# How does 2 dimensional space transforms in 3-harmonic-oscillators problem?

I'm reading Lie Algebras in Particle Physics by Howard Georgi. When talking about an example of harmonic oscillator (on Page 27), it says

"The 2 dimensional space transforms by the representation $$D_2$$ below"

and then gives a unitary irreducible representation of $$S_3$$. I'm puzzled about what happened here. How does this representation come out? Is that a subgroup of $$SO(2)$$?

• Is that a subgroup of $SO(2)$? Do you know what 2D rotation matrices look like? Do all of these have that form? Nov 3, 2023 at 5:01
• Sorry for my carelessness. What I'm thinking is whether these first 3 representations are connected with some rotation transformations. And it seems that these last 3 are something to do with rotation+reflection. Nov 8, 2023 at 7:54

Invoke Cayley's Theorem, which states that every group is isomorphic to a permutation group. Following that, if we consider the symmetry group of an equilateral triangle, which is isomorphic to $$S_{3}$$, the permutation group of order three. For a chemist $$S_{3}$$ is the symmetry group of the Ammonia.
The group elements are $$\{E,R_{1},R_{2},R_{3},R_{4},R_{4}\}$$, where $$E$$ is the identity element, $$R_{1}$$ and $$R_{2}$$ rotate the triangle by the angle $$2\pi/3$$ and $$4\pi/3$$ about the $$z-$$ axis, respectively. The remaining operations $$R_3$$, $$R_4$$, and $$R_5$$ denote reflections at the axes. Pictorially,
One can see immediately that the action of $$R_{1}$$ is given by $$\begin{eqnarray} D(R_{1})\mathbf{e}_{x} & = & -\frac{1}{2} \mathbf{e}_{x} + \frac{\sqrt{3}}{2} \mathbf{e}_{y} ,\quad D(R_{1})\mathbf{e}_{y} & = & -\frac{\sqrt{3}}{2} \mathbf{e}_{x} - \frac{1}{2} \mathbf{e}_{y} ,\quad D(R_{1})\mathbf{e}_{z} & = & \mathbf{e}_{z} \end{eqnarray}$$ Or $$\begin{eqnarray} D(R_{1}) = \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{eqnarray}$$
This representation is reducible (look into the $$2\times 2$$ block). Do the same for the rest of the transformations, which will immediately give the matrices you are asking.