The $k/n$ representation of the Dihedral group of order $2n$ in $GL(2,\mathbb{C})$ is induced by mapping the rotation element of $D_n$ to the Rotation Matrix $R(\frac{2\pi k}{n})$, and the reflection element of $D_n$ to the matrix $ \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$. As mentioned in the first hyperlink, this representation is irreducible unless $n$ is even and $k=n/2$. The article also gives a brief justification for this, however I don't immediately understand why this is the case. Thanks in advance, and apologies for my inexperience formatting these posts!

  • $\begingroup$ What part of the argument are you more specifically struggling with? $\endgroup$ – Ben Sheller Nov 2 '15 at 22:50
  • $\begingroup$ The article says that the fact that this representation is irreducible unless $n=2k$, and that this "can be seen by computing its character norm"...this quoted phrase is what I don't understand. Thanks for the edits, now I see it's actually quite simple to correctly format these posts. $\endgroup$ – L. T. P. L. Nov 2 '15 at 23:00
  • $\begingroup$ It's more or less clear to me that in the case $n=2k$, this representation is not irreducible. However, I do not understand why: if we take $0<k<n$ with $n$ not equal to $2k$, this representation should be irreducible. $\endgroup$ – L. T. P. L. Nov 3 '15 at 3:31

Let's suppose we are OK with the case when $n=2k$ (in this case the rotation matrix simplifies significantly and it becomes easy to find an invariant subspace).

Call the representation $\rho$ and let $\chi_\rho$ denote the character of this representation.

In the case $n=2k$ the representation gives rise to an invariant subspace - the rotation matrix will simplify to a diagonal matrix, and then any product of these two diagonal matrices will also be diagonal. This allows us to easily see that the subspace of $V=\mathbb{C}^2$ given by $V'=\{(z,0)\in\mathbb{C}^2\}$ is invariant.

Now let $n\neq2k$. To show irreducibility we wish to prove that the character norm $<\chi_\rho,\chi_\rho>$ is equal to 1.$

$<\chi_\rho,\chi_\rho>=\frac{1}{2n}\sum_{g\in D_n}\chi_\rho(g)\chi_\rho(g^{-1})$

$\implies<\chi_\rho,\chi_\rho>=\frac{1}{2n}[(2cos(\frac{2\pi k}{n}))^2+(2cos(2\frac{2\pi k}{n}))^2\cdots+2cos(n\frac{2\pi k}{n})+(0+0)+(cos\frac{2\pi k}{n}-cos\frac{2\pi k}{n})+...+(cos(n\frac{2\pi k}{n})-cos(n\frac{2\pi k}{n}))]$

We now apply the identity $2cos^2\theta=1+cos(2\theta)$ (we skip a few substeps) to obtain that

$\frac{1}{n}[(1+cos(2\frac{2\pi k}{n}))+(1+cos(2\frac{4\pi k}{n}))+\cdots+(1+cos(2\frac{2n\pi k}{n}))]$

$=\frac{1}{n}[n+cos(\frac{4\pi k}{n})+cos(2\frac{4\pi k}{n})+\cdots+cos(n\frac{4\pi k}{n})]$.

Now we use another trigonometric result, namely:

$cos(\alpha)+cos(2\alpha)+\cdots+cos(n\alpha)=\frac{cos(\alpha+\frac{n-1}{2}\alpha)sin(\frac{n\alpha}{2})}{sin(\frac{\alpha}{2})}$ with $\alpha=\frac{4\pi k}{n}$.

[note: this is the moment where we lose the case where $n=2k$, since it would give a zero in this denominator.]

It is easy to see now that in our case this sum is zero. Since $sin(\frac{n\alpha}{2})=sin(\frac{4\pi k}{n}\frac{n}{2})=sin(2\pi k)=0$ for all $k$.

Finally, $<\chi_\rho,\chi_\rho>=\frac{1}{n}n=1$ and the representation is irreducible.


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