Can an irreducible representation have a zero character?

I asked the following question on Stackexchange,

https://math.stackexchange.com/questions/1978407/can-an-irreducible-representation-have-a-zero-character

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

• Regarding "but it got no answer": You got 10 upvotes, which indicates much interest from the readers, but more importantly you only asked this yesterday. I think that it is good practice to wait at least three days (actually, I would suggest one weak), so that the mathematics sites don't have unnecessary duplicates. By the way, that's an interesting question. Oct 23, 2016 at 7:38
• The cross-post immediately gave rise to useful answers, so it sounds fine to me.
– YCor
Oct 23, 2016 at 19:17

Well, it seems that the character of an irreducible representation of a finite group over whatever field is always nonzero. I find this statement in I.M. Isaacs, Character Theory of Finite Groups, coroll 9.22, Dover, p. 155.

• This is a precise reference for the result (and more). The corollary itself depends heavily on the earlier Lemma (9.12) where the field is assumed to be a splitting field for the group: here the representing matrices span a full matrix algebra, where the character can take value 1. [I suspect the general result goes back to Burnside's work, though Isaacs' book does not emphasize the history.] Oct 23, 2016 at 18:35

Representations of semigroups whose characters are identically zero have been classified in

A. Freedman, R. N. Gupta, and R. M. Guralnick, Shirshov’s theorem and representations of semigroups, Pacific J. Math. (1997), no. 181, 159--176.

Their result is the following, see Theorem 2.5 in the aforementioned paper:

Theorem. Let $\rho \colon S \to \mathrm{End}(V )$ be a finite irreducible representation of the semigroup $S$ and let $A$ be the $F$-algebra generated by $\rho(S) \subseteq \mathrm{End}(V )$.

Then the character $\chi$ of $\rho$ is identically $0$ if and only if $A \simeq M_r(D)$, where $D$ is division ring such that either $Z = Z(D)$ is not separable over $F$ or $\mathrm{dim}_Z(D)$ is a multiple of the characteristic. In particular, if $F$ is perfect, the character in nonzero.

${}$ Edit. As explained by B. Steinberg in the comments below, none of the conditions in the previous Theorem can occur when $S$ is a finite semigroup. In particular, when $G$ is a finite group over any field $F$, the character of any irreducible $F$-representation of $G$ is non-zero. This agrees with the result in Isaac's book quoted in Panurge'a answer.

• Thanks. I'm learning Group Theory with Robinson's book as textbook and I must confess that your answer is difficult to understand for me. May I ask if I understood correctly the satement of I.M. Isaacs that I quoted in my answer and if it agrees with the result quoted by you ? Oct 23, 2016 at 10:10
• I checked Isaac's result, and it seems to me that you quoted it correctly. So it appears that none of the situations in Friedman-Gupta-Guralnick theorem can occur when the semigroup $S$ is actually a group. However I'm not a a specialist in the area, so I do not see immediately why this is true. Maybe some more expert user can explain this. Oct 23, 2016 at 11:02
• It can occur for infinite groups and semigroups but not finite ones. Look at my question math.stackexchange.com/questions/819466/… which proves it doesn't happen for finite semigroups. Oct 23, 2016 at 11:07
• The paper you cite gives infinite groups. The multiplicative group of a not "nice" field extension. Oct 23, 2016 at 11:08
• @BenjaminSteinberg: So this agrees with Isaac's claim, good. Thank you. Oct 23, 2016 at 11:11