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Apr 16, 2019 at 16:20 vote accept Nadia SUSY
Apr 16, 2019 at 15:48 answer added spin timeline score: 11
Apr 16, 2019 at 15:17 comment added Victor Protsak In order to get a faithful representation of the spinor group, you must use a spinor representation somehow. So contrary to the impression you got from that comment, the standard representation is not atypical when viewed among all simple modules. What it is true that you can build all non-spinor fundamental weights by tensor products and decomposing.
Apr 16, 2019 at 15:08 answer added Victor Protsak timeline score: 8
Apr 16, 2019 at 14:08 history edited Victor Protsak CC BY-SA 4.0
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Apr 16, 2019 at 13:19 comment added Nadia SUSY @Venkataramana: What does this look like dually on the Lie algebra side?
Apr 16, 2019 at 13:17 comment added Venkataramana these are exactly the irreducible representations with highest weight (which is a character on a maximal torus $T$ of the simply connected compact group $G$), which is trivial restricted to the centre of G .
Apr 16, 2019 at 12:47 history edited Nadia SUSY CC BY-SA 4.0
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Apr 16, 2019 at 12:36 history edited Nadia SUSY CC BY-SA 4.0
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Apr 16, 2019 at 12:19 comment added Christoph Mark AFAIK, all finite dimensional irreducible representations of $\mathfrak{g}$ are integrable. So, no need to worry about the difference between $\mathfrak{g}$ and $G$. AFAIK, for a integral positive weigth $\omega$, it will not give rise to a faithful representation if $(\omega,\beta^\vee)=0$ for a simple root $\beta$ (which seems equivalent to me).
Apr 16, 2019 at 11:40 review Close votes
Apr 16, 2019 at 15:00
Apr 16, 2019 at 11:36 history edited Nadia SUSY CC BY-SA 4.0
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Apr 16, 2019 at 11:36 history undeleted Nadia SUSY
Apr 15, 2019 at 17:08 history deleted Nadia SUSY via Vote
Apr 15, 2019 at 16:30 history asked Nadia SUSY CC BY-SA 4.0