Idempotent of the complex group ring
Idempotent primitivity criterion
An idempotent
Proof
Let
be the minimal left ideΓ€l generated by primitive idempotent πΏ π πΌ . Since for all π π πΌ π , π β β [ πΊ ] P β [ πΊ ] ( π π πΌ β π β π π πΌ ) Ξ β [ πΊ ] ( π ) = Ξ β [ πΊ ] ( π ) P β [ πΊ ] ( π π πΌ β π β π π πΌ ) and
transforms in an irrep πΏ π πΌ , in particular Ξ π P β [ πΊ ] ( π π πΌ β π β π π πΌ ) Ξ π πΌ ( π ) = Ξ π πΌ ( π ) P β [ πΊ ] ( π π πΌ β π β π π πΌ ) for all
and thus by Schurβs lemma π β πΊ is P β [ πΊ ] ( π π πΌ β π β π π πΌ ) in π π π and zero everywhere else, i.e. πΏ π πΌ . π π πΌ β π β π π πΌ = π π π π πΌ For the converse, assume
is non-primitive and for every π π there exists a scalar π β β [ πΊ ] such that π π β β . From non-primitivity π π β π β π π = π π π π for nonzero idempotents with π π = π 1 + π 2 . Then on the one hand π 1 β π 2 = 0 = π 2 β π 1 π π β π 1 β π π = ( π 1 + π 2 ) β π 1 β ( π 1 + π 2 ) = π 1 but on the other hand
π π β π 1 β π π = π π π so
. But this is a contradiction, since it implies π π π = π 1 π 1 β π 1 = π 2 π π β π π = π 2 π π = π 1 = π π π and thus
and π = π 2 . Hence the assumption is false.1 π β 0
Footnotes
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2023, Groups and representations, p. 58 β©