Can anyone tell me whether the weight diagram associated with a given semi-simple Lie algebra is unique to that algebra? I feel morally certain that it is but I just can't seem to get it out. Any pointers gratefully received.
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$\begingroup$ I googled for "weight diagram" and (after a first result which is a link to your question (it is amazing how fast Google indexes MO!) comes a link math.oregonstate.edu/~tevian/JOMA/joma_paper_softlinks.pdf to a paper by Wangberg and Dray which, if I understand correctly, says the answer to your question is yes. $\endgroup$– Mariano Suárez-ÁlvarezCommented Apr 7, 2011 at 18:04
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1$\begingroup$ Are you referring to the weight diagram of a given representation, or to the entire weight lattice of the algebra? $\endgroup$– ARupinskiCommented Apr 7, 2011 at 19:01
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$\begingroup$ Thanks! To ARupinski, what I'm asking is whether, given a weight diagram corresponding to an arbitrary representation of an algebra A, that weight diagram is a weight diagram of a representation of the algebra A only (ie, no other algebra). Does that help? To Mariano - thank you so much for the link! Awesome. Can I ask you what section (to your mind) implies that the answer is yes? (My math is a few orders of magnitude beneath these guys [and doubtless you guys].) $\endgroup$– K McKenzieCommented Apr 7, 2011 at 19:53
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$\begingroup$ Then the answer is no. You cannot differentiate $B_2$ and $C_2$. $\endgroup$– David HillCommented Apr 7, 2011 at 21:35
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1$\begingroup$ @David: $B_2$ and $C_2$ define the same Lie Algebra. Consequently one usually specifies that the $B_n$ series begins at $n=2$ and the $C_n$-series begins with $n=3$ (although occasionally the reverse is also used) $\endgroup$– ARupinskiCommented Apr 7, 2011 at 22:54
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According to the comments in this recent MO question: Can A Simple Lie Algebra Be Determined By Weights of its Representation, the answer to your question is no with simple examples afforded by the defining 2n-dimensional representations of Lie algebras of types $C_n$ and $D_n$.
However, unless I am overlooking some very subtle point, this result is only dependent on the geometry of the weights (i.e. the angles between the different weights within the weight lattice). If instead one looks at the set of weights of a given representation as expressed in terms of the fundamental weights then the set of weights does determine the corresponding algebra (from the set of weights one can reconstruct the Cartan Matrix and hence the algebra).
For example, the defining 10-dimensional representation of $C_5$ (with highest weight $\omega_1$) has the following weights:
$\pm\omega_1,\pm(\omega_2-\omega_1), \pm(\omega_3-\omega_2), \pm(\omega_4-\omega_3)$, and $\pm(\omega_5-\omega_4)$
On the other hand the defining 10-dimensional representation of $D_5$ (with highest weight $\omega_1$) has the following weights:
$\pm\omega_1, \pm(\omega_2-\omega_1), \pm(\omega_3-\omega_2), \pm(\omega_4+\omega_5-\omega_3)$, and $\pm(\omega_5-\omega_4)$
In the above I am using the conventions used by LiE for indexing the fundamental weights. When expressed in terms of the fundamental weights, one sees that the sets of weights do distinguish between $C_5$ and $D_5$ even though geometrically they are indistinguishable.
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$\begingroup$ 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. (...) $\endgroup$ Commented Apr 8, 2011 at 9:22
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$\begingroup$ 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! $\endgroup$ Commented Apr 8, 2011 at 9:28
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$\begingroup$ 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. $\endgroup$ Commented Apr 11, 2011 at 22:38
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$\begingroup$ 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. $\endgroup$ Commented Apr 11, 2011 at 22:45
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$\begingroup$ 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! $\endgroup$ Commented Apr 12, 2011 at 0:10