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Fully reducible rep-invariant subspace

  1. Feb 2, 2014 #1

    ChrisVer

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    How could I show that a representation [itex]D[/itex] is fully reducible if and only if for every invariant subspace [itex]V_{1} \in V[/itex] then also [itex]V_{1}^{T}[/itex] (meaning orthogonal to V1) is also invariant?
    http://www.crystallography.fr/mathcryst/pdf/nancy2010/Souvignier_irrep_syllabus.pdf [Broken]
    (Lemma 1.6.4)
    in fact I don't understand the inaffected scalar product it mentions...
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 3, 2014 #2

    ChrisVer

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    It's OK I think I came into the answer....
    if we take [itex]v \in V_{1}[/itex] and [itex]w \in V_{1}^{T}[/itex]
    then [itex] <v,w>=0[/itex]
    but v doesn't change subspace under the action of the repr
    [itex] <D(g)v, w>=0= <v,D^{t}(g) w>=<v,D(g^{-1}) w>=<v,D(g) w>[/itex]
    However this needs the unitarity for the [itex]g^{-1}[/itex] step
     
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