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Homework Help: 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


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    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


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    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
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