MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

I wonder if anyone can help me with this question regarding algebras and multiplets. In a nice review paper (McVoy, Rev Mod Phys 37(1)) the author states the following theorem: Given any set of operators which satisfy the [Lie algebra commutation relations], there exists a Lie group which has these operators as its generators. Its multiplets are uniquely determined by the structure constants.'' He then goes on to say that it follows thatgiven the generator commutators, and nothing else, we can directly work out the multiplets of the group and all their properties, with no further assistance.''

But I'm confused about this, for the following reason. As a general rule, there are a number of different groups corresponding to the same Lie algebra (which are identical locally but which may differ globally). SU(3) and SU(3)/Z3 is a case in point. But (as anyone who's familiar with the history of the Eightfold Way and subsequent quark model will be aware) the triplet irrep is an irrep of SU(3), but not of SU(3)/Z3. (The lowest-dimensional non-trivial representation of SU(3)/Z3 is the 8.) So how can the algebra determine the multiplets of a group corresponding to it, if different groups with the same algebra will in general have different multiplets?

Any help much appreciated!

share|cite|improve this question
I think that the author refers to the Lie group generated in the same operator algebra, by the 1-parameter groups associated with the given operators (i.e. the $\exp(tH)$, where $t\in\mathbb{R}$ and $H$ is one of the given operators). – Alain Valette Aug 31 '11 at 15:17
Also posted at – Simon Sep 4 '11 at 9:00
@k mc , it's just physicists usually doesn't distinguish Lie algebras and Lie groups. – Yuji Tachikawa Sep 10 '11 at 14:47

The question is not really a question of research in mathematics. The answer is that McVoy is correct that there is a Lie group with any given finite dimensional algebra. But, as pointed out in asking the question, there are actually several Lie groups with any given Lie algebra. In fact, if we allow disconnected Lie groups, then there are uncountably many. In gauge theory, the physicists (almost always) need compact Lie groups, and usually assume that the Lie groups are connected. Up to isomorphism, there is a unique connected and simply connected real Lie group with any finite dimensional real Lie algebra. However, that Lie group may be the universal covering space of numerous other Lie groups. Assuming that the Lie group $G$ is compact and connected and simply connected, its center $Z(G)$ is a discrete abelian subgroup (for example, the center of $SU(3)$ is $Z_3$), and every Lie group with the same Lie algbra has the form $G/\Gamma$ where $\Gamma$ can be any subgroup of $Z(G)$. See any textbook on Lie groups (for example, Onishchik and Vinberg, or Brocker and tom Dieck or Fulton and Harris) for the full story. McVoy is almost certainly taking the connected and simply connected Lie group with the given Lie algebra.

share|cite|improve this answer
Everything here is spot on. To supplement: any connected Lie group is generated by its Lie algebra, in the appropriate sense. The simply-connected connected Lie group for a given Lie algebra is special in many ways. One of them is that any (finite-dimensional!) representation (multiplet) of the Lie algebra does extend to a representation of the group. Note that this fact alone does not distinguish the simply-connected group. The standard example is SL(2,R), which has the homotopy type of a circle, but acts on all finite-dimensional sl(2,R) representations. – Theo Johnson-Freyd Sep 11 '11 at 23:07
Suppose that $G$ is a compact connected and simply connected Lie group. Then any other connected Lie group with the same Lie algebra is a quotient $G/\Gamma$. If every Lie algebra representation of $G$ descends to a representation of $G/\Gamma$, then $\Gamma=\{1\}$. This is a consequence of the splitting of the square integrable functions on $G$ into a sum of irreducible finite dimensional representations. – Ben McKay Sep 13 '11 at 9:51

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.