Complete set of irreducible representations of small groups (2)

Posted: February 10, 2011 in Representations of Finite Groups
Tags: ,

Example 3. Irreducible Representations of D_8 : In part 5) in here we showed that D_8, the dihedral group of order 8, has five non-equivalent irreducible representations. Four of them have degree one and one of them has degree two. So let D_8 = \langle g_1, g_2 : \ g_1^2=g_2^4=(g_1g_2)^2=1 \rangle. Every element of D_8 can be written uniquely as g_1^j g_2^k, where 0 \leq j \leq 1 and 0 \leq k \leq 3.

1) Representations of degree one. These have been completely described in Example 2 in this post. In our case m=4, which is an even number. So there are four representations of degree one, say \rho_1, \rho_2, \rho_3, \rho_4. They are defined on D_8 by

\rho_1(g)=1, \ \rho_2(g)=(-1)^k, \ \rho_3(g)= (-1)^j, \ \rho_4(g)= (-1)^{j+k},

where g = g_1^j g_2^k, \ 0 \leq j \leq 1, \ 0 \leq k \leq 3.

2) Representation of degree two. Here we described m representations of degree two for the dihedral group D_{2m}. We proved in here that all except those corresponding to \zeta = \pm 1, \ m \geq 3, \rho_0 are irreducible. In our case m=4. So there are two irreducible representations and we’ll pick the one corresponding to \zeta = i. Let’s call it \rho. So if  v \in \mathbb{C}^2, then \rho is defined on D_8 by

\rho(g_1^jg_2^k)(v)= \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}^j \begin{pmatrix} \zeta & 0 \\ 0 & \zeta^{-1} \end{pmatrix}^k v,

for all 0 \leq j \leq 1 and 0 \leq k \leq 3. I’ll leave it to the reader to find an explicit formula for \rho(g_1^j g_2^k)(v).

Example 4. Irreducible Representations of Q_8 : In part 4) in here we showed that Q_8, the quaternion group of order 8, has five non-equivalent irreducible representations. Four of them have degree one and one of them has degree two.

1) Representations of degree one. These were fully described in Example 1 in this post. We showed that if \rho is a degree  one representation of Q_8, then \rho(1)=\rho(-1)=1, \ \rho(i)=\pm 1, \ \rho(j)=\pm 1. Therefore  \rho(-i)=\rho(-1)\rho(i)=\rho(i) and similarly \rho(-j)=\rho(j). Finally, since k=ij, we  have \rho(k)=\rho(i)\rho(j) and \rho(-k)=\rho(k).

2) The representation of degree two. So we need 2 \times 2 matrices with entries in \mathbb{C} which satisfy the same relations as i,j, k do. Let A = \begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix} and B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. Let C=AB. Now if v \in \mathbb{C}^2, we define \rho(i)(v)=Av and \rho(j)(v)=Bv. Since A,B satisfy the same relations as i,j do, \rho will give us a group homomorphism from Q_8 to \text{GL}(\mathbb{C}^2) and hence it is a representation of degree two. So, for example \rho(-1)(v)=\rho(i^2)(v)=\rho(i) \rho(i)(v)=A^2v, etc. It is easy to see that A,B have no common eigenvector and thus \rho is irreducible by the theorem we proved in here.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s