# The Character Ring

Proposition 3.1.1: The sum or product of two characters of the finite group $G$ is a character.

Proof. (Click to Expand/Collapse)

If $(\pi, V)$ and $(\pi', V')$ are representations, let us choose bases for $V$ and $V'$, so that we can identify $V =\mathbb{C}^d$ and $V' =\mathbb{C}^{d'}$ for $d = \dim (V)$, $d' = \dim (V')$. Then $\pi (g)$ and $\pi' (g)$ may be identified with their matrices $\Pi (g) = \left(\begin{array}{cc} \pi (g) & 0\\ 0 & \pi' (g) \end{array}\right) \hspace{2em} \text{(block matrix)}$ gives a representation with character $\chi_{\pi} + \chi_{\pi'}$. This shows that a sum of characters is a character.

On the other hand, let $W$ be the $d d'$-dimensional vector space of $d \times d'$ rectangular matrices. Define an action $\Pi' : G \longrightarrow \operatorname{GL} (W)$ by
 $\Pi' (g) X = \pi (g) X \pi' (g)^t, \hspace{2em} X \in W,$ (3.1.1)

where as usual $\pi' (g)^t$ is the transpose matrix. If $X$ has entries $x_{i j}$, $\pi (g)$ has entries $a_{i j}$ and $\pi' (g)$ has entries $b_{i j}$, then $\Pi' (g) X$ has $i, j$-th entry $\sum_{k, l} a_{i k} x_{k l} b_{j l} = \sum_{k, l} \left( a_{i k} b_{j l} \right) x_{k l}$ where the left-hand side comes from simple matrix multiplication remembering to take the transpose of $\pi' (g)$. The right-hand side is obviously the same, but we have rewritten it in order to facilitate thinking about the matrix representing this linear transformation. If we refer this to a matrix by taking the $d d'$ entries $x_{i, j}$ in some order, the entries in this matrix will be the $(d d')^2$ coefficients $a_{i k} b_{j l}$, and the diagonal terms will be the contribution of $i = k$ and $j = l$, that is $\operatorname{tr} \; \Pi' (g) = \sum_{i, j} a_{i i} b_{j j} = \left( \sum_i a_{i i} \right) \left( \sum_j b_{j j} \right) = \chi_{\pi} (g) \chi_{\pi'} (g) .$ This shows that the product of two characters is a character.

It is not true that the characters form a ring, since although they are closed under addition and multiplication, they are not closed under subtraction. However generalized characters do form a ring. We recall that a generalized character is a difference of two characters. If $f$ is any class function on a finite group $G$, and if $\chi_1, \cdots, \chi_h$ are the irreducible characters, we know that $f$ can be written uniquely as $\sum c_i \chi_i$. Clearly:

• The class function $f$ is a character if and only if the $c_i$ are nonnegative integers.
• The class function $f$ is a generalized character if and only if the $c_i$ are integers.

Proposition 3.1.2: The generalized characters form a ring, called the character ring.

Proof. (Click to Expand/Collapse)

It is clear from Proposition 3.1.1 that the sum or product of two generalized characters is a generalized character; since the difference of two generalized characters is a generalized character, they are indeed closed under the ring operations and form a subring of the ring of class functions.

Exercise 3.1.1: Let $G = S_3$, and let $\chi_1, \chi_2, \chi_3$ be the irreducible characters, so $\chi_3 (1) = 2$. Determine the decomposition of $\chi_3^n$ into irreducibles when $n \le 10$. That is, write $\chi_3^n = a_n \chi_1 + b_n \chi_2 + c_n \chi_3$ and compute $a_n, b_n$ and $c_n$.

Exercise 3.1.2: Let $G$ and $H$ be groups, and let $\chi, \sigma$ be characters of $G$ and $H$. Prove that $\chi \otimes \sigma$ is a character, where $\chi \otimes \sigma$ is defined by $(\chi \otimes \sigma) (g, h) = \chi (g) \sigma (h) .$ Hint: Let $\pi : G \longrightarrow \operatorname{GL} (n, \mathbb{C})$ and $\pi' : H \longrightarrow \operatorname{GL} (m, \mathbb{C})$ be representations with characters $\chi$ and $\sigma$. Let $W$ be the vector space of $n \times m$ rectangular complex matrices, and define $\Pi : G \times H \longrightarrow \operatorname{GL} (W)$ by $\Pi (g, h) X = \pi (g) X \pi' (h)^t, \hspace{2em} (X \in W) .$ Compute the character.

Exercise 3.1.3: In the last exercise, if $\chi$ and $\sigma$ are irreducible characters of $G$ and $H$, prove that $\chi \otimes \sigma$ is an irreducible character of $G \times H$. (Hint: one way is to compute $\left\langle \chi \otimes \sigma, \chi \otimes \sigma \right\rangle_{G \times H}$.

Exercise 3.1.4: Prove that if $\chi_1, \cdots, \chi_h$ are distinct irreducible characters of $G$, and $\sum_i \chi_i (1)^2 = |G|$ then every irreducible character is one of the $\chi_i$.

Exercise 3.1.5: Prove that every irreducible character of $G \times H$ is of the form $\chi \otimes \sigma$ for some irreducible characters $\chi$ and $\sigma$ of $G$ and $H$. (Hint: use the criterion in the last exercise.)