Timeline for The normalised form for the twisted Kac-Moody algebra
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12 events
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| May 25, 2017 at 10:18 | answer | added | Paul Levy | timeline score: 2 | |
| May 25, 2017 at 9:22 | comment | added | Z.A.Z.Z | $\widehat{sl}_n$ is the untwisted affine Kac-Moody algebra. I don't see why they are isomorphic. By Proposition 8.5 and Remak 8.5 in Kac's book. It is given a condition when they are isomorphic. In fac they if there exists a Lie algebra automorphism $f(t)$ of $\mathfrak g$ such that $$\tau=f(t)\circ \sigma\circ f(-t)^{-1}$$ | |
| May 24, 2017 at 22:07 | comment | added | Paul Levy | ... $\mathfrak{so}_n$ and $\mathfrak{sp}_n$ (when $n$ is even) respectively. These are both outer so we obtain isomorphic algebras ${\mathcal L}(\mathfrak{sl}_{2n},\sigma)$ and ${\mathcal L}(\mathfrak{sl}_{2n},\tau)$. Its Dynkin diagram is obtained by glueing together the type $D$ and type $C$ diagrams (they share $n-1$ vertices). The canonical central element belongs to the span of the simple co-roots in this diagram - I can't see how you would relate it to the span of the co-roots in the untwisted affine type $A$ diagram. | |
| May 24, 2017 at 21:56 | comment | added | Paul Levy | I'm not familiar with the way you are writing things here - in the formulation in Kac's book (a good place to start), you pick a periodic automorphism of the finite-dimensional Lie algebra and associate a Kac-Moody Lie algebra to that - for any outer automorphism of $\mathfrak{sl}_n$ ($n>2$) we obtain the same twisted affine type Kac-Moody algebra, which won't have much to do with $\mathfrak{sl}_n({\mathbb C}((t)))$. (I don't know what you mean by $\widehat{\mathfrak{sl}_n}$.) I can only assume that by $\sigma$ and $\tau$ you want to take involutions of $\mathfrak{sl}_n$ with fixed points... | |
| May 24, 2017 at 19:08 | comment | added | Z.A.Z.Z | yes.....correct | |
| May 24, 2017 at 19:07 | comment | added | Paul Levy | I assume you mean $c/2$ and not $1/2c$ in the latter case then... | |
| May 24, 2017 at 19:06 | comment | added | Z.A.Z.Z | If $c$ is the central element of $\mathfrak g$, then the canonical central element for $\mathcal L(\sigma)$ is again $c$ and for $\mathcal L(\tau)$ is $c/2$. I expect this from diffrent geometric approch. | |
| May 24, 2017 at 18:35 | comment | added | Paul Levy | I don't understand your final sentence (before the "Thanks"). Can you state a bit more precisely what you mean here? | |
| May 24, 2017 at 14:23 | history | edited | Z.A.Z.Z | CC BY-SA 3.0 |
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| May 24, 2017 at 13:50 | history | edited | Z.A.Z.Z | CC BY-SA 3.0 |
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| May 24, 2017 at 13:44 | history | edited | Z.A.Z.Z | CC BY-SA 3.0 |
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| May 24, 2017 at 13:33 | history | asked | Z.A.Z.Z | CC BY-SA 3.0 |