8
$\begingroup$

The representations of a finite group can be understood by their irreducible characters. A class function is a function from the group to the complex numbers that is constant on the conjugacy classes.

I know that any linear combination of the irreducible characters is the character of some representation. I also know that not all class functions are characters of a representation.

Let's say that I don't know all the irreducible characters of a group, but I come across a class function whose inner product with itself is 1. My question is: How do I know whether this function is actually the character of an irreducible representation?

More generally: How do I know whether a given class function is the character of some representation of a group without knowing all the irreducible representations?

EDIT: I see this question with answers: Class function as a character. This almost answers my question. To clarify what I am specifically interested in knowing, if I have found some irreducible representations of a group $G$. Say I have $\chi_1, \dots, \chi_m$. I know I haven't found all of them because I know the number of conjugacy classes. Then, say, I some other non-irreducible character $\chi$ and I know, say, that this is the character of some representation. Then I subtract a linear combination of $\chi_1, \dots, \chi_m$, and define the class function $\psi = \chi - (a_1\chi_1 + \dots + a_m\chi_m)$. How do I know whether this $\psi$ is the character of some representation?

$\endgroup$
  • $\begingroup$ First step is to find all irreducible characters. You are essentially asking how to avoid this first step. But if you could do what you asked, then you could just use that to find all irreducible characters. $\endgroup$ – Somos Mar 27 '19 at 19:13
  • $\begingroup$ Not any linear combination of characters. The coefficiens have to be positive integers. Otherwise the zero function would be character of a representation. $\endgroup$ – Somos Mar 27 '19 at 19:23
  • 1
    $\begingroup$ The zero function is the character of a representation - the zero-dimensional representation. $\endgroup$ – Qiaochu Yuan Mar 27 '19 at 20:08
  • $\begingroup$ Also if the class function $c(x) = c( g^{-1} x g)$ is $|G|\Bbb{Z}$-valued and $c(1)$ is large enough compared to $\sup_{g \ne 1} |c(g)|$ then $c$ is a character. $\endgroup$ – reuns Mar 28 '19 at 15:20
8
$\begingroup$

To clarify what I am specifically interested in knowing, if I have found some irreducible representations of a group $G$. Say I have $\chi_1, \dots, \chi_m$. I know I haven't found all of them because I know the number of conjugacy classes. Then, say, I some other non-irreducible character $\chi$ and I know, say, that this is the character of some representation. Then I subtract a linear combination of $\chi_1, \dots, \chi_m$, and define the class function $\psi = \chi - (a_1\chi_1 + \dots + a_m\chi_m)$. How do I know whether this $\psi$ is the character of some representation?

This question is much easier than your general question; assuming the $a_i$ are nonnegative integers, the answer is if and only if $\langle \chi, \chi_i \rangle \ge a_i$ for all $i$. This follows from:

Lemma: A class function $\chi$ is the character of a representation iff for every irreducible character $\chi_i$, $\langle \chi, \chi_i \rangle$ is a nonnegative integer.

Proof. If $\chi$ is the character of a representation $V$ then $\langle \chi, \chi_i \rangle$ is the multiplicity of the irreducible representation $V_i$ corresponding to $\chi_i$ in $V$, so this condition is clearly necessary. On the other hand, if this condition holds, then $\chi = \sum \langle \chi, \chi_i \rangle \chi_i$, and hence the direct sum of $\langle \chi, \chi_i \rangle$ copies of $V_i$ is a representation with character $\chi$. $\Box$

If you set $a_i = \langle \chi, \chi_i \rangle$ then you've removed the components of the representation corresponding to $\chi$ which correspond to the irreducibles with character $\chi_i$. So all you're left with is the components corresponding to the irreducibles you haven't found yet.

$\endgroup$
2
$\begingroup$

There is another angle at this. As so well-explained by @Qiaochu Yuan, if one knows the irreducible characters $\chi$ of the finite group $G$, then for some class function $\varphi$ you need to compute all inproducts $[\varphi,\chi]$ with these irreducibles and check if all the multiplicities are non-negative integers.

A different situation is that one has information about a set of subgroups $\mathcal{H}$, in the sense that $\varphi_H$ is indeed a character for all subgroups $H \in \mathcal{H}$. There is an old and famous result of Richard Brauer (sometimes referred to as characterization of characters) showing that indeed certain sets of subgroups $\mathcal{H}$ will guarantee that $\varphi$ is a character of $G$. This works for example if $\mathcal{H}=\{\text{all nilpotent subgroups of G}\}$. However, this "testing set" can be made smaller to what is called $p$-elementary subgroups: a $p$-elementary group is a direct product of a $p$-group and a cyclic $p'$-group.

$\endgroup$

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.