Timeline for Does a finite-dimensional Lie algebra always exponentiate into a universal covering group
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9 events
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| Oct 12, 2010 at 4:49 | comment | added | Victor Galitski | Ooops, somehow I erased the comment to which you just responded. Thanks. I think I clarified for myself the key issue. | |
| Oct 12, 2010 at 4:40 | comment | added | Theo Johnson-Freyd | However, in general the exponential map is neither injective nor surjective. It is surjective if the group is compact, which seems to be the case you care about. But then it definitely cannot also be injective: the Lie algebra, begin a real vector space, is definitely not compact as a manifold. (In very special cases --- in particular, for connected simply-connected nilpotent groups --- the exponential map is a global diffeomorphism.) | |
| Oct 12, 2010 at 4:37 | comment | added | Theo Johnson-Freyd | Yes. Lie's Third Theorem says that every Lie algebra over R (i.e. real vector space with antisymmetric bracket satisfying Jacobi) exponentiates to some Lie group (i.e. smooth manifold with smooth group multiplication). Here by "exponentiates" I simply mean that the Lie algebra is naturally isomorphic to, for example, the algebra of left-invariant vector fields on the group. (Equivalently, the algebra is the tangent space to the identity of the group, with the bracket given as the one-jet of conjugation.) Moreover, there is an "exponential map" from the algebra to the group. | |
| Oct 10, 2010 at 20:26 | comment | added | Theo Johnson-Freyd | @Dick Palais: I took "Lie algebra" to mean "real vector space with anti-symmetric Jacobi-satisfying bracket". Then it is a nontrivial theorem that such a thing can always be exponentiated to an "abstract" group, and even more nontrivial that it can be faithfully represented as a matrix Lie algebra. It's reasonable to say that this is not what OP wanted. In particular, not every real (in the mathematicians' sense) Lie group has a faithful representation; covers of SL(2,R) being the classic examples. So the maximal matrix group integrating sl(2,R) is not the same as the simply-connected cover. | |
| Oct 10, 2010 at 20:22 | comment | added | Theo Johnson-Freyd | It's likely that I misinterpreted part of the question: mathematicians do consider sl(2,R) to be a "real Lie group" and its defining representation to be a "real representation". I agree that physicists are generally interested only in unitary representations, and sl(2,R) has no nontrivial finite-dimensional unitary representations, whereas su(2) does. | |
| Oct 10, 2010 at 5:26 | comment | added | Victor Galitski | Dr. Palais and Theo: I'd like to stress that I am not only interested to know what exp(i A) is NOT, but rather to figure out what it is? Also, yes, I do assume the existence of a faithful matrix repres. of A, but my main question was independent of it. In question 4 I tried to ask the following: Let T[A] be a faithful representation of A of minimal dimension [e.g., d=2 for su(2)]. Form all possible matrix exponents exp(i B^k T[X_k]) and their products. Ok, if it's not always a universal cover, then what is it and what are the conditions for A to map globally on G[A]? Thanks for your time. | |
| Oct 10, 2010 at 4:56 | comment | added | Victor Galitski | Hi Theo -- Thanks. I looked into your notes (very impressive!) and also noticed your paper on evolution operator in QM, which is in fact related to my question. Your example of sl(2,R) however doesn't apply to my specific question, because, I am interested in a REAL form arising from the algebra [sl(2,R) can't possibly describe a physical Hamiltonian, but, e.g., su(2) can]. I am not sure how to formulate it in math terms precisely, but what ACTUALLY happens in physics: structure constants, in [X_k,X_p] = i f_{kp}^q X_q, are all pure imaginary. Is exp(iA) guaranteed to be a group then? -Victor | |
| Oct 10, 2010 at 4:42 | comment | added | Dick Palais | Theo, I agree that everything you say is correct if you interpret a Lie algebra to mean the Lie algebra of some abstract Lie group (e.g., the left invariant vector fields) and exponential to mean the 1-parameter subgroup generated by---however I don't think it really addresses the OP's question. At least as I understand his question, he is asking about matrix Lie algebras and when he talks about the exponential of a matrix I believe he means the usual exponential of a matrix. Victor, please clarify this point. | |
| Oct 10, 2010 at 4:19 | history | answered | Theo Johnson-Freyd | CC BY-SA 2.5 |