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Is the following statement true?

Every finite group $G$ has a non-trivial irreducible representation of dimension $O(\lvert G\rvert^{1/3})$.

Context: Groups with no small irreducible representations are called quasirandom in Gowers - Quasirandom groups as these groups always have good mixing properties. The standard example of a quasirandom group used everywhere is $\operatorname{PSL}(2,q)$, which has $\sim q^3/2$ elements and no irreducible representation of dimension less than $(q-1)/2$.

Is $\operatorname{PSL}(2,q)$ known to be quantitatively the best example?

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  • $\begingroup$ Similar question on math stackexchange, although with no real answers: math.stackexchange.com/questions/3574456/… $\endgroup$
    – Sam Hopkins
    Commented Aug 2, 2021 at 18:04
  • $\begingroup$ @GeoffRobinson If I'm understanding right, I think you have the sense of the question wrong? Your dihedral group example would have all those characters of order $O(|G|^{1/3})$, since they're much smaller than that size. What I believe Yufei is effectively asking is whether there are any groups whose lowest-degree irreps are larger than those of $PSL(2,q)$ (which are of size $\Theta(|G|^{1/3})$), or whether those groups are (among) the best. $\endgroup$
    – Steven Stadnicki
    Commented Aug 2, 2021 at 18:46
  • $\begingroup$ Another similar question from math stackexchange without an answer: math.stackexchange.com/questions/2804537/… $\endgroup$
    – Carl Schildkraut
    Commented Aug 2, 2021 at 19:32
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    $\begingroup$ An amusing example is the sporadic simple group $J_1$, of order $175560=55.56.57$. The lowest degree among its nonprincipal irreducible complex characters is $56$. $\endgroup$
    – Richard Lyons
    Commented Aug 2, 2021 at 22:01
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    $\begingroup$ @LSpice: In this case I think it's easy enough to say, there is some absolute constant $C$ so that every finite group $G$ has a nontrivial irreducible representation of dimension $\leq C |G|^{1/3}$. $\endgroup$
    – Sam Hopkins
    Commented Aug 3, 2021 at 3:21

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The answer is likely yes. Here is a partial proof:

Every finite group has a maximal proper normal subgroup, and the quotient by this gives a surjection to a finite simple group. Any nontrivial irreducible representation of that finite simple quotient gives a nontrivial irreducible representation of the original group. So it suffices to prove such a bound for finite simple groups.

Next it is natural to invoke the classification of finite simple groups.

Cyclic and alternating groups are obviously fine, having irreps of much lower dimension. We can absorb the sporadics into the big $O$. So we really only have to consider finite simple groups of Lie type.

Let's consider for simplicity the finite simple groups of Lie type arising from split algebraic groups $G$ over finite fields $\mathbb F_q$. These have dimension roughly $q^{ \dim G}$ and we can construct a nontrivial permutation representation of size roughly $q^n$ where $n$ is the dimension of any flag variety of $G$. So it suffices to find such a flag variety where $\frac{n}{ \dim G} \leq \frac{1}{3}$.

For $G = SL_n$, of dimension $n^2-1$, projective space is a flag variety of dimension $n-1$, for a ratio of $\frac{1}{n+1} \leq \frac{1}{3}$.

For $G = Sp_{2n}$, of dimension $n (2n+1)$, projective space is a flag variety of dimension $2n-1$, for a ratio of $\frac{2n-1}{ n (2n+1)} \leq \frac{1}{ n+1} \leq \frac{1}{3}$ since $\frac{2n-1}{2n+1} \leq \frac{2n-2}{2n}$.

For $G= SO_n$, of dimension $n (n-1)/2$, the set of rank one maximal isotropic subspaces is a flag variety of dimension $n-2$, for a ration of $\frac{2n-4}{n (n-1)} = \frac{1}{3} - \frac{(n-3)(n-4)}{ 3 n(n-1)} \leq \frac{1}{3}$

For $G = G_2, F_4, E_6, E_7, E_8$, the dimension is $14, 52, 78, 133, 248$ and they have (lowest-dimensional) flag varieties of dimension $5, 15, 16, 33, 78 $. So each one satisfies the inequality except $G_2$.

Unless I have misread the tables from Character Degrees and their Multiplicities for some Groups of Lie Type of Rank < 9, regardless of the congruence class of $q$ mod $6$, $G_2(q)$ has an irreducible representation of degree $O(q^3)$ or $O(q^4)$ and thus does satisfy this inequality. (In fact, there are always exactly one or two irreducible representations in that range.)

So one is reduced to considering the various non-split groups...

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    $\begingroup$ I think Table 4 of this paper -- arxiv.org/pdf/1301.5166.pdf -- gives enough information to complete the proof... $\endgroup$
    – Nick Gill
    Commented Aug 3, 2021 at 16:02

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