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Xyx^-1y^-1 a Lie group homomorphism?

  1. May 6, 2012 #1
    Hi! I was just going through this script on Lie groups: http://www.mit.edu/~ssam/repthy.pdf

    At one point the following is said:

    (see attachment)

    I've spent multiple hours trying to figure out why this is a group homomorphism. Sure, once you know the theorem is correct, this follows. But without knowing this, I can't figure out why it should be.

    I want to see that:

    [itex]\varphi_y(x_1x_2) = \varphi_y(x_1)\varphi_y(x_2) \Leftrightarrow x_1x_2yx_2^{-1}x_1^{-1}y^{-1} = x_1yx_1^{-1}y^{-1}x_2yx_2^{-1}y^{-1}[/itex]


    If I write [itex]x_1y = \bar y_1x_1, x_2y = \bar y_2x_2[/itex], then the above equation is equivalent to

    [itex]x_1\bar y_2x_1^{-1} = \bar y_1y^{-1}\bar y_2[/itex]

    but I don't get any further from here.
     

    Attached Files:

  2. jcsd
  3. May 6, 2012 #2
    Anyways, I think whether or not this is a homomorphism is not necessary for this proof. One can just argue that this map is smooth since G is a Lie group, therefore the image under G is connected, therefore it is a one point set and since eye-1y-1 = e one gets that this set must be exactly {e}.
     
  4. May 6, 2012 #3


    I agree with you: I can't either see how that map is a homomorphism in a general case, but it is unnecessary to consider it that way.

    DonAntonio
     
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