Timeline for Killing form vs its counterpart in a given represenation
Current License: CC BY-SA 2.5
12 events
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| Dec 18, 2018 at 2:39 | comment | added | Arturo don Juan | @BugsBunny Regarding your original answer, could you (or anybody else) provide me with a reference to a more explicit proof of what you said? (wow, this answer is a little over 8 years old) | |
| Jan 4, 2011 at 16:31 | comment | added | Pasha Zusmanovich | @Jose: The following seems an easy exercise to me: for a simple Lie algebra over non-algebraically closed field, every two symmetric invariant nondegenerate forms are proportional up to an element of the centroid. This is also proved, as far as I understand (but I haven't seen the paper) in: W.C. Waterhouse, Invariant bilinear forms on semisimple Lie algebras, Algebras Groups Geom. 9 (1992), No.1, 49-52. | |
| Dec 15, 2010 at 7:15 | comment | added | Bugs Bunny | @ Jose $so(3,1)$ turns to $so(4,C)$ after complexification, thus it is not absolutely irreducible. The question, indeed, remains for this one and requires an explicit computation. The endomorphism ring of the adjoint representation is $C$ and I bet that $K_\phi$ and $K$ are not, in general, proportional ove $R$... | |
| Dec 15, 2010 at 7:04 | comment | added | Bugs Bunny | @ Michal If $g$ is semisimple but not simple, the proportionality is difficult to achieve on an irreducible representation. Such representation is annihilated by all but one simple components, i.e. $K_\phi$ has on all but one simple components in its kernel. It follows that $\phi$ must be trivial to have proportionality... The point is that $\phi$ needs to be faithful to have anything interesting. | |
| Dec 15, 2010 at 5:11 | comment | added | David Hill | Michal-It is fine for the constants to be the same, but there is no reason why this has to be the case. | |
| Dec 15, 2010 at 0:07 | comment | added | Michał Oszmaniec | Thanks a lot for your answers. I find them very inspiring. I'm a newbie in this field and I still have to learn A LOT :) Just a little question: What if (for semi-simple case) all those constants of proportionality that correspond to a different simple Lie algebras (on which $\mathfrak{g}$ decomposes) turn out to be the same? Is there a reason for which it should not be the case? | |
| Dec 14, 2010 at 23:42 | comment | added | José Figueroa-O'Farrill | OK thanks -- so the question remains whether for, say, $\mathfrak{so}(3,1)$ the two forms are always proportional. Is that obvious? Is that even true? | |
| Dec 14, 2010 at 23:20 | vote | accept | Michał Oszmaniec | ||
| Dec 14, 2010 at 23:06 | comment | added | Bugs Bunny | You need endomorphisms of the adjoint representation to be $R$. This means the adjoint representations is absolutely irreducible. I guess it is equivalent to the complexification of $g$ being simple... | |
| Dec 14, 2010 at 23:01 | vote | accept | Michał Oszmaniec | ||
| Dec 14, 2010 at 23:20 | |||||
| Dec 14, 2010 at 22:57 | comment | added | José Figueroa-O'Farrill | This is true over $\mathbb{C}$, but is it true over $\mathbb{R}$, say? Look at $\mathfrak{so}(3,1)$, say. It possesses a two-parameter family of invariant nondegenerate bilinear form, yet the adjoint action is irreducible. | |
| Dec 14, 2010 at 22:54 | history | answered | Bugs Bunny | CC BY-SA 2.5 |