most recent 30 from http://mathoverflow.net 2013-05-19T17:39:03Z http://mathoverflow.net/feeds/question/74163 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74163/a-question-on-lie-algebras-lie-groups-and-multiplets k mc 2011-08-31T14:39:37Z 2011-09-10T13:43:31Z

<p>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: <code>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 that</code>given the generator commutators, and nothing else, we can directly work out the multiplets of the group and all their properties, with no further assistance.''</p> <p>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? </p> <p>Any help much appreciated! </p>

http://mathoverflow.net/questions/74163/a-question-on-lie-algebras-lie-groups-and-multiplets/75097#75097 Ben McKay 2011-09-10T13:43:31Z 2011-09-10T13:43:31Z

<p>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.</p>