2
$\begingroup$

Excercise 3.2. from Serre's book asks to prove a bound for the degree $n$ of an irreducible representation of a finite group $G$. Indeed, if $Z(G)$ denotes the center of $G$, then $n^2\leq\frac{|G|}{|Z(G)|}$.

The intended proof uses the fact that by Schur's lemma $|\chi(g)|=n$, whenever $g\in Z(G)$ (here $\chi$ is the character of our representation). Combined with the observation that, by the orthogonality relations: $$\sum_{g\in G}|\chi(g)|^2 \;=\; \sum_{g\in G}\chi(g)\overline{\chi(g)}=(\chi,\chi)|G|=|G|$$

Indeed, $$n^2|Z(G)| = \sum_{g\in Z(G)}|\chi(g)|^2 \leq \sum_{g\in G}|\chi(g)|^2 =|G|$$ which implies the result.

My personal gripe is that, thus far, all statements and arguments involving the inner product on class functions, could be rewritten in a way that avoids referencing the complex numbers. This is done by considering the symmetric bilinear form: $$\langle\chi,\psi\rangle:=\frac{1}{|G|}\sum_{g\in G}\chi(g)\psi(g^{-1})$$

Which is equal to the inner product on characters of $G$, as $\overline{\chi(g)}=\chi(g^{-1})$. The proof above does not have a straightforward generalization to this approach.

My question is: does anyone know a proof of the above statement which does not reference $\mathbb{C}$ directly, and thus applies to representations over arbitrary algebraically closed fields of characteristic 0?

$\endgroup$
2
  • $\begingroup$ But $\chi(g)$ is always a sum of roots of unity ie. in characteristic $0$ of complex numbers (this is because $\rho(g)^{|G|}=I$ implies the characteristic polynomial divides $X^{|G|}-1$) $\endgroup$
    – reuns
    Jan 10, 2020 at 0:43
  • $\begingroup$ @reuns: the minimal polynomial will be the one to surely divide $X^{|G|}-1_K$ over arbitrary commutative field $K$, I don't think it will follow in general that the characteristic polynomial also has this property. For instance, it could very well be that $|G|<\mathrm{dim}_K V$. $\endgroup$
    – ΑΘΩ
    Jan 10, 2020 at 0:54

1 Answer 1

1
$\begingroup$

The idea here is that an arbitrary algebraically closed field of characteristic $0$ can be endowed with a structure that emulates the behaviour of $\mathbb{C}$ together with the complex conjugation automorphism and the absolute value. By this I specifically mean the following

Theorem. Let $K$ be algebraically closed of characteristic $0$. Then there exists an involutive field automorphism $\iota \in \mathrm{Aut}_{\mathbb{Q}}K$ such that the fixed subfield $E:={}^{\iota}K$ is orderable by a certain total order $R$ and such that the structure $(E, +, \cdot, R)$ be a formally real field.

Sketch of proof: since it is of characteristic $0$, $K$ has a natural $\mathbb{Q}$-algebra structure; considering the extension $K/\mathbb{Q}$, let us fix a certain transcendence basis $B \subseteq K$.

It is known that free commutative monoids over any set (of arbitrary cardinality) are totally orderable (in the sense of admitting total orders compatible with the monoid structure) and that consequently polynomial rings (in arbitrarily many indeterminates) over (totally) ordered rings are (totally) orderable.

Hence in particular the integral domain $\mathbb{Q}[X_t]_{t \in B}$ is totally orderable and thus this order structure can be extended to the rational fraction field $\mathbb{Q}(X_t)_{t \in B}$, rendering it into a totally ordered field. Consider now a real closure $E$ of $\mathbb{Q}(X_t)_{t \in B}$ and furthermore an algebraic closure $F$ of $E$, which by the famous Euler-Lagrange theorem is actually a quaint quadratic extension of $E$ obtained by adjoining a square root of $-1_E$.

Clearly, $F$ and $K$ are by construction both algebraic closures of $\mathbb{Q}(X_t)_{t \in B}$ (to be specific with the details, $K$ is an algebraic closure of its subfield $\mathbb{Q}(B)$, the latter being canonically isomorphic to $\mathbb{Q}(X_t)_{t \in B}$), so they must be isomorphic $\mathbb{Q}(X_t)_{t \in B}$-algebras, via say an isomorphism $\varphi$.

By the general theory surrounding the Euler-Lagrange theorem, $F$ will naturally come equipped with a conjugation $\gamma$, which is none other than the automorphism fixing $E$ and taking $i$ to $-i$, where we have taken the liberty to denote by $i$ a certain fixed square root of $-1_E$. Transporting this entire structure via the isomorphism $\varphi$ introduced above is what establishes the existence of $\iota$ such that its fixed subfield be orderable with the structure of a really closed field. $\Box$

Having equipped $K$ with this structure, where we agree to denote the fixed subfield ${}^{\iota}K=E$, one can introduce the absolute value map on $K$ given by

$$| \bullet|: K \to K, \\ |z|=\sqrt{z \iota(z)}$$

since $z\iota(z)$, being fixed by $\iota$, will necessarily belong to $E$ and since really closed fields admit radicals of all orders (every positive element will have a unique positive root of order $n$ for any $n \in \mathbb{N}^*$).

The ''conjugation'' $\iota$ and the absolute value map $| \bullet |$ thus introduced exhibit exactly the same behaviour as the standard complex ones, and it is with this equipment that you can generalize results such as the one you discuss to the general setting of characteristic $0$ (with the proviso of algebraic closure, of course).

As a final remark, the ''conjugation'' introduced above is far from being a canonical object (somewhat unlike the case of complex conjugation, although ultimately one could argue that depending on the axiomatic system used to formalize mathematics not even the natural number $1$ is truly canonically fixed, in the sense of there being some arbitrariness behind its choice; but I digress with philosophical contemplations). However, the only important aspect is the existence of such structures, for they suffice to allow one to carry on the same kind of arguments and obtain the same type of inequalities/bounds as in the standard complex case.

$\endgroup$
3
  • $\begingroup$ Thank you for this answer, it is quite insightful. I admit I was hoping for a more direct proof, that didn't use a conjugation at all, but at the very least this settles my philosophical concerns. $\endgroup$
    – rridder
    Jan 10, 2020 at 10:45
  • $\begingroup$ @rridder: complex conjugation and the behaviour of the absolute value map (especially in relation to roots of unity) is exactly what enables you derive such results in the complex case, by enabling a ''translation'' of inverting elements in $G$ into taking conjugates in $\mathbb{C}$; $\endgroup$
    – ΑΘΩ
    Jan 10, 2020 at 10:54
  • $\begingroup$ in the general case I doubt that you could expect to find an essentially different way of arguing for such results, which is why this general theory allowing you to actually equip any algebraically closed field of characteristic $0$ with a conjugation is I would say the most direct of all and perhaps even the only tool you have available! Not to mention it is a lovely and remarkable result in and of itself as far the structure theory of (ordered) commutative fields is concerned. Regardless, I am glad to hear you found it helpful. $\endgroup$
    – ΑΘΩ
    Jan 10, 2020 at 10:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .