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This might be a silly question, but I need help understanding an example of a representation of a group ,the following example is from Yvette Kosmann-Scwartbach's book called Groups and Symmetries:

''Let $t\in S_3 $ be the transposition $123\to 132$ and $c$ the cyclic permutation $123\to 231$ that generates $S_3$.We set $j=e^{2i\pi/3}$ ,so that $j^2+j+1=0$.We can represent $S_3$ on $\mathbb{C}^2 $ by defining $ρ(e)=I_2$,$ρ(t)=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ and $ρ(c)=\begin{pmatrix} j & 0 \\ 0 & j^2 \\ \end{pmatrix}$''.

My understanding is that $ρ_g , (ρ_g=ρ(g))$ is defined by the action of $S_3$ on $\mathbb{C^2}$ and so for $ρ(e)=I_2$ we get $ρ(e)=[a_{e(1)},a_{e(2)}]$,where $e$ the ''neutral'' permutation and $a=\{a_1,a_2\}=\{(1,0),(0,1)\}$ a basis of $\mathbb{C^2}$ and thus $ρ(e)=[a_1,a_2]=I_2 $ ,since $e:123 \to 123$.
That kind of logic doens't seem to work on the rest.What am I missing and how does the author introduce $j$ on the matrices?
Thank you in advance !

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    $\begingroup$ I'm not sure I understand your question. You need to show that you have a homomorphism from $S_3$ to $GL_2(\mathbb C)$ compatible with those two datapoints $\rho(c)$ and $\rho(t)$. The $j$ is coming from the fact that since $t^3=e$, $\rho(t)^3=I_2$, and so the only possible eigenvalues of $\rho(t)$ are 1, j, and $j^2$. All the matrices in a reputation are diagonalizable, and there is no loss in generality picking a basis such that one particular matrix ends up diagonal. If $\rho(t)$ is the matrix you make diagonal, there aren't many choices. $\endgroup$
    – Aaron
    Oct 9, 2022 at 15:14
  • $\begingroup$ I,m sry $ρ(c)=\begin{pmatrix} j & 0 \\ 0 & j^2 \\ \end{pmatrix}$,not ρ(t),i just edited. $\endgroup$
    – GGG
    Oct 9, 2022 at 15:39
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    $\begingroup$ Regardless, any relation that holds in the original group holds in the representation, and any collection of matrices that satisfies all the relations between all of the generators will be a representation. The j is coming in because you have a generator with order 3. However, there are many other choices. If you want a real 2 dimensional representation, take a reflection and a rotation by 120 degrees. $\endgroup$
    – Aaron
    Oct 9, 2022 at 15:43
  • $\begingroup$ Ah I see ,I see .Thank you Aaron ! $\endgroup$
    – GGG
    Oct 9, 2022 at 15:47

1 Answer 1

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Here is one way to understand this. For each $\sigma\in S_3$ you have $\rho(\sigma): \mathbb{C}^2 \to \mathbb{C}^2$ and this is simply given as matrix multiplication. So, for example $$ \rho(t)\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}y \\ x \end{pmatrix}. $$

The point is that since $S_3$ is generated by $t$ and $c$, the three (really only need $2$ equations you have is enough to fully define $\rho$. You can't just select any matrices for the value of $\rho$ at $t$ and $c$. So, there are things to check.

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  • $\begingroup$ And so if every element of $S_3$ is a ''power'' of c or t ,then every ρ is a power of ρ(t) and (or ) ρ(c) $\endgroup$
    – GGG
    Oct 9, 2022 at 15:50
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    $\begingroup$ @GGG It isn't that every element in $S_3$ is a power of $x$ and $t$, you can also have products. For example, to find $\rho$ at $(1\;3)$, you can use $(1\;3) = tc$ (assuming I understand your notation). So $\rho(1\;3) = \rho(tc) = \rho(t)\rho(c)$. $\endgroup$
    – Thomas
    Oct 9, 2022 at 16:21
  • $\begingroup$ Yeah, of course you can ! Thank you for your time Thomas ! $\endgroup$
    – GGG
    Oct 9, 2022 at 16:35

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