9
$\begingroup$

It is well-known that the entries of the character table of a finite group are sums of roots of unity.

Question: Is the converse true? Explicitly, given $z\in \mathbb{Z}[\mu_\infty]$, can I find a finite group $G$ and irreducible character $\chi$ with $z=\chi(g)$ for some $g\in G$?

It's certainly true if I dropped the irreducibility; in this case one can take $G$ to be abelian, and just hit each summand of $z$ one at a time and take a direct sum. Also noteworthy is that if one has a single $G$ one can compose with a surjection to $G$, so it will be true for infinitely-many groups if it is for one.

A related question (possibly less trivial): If it's YES, is there a natural "minimal" subclass of finite groups that suffice for this purpose?

If it's "NO," are the obstructions completely understood?

This is partially meant as a (hopefully) easier relative to a previous question: A Realization Problem for Character Tables. I apologize in advance if it is trivial (one way or the other), as I fear it may be.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
11
$\begingroup$

I think you can take your idea and make it into an irreducible representation.

Take $m\in\mathbb{Z}$ large enough that all the summands are $m$th roots of unity. Let $n$ be the number of such summands. Represent $G=(\mathbb{Z}/m\mathbb{Z})^n$ on $\mathbb{C}^n$ by diagonal matrices, with the $i$th generator of the group acting by multiplication by the primitive $m$th root of unity on the $i$th coordinate. Your $z$ appears as a value of the character, but it's not irreducible.

To make it irreducible, you just need to intertwine the coordinates. So let the permutation group $S_n$ act by automorphisms on $G$ by permuting the generators. Likewise, represent it on $\mathbb{C}^n$ by permuting the coordinates. This defines a covariant representation, and hence a representation of the semidirect product $S_n\ltimes G$. It should be irreducible for $m>1$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ All the irreducible representations of $S_n\ltimes G$ are obtained as follows: start with a one-dimensional representation $\chi$ of $G$, extend it trivially to $S\ltimes G$, where $S$ is the stabiliser of $\chi$ in $S_n$, take any irreducible representation $\rho$ of $S$, viewed as a representation of $S\ltimes G$. Then ${\rm Ind}^{S_n\ltimes G}\chi\otimes \rho$ is an irreducible representation, and they all arise in this way. Is the representation you have constructed somewhere among these? I don't quite see that right now. $\endgroup$
    – Alex B.
    Mar 6, 2013 at 10:31
  • $\begingroup$ So here is the problem, I think: suppose that among the $m$-th roots of unity, the first two are 1, and the remaining ones are distinct primitive $m$-th roots of unity. Then the subgroup whose $S_n$-component is trivial and whose $G$-component is of the form $(x,y,1,1,\ldots, 1)\in (\mathbb{Z}/m\mathbb{Z})^n$ is the kernel of the representation you are constructing. But this subgroup is not normal. So, unless I am mistaken, what you have constructed is not a representation. $\endgroup$
    – Alex B.
    Mar 6, 2013 at 10:38
  • $\begingroup$ But where is the mistake in his proof? He has provided an explicit construction of an irreducible imprimitive subgroup of ${\rm GL}(n,{\mathbb C})$ that is isomorphic to the wreath product of a cyclic group of order $m$ by $S_n$. $\endgroup$
    – Derek Holt
    Mar 6, 2013 at 11:00
  • 1
    $\begingroup$ Alex, I think you have misunderstood Bob's construction. In your notation, $\chi$ is the representation sending each generator to a primitive $m$th root of unity (the same one), so your $S$ is $S_n$, $\rho$ is the natural $n$-dimensional representation of $S$, and Bob's your uncle. $\endgroup$
    – user30035
    Mar 6, 2013 at 11:03
  • $\begingroup$ The image of $S_n \ltimes G$ in $\mathrm{GL}(n, \mathbb{C})$ is the set of all monomial matrices having $m$-th roots of unity as nonzero entries. (By monomial I mean: having exactly one nonzero entry in each row and column.) In fact, every $z$ that can be written as a sum of $n$ or fewer $m$-th roots of unity occurs as trace of such a matrix. $\endgroup$
    – Frieder Ladisch
    Mar 6, 2013 at 14:20
6
$\begingroup$

I think that a clearer way to construct the desired example is not to let the acting group be $S_n$, but take the cyclic group $C = C_n$ instead. Thus, start with the group $G = (C_m)^n$, as before, and let $C$ act by cycling the direct factors of $G$. Let $\lambda$ be a linear character of $G$ whose kernel is exactly the product of all but the first cyclic factor of $G$ (so the restriction of $\lambda$ to the first factor of $G$ is faithful). Only the identity of $C$ stabilizes $\lambda$, so working in the semidirect product $\Gamma = GC$, the stabilizer of $\lambda$ is $G$, and it follows that the induced character $\chi = \lambda^\Gamma$ is irreducible.

Now the restriction $\chi_G$ is a sum of $m$ linear characters $\lambda_i$, where the kernel of $\lambda_i$ is the product of all but the $i$th cyclic factor of $G$. Now given any list of $n$ $m$th roots of unity $\varepsilon_i$, we can choose an element $g$ of $G$ such that $\lambda_i(g) = \varepsilon_i$, and thus $\chi(g) = \sum\varepsilon_i$, as wanted.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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