by ericc
Last Updated October 09, 2019 17:20 PM

I know that up to isomorphism there is only one irreducible representation of $D_3$ of dimension 2. Write $D_3=\{1, x, x^2, xy, x^2y, y\}$, where $x^3 = 1$, $y^2 = 1$, $xy = yx^2$. Given any $\theta\in[0, 2\pi)$, I think the following is one of such irreducible representation:

$R_y = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right],$

$R_x = \left[ \begin{array}{cc} -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}e^{-i\theta} \\ \dfrac{\sqrt{3}}{2}e^{i\theta} & -\dfrac{1}{2} \end{array} \right].$

Am I correct? If the answer is yes, then what is the isomorphism $T:\mathbb{C^2}\rightarrow\mathbb{C^2}$ between this representation and the standard representation

$R'_y = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right],$

$R'_x = \left[ \begin{array}{cc} -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \end{array} \right].$

On the other hand, if the answer is no, then what forces $\theta$ to be $0$?

Thanks!

You have $R_xA=AR'_x$, where $$ A=\begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix} $$ So the two representations are equivalent.

Updated April 20, 2015 09:08 AM

Updated April 04, 2017 17:20 PM

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