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Euclidean Reflection Groups _ Kane's text

  1. Mar 28, 2012 #1
    I am reading Kane - Reflection Groups and Invariant Theory and need help with two 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 [itex] \ell [/itex] dimensional Euclidean space ie [itex] E = R^{\ell}[/itex] where [itex] R^{\ell}[/itex] has the usual inner product (x,y).

    In defining reflections with respect to vectors Kane writes:

    " Given [itex] 0 \ne \alpha \in E [/itex] let [itex] H_{\alpha}\subset E [/itex] be the hyperplane

    [itex] H_{\alpha} = \{ x | (x, \alpha ) = 0 \} [/itex]

    We then define the reflection [itex] s_{\alpha} : E \longrightarrow E [/itex] by the rules

    [itex] s_{\alpha} \cdot x = x [/itex] if [itex] x \in H_{\alpha} [/itex]

    [itex] s_{\alpha} \cdot \alpha = - \alpha [/itex] "

    Then Kane states that the following two properties follow:

    (1) [itex] s_{\alpha} \cdot x = x - [2 ( x, \alpha) / (\alpha, \alpha)] \alpha [/itex] for all [itex] x \in E [/itex]

    (2) [itex] s_{\alpha} [/itex] is orthogonal, ie [itex] ( s_{\alpha} \cdot x , s_{\alpha} \cdot y ) = (x,y) [/itex] for all [itex] x, y \in E [/itex]


    I would appreciate help to show (1) and (2) above.

    Peter
     
  2. jcsd
  3. Mar 28, 2012 #2

    morphism

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    Note that ##H_\alpha## is codim 1 in E with perp spanned by ##\alpha##. So any linear operator defined on E is completely determined by its action on ##H_\alpha## and ##\alpha##. In particular, there is a unique linear operator satisfying the defining conditions for ##s_\alpha##. Since the mapping defined by the RHS of (1) is linear and satisfies these conditions as well, it must be ##x \mapsto s_\alpha x##.

    Of course what's going on geometrically is that ##s_\alpha## is simply reflection in the hyperplane ##H_\alpha##. Try this out in ##R^2##, with ##H_\alpha## a line through the origin that is perpendicular to ##\alpha##: what is the formula for reflecting about ##H_\alpha##?

    As for (2), it suffices to check that it holds in the 3 cases: (a) ##x,y \in H_\alpha##; (b) ##x\in H_\alpha## and ##y=\alpha##; and (c) ##x=y=\alpha##. Or you could use (1).
     
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