V denotes the direct sum of three 3-dimensional irreducible $\mathfrak{sl}_2$-modules then V is isomorphic to the direct sum of a trivial representation and V itself. Given that the trivial representation is one-dimensional, how is this possible? | |
| Mar 25, 2015 at 22:40 | history | answered | Dave Witte Morris | CC BY-SA 3.0 |