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?
Best Answer
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:
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.