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Finite Reflection Groups in Two Dimensions - R2

  1. Apr 22, 2012 #1
    I am seeking to understand reflection groups and am reading Grove and Benson: Finite Reflection Groups

    On page 6 (see attachment - pages 5 -6 Grove and Benson) we find the following statement:

    "It is easy to verify (Exercise 2.1) that the vector [itex] x_1 = (cos \ \theta /2, sin \ \theta /2 ) [/itex] is an eigenvector having eigenvalue 1 for T, so that the line
    [itex] L = \{ \lambda x_1 : \lambda \in \mathbb{R} \} [/itex] is left pointwise fixed by T."

    I am struggling to se why it follows that L above is left pointwise fixed by T (whatever that means exactly).

    Can someone please help - I am hoping to be able to formally and explicitly justify the statement.

    The preamble to the above statement is given in the attachment, including the definition of T

    Notes (see attachment)

    1. T belongs to the group of all orthogonal transformations, [itex] O ( \mathbb{R} ) [/itex].

    2. Det T = -1

    For other details see attachment

    Peter
     

    Attached Files:

    Last edited: Apr 22, 2012
  2. jcsd
  3. Apr 22, 2012 #2


    "[itex]\,x_1[/itex] is an eigenvector of [itex]\,T[/itex] with eigenvalue [itex]\,1\,\,[/itex]" means [itex]\,Tx_1=x_1[/itex]

    To leave a subspace L fixed pointwise means [itex]\,\,Tv=v\,,\,\,\forall v\in L[/itex]

    Can you now apply the above to see the statement in that book is trivial?

    DonAntonio
     
  4. Apr 22, 2012 #3
    Thanks for the help - thinking that over

    What is meant by (or what is the significance of) 'left' pointwise fixed

    Peter
     
  5. Apr 22, 2012 #4


    In this case "left" is the past participle of "leave", and NOT the left of "side", as opposite to "right"...:P)

    DonAntonio
     
  6. Apr 22, 2012 #5
    Oh ... right ... completely misunderstood the text :-(

    Thanks

    Peter
     
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