It suffices just to consider a linear transformation $f$ such that $f^n=id$ and require $V$ to have no proper subspace invariant under $f$. But I still don't have a picture of what's going on.
What are the irreducible representations of the cyclic group $C_n$ over a real vector space $V$?
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1@JulianKuelshammer Done. – 2015-11-08
1 Answers
Let $\rho : C_n \to \mathrm{GL}(V)$ be a representation, equivalently let $f : V \to V$ be an automorphism such that $f^n = \operatorname{id}_V$. Then $f$ is a root of the polynomial $X^n - 1$, hence it is diagonalizable over $\mathbb{C}$ and its complex eigenvalues are $n$th roots of unity. These roots are either real or come in pair of complex conjugate numbers: let $A \in M_d(\mathbb{R})$ be the matrix associated to $f$; then if $\lambda \in \mathbb{C} \setminus \mathbb{R}$ is an eigenvalue of $A$, i.e. $Av = \lambda v$ for nonzero $v$, then $\overline{Av} = \overline{\lambda v} \implies A \bar{v} = \bar{\lambda} \bar{v}$ hence $\bar{\lambda}$ is also an eigenvalue of $A$.
Now you can pair up $\lambda = e^{2ik\pi/n}$ and $\bar{\lambda} = e^{-2ik\pi/n}$. Let $\theta = 2k\pi/n$, such that $n \theta \equiv 0 \pmod{2\pi}$. The two following matrices are similar: $\begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} \sim \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$
Conclusion: The irreducible real representations of $C_n$ are either
- 1-dimensional, with matrix a real $n$th root of unity;
- 2-dimensional, with matrix $\left(\begin{smallmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{smallmatrix}\right)$ where $n\theta \equiv 0 \pmod{2\pi}$ but $\sin\theta \neq 0$.