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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 $\displaystyle \phi $ is an orthogonal automorphism of E then

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

    $\displaystyle \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 $\displaystyle H_{\phi \cdot \alpha} $ but I assume

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

    And I assume that

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

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


    Last edited by Bernhard; Mar 30th 2012 at 11:56 PM.
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  2. #2
    MHF Contributor

    Mar 2011

    Re: Reflections - Kane's Book

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

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

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

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

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

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

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

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

    which shows that $\displaystyle H_{\phi(\alpha)} \subseteq \phi(H_\alpha)$, 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 $\displaystyle E$.

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

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

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

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

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

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