Timeline for Recovering a Lie algebra from its affine Lie algebra
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
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May 11, 2013 at 9:45 | answer | added | Pasha Zusmanovich | timeline score: 1 | |
May 25, 2012 at 3:57 | answer | added | S. Carnahan♦ | timeline score: 2 | |
May 24, 2012 at 20:07 | comment | added | H. Arponen | I'm assuming you just have the Lie algebra as commutation relations, maybe due to some symmetries of PDEs or such, and you're trying to determine if it corresponds to an affine Lie algebra? | |
May 24, 2012 at 16:34 | history | edited | Eric O. Korman | CC BY-SA 3.0 |
added 35 characters in body
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May 24, 2012 at 15:34 | answer | added | Jim Humphreys | timeline score: 0 | |
May 24, 2012 at 15:30 | comment | added | Bill Cook | However, an affine algebra $\hat{\mathfrak{g}}$ carries more information than just its Lie algebra structure. It has a built in gradation: $\hat{\mathfrak{g}}(n) = \mathfrak{g} \otimes t^n$. If you have this information, then your problem is trivial since $\hat{\mathfrak{g}}(0) = \mathfrak{g}$. | |
May 24, 2012 at 15:29 | comment | added | Bill Cook | This is what comes to mind: If you can find a cartan subalgebra (CSA), then $\hat{\mathfrak{g}}$ decomposes into root spaces. Take a maximal collection $\Delta$ of real roots such that the sum of any two elements of $\Delta$ is either in $\Delta$, 0, or not a root. Then the direct sum of the root spaces associated with the roots in $\Delta$ along with the CSA will give you $\mathfrak{g}$. | |
May 24, 2012 at 14:59 | history | asked | Eric O. Korman | CC BY-SA 3.0 |