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A complex character is irreducible if and only if its self-inner-product is
Statement
Let be a finite group and let be a complex character of . Then is irreducible if and only if .
Facts & Assumptions
Given: A finite group and a complex character of a finite-dimensional representation , completely reduced as .
The multiplicity of in is (The multiplicity of an irreducible summand is a character inner product).
Characters add on direct sums, so the decomposition gives (Characters add on direct sums, multiply on tensor products, and conjugate on duals).
If and , then conjugate-symmetry and linearity in the first argument give , because the multiplicities are nonnegative integers.
A representation is irreducible exactly when it has one irreducible summand with multiplicity and no others.
Proof
Completely reduce with integers. By [F2] this decomposition gives , and by [F1] each ; therefore [A1] gives .
Assume is irreducible. Then by [A2] exactly one multiplicity equals and the rest are , so the sum in step 1.1 is ; hence .
Conversely, assume . The sum of nonnegative integers in step 1.1 can equal only when exactly one equals and all the others are . By [A2], is irreducible, so is irreducible.
Steps 2.1 and 2.2 prove the two implications, hence the biconditional.
Depends on
Used by
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Sources
- Peter Webb, A Course in Finite Group Representation Theory, Corollary 3.3.4 (standard reference, not scraped)
- Pavel Etingof et al., Introduction to Representation Theory, Section 3.5 (standard reference, not scraped)