Timeline for Weight diagrams and semi-simple Lie algebras
Current License: CC BY-SA 2.5
9 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 12, 2011 at 0:10 | comment | added | K McKenzie | Thanks very much - you reassure me that at least the problem has hope of being solvable! From the dimension of the weight diagram one can at least immediately infer what the dimension of the (semi-simple) algebra is, and there are only finitely many of each - so one could at least go and check. But of course the question is whether it's true! (p4 of the reference in the first comment above suggests that it is, but it doesn't go into it further.) I should have also specified that the representations for the weight diagrams I'm concerned with are indeed irreducible. I'll have a think! | |
| Apr 11, 2011 at 22:45 | comment | added | ARupinski | Essentially this finiteness result arises from looking at the possible weight lattices and checking whether the given configuration of weights can occur in each possible lattice of the correct dimension; there are only finitely many lattices in each dimension. For those lattices which contain your configuration, you can then look at which combinations of irreps of the corresponding algebras add up to the required configuration. Since there are only finitely many irreps with at most a given number of weights, one therefore only has finitely many possibilities. | |
| Apr 11, 2011 at 22:38 | comment | added | ARupinski | I guess from the viewpoint of only knowing the weight diagram without the actual weights your problem is at the very least difficult, unless perhaps you know that the representations are irreducible (although the $C_n$/$D_n$ example shows even this is not enough). On the other hand, knowing a weight diagram (including the multiplicities of the various weights) does allow you to narrow the possible algebras and corresponding decomposition into irreps into a finite number of possibilities from which you might be able to guess the correct algebra. | |
| Apr 8, 2011 at 9:28 | comment | added | K McKenzie | This is because I'm trying to figure out an issue in particle physics, where we only have direct access to the particles themselves - ie, the weights of some representation of the algebra that governs their interaction, and not necessarily the fundamental rep. So may I ask: is there a general procedure in which one can deduce from a given weight diagram the corresponding algebra, without already knowing the fundamental weights? (Or equivalently, a procedure by which one can deduce the set of fundamental weights associated with that weight diagram?) I've failed so far, but hope is strong! | |
| Apr 8, 2011 at 9:22 | comment | added | K McKenzie | Thanks for this (yes, as far as I was aware B2 and C2 were isomorphic, so I was a bit confused by that. For the record, all I'm interested in for now is whether a weight diagram determines an algebra up to isomorphism.) So to Rupinski: I know how to determine the Cartan matrix given the full set of fundamental weights. But what I'm interested in is whether one can determine the algebra from the weight diagram of an arbitrary representation without prior knowledge of what the fundamental weights are. (...) | |
| Apr 8, 2011 at 9:18 | vote | accept | K McKenzie | ||
| Apr 7, 2011 at 23:43 | history | edited | ARupinski | CC BY-SA 2.5 |
edited body
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| Apr 7, 2011 at 22:50 | history | answered | ARupinski | CC BY-SA 2.5 |