33
$\begingroup$

It is a theorem in basic representation theory that the degree of an irreducible representation on $G$ over $\mathbb{C}$ divides the order of $G$. The usual proof of this fact involves algebraic integers (see for example Fulton & Harris' Representation Theory, Serre's Linear representations of finite groups, or Simon's Representations of finite and compact groups). However I find this proof somewhat unsatisfying precisely because it uses algebraic integers, which don't show up much elsewhere in basic representation theory, and it is not at all evident why algebraic integers should be used. I feel that there has to be another proof of this theorem that uses techniques of group theory and representation theory, but the only other proof I know is one by Kopp and Wiltshire-Gordon, but that proof seems to use even more complicated ideas if not machinery!

What are some other proofs of this theorem?

$\endgroup$
6
  • $\begingroup$ Actually, algebraic integers can be considered beasts from (Galois) group theory. $\endgroup$ Nov 23, 2012 at 15:23
  • $\begingroup$ Is the fulton and harris proof the character theoretic one which uses $\langle \chi,\chi \rangle$ to show $|G|/\chi(1)$ is an algebraic integer? $\endgroup$
    – Alexander Gruber
    Nov 23, 2012 at 20:12
  • $\begingroup$ @Alexander More or less. Fulton and Harris presents this theorem as an exercise (Problem 2.38) and works it out at the back of the book. $\endgroup$
    – JHF
    Nov 23, 2012 at 22:14
  • $\begingroup$ “The proof of Kopp and Wiltshire-Gordon seems to use more complicated ideas” ? Their abstract claims otherwise : “the only ingredients are Schur’s lemma, basic counting, and a divisibility argument.” $\endgroup$ Dec 1, 2012 at 6:20
  • 1
    $\begingroup$ @Ewan: While the calculations might not be very sophisticated, Remark 6.4 of the Kopp and Wilton-Gordon indicates that the proof depends on corollary 4.2. Certainly the involvement of surfaces defined by words or equation (52) is a surprising idea, no? So I'm wondering if there's a more group-theoretic proof. $\endgroup$
    – JHF
    Dec 1, 2012 at 23:24

4 Answers 4

30
+50
$\begingroup$

It seems to me that Etingof et al (pages 51-52) and Kopp and Wiltshire-Gordon give very similar proofs. I'll write them out in as elementary a way as I can, and also give a third proof of my own along similar lines.

Notation: Let $G$ be our finite group, $V$ an irrep over the complex numbers, $\rho_V$ the map $G \to GL(V)$ and $\chi_V$ the character of $V$. We write $\mathrm{Id}_V$ for the identity map $V \to V$. Let $g_1$, $g_2$, ..., $g_c$ be representatives for the conjugacy classes of $G$, and let $C(g)$ be the conjugacy class of $g$.

Lemma: For any $g \in G$, $$\sum_{h \in C(g)} \rho_V(h) = \frac{|C(g)| \chi_V(g)}{\dim V} \mathrm{Id}_V \quad\quad (\ast)$$

Proof: For any $f$ in $G$, we have $$\rho_V(f) \cdot \left( \sum_{h \in C(g)} \rho_V(h) \right) = \sum_{h \in C(g)} \rho_V(fh) =$$ $$\left( \sum_{h \in C(g)} \rho_V(fhf^{-1}) \right) \cdot \rho_V(f) = \left( \sum_{h \in C(g)} \rho_V(h) \right) \cdot \rho_V(f).$$

So the left hand side of $(\ast)$ commutes with every $\rho_V(f)$. By Schur's lemma, this means that the left hand side of $(\ast)$ is $a \mathrm{Id}_V$ for some scalar $a$. Taking traces, we compute that $|C(g)| \chi_V(g) = a \dim V$, so $a = |C(g)| \chi_V(g)/\dim V$ as required. $\square$

Let's define $$P(g) = \sum_{h \in C(g)} h.$$ This is an element in $\mathbb{Z}[G]$. Let $$P_V(g) = \sum_{h \in C(g)} \rho_V(h).$$ This is how $P(g)$ acts on $V$. So the above Lemma shows that $P_V(g) = |C(g)| \chi_V(g)/ \dim V \cdot \mathrm{Id}_V$.

Both the papers I cite want to sum up $P_V(g_i)$ in some way, use the identity $$\sum_{i} |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) = |G| \quad \quad (\ast \ast)$$ and wind up with $\dim V$ in the denominator of something which they can prove, by other means, is an integer. Note that you may recognize $(\ast \ast)$ better in the form $\sum_{g \in G} \chi_V(g) \overline{\chi_V(g)} = |G|$; we get identity $(\ast \ast)$ by grouping together the terms in the same conjugacy class and using $\chi_V(g^{-1}) = \overline{\chi_V(g)}$.

Etingof et al's proof: Consider $$Q_V = \sum_i P_V(g_i) \chi_V(g_i^{-1}).$$ On the one hand, using the Lemma, $$Q_V = \sum_i |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) \frac{1}{\dim V} \mathrm{Id}_V = \frac{|G|}{\dim V} \mathrm{Id}_V$$ using $(\ast\ast)$.

On the other hand if you expand out $Q_V$ in $\mathbb{C}[G]$, you'll see that the coefficient of every group element is an algebraic integer. So $|G|/\dim V$ is an algebraic integer and, since it is rational, must be an integer. $\square$.

Kopp and Wiltshire-Gordon's proof: Set $$R = \sum_{i} \frac{|G|}{|C(g)|} P(g_i) P(g_i^{-1}).$$ Let $R_V = \rho_V(R)$.

Using the Lemma, $$R_V = \sum_i \frac{|G|}{(\dim V)^2} |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) \mathrm{Id} = \frac{|G|^2}{(\dim V)^2} \mathrm{Id}_V.$$

So, similarly, $$\rho_V(R^k) = \frac{|G|^{2k}}{(\dim V)^{2k}} \mathrm{Id}_V.$$

Let $U$ be the regular representation of $G$. So $\chi_U = \sum (\dim V) \chi_V$. We deduce that $$\chi_U(R^k) = \sum_V (\dim V) \frac{|G|^{2k}}{(\dim V)^{2k}} \mathrm{Tr}(\mathrm{Id}_V) = \sum_V (\dim V)^2 \frac{|G|^{2k}}{(\dim V)^{2k}}.$$

One the other hand, $R$ is clearly in $\mathbb{Z}[G]$. (Recall that $|C(g)|$ divides $|G|$ because $|C(g)|$ has a transitive action of $G$ by conjugation.) So $R^k$ is in $\mathbb{Z}[G]$. And the trace of any element of $\mathbb{Z}[G]$ acting on the regular rep is an integer (in fact, it is $|G|$ times the coefficient of the identity). So $\sum_V (\dim V)^2 \frac{|G|^{2k}}{(\dim V)^{2k}}$ is an integer (in fact, one divisible by $|G|$) for all $k$. In Lemma 6.2, the authors show that this forces each $|G|/\dim V$ to be an integer. $\square$

Remark Other parts of the paper make use of the pleasant identity $R = \sum_{g \in g} \sum_{h \in G} ghg^{-1} h^{-1}$. But this doesn't seem to be important if our goal is solely to get this fact.

Variant on the second proof If one considers the action of $R$ on $\mathbb{Z} G$, it is clearly given by a matrix with integer entries. Let $f(\lambda)$ be the characterisitc polynomial of that matrix, so $f$ is a monic polynomial with rational coefficients.

We showed that $R$ acts on the subspace $V^{\oplus \dim V}$ of $\mathbb{C} G$ by $(|G|/\dim V)^2$. So $(|G|/\dim V)^2$ is an eigenvalue of the above integer matrix. So $(|G|/\dim V)^2$ is a root of $f$. By the rational root theorem, we deduce that $(\dim V)^2$ divides $|G|^2$, so $\dim V$ divides $|G|$. $\square$

$\endgroup$
5
  • $\begingroup$ Thanks for the detailed explanation @DavidSpeyer, I think this is very nice. $\endgroup$
    – Alexander Gruber
    Dec 8, 2012 at 18:32
  • 2
    $\begingroup$ Nice survey! But I'm afraid you got Etingof's proof slightly wrong: What do you mean by "if you expand out $Q_V$ in $\mathbb C[G]$, you'll see that the coefficient of every group element is an algebraic integer"? I don't think there is a "sum of algebraic integers is an algebraic integer" result for noncommutative rings. I think this is the reason Etingof works with numbers rather than elements of $\mathbb C[G]$ or $\mathrm{End}V$. $\endgroup$ Apr 10, 2013 at 2:37
  • $\begingroup$ @darijgrinberg I think you're right. I'm probably not going to get to this soon; feel free to fix it for me if you want to. You're right that sum of algebraic integers need not be an algebraic integer in a noncommutative ring; consider $\left( \begin{smallmatrix} 0 & p \\ 1/p & 0 \end{smallmatrix} \right) + \left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$ in $\mathrm{Mat_2}(\mathbb{Q})$. $\endgroup$ Apr 10, 2013 at 3:06
  • $\begingroup$ @DavidESpeyer I'm wondering whether one could go anywhere toward an elementary proof with the following naive idea: Take a nonzero vector $v \in V$ with $V$ an irrep. Consider the finite set of $\{gv\}$ for $g\in G$. The span of these vectors is apparently $G$-invariant and not the empty set so must be $V$. By the Orbit-Stabilizer theorem $|G|=|Stab(v)||Orb(v)|$. It seems plausible that the finite set $Orb(v)$ could now be a basis of $V$ and thus prove the result, but I was curious whether someone with more rep theory experience had an example to contradict this before I try to proceed. $\endgroup$ Mar 6, 2018 at 16:34
  • $\begingroup$ @BrianKlatt : Note that the only representations of $G$ for which $Orb(v)$ is a basis for some $v$ are the permutation representations (linear extensions of group actions). These representations are never irreducible, since the sum of the elements of $Orb(v)$ is a non-zero element which is $G$-invariant. $\endgroup$ Sep 20, 2018 at 12:00
7
$\begingroup$

I was thinking about something else the other day, and came up with another proof. It's heavier on the number theory, but lighter on the character theory, then the other proofs so far.

Lemma Let $V$ be an irrep of $G$, with representation map $\rho: G \to GL(V)$. Let $A$ be any element of $\mathrm{End}(V)$. Then $$\sum_{g \in G} \rho(g) A \rho(g^{-1}) = \frac{\mathrm{Tr}(A) \cdot |G|}{\dim V} \mathrm{Id}.$$

Proof We have $$\rho(h) \left( \sum_{g \in G} \rho(g) A \rho(g^{-1}) \right) = \sum_{g \in G} \rho(hg) A \rho(g^{-1}) = \left( \sum_{f \in G} \rho(f) A \rho(f^{-1}) \right) \rho(h)$$ by the change of variable $f=hg$. So the left hand side commutes with the $G$-action and is thus a scalar. Taking traces of both sides shows us which scalar it is. $\square$

Lemma With notation as above, we can choose a basis for $V$ so that all the entries of $\rho(g)$ are algebraic integers.

Proof See here. $\square$

Take such a basis, and let $A$ be a matrix with integer entries and trace $1$. Then the first lemma shows that $|G|/\dim V$ is a sum of algebraic integers, hence an algebraic integer, hence an integer.

$\endgroup$
1
  • 1
    $\begingroup$ You don't need to prove that $rho(g) $ can be made to have algebraic integer entries. You only need to work p-adically for every prime separately and over a local field, it's easy to see that you can find an integral basis. $\endgroup$
    – Asvin
    Nov 16, 2021 at 5:36
3
$\begingroup$

Consider the following Lemma: Let $G$ be a finite group, $g \in G$ and $w$ be a given word on $G$. Let $\gamma_G^g(w) = |\{\vec{g} \in G^n: w(\vec{g})=g\}|$. Then $\gamma_G^g(w)=\sum_{\chi} \frac{\overline{\chi(g)}}{|G|} \sum_{\vec{g} \in G^n} \chi(w(\vec{g}))$.

Proof: Let $f: G \to \mathbb{C}$ be defined in the following way: $f(x)=|G|$, if $x \in g^G$ (that is, if $x$ is in the conjugacy class of $g$) and $f(x)=0$ otherwise. Clearly, $f$ is a class function. Now let $A=\sum_{\vec{g} \in G^n} f(w(\vec{g}))$. Since $f(w(\vec{g}))$ is nonzero if and only if $w(\vec{g}) \in g^G$, and since $f$ is a class function, it follows that $A=|g^G| |G| \gamma_G^g(w)$, so $\gamma_G^g(w)=\frac{A}{|g^G||G|}$.

As $f$ is a class function, we have that $f=\sum_{\chi} \langle f,\chi \rangle \chi$, where the sum is taken over all the irreducible characters of $G$. But $\langle f, \chi \rangle = \frac{1}{|G|} \sum_{x \in G} f(x) \overline{\chi(x)} =\frac{1}{|G|} |g^G| |G| \overline{\chi(g)} = |g^G| \overline{\chi(g)}$ (using only the definition of $f$, so $f=\sum_{\chi} |g^G| \overline{\chi(g)} \chi$) and thus

\begin{equation} A= |g^G|\sum_{\vec{g} \in G^n} \sum_{\chi} \overline{\chi(g)}\chi(w(\vec{g}))=|g^G| \sum_{\chi} \overline{\chi(g)} \sum_{\vec{g} \in G^n} \chi(w(\vec{g})) \end{equation}

Since $\gamma_G^g(w)=\frac{A}{|G||g^G|}$, we get $\gamma_G^g(w)=\sum_{\chi}\frac{\overline{\chi(g)}}{|G|}\sum_{\vec{g} \in G^n}\chi(w(\vec{g}))$.

$\square$

Now apply this lemma to the word which is a product of $k+1$ commutators. Using lemmas 3.2 and 3.3 from Kopp and Wiltshire-Gordon (which rely only on Schur's Lemma and are simple, as the group is finite, so you get sums instead of integrals), you'll recover Kopp's and Wiltshire-Gordon's equation 70, bypassing the topological arguments. From that point, you just need the basic number theory he uses in section 6 to finish the proof.

$\endgroup$
3
$\begingroup$

Here's my attempt at an "elementary" proof, which is basicially a geodesic path through the proof that Serre gives in Linear representations of finite groups.

Theorem. Let $\rho$ be an irreducible representation of $G$. Then $d=\dim \rho$ divides $n=\lvert G\rvert$.

For any $x\in G$, consider the operator $$\def\Cl{\operatorname{Cl}}T_x := \sum_{g\in \Cl(x)} \rho_g.$$ Note that this only depends on the conjugacy class of $x$. We have $\rho_a T_x\rho_{a^{-1}} = T_x$ for any $a\in G$, so by Schur's lemma, $T_x=\lambda_x I$ for some $\def\C{\mathbb{C}}\lambda_x\in \C$. We can actually compute $\lambda_x$ by taking traces: $$\lambda_x = \frac{\lvert\Cl(x)\rvert}{d} \chi(x)$$ where $\chi$ is the character of $\rho$. In particular, $\lambda_e=1$.

By taking the trace of $\sum_{g\in G} \chi(g^{-1})\rho_g = \sum_{x_i} \chi(x_i^{-1}) T_{x_i}$ (sum over a set $\{x_i\}$ of representatives of conjugacy classes), and using the orthogonality relation $\langle\chi,\,\chi\rangle=1$, we get an identity $$\frac{n}{d} = \sum_{x_i}\chi(x_i^{-1})\lambda_{x_i}.$$

Let $R=$ the abelian subgroup of $\C$ generated under addition by the finite set of elements $$\zeta^k \lambda_x, \qquad 0\leq k<n,\quad x\in G,$$ where $\zeta=e^{2\pi i/n}$. Since $\chi(g)$ is a sum of $n$-th roots of unity, we have from the above identity that $n/d\in R$.

Claim. For any $g,h\in G$, there is a formula $$T_gT_h = \sum_{x_i} m_{x_i} T_{x_i}\qquad \text{and therefore}\qquad \lambda_g\lambda_h = \sum_{x_i} m_{x_i} \lambda_{x_i},$$ where the sum is over $\{x_i\}$ a set of representatives of conjugacy classes, and $m_x=$ size of $\{(u,v)\in \Cl(g)\times \Cl(h)\;|\;uv=x\}$, and thus $\def\Z{\mathbb{Z}}m_x\in \Z$. Proof. Straightforward.

By the above claim, $R$ is in fact a subring of $\C$, so multiplication by any element of $R$ such as $n/d$ defines an endomorphism $\phi\colon R\to R$ of the underlying abelian group.

As an abelian group $R$ is certainly finitely generated, and since $R\subseteq \C$ it is torsion free. So by the classification of finitely generated abelian group it is isomorphic to $\Z^N$ for some $N\geq 1$. Choose a basis for $R$ over $\Z$, and let $A$ be the integer matrix representing $\phi$ in this basis.

Since $\def\Q{\mathbb{Q}}n/d\in \Q$, for any $r\in R$ we must have $d\phi(r)=nr$, so $dA=nI$, i.e., $A=(n/d)I$. But $A$ is an $N\times N$ integer matrix with $N\geq1$, so $n/d\in \Z$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.