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Using the following presentation of the dihedral group $D_{3}$ \begin{equation} D_{3} = \left\langle r,s \mid r^{2} = s^{2} = (rs)^{3} = e \right\rangle \end{equation} There is one (up to isomorphism) irreducible 2-dimensional complex representation \begin{equation*} \begin{matrix} \rho : D_{3} \to \operatorname{GL}(2, \mathbb{C}) \\ r \mapsto \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ s \mapsto \begin{pmatrix} 0 & e^{\frac{-2\pi i}{3}} \\ e^{\frac{2\pi i}{3}} & 0 \end{pmatrix} \end{matrix} \end{equation*} is there an isomorphic complex irreducible representation to this one such that the entries in the matrices are all in $\mathbb{Z}$? I think there should be since $D_{3} \cong S_{3}$.

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  • $\begingroup$ would you like to see one in $SL_2({\Bbb{Z}}_2)$? $\endgroup$
    – janmarqz
    Mar 20, 2019 at 2:26
  • $\begingroup$ I commented cuz $\Bbb Z$ hasn't finite subgroups. $\endgroup$
    – janmarqz
    Mar 20, 2019 at 2:29
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    $\begingroup$ Nitpicking: direct sum of two trivial representations is also two-dimensional, which is inequivalent to $\rho$. You need an extra adjective irreducible. $\endgroup$
    – Orat
    Mar 20, 2019 at 6:25
  • $\begingroup$ @Orat I have edited the question to specify that the representations should be simple. Thanks for your detailed and intuitive answer! $\endgroup$
    – Anfänger
    Mar 20, 2019 at 13:16

1 Answer 1

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Consider the following assignment: $$\sigma \colon r \mapsto \begin{bmatrix}-1 & 0\\ 1 & 1\end{bmatrix},\qquad s \mapsto \begin{bmatrix}1 & 1\\ 0 & -1\end{bmatrix}.$$ You can check that it is indeed a representation of $D_3$ that is equivalent to $\rho$ by comparing their characters.

Geometric meaning. Let $e_1, e_2, e_3$ be the standard basis of $\mathbb{R}^3$. Take $\alpha = e_1 - e_2, \beta = e_2 - e_3$ and consider $2$-dimensional subspace $V = \operatorname{Span}_\mathbb{R}\{\alpha, \beta\} = \{\, x_1e_1 + x_2e_2 + x_3e_3 \mid x_1 + x_2 + x_3 = 0 \,\}$. Define a linear transformation $\sigma_\alpha$ on $V$ as a reflection with respect to $\alpha^\perp = \{\, v \in V \mid \langle \alpha, v \rangle = 0 \,\}$: $$ \sigma_\alpha(v) = v - 2\frac{\langle \alpha, v \rangle}{\langle \alpha, \alpha\rangle}\alpha \qquad (v \in V).$$ The representation matrix of $\sigma_\alpha$ is $\left[\begin{smallmatrix}-1 & 0\\ 1 & 1\end{smallmatrix}\right]$ as $$ \sigma_\alpha\begin{bmatrix}\alpha\\ \beta\end{bmatrix} =\begin{bmatrix}-1 & 0\\ 1 & 1\end{bmatrix}\begin{bmatrix}\alpha\\ \beta\end{bmatrix}.$$ Similarly, the representation matrix of $\sigma_\beta$ is $\left[\begin{smallmatrix}1 & 1\\ 0 & -1\end{smallmatrix}\right]$.

Root system of type A2

Basically, they are symmetry of an equilateral triangle which $S_3$ also acts on.

Symmetry of an equilateral triangle

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    $\begingroup$ For anyone curious about the more general setup where this approach applies, they should look up root systems and Weyl groups (and more generally Coxeter groups and reflection groups). $\endgroup$ Mar 20, 2019 at 13:22

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