Tuesday, July 5, 2016

linear algebra - If Brauer characters are barmathbbQ-linearly independent, why are they mathbbC-linearly independent?



If Brauer characters are ˉQ-linearly independent, why are they C-linearly independent?



I think this is a linear algebra fact showing up when proving the irreducible Brauer characters on a finite group are linearly independent over C. The proof I've seen observes that the characters take values in the ring of algebraic integers, and then proves linear independence over ˉQ.




Why is it sufficient to only check linear independence over ˉQ? It seems like something could go wrong when extending the field all the way up to C.



The proof I'm reading is Theorem 15.5 in Isaacs' Character Theory of Finite Groups.



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Answer



If E/F is a field extension, we have FnEn, and if a subset of Fn is F-linearly independent, then it is also E-linearly independent. A nice, super easy way to see it: extend the subset to a basis for Fn. Form the matrix whose columns are elements of this basis. Its determinant is nonzero. But this shows that the columns form a basis for En since the determinant has the same formula regardless of the field you work over.



The space of class functions of a finite group can be identified with Fn in an obvious way (n= number of conjugacy classes).


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