Is there an example of the following situation : FFF is a field, GGG is a finite group, ρ\rho is an irreducible FF-representation of GG and the character of this representation takes the zero value at every element of GG ? If I’m not wrong, FF cannot be algebraically closed (Robinson, A Course in the Theory of Groups, 8.1.9, p. 220) and must have a nonzero characteristic pp such that the degree of the representation is divisible by pp and such that GG is not a pp-group (Robinson, exerc. 8.1.5, p. 222). But that doesn’t solve the problem. Thanks in advance for the answers.

(Edit. Since I didn’get answers here, I asked the question on Mathoverflow :

http://mathoverflow.net/questions/252855/can-an-irreducible-representation-have-a-zero-character )

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Won’t ρ(1)=I\rho(1) = I the identity of GLn(F)GL_n(F) ?

– user1952009

20 hours ago

4

Yes, so the value of the character at 1 is the trace of the identity and this trace is zero if the degree of ρ\rho is divisible by pp (the characteristic of FF).

– Panurge

20 hours ago

SImple remark: the FF-algebra generated by ρ(G)\rho(G) will also have trace identically zero, and will be a simple FF-algebra.

– YCor

20 hours ago

@YCor Why is it clear that the trace is zero? If that is true, we are done, because I can reduce to the case that FF is a finite field. All simple algebras over a finite field are separable field extensions (Wedderburn’s little theorem + finite fields are perfect) and the trace of a separable field extension is nonzero.

– David Speyer

12 hours ago

@DavidSpeyer because the FF-subspace generated by a subgroup of GLn(F)GL_n(F) is stable under multiplication, and hence equals the FF-subalgebra it generates.

– YCor

12 hours ago

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