Class function and character of $S_3$ representation

Question

I particularly need help with question 2. The $\textit{character table}$ for $S_3$ is given as follows:$$\begin{array}{c|c|c|} & \text{1} & \text{(12)} & \text{(123)} \\ \hline \text{$X_1$} & 1 & 1 & 1 \\ \hline \text{$X_2$} & 1 & -1 & 1 \\ \hline \text{$X_3$} & 2 & 0 & -1 \\ \hline \end{array}$$

and the $\textit{class function, f}$, whose values on the $\textit{conjugacy classes}$ are given by: $$\begin{array}{c|c|c|} & \text{1} & \text{(12)} & \text{(123)} \\ \hline \text{f} & 19 & -1 & -2 \\ \hline \end{array}$$

$\textbf{Question 1}$ Express f as a linear combination of the characters $X_1, X_2$ and $X_3$.

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I calculated: $(f, X_1)=8/3$, $(f, X_2)=3$ and $(f, X_3)=20/3$. So: $f={8/3}{X_1}+3{X_2}+20/3{X_3}$

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$\textbf{Question 2}$ Is f the $\textit{character}$ of a representation?

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I know that the character is related to the trace of the representation but I cannot think how to apply this. Thanks

Answer 1

Your computations are wrong. For instance, we find that $$(f,X_1) = \frac{1}{6}\left(1\cdot 19 + \color{red}3\cdot 1 \cdot (-1) + \color{red}2\cdot 1 \cdot (-2)\right) = 2.$$ (There are $\color{red}3$ elements in the conjugacy class of $(12)$ and $\color{red}2$ elements in the conjugacy class of $(123)$.)

In the end we get $f = 2X_1 + 3X_2 + 7X_3$.

For the second question, let $\rho_i$ be the irreducible representation with character $X_i$. Then $2\rho_1 \oplus 3\rho_2 \oplus 7\rho_3$ has character $f$.

Answer 2

A class function is a character if and only if it is a nonnegative integral combination of irreducible characters, so no. The key reason for that is complete reducibility: every $G$-module is isomorphic to a direct sum of irreducible modules.

Answer 3

A representation is a direct sum of irreducible representations. If $f$ was the character of a representation, your result would get you that it could be written as an integer valued sum of the irreducible representations.

Answer 4

HINT:

You got the inner products with $f$ wrong ( there are some weights, coming from the size of the conjugacy classes appearing there).

Another approach: you need to find $a$, $b$, $c$ so that $$a(1,1,1) + b(1,-1,1) + c(2,1,0) = (19, -1,-2)$$

which gets the (unique) solution $a=2$, $b=3$, $c=7$. Since these are natural numbers, $f$ is the character of a representation.