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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.

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  • $\begingroup$ Hi @nik do you mean $V$ must be less than 2 dimensional? $\endgroup$
    – Spook
    Oct 23, 2012 at 13:37
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    $\begingroup$ It depends on $n$. If $n = 1,2$, then you can easily find a representation of degree 1; otherwise, if you want to find all the representations, for some of them you will need $V$ to be of dimension at least 2 (and in fact 2 is enough). $\endgroup$ Oct 23, 2012 at 13:40
  • $\begingroup$ You mean if the dimension is greater than 2 I can always decompose into direct sums of 1 or 2 dimensional subspaces? Why $\endgroup$
    – Spook
    Oct 23, 2012 at 13:45
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    $\begingroup$ @NajibIdrissi Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ Nov 7, 2015 at 20:06

1 Answer 1

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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 $X^n - 1$. This polynomial splits as a product of linear terms over $\mathbb{C}$. Therefore $f$ is diagonalizable over $\mathbb{C}$. It follows that $V$ splits as a direct sum of the sub-representations (over $\mathbb{C}$):

$$V \otimes_{\mathbb{R}} \mathbb{C} = \bigoplus_{i=1}^k \ker(f - \lambda_i \operatorname{id}_V),$$

where the $\lambda_i$ are the complex eigenvalues of $f$. These are $n$th roots of unity since $f^n = \operatorname{id}$.

The root $\lambda_i$ is either:

  • Real, in which case $\ker(f - \lambda_i \operatorname{id})$ is also a sub-representation. It splits as a direct sum of one-dimensional representations (simply choose a basis).
  • Or it comes in a pair of complex conjugate numbers. Indeed, let $A \in M_d(\mathbb{R})$ be the matrix associated to $f$. Then $Av = \lambda_i v$ for a nonzero $v$ in the eigenspace, and therefore: $$\overline{Av} = \overline{\lambda_i v} \implies A \bar{v} = \bar{\lambda}_i \bar{v},$$ and therefore $\bar{\lambda}_i$ 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 $\sin\theta \neq 0$.
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    $\begingroup$ I'm sorry, but I don't really see where the argument here is that there are no higher dimensional representations that cannot be decomposed into the two stated. Could you maybe make this a bit more explicit? $\endgroup$
    – Sito
    Apr 21, 2020 at 15:34
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    $\begingroup$ @Sito Let $C_n$ be generated by $a$. Being $\rho$ a homomorphism, to know the irreducible subspaces of the representation it is enough to know the subspaces stabilised by $\rho(a)$. As shown in the answer, the real eigenspaces of $\rho(a)$ are all one- or two-dimensional, with the two-dimensional ones being irreducible. It follows that the irreducible subspaces of $\rho$ are all either one- or two-dimensional. Note that this argument doesn't make any assumption on the dimension of $\rho(a)$: you can have $n=2$ but $\rho(a)$ of dimension $6\times 6$, etc. $\endgroup$
    – glS
    Jul 11, 2021 at 12:06
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    $\begingroup$ @Sito Sorry, I hadn't seen your comments. I added some more details. $\endgroup$ Jul 12, 2021 at 7:53
  • $\begingroup$ Thanks a lot for the additional comments! $\endgroup$
    – Sito
    Jul 13, 2021 at 17:55

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