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Thread: Reflections - Kane's Book

  1. #1
    Super Member Bernhard's Avatar
    Jan 2010
    Hobart, Tasmania, Australia

    Reflections - Kane's Book

    I am reading Kane - Reflection Groups and Invariant Theory and need help with property A-4 of the properties of reflections stated on page 7

    (see attachment - Kane _ Reflection Groups and Invariant Theory - pages 6-7)

    On page 6 Kane mentions he is working in dimensional Euclidean space ie where has the usual inner product (x,y).

    In defining reflections with respect to vectors Kane writes:

    " Given let be the hyperplane

    We then define the reflection by the rules



    Then Kane states that the following property follows:

    (A-4) If  \phi is an orthogonal automorphism of E then

     \phi \cdot H_{\alpha} = H_{\phi \cdot \alpha}

     \phi s_{\alpha} {\phi}^{-1} = s_{\phi \cdot \alpha

    Can someone explicitly and formally demonstrate A-4?

    I am not really sure of the meaning of  H_{\phi \cdot \alpha}  but I assume

     H_{\phi \cdot \alpha} = \{ x  \ | \ (x, {\phi \cdot \alpha ) = 0   \}

    And I assume that

     \phi \cdot H_{\alpha} = \{ \phi \cdot x \ | \ (x, \alpha \} = 0 \}

    Can anyone help me formulate an explicit and detailed proof of A-4


    Last edited by Bernhard; Mar 31st 2012 at 12:56 AM.
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  2. #2
    MHF Contributor

    Mar 2011

    Re: Reflections - Kane's Book

    first of all, realize that \phi(H_\alpha) = H_{\phi(\alpha)} is an equality of two sets.

    so suppose y \in \phi(H_\alpha). then  y = \phi(x) for some x \in H_\alpha.

    so (x,\alpha) = 0 and because \phi is orthogonal, (\phi(x),\phi(\alpha)) = 0, that is (y,\phi(\alpha)) = 0

    so y \in H_{\phi(\alpha)}, so \phi(H_\alpha) \subseteq H_{\phi(\alpha)}.

    now suppose we have z \in H_{\phi(\alpha)}, so that (z,\phi(\alpha)) = 0.

    since \phi is an orthogonal automorphism, it is bijective, and its inverse \phi^{-1} is also orthogonal.

    so (\phi^{-1}(z),\phi^{-1}(\phi(\alpha)) = (z,\phi(\alpha)) = 0. but (\phi^{-1}(z),\phi^{-1}(\phi(\alpha)) = (\phi^{-1}(z),\alpha).

    this means that \phi^{-1}(z) \in H_\alpha so z = \phi(\phi^{-1}(z)) \in \phi(H_\alpha),

    which shows that H_{\phi(\alpha)} \subseteq \phi(H_\alpha), thus the two sets are equal.

    the second equality, is an equality of functions, so we must show that the two functions have the same value at every point of E.

    so let v \in E. since \phi is surjective, we can write v = \phi(w) for some w = x + k\alpha,

    where x \in H_\alpha ,k \in \mathbb{R}. so then we have:

    \phi s_\alpha \phi^{-1}(v) = \phi s_\alpha \phi^{-1}(\phi(w)) = \phi s_\alpha(w) = \phi s_\alpha(x + k\alpha) = \phi(s_\alpha(x) + ks_\alpha(\alpha))

    = \phi(x - k\alpha) = \phi(x) - k\phi(\alpha) = s_{\phi(\alpha)}(\phi(x) + k\phi(\alpha)) = s_{\phi(\alpha)}(\phi(x+k\alpha))

    = s_{\phi(\alpha)}(\phi(w)) = s_{\phi(\alpha)}(v) for all v \in E.

    this shows the two functions are indeed equal on all of E.
    Thanks from Bernhard
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