Timeline for Killing form vs its counterpart in a given represenation
Current License: CC BY-SA 2.5
9 events
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| Dec 19, 2018 at 1:20 | comment | added | Arturo don Juan | I cannot find any proof of this fact. In fact even with the schematic answer given by @BugsBunny below, I cannot prove that for simple $\mathfrak g$ the form $K_\phi$ is proportional to $K_{\text{Ad}}$. Could somebody direct me to one? | |
| Dec 17, 2010 at 14:34 | comment | added | Jim Humphreys |
This question comes up in both mathematics and physics literature. For example in section 22.1 of my 1972 Springer Graduate Text on Lie algebras I gave a purely mathematical treatment for a semisimple Lie algebra over a field such as $\mathbb{C}$. Here a "Casimir element" attached to a representation and trace function is contrasted with a "universal Casimir element" defined using the Killing form. To treat semisimple (or reductive) rather than just simple Lie algebras, you just need to be careful about the ideals acting as zero in a representation.
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| Dec 14, 2010 at 23:20 | vote | accept | Michał Oszmaniec | ||
| Dec 14, 2010 at 23:01 | vote | accept | Michał Oszmaniec | ||
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| Dec 14, 2010 at 22:54 | answer | added | Bugs Bunny | timeline score: 5 | |
| Dec 14, 2010 at 22:53 | comment | added | José Figueroa-O'Farrill | Small nitpick: $\mathfrak{gl}(N,\mathbb{C}) is *not* semisimple. $$ $$ No matter, over the complex numbers, if the Lie algebra is simple then they are proportional, since any two invariant bilinear forms are proportional. (Over the reals, for example, the latter statement is false: c.g., $\mathfrak{so}(3,1)$, but I don't recall whether the two forms in the answer are still proportional.) $$ $$ For semisimple Lie algebras which are not simple, I don't think they have to be proportional. | |
| Dec 14, 2010 at 22:52 | answer | added | David Hill | timeline score: 0 | |
| Dec 14, 2010 at 22:44 | history | edited | Michał Oszmaniec | CC BY-SA 2.5 |
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| Dec 14, 2010 at 22:36 | history | asked | Michał Oszmaniec | CC BY-SA 2.5 |