Fully reducible rep-invariant subspace

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The discussion centers on proving that a representation D is fully reducible if and only if every invariant subspace V₁ also has its orthogonal complement V₁ᵀ as invariant. The key argument involves demonstrating that for vectors v in V₁ and w in V₁ᵀ, the inner product equals zero, indicating orthogonality. The proof requires the unitarity of the representation, particularly in the context of the transformation D(g) and its inverse D(g⁻¹).

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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
(Lemma 1.6.4)
in fact I don't understand the inaffected scalar product it mentions...
 
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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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