MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a simple lie algebra (over ${\mathbb C}$ or ${\mathbb R}$). What is the number of operators such that their eigenvalues sufficiently label all state vectors in the algebra's representation space.

According to this paper (page 24), if $d$ is the dimension of the algebra and $n$ is its rank then the number is $(d+n)/2$.

I am not familiar with this result; a reference or explanation would be appreciated.

(adding additional background information from the paper above; this was too large to include in a comment) :

Although the setting in the paper is the lie group $SO(4,2) \otimes SU(2)$ I moved it to a lie algebra setting : $so(6) \oplus su(2)$ or $A_3 \oplus A_1$ for simplicity.

What the author is saying is that it takes (15+3)/2=9 operators to label an $A_3$ state and (3+1)/2=1 operators to label an $A_1$ state. A state here would be any element in the basis of a finite dimensional irrep. For $A_3$ the 9 operators would include 3 casimir elements and 3 cartan generators. Quoting the author : "According to a (not very well-known) result popularised by Racah, we need to find $(d-3n)/2=3$ additional operators in order to complete the set of the six preceding operators". This is what motivated the question. So it seems that these 9 operators would correspond to 9 commuting elements of the universal eneveloping algebra of $A_3$; I would expect that they are linearly independant.

share|cite|improve this question
This question is really unclear. What do you mean by representation space (a space of representations, or a space which is a representation)? I'm also not sure what "the number of operators such their eigenvalues sufficiently label all state vectors" is supposed to mean. If you have a collection of operators whose joint spectrum on a finite dimensional space is simple (my best guess for what "sufficiently label all state vectors" means), then any generic linear combination of them also has simple spectrum. – Ben Webster Feb 20 '14 at 9:24
Though I think I understand what "repreentation space" means here, Ben's comments point to a standard problem of interpreting in mathematical language what goes on in mathematical physics. Most of the work here is probably involved in translating the problem carefully into mathematics, where much is known about finite dimensional representations. (However, the real Lie groups add an extra layer of complication.) – Jim Humphreys Feb 20 '14 at 14:01
I was following the authors terminology; which I agree is not precise. I'm overlooking this imprecision with the assumption that there's something useful here to learn. I added some additional background from the paper above; hopefully that's of value. – Y Macdisi Feb 20 '14 at 20:01
I found a reference...thanks to all who responded for their useful comments and valuable insights – Y Macdisi Feb 23 '14 at 18:52

After much looking I found the reference I was looking for : Here it is for those interested The terminilogy in the original question isn't far off what seems to be common lingo (maybe not around here).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.