Finite Group D3

The Dihedral Group $D_3$ is one of the two groups of Order 6. It is the non-Abelian group of smallest Order. Examples of $D_3$ include the Point Groups known as $C_{3h}$, $C_{3v}$, $S_3$, $D_3$, the symmetry group of the Equilateral Triangle, and the group of permutation of three objects. Its elements $A_i$ satisfy ${A_i}^3=1$, and four of its elements satisfy ${A_i}^2=1$, where 1 is the Identity Element. The Cycle Graph is shown above, and the Multiplication Table is given below.

$D_3$	1	$A$	$B$	$C$	$D$	$E$
1	1	$A$	$B$	$C$	$D$	$E$
$A$	$A$	1	$D$	$E$	$B$	$C$
$B$	$B$	$E$	1	$D$	$C$	$A$
$C$	$C$	$D$	$E$	1	$A$	$B$
$D$	$D$	$C$	$A$	$B$	$E$	1
$E$	$E$	$B$	$C$	$A$	1	$D$

The Conjugacy Classes are $\{1\}$, $\{A, B, C\}$

$$A^{-1}AA = A$$ (1)

$$B^{-1}AB = C$$ (2)

$$C^{-1}AC = B$$ (3)

$$D^{-1}AD = C$$ (4)

$$E^{-1}AE = B,$$ (5)

and $\{D$, $E\}$,

$$DA^{-1}D = E$$ (6)

$$B^{-1}DB = D.$$ (7)

A reducible 2-D representation using Real Matrices can be found by performing the spatial rotations corresponding to the symmetry elements of $C_{3v}$. Take the z-Axis along the $C_3$ axis.

$$I = R_z(0) = \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$$ (8)

$$A = R_z({\textstyle{2\over 3}}\pi) = \left[\begin{array}{cc}\cos({\te... ...({\textstyle{2\over 3}}\pi) & \cos({\textstyle{2\over 3}}\pi)\end{array}\right]$$

$$= \left[\begin{array}{cc}-{\textstyle{1\over 2}}& -{\textstyle{1\ov... ...}\\ {\textstyle{1\over 2}}\sqrt{3} & -{\textstyle{1\over 2}}\end{array}\right]$$ (9)

$$B = R_z({\textstyle{4\over 3}}\pi) = \left[\begin{array}{cc}-{\textst... ...\\ -{\textstyle{1\over 2}}\sqrt{3} & -{\textstyle{1\over 2}}\end{array}\right]$$ (10)

$$C = R_C(\pi) = \left[\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right]$$ (11)

$$D = R_D(\pi) = CB = \left[\begin{array}{cc}{\textstyle{1\over 2}}& -{... ...\\ -{\textstyle{1\over 2}}\sqrt{3} & -{\textstyle{1\over 2}}\end{array}\right]$$ (12)

$$E = R_E(\pi) = CA = \left[\begin{array}{cc}{\textstyle{1\over 2}}& {\... ...\\ {\textstyle{1\over 2}}\sqrt{3} & -{\textstyle{1\over 2}}\end{array}\right].$$ (13)

To find the irreducible representation, note that there are three Conjugacy Classes. Rule 5 requires that there be three irreducible representations satisfying

$$h={l_1}^2+{l_2}^2+{l_3}^2 = 6,$$ (14)

so it must be true that

$$l_1 = l_2 = 1, l_3 = 2.$$ (15)

By rule 6, we can let the first representation have all 1s.

${\mathcal D}_3$	1	$A$	$B$	$C$	$D$	$E$
$\Gamma_1$	1	1	1	1	1	1

To find a representation orthogonal to the totally symmetric representation, we must have three $+1$ and three $-1$ Characters. We can also add the constraint that the components of the Identity Element 1 be positive. The three Conjugacy Classes have 1, 2, and 3 elements. Since we need a total of three $+1$s and we have required that a $+1$ occur for the Conjugacy Class of Order 1, the remaining +1s must be used for the elements of the Conjugacy Class of Order 2, i.e., $A$ and $B$.

$D_3$	1	$A$	$B$	$C$	$D$	$E$
$\Gamma_1$	1	1	1	1	1	1
$\Gamma_2$	1	1	1	$-1$	$-1$	$-1$

Using Group rule 1, we see that

$$1^2+1^2+{\chi_3}^2(1) = 6,$$ (16)

so the final representation for 1 has Character 2. Orthogonality with the first two representations (rule 3) then yields the following constraints:

$$1 \cdot 1 \cdot 2+1 \cdot 2 \cdot \chi_2+1 \cdot 3 \cdot \chi_3 = 2+2\chi_2+3\chi_3 = 0$$ (17)

$$1 \cdot 1 \cdot 2+1 \cdot 2 \cdot \chi_2+(-1) \cdot 3 \cdot \chi_3 = 2+2\chi_2-3\chi_3 = 0.$$ (18)

Solving these simultaneous equations by adding and subtracting (18) from (17), we obtain $\chi_2=-1$, $\chi_3=0$. The full Character Table is then

$D_3$	1	$A$	$B$	$C$	$D$	$E$
$\Gamma_1$	1	1	1	1	1	1
$\Gamma_2$	1	1	1	$-1$	$-1$	$-1$
$\Gamma_3$	2	$-1$	$-1$	0	0	0

Since there are only three Conjugacy Classes, this table is conventionally written simply as

$D_3$	1	$A=B$	$C=D=E$
$\Gamma_1$	1	1	1
$\Gamma_2$	1	1	$-1$
$\Gamma_3$	2	$-1$	0

Writing the irreducible representations in matrix form then yields

$$1 = \left[\begin{array}{cccc}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$$ (19)

$$A = \left[\begin{array}{cccc}-{\textstyle{1\over 2}}& -{\textstyle{1\... ...extstyle{1\over 2}}& 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$$ (20)

$$B = \left[\begin{array}{cccc}-{\textstyle{1\over 2}}& {\textstyle{1\o... ...extstyle{1\over 2}}& 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$$ (21)

$$C = \left[\begin{array}{cccc}-1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{array}\right]$$ (22)

$$D = \left[\begin{array}{cccc}{\textstyle{1\over 2}}& -{\textstyle{1\o... ...xtstyle{1\over 2}}& 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{array}\right]$$ (23)

$$E = \left[\begin{array}{cccc}{\textstyle{1\over 2}}& {\textstyle{1\ov... ...tstyle{1\over 2}}& 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{array}\right].$$ (24)

See also Dihedral Group, Finite Group D4, Finite Group Z6