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Fix a finite dimensional representation $\rho: G \longrightarrow GL(V)$ of $G$. Its trace is defined as the function $tr:G \longrightarrow F$ defined by $tr(g) = tr(\rho(g))$. Explicitly compute the trace for the tautological representation of $D_4$ of $\mathbb{R}^2$.

Here's my answer, can someone please verify it? Also, is this the same thing that people call the character of a representation? Note: This is NOT homework!

Let $r_1$ denote the counterclockwise rotation by $\frac{\pi}{2}$, $r_2$ the counterclockwise rotation by $\pi$, $r_3$ the counterclockwise rotation by $\frac{3 \pi}{2}$, $H$ the reflection across the horizontal axis, $D$ the reflection across the main diagonal (joining the top left vertex with the bottom right), and $A$ the reflection across the off diagonal (joining the top right vertex with the bottom left), and $id$ the identity element. Then, we have:

$$\rho(r_1) = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$

$$\rho(r_2) = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$$

$$\rho(r_3) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$

$$\rho(id) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

$$\rho(H) = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$

$$\rho(V) = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$$

$$\rho(D) = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$$

$$\rho(A) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

Therefore $tr(r_1) = 0$, $tr(r_2) = -2$, $tr(r_3) = 0$, $tr(id) = 2$, $tr(H) = 0$, $tr(D) = 0$, $tr(A) = 0$, $tr(V) = 0$.

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Looks good to me. As a further check you can do the following. The trace of a 2D rotation matrix is $2\cos\theta$ where $\theta$ is the angle of rotation, because its eigenvalues are $e^{\pm i\theta}$ (if you next need the trace of a 3D rotation, then you need to add $+1$ for the eigenvalue $+1$ from the axis of rotation). The trace of an nD reflection is $(n-1)-1=n-2$ because a reflection has eigenvalues $+1$ with multiplicity $(n-1)$ and $-1$ with multiplicity $1$.

This is, indeed, the character of the representation.

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