Related to an answer to a previous question. The answer assume the following result:

Let $G$ be a finite group and $\rho : G \rightarrow \text{GL}(\mathbb{C}, n)$ be a faithful representation of $G$ (ie. $\text{Ker}(\rho) = 1_G$). Let $\chi$ be the character associated to $\rho$. Then, for all $g \in G$ such that $g \not= 1_G$ we have $|\chi(g)| < n$.

Is this true? If yes, why? I couldn't find any proof and I can't understand the small justification given in the previous answer.

  • 2
    $\begingroup$ This is false; take rho to be a direct sum of copies of a nontrivial one-dimensional representation of Z/nZ on C*. However, we can say the following: every element of a finite group has finite order, so the eigenvalues of rho(g) must be roots of unity. It follows that chi(g) is the sum of n roots of unity, which has absolute value less than or equal to n, with equality if and only if they're all the same by the triangle inequality. $\endgroup$ – Qiaochu Yuan Mar 14 '10 at 3:14

The other result in this direction is that if $\rho$ is irreducible, then $|\chi(g)| = n$ if and only if g is in the center of G/Ker. The proof is to start wiht what Qiaochu said, namely that by the triangle inequality and the fact that the eigenvalues are roots of unity you get that $|\chi(g)|=n$ if and only if g is a scalar matrix. Thus they commute with everything in End(V), and hence lie in the center of G/Ker. Conversely, use irreducibility to show that commuting with everything in G/Ker implies that you commute with everything in End(V).

When I took representation theory with Lenstra this argument was very memorable. He had started out in complete generality (arbitrary fields etc.) and as the course went on we needed more and more assumptions (algebraically closed, characteristic prime to the size of the group, etc.). When he got to this argument he said "Now this is the only time that we need to assume that the field is the complex numbers. This argument doesn't work over an arbitrary algebraically closed field of characteristic zero. (Although it's still true for such fields by model theoretic arguments.)"

A good related theorem to try to prove when you're thinking about the question you asked is that a representation is faithful if and only if every representation appears inside one of its tensor powers.

  • 1
    $\begingroup$ @Noah: your last statement is false, as explained by Richard Stanley here: mathoverflow.net/questions/10487/… - what you mean to say is that every irreducible occurs as a summand in a tensor power (that was discussed here: mathoverflow.net/questions/10126/…), which is not literally the same. $\endgroup$ – Vladimir Dotsenko Mar 14 '10 at 12:16
  • $\begingroup$ Indeed, and I should know better. I made a similar mistake on a homework set when taking grad algebra. That mistake was at least productive as my TF decided to actually prove I was wrong, which led to this paper: arxiv.org/abs/math/0002106 $\endgroup$ – Noah Snyder Mar 14 '10 at 15:24
  • $\begingroup$ The false statement which Vladimir refers to, now edited away, is that a representation is faithful if and only if the powers of its character span the whole character ring. $\endgroup$ – David E Speyer Mar 14 '10 at 15:35
  • $\begingroup$ The following result is due to Burnside-Brauer. Let a faithful character $\chi$ of a group $G$ takes exactly $N$ values. Then the sum $1_G+\chi+\dots+\chi^{N-1}$ contains all irreducible characters of $G$. The following dual result is due to S. Garrison: Let $K$ be a $G$-class and $\langle K\rangle=G$, $k$ the corresponding class sum and $m$ is the number of distinct values of the function $\chi\to\frac{\chi(k)}{\chi(1)}$, where $\chi\in\text{Irr}(G)$. Then every element of $G$ can be written as a product of less than $m$ elements of $K$. $\endgroup$ – yakov Jul 8 '16 at 14:25

It is a well known fact that $Ker(\rho)$ is the set of elements $g$ such that $\chi(g)=\chi(1) $ -this can be found in Isaacs character theory or any other book in character theory- Notice that, as pointed out in the above example, $\chi(g)=\chi(1) $ is not the same as $|\chi(g)|=\chi(1)$. So the point of the previous answer is that for faithful $\rho$ one has that $\chi(g)=n$ if and only if $g=1$.

Edit: See Darij's comment below.

The proof I know is algebraic(I think). Let $\alpha$ be the arithmetic mean of the root of units in question. Then, for all $\beta$ which is conjugated to $\alpha$ over $\mathbb{Q}$ we have that $|\beta| \leq 1$. In particular the product of all such $\beta$'s has absolute value less than equal to $1$. On the other hand the product must be an integer, by the hypothesis on $\alpha$, hence it is either $1$ or $0$. If it is non-zero then each term in the product must be equal to $1$, hence $\alpha =1$. The last can only happen if all the root's of unity are the same(we have equality in the triangle inequality.)

  • $\begingroup$ I would really like to know a non-analysis proof of this fact. Or, better, the more general fact that the arithmetic mean of some roots of unity is an algebraic integer if and only if these roots are either all equal or the arithmetic mean is zero. $\endgroup$ – darij grinberg Mar 14 '10 at 15:47
  • $\begingroup$ Well, the triangle inequality is not quite algebraic in my opinion ;) - in fact, the problem is that you use the fundamental theorem of algebra, which is analysis (despite the name). Most proofs that use complex numbers would work well in an abstract field extension of Q, but here you explicitly use that the roots of unity lie in C. $\endgroup$ – darij grinberg Mar 16 '10 at 12:56
  • $\begingroup$ I was afraid you were going to say that about the triangle inequality, and I sort of see your point. Now, from the point of view of a number theorist $\mathbb{Q}_p$'s -where $p$ includes $\infty$- are just algebraic entities that are attached to the arithmetic of $\mathbb{Q}$. Now for some, $\Z_p$ might be a very analytical object and for some others like me it is an algebraic one. In particular, Cauchy-Schwarz inequality on $\bar{Q}_p$ is an algebraic phenomenon for me while it is not for you. $\endgroup$ – Guillermo Mantilla Mar 17 '10 at 2:57

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