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 Algebras: [H1, H2] = 0 => h1h2 = h2h1

  1. May 4, 2012 #1
    Hi! I have the following problem:

    I tried to solve this by using the adjoint represantation and although it looks very promising, I can't really get it to work.

    If I define [itex]C_x(y) := xyx^{-1}[/itex] as the conjugation, then what needs to be shown is: [itex]C_{h_1}h_2 = h_2[/itex].

    I'm using these symbols: [itex]Ad: G \rightarrow GL(\mathfrak g), Ad(g) := C_{g*e}[/itex] and [itex]ad_X Y = [X, Y] [/itex].

    My idea is as follows: [itex][\mathfrak h_1, \mathfrak h_2] = 0[/itex] implies [itex]ad_{h_1}ad_{h_2}=ad_{h_2}ad_{h_1} \Rightarrow Ad_{*e}(h_1)Ad_{*e}(h_2) = Ad_{*e}(h_2)Ad_{*e}(h_1) \Rightarrow (Ad(h_1)Ad(h_2))_{*e} = (Ad(h_2)Ad(h_1))_{*e}[/itex]

    [itex]\Rightarrow (Ad(h_1h_2))_{*e} = (Ad(h_2h_1))_{*e}[/itex]. Since [itex]G[/itex] is connected this implies [itex]Ad(h_1h_2) = Ad(h_2h_1) \Rightarrow C_{h_1h_2*e} = C_{h_2h_1*e} \Rightarrow C_{h_1h_2}= C_{h_2h_1} \Leftrightarrow C_{h_1} = C_{h_2h_1h_2^{-1}}[/itex] but I'm not getting any further from here.

    Maybe this is the wrong path altogether but I don't see how else I could use the connectedness of G as well as the information on the Lie bracket to make a statement on the group elements themselves.
     
  2. jcsd
  3. May 4, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    How do you define the Lie subgroup associated the Lie subalgebra h1? I'm going to guess that you want to use the exponential map here.
     
  4. May 5, 2012 #3
    "If [itex]\mathfrak h \subset \mathfrak g[/itex] is a Lie subalgebra, there exists a unique connected Lie subgroup [itex]H \subset G[/itex] with Lie algebra [itex]\mathfrak h[/itex]."

    I'm sure this unique connected subgroup is meant in the assignment.
     
  5. May 5, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I'm sure it is too. Look at how they proved that statement. I'm guessing you can use exp(A)exp(B)=exp(B)exp(A) if [A,B]=0.
     
  6. May 5, 2012 #5
    Hi! Thank you for your response.

    The proof for this theorem uses distributions and leafs. But it is also shown that exp(A)exp(B) = exp(B)exp(A) if [A, B] = 0 at a different point in our script. But are you sure this fact can be used here? Is the exponential map necessarily surjective on a connected Lie group?

    I'm a little confused. I tried to look this up and while I found two sources stating

    "If G is a Lie group and G0 is the connected component of 1, then G0 is generated by exp(g)."

    which would apply to a connected Lie group (right?), I found other sources stating:

    "Not even on connected Lie groups is the exponential map necessarily surjective. Example: SL(2,IR)."

    And if it's not surjective, how I could I use your idea if [itex]h_1[/itex] or [itex]h_2[/itex] are not included in the image of exp?

    edit: Ok, I think I get it now. The first statement is just about a generator of the Lie group. I'll take a closer look at it now.

    edit 2: Thank you, I think this works!
     
    Last edited: May 5, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lie Algebras: [H1, H2] = 0 => h1h2 = h2h1
  1. Lie Algebras (Replies: 45)

  2. Derived lie algebra (Replies: 0)

  3. Help with Lie Algebra (Replies: 2)

Loading...