# Class function as a character

Suppose $G$ is a finite group, and $\{ \chi_1,\chi_2,\cdots,\chi_k \}$ be the complete set of irreducible complex characters of $G$. If $\theta$ is a class function on $G$, i.e. function from $G$ to $\mathbb{C}$ which takes same values on conjugacy classes of $G$,

Question: When can we say that $\theta$ is a character of $G$?

If we take characters $\chi_i$ and $\chi_j$, then their sum $\psi=\chi_i + \chi_j$ is also a character of $G$. Looking in other way, the character $\chi_j$ is difference of two characters, $\psi -\chi_i$. So a difference of two characters may be a character of $G$. So if a class function $\theta$ when expressed as an integral linear combination of $\chi_i$'s, then it will be a character provided $\theta(1)>0$.

But in the online notes on "Group Representation Theory", by Daniel Bump (here), he has proved that the class function (namely $\theta^G$ in Theorem 2.5.1 in the notes) is character by showing that it is non-negative integral linear combination of irreducible representations $\chi_i$'s. Why should we show that it is non-negative integral linear combination?

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Every character is the trace of a (finite-dimensional) representation; by Maschke's theorem, every such representation is a direct sum of irreducible ones, and therefore the original character is equal to the sum of the traces of these irreducible representations, possibly with multiplicity. The multiplicities are all positive integers, because the answer "how many times does a copy of the irrep $V$ appear inside the representation $W$?" is always a non-negative integer.

Conversely, if a class function $\psi$ is equal to $\sum_{i=1}^m c_i \chi_i$ where each $c_i$ is a positive integer, then $\psi$ is the trace of the representation $\bigoplus_{i=1}^k c_i \pi_i$, where we choose each $\pi_i$ to be an irreducible representation whose trace is $\chi_i$.

That was my attempt to answer the second question. To answer the first question: it is a character when it is a character.

Less facetiously, take the inner product of your class function with each of the irreducible characters, using the normalization if the inner product which makes the irreducible characters an orthonormal basis for $\ell^2(G)$. If each of the resulting numbers is a non-negative integer, this expresses your class function as a non-negative integral combination of irreducible characters.

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if $\chi$ is any character than $-\chi$ is a class function, but not a character. So my question is when a class function is character? I agree that "any character can be expressed as non-negative integral linear combination of irreducible characters. –  Marshal Kurosh Sep 22 '11 at 9:55
Marshal, I was answering your last question. I am not sure how to answer your repeated question: it is a character when it is a character. –  user16299 Sep 22 '11 at 9:58

While one often proves that θ is a character by showing θ(1) > 0, this is not sufficient, even for differences of characters.

For instance, consider the non-abelian group of order 6 with classes 1a, 2a, 3a. It has a two-dimensional character (2,0,−1), and subtracting the trivial character (1,1,1) one gets the virtual character (integral linear combination of irreducible characters) (1,−1,−2). This cannot possibly be a character since a one-dimensional character is in fact a homomorphism, but −2 has infinite order in $\mathbb{C}^\times$. Alternatively, one already knows the one-dimensional characters: (1,1,1) and (1,−1,1).

The subset of characters within the abelian group of virtual characters is a little funny. It is not like positive and negative integers where either θ or −θ is a character. It is more like the X–Y plane, where there is the first quadrant (X>0, Y>0) and its opposite the third quadrant (X<0, Y<0), but also the second and fourth quadrants that are just weird mixtures.

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