Throughout $G$ is a finite, **non-abelian** group.
$\DeclareMathOperator\Irr{Irr}\DeclareMathOperator\AD{AD}\DeclareMathOperator\cp{cp}\newcommand\card[1]{\lvert#1\rvert}$

Let $\Irr(G)$ be the set of irreducible characters (working over the complex ground field). Let $$ \AD(G) = \card G^{-1} \sum_{\chi\in\Irr(G)} (d_\chi)^3 \;. $$ (The invariant $\AD(G)$ arose under a different name in work of Johnson on amenablity constants of Fourier algebras, and is a special case of something more general that can be defined for countable virtually abelian groups in terms of Plancherel measure for such groups. But this question is only concerned with character theory of finite groups.)

**Question.** If $\AD(G)\leq 2$, does this imply that $d_\chi\leq 2$ for all $\chi\in\Irr(G)$?

Here is my naive reasoning to support my guess that the answer is positive. Let $\Irr_n(G)=\{\chi\in\Irr(G) \mid d_\chi =n\}$. Then $$ 2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} +\card G\AD(G) = \sum_{\chi\in\Irr_1(G)} 3 + \sum_{\chi\in\Irr_2(G)} 12 + \sum_{n\geq 3} \sum_{\chi\in\Irr_n(G)} (d_\chi)^3 $$ and therefore, since $\card G=\sum_{n\geq 1} \sum_{\chi\in \Irr_n(G)} (d_\chi)^2$, $$ 2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} +\card G\AD(G) \geq 3\card G. $$ If we impose the condition that $\AD(G)\leq 2$, this yields $$ 2\card{\Irr_1(G)} + 4\card{\Irr_2(G)} \geq \card G $$ which seems difficult to achieve if there are irreps of degree $\geq 3$.

**EDIT:** the original version of this post asked if there existed any $G$ with irreps of degree $\geq 3$ that satisfied this inequality, with the naïve hope that no such $G$ existed. Victor Ostrik has pointed out in a comment that an extraspecial $2$-group of order $32$ has $16$ linear characters and a single nonlinear character which has degree $4$, so that the inequality above is indeed satisfied. On the other hand the AD of this group is $(16+64) / 32 = 5/2$.

Some other background facts which may or may not be useful.

If $H\leq G$ then $\AD(H)\leq\AD(G)$. I am not sure if this has a simple proof just using the definition above, but it follows from the alternative description in tems of "amenability constants of Fourier algebras".

There is some relationship between $\AD(G)$ and the

*commuting probability*$$ \cp(G) = \frac{\# \hbox{conjugacy classes of $G$}}{\card G} = \frac{\card{\Irr(G)}}{\card G} $$ which shows that for the former to be “small” the latter should be “large”. To be precise, note that Hölder with conjugate exponents 3 and 3/2 gives $$ \sum_{\chi\in\Irr(G)} (d_\chi)^2 \leq \left( \sum_{\chi\in\Irr(G)} 1^3 \right)^{1/3} \left( \sum_{\chi\in\Irr(G)} (d_\chi)^3\right)^{2/3} $$ so that $$ 1 \leq \cp(G) \AD(G)^2. $$ Moreover, equality is strict since we assume $G$ is non-abelian. In particular, if $\AD(G)\leq 2$ this implies $\cp(G) > 1/4$. Can this be leveraged, perhaps by using the good behaviour of $\cp$ with respect to quotients? I had a quick look in the 2006 paper On the commutating probability in finite groups of Guralnick–Robinson and I suspect that results there could be useful, but as I am not a specialist in finite group theory I couldn't see how to make progress.

`\newcommand\AD{\operatorname{AD}}`

is the same as`\DeclareMathOperator\AD{AD}`

. $\endgroup$ – LSpice Apr 20 at 22:561more comment