1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lie groups and Lie algebras

  1. May 16, 2014 #1
    Why is it that several lie groups can have the same Lie algebra? could it have to do with the space where they act transitively? Could two different Lie groups acting transitively on the same space have the same Lie algebra?
     
  2. jcsd
  3. May 16, 2014 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    For this, one needs to understand the relation between a Lie Algebra and a Lie group. A Lie algebra, as conceptualized by Sophus Lie himself, are those group members "infinitesimally close" to the identity element of the Lie group. Perhaps in more sophisticated/rigorous terms, the Lie Algebra is the tangent space of the identity element, with the Lie bracket being the matrix commutator for matrix Lie groups (which are the largest class of Lie groups, let's not worry about the exceptional groups for now).

    This should tell you that two groups will have the same Lie algebra if their neighborhoods of the identity are the same. In other words, near the identity of two Lie groups which share a Lie algebra (say SO(3) and SU(2)), the two Lie groups look identical. It is the global topology of these groups that differentiate between them. In our example, SU(2), which is isomorphic to the three-sphere, forms a double covering of SO(3) with radially opposite points on the three sphere identified. Also SO(3) is not simple, nor simply connected, where SU(2) is both. Notice though that because SO(3) is SU(2) with antipodal points identified, its neighborhood of the identity (which cannot see the global structure of the group) is the same as SU(2). This has nothing to do with the space upon which they act (which is called a realization, or representation of the group).
     
  4. May 16, 2014 #3
    So, if we in a neighboorhood of a particular point on a 2-sphere, approximate that neighborhood by a plane and perform a small rotation of that point by multiplying by elements near the identity of the group SO(3). That would equal a small linear translation generated by T(3) (translation group in 3 dimensions). And so they will have the same Lie algebra.
     
  5. May 16, 2014 #4

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    No, the translations are commutative. This is a fundamentally different structure than the rotations. As such, the Lie algebra of the translations must be the trivial one ##[T_i,T_j]=0##. In the group lingo, the translation group is Abelian.

    You are still thinking about the space on which the group acts. This is fundamentally different than the group itself.

    For example, the group SU(2) acts fundamentally on the space of complex 2-vectors (vectors with 2 components over the complex field). This is a flat space ##\mathbb{C^2}##. However, SU(2), the group, is a three-sphere ##S^3## which is a curved space!

    What I was talking about in my post was about the group space itself. The groups which have isomorphic Lie algebras will have neighborhoods of the identity which are equal to each other.
     
  6. May 16, 2014 #5

    Matterwave

    User Avatar
    Science Advisor
    Gold Member


    Sorry I made one erroneous statement in the above. SO(3) is certainly a simple group, but it's true that it's not simply connected like SU(2), and SU(2) is technically not a simple group (its Lie Algebra, however, is simple).
     
  7. May 17, 2014 #6
    All Bianchi universes are spatially homogeneous cosmological models. But that doesn't mean that the 3-sphere is translation invariant. It just means that the group acts transitively on the 3-sphere? But what does it mean when SO(3) is the isometry group acting transitively on the 3-sphere? What does that imply about the geometry of spacetime?
     
    Last edited: May 17, 2014
  8. May 17, 2014 #7

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    I'm sorry, I'm not familiar with the term "Bianchi universe". So I'm not sure where you're going with this. For a spatially homogeneous universe, the group of symmetries should be the translations. For the group of rotations SO(3) to be the group of symmetries as well, you need isotropy in your universe.

    I'm also not sure where the 3-sphere is coming into play in your question. My mention of the 3-sphere earlier was as the manifold diffeomorphic to SU(2). The group SU(2) itself is the 3-sphere. We weren't talking about universes earlier.

    SO(3), being the group of 3x3 orthogonal matrices of determinant 1, most naturally acts on the 3-space ##\mathbb{R}^3## and not on the 3-sphere ##S^3##. If you want to ask about its action on the 3-sphere, you first have to define the action. Perhaps you've gone beyond my ability to answer your questions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lie groups and Lie algebras
Loading...