Timeline for Killing form vs its counterpart in a given represenation
Current License: CC BY-SA 2.5
8 events
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Dec 14, 2010 at 23:18 | comment | added | Keerthi Madapusi | Okay, so that's the obstruction: lack of absolute simplicity. Thanks for the clarifications! | |
Dec 14, 2010 at 23:15 | comment | added | David Hill | The problem is that you might start with a simple Lie algebra over $\mathbb{R}$ which is no longer simple after extending scalars to $\mathbb{C}$ (e.g. $\mathfrac{so}(3,1)$). So the forms are not necessarily proportional over $\mathbb{C}$ and therefore, by your argument, they are not necessarily proportional over $\mathbb{R}$. | |
Dec 14, 2010 at 23:07 | comment | added | Keerthi Madapusi | Yes, but formally speaking: the two invariant forms are elements of some vector space over the base field. It seems from your argument, that, once I tensor over $\mathbb{C}$, they are proportional. So shouldn't they already be proportional over $\mathbb{R}$? Or am I just being stupid? | |
Dec 14, 2010 at 23:04 | comment | added | David Hill | I don't think this is a characteristic $p$ issue. The point is that I assummed we were working over an algebraically closed field. Schur's lemma doesn't work over the reals. | |
Dec 14, 2010 at 23:02 | comment | added | David Hill | You are right, I was assuming the base field was $\mathbb{C}$, and had Schur's lemma in mind. | |
Dec 14, 2010 at 23:00 | comment | added | Keerthi Madapusi | If they are both defined over the same field, then they can be proportional over a field extension if and only if they're already proportional over a base field. No? The issue is probably one of characteristic $p$ versus characteristic $0$. See <a href=mathoverflow.net/questions/46813/…> | |
Dec 14, 2010 at 22:55 | comment | added | José Figueroa-O'Farrill | Is this still true over the reals when the complexification of the algebra is not semisimple? e.g., \mathfrak{so}(3,1) | |
Dec 14, 2010 at 22:52 | history | answered | David Hill | CC BY-SA 2.5 |