Multiplicativity, where art thou?

The inequality I was pondering during my recent vacations is another result of Burnside: for a finite cyclic group $G=\mathbf{Z}/m\mathbf{Z}$ , denoting by $S$ the set of generators of $G$ , we have
$\sum_{x\in S}|\chi_{\rho}(x)|^2\geq |S|=\varphi(m)$
for any finite-dimensional representation $\rho$ of $G$ with character $\chi_{\rho}$ , provided the latter does not vanish identically on $S$ (this may happen, e.g., for the regular representation of $G$ ). This was used by Burnside (and still appears in many books) in order to prove that if an irreducible representation $\rho$ of a finite group has degree at least $2$ , its character must be zero on some elements of the group. (The cyclic groups $G$ used are those generated by non-trivial elements of the group, and the representations are the restrictions of $\rho$ to those cyclic groups; thus it is an interesting application of reducible representations of abelian groups to irreducible representations of non-abelian ones… the lion and the mouse come to mind.)

This inequality can not be considered hard: it is dispatched in two lines invoking Galois theory to argue that
$\prod_{x\in S}{|\chi_{\rho}(x)|^2}$
is a (non-negative) integer, and the arithmetic-geometric inequality to deduce that
$\frac{1}{|S|}\sum_{x\in S}{|\chi_{\rho}(x)|^2}\geq \Bigl(\prod_{x\in S}{|\chi_{\rho}(x)|^2}\Bigr)^{1/|S|}\geq 1$
unless this product is zero.

However, if we view this as a property of representations of a finite cyclic group, one (or, at least, I) can’t help wonder whether there is a “direct” proof, which only involves the formal properties of these representations, and does not refer to Galois theory. This is what I was trying to understand while basking in the italian spring and gardens. As it turns out, there is is in fact quite a beautiful argument, which is undoubtedly too complicated to seriously replace the Galois-theoretic one, but is directly related to very interesting facts which I didn’t know before — I therefore consider well-spent the time spent thinking about this inequalitet. (Philological query: what is the diminutive of “inequality”? or of “formula”?)

The idea is that a representation $\rho$ of $G$ is a direct sum
$\rho=\bigoplus_{a}{n(a)\chi_a},$
where $a$ runs over $\mathbf{Z}/m\mathbf{Z}$ , and $\chi_a$ is the character
$x\mapsto e(ax/m),$
for some integral multiplicities $n(a)$ . Hence the left-hand side of Burnside’s inequality, namely
$\sum_{x\in S}{|\chi_{\rho}(x)|^2}$
can be seen as a quadratic form in $m$ integral variables, and the question we ask is: what is the minimal non-zero value it can take? (Note that the example of the regular representation shows that this quadratic form, though obviously non-negative, is not definite positive.)

Now the funny part of the story is that this quadratic form, say $Q_m$ , is naturally a tensor product
$Q_m=\bigotimes_{p^{k_p}\mid\mid m}{Q_{p^{k_p}}}$
of the corresponding quadratic forms arising from the primary factors of $m$ . Hence we have a problem about a multiplicatively defined quadratic form, and the answer (given by Burnside) is rhe Euler function $\varphi(m)$ , which is also multiplicative; should it not then feel natural to treat the case of $m=p^k$ for some prime $p$ , and claim that the general case follows by multiplicativity?

Alas, a second’s thought shows that it is by no means clear that the minimum of a tensor product of (non-negative, integral) quadratic forms should be the product of the minima of the factors! Denoting by $s(Q)$ the minimum of $Q$ (among non-zero values), we certainly have
$s(Q_1\otimes S_2)\leq s(Q_1)s(Q_2),$
but the converse inequality should immediately feel doubful when the forms are not diagonalized. And indeed, in general, this is not true! I found examples in the book of Milnor and Husemoller on bilinear forms (which I had not looked at for a long time, and I am very happy to have been motivated to look at it again!); they are attributed to Steinberg, through a result of Conway and Thompson, itself based on the famous Siegel formula for representation numbers of quadratic forms…

However, despite this fact about general forms, it turns out that the quadratic forms $Q_{p^k}$ have the property that
$s(Q_{p^k}\otimes Q')=s(Q_{p^k})s(Q')$
for all other quadratic forms $Q'$ (always non-negative integral). This I found explained by Kitaoka. Here the nice thing is that it depends on writing $Q_p$ as (essentially) a variance
$Q_p(n)=p\sum_{a}{n(a)^2}-\Bigl(\sum_{b}{n(b)}\Bigr)^2$
of the coefficients $n(a)$ , and using the alternate formula
$Q_p(n)=\frac{1}{2}\sum_{a,b}{(n(a)-n(b))^2}\quad\quad\quad\quad\quad\quad (\star)$
which corresponds, in probability, to the formula
$\mathbf{V}(X)=\frac{1}{2}\mathbf{E}((X-Y)^2)$
for the variance of a random variable $X$ , using an independent “copy” $Y$ of $X$ (i.e., $Y$ is has the same probability distribution as $X$ , but is independent.) This probabilistic formula, I have to admit, I didn’t know — or had forgotten –, though it is very useful here! Indeed, the reader may try to prove directly that $Q_p(n)\geq p-1$ unless it is $0$ , using any of these formulas for $Q_p(n)$ ; I succeeded with the other formula
$Q_p(n)=p\sum_{a}{\Bigl(n(a)-\frac{1}{p}\sum_{b}{n(b)}\Bigr)^2}$
but the argument was much uglier than the one with $(\star)$ above! (It is also clearer from the latter that $Q_p$ takes integral values for integral arguments, despite the $1/p$ which crop up here and there…)