Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. 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

The question is regarding SU(2) group and SU(2) algebra. The SU(2) group can be generated by exponentiating the generators of SU(2) algebra $X_a$ as $exp(i t_a X_a )$ with $t_a$ being three parameters. Genreally we use half the Pauli matrices as $X_a$ when discussing SU(2) algebra as well as the group. But if we use the spin-1 representation of $X_a$ (which can surely be done at the Lie Alebra level) can they be exponentiated to give another rep. of the SU(2) group (which would be 3 dimensional)?

share|cite|improve this question
By the way, the group you obtain in this case is isomorphic to SO(3). If you choose infinitesimal rotations as the spin-1 representation, you obtain the SO(3) matrices directly. – The Vee Sep 20 '12 at 19:42

Yes. A representation of a Lie algebra always exponentiates to a representation of the simply connected Lie group corresponding to it. The representations of each are in bijective correspondence this way.

I recommend the free online book by Kirillov ( ) as a reference for this business.

share|cite|improve this answer
It exponentiates if the Lie algebra is finite-dimensional (as in case of SU(2)), but not necessarily in infinite dimensions. – Arnold Neumaier Sep 5 '12 at 8:40
I second a vote for Kirillov's Lie Group notes! Download them and keep a copy lest they should disappear. The university sector's shrinking faster than a pierced balloon in my home land and, although Kirillov's present home land is not mine, it is also infected with the English language bacterium which seems to be an effective economic "rationalist" virus's vector organism! Also see my answer on how several different exponentiations can happen in finite dimensions, and I also recommend Chapter 8 of John Stillwell's "Naive Lie Theory". – WetSavannaAnimal aka Rod Vance Sep 27 '13 at 1:49

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.