# Fully reducible rep-invariant subspace

1. Feb 2, 2014

### ChrisVer

How could I show that a representation $D$ is fully reducible if and only if for every invariant subspace $V_{1} \in V$ then also $V_{1}^{T}$ (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. Feb 3, 2014

### ChrisVer

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