So I'm pretty new into Representation Theory having so far covered only a couple of example sheets. I'm thinking about the following question:

Suppose we have the group $D_{2n}$ (for clarity this is the dihedral group of order $2n$, as notation can differ between texts).

We can describe this group as follows:

$$\langle \sigma, \tau | \sigma ^n=1, \tau ^2=1, \tau \sigma \tau = \sigma ^{-1} \rangle$$

We know this is isomorphic to the symmetries of the regular $n$-gon.

Question How can we construct a two-dimensional representation $\psi: D_{2n} \rightarrow \text{GL}_2(\mathbb{R})$?

Thoughts Do we simply set up a map that takes $\sigma$ to the standard $2 \times 2$ rotation matrix, with $\theta$ being $\frac{2 \pi}{n}$ and $\tau$ to something like the matrix $\left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)$? If this is the case, is there any simple way to show it is two-dimensional?

  • 2
    $\begingroup$ There is not really any trick to showing that it is $2$-dimensional. It is a map to $GL_2(\mathbb{R})$ and therefore a $2$-dimensional (real) representation. $\endgroup$ – Tobias Kildetoft Oct 30 '13 at 14:26

As stated in the comments, once you show that your map is a homomorphism $D_{2n}\to \operatorname{GL}_2(\mathbf{R})$, then you have a 2-dimensional representation (since the matrices are 2x2). Thus all you need to show is that the matrices for $\sigma$ and $\tau$ satisfy the relations in the group presentation.

Here is another way to get the 2-dimensional reps using induced representations, just for fun.

There is an obvious subgroup $\{1, \sigma\, \ldots, \sigma^{n-1}\}$ which is a cyclic group of order $n$, let's call it $C_n < D_{2n}$. Since $C_n$ is abelian, it has $n$ irreducible 1-dimensional representations over $\mathbf{C}$, namely

$$\sigma\mapsto e^{2\pi ki/n}\qquad 0\leq k < n$$ which captures the idea of rotating by an angle of $2\pi k/n$. The trick is to induce these easily-described representations to $D_{2n}$ in order to find some possibly new representations.

Recall that if we have a representation $W$ of a subgroup $H\leq G$ (i.e. an H-linear action on $W$), the induced representation of $W$ is $$\bigoplus_{g\in G/H}g\cdot W$$ where $g$ ranges over a set of representatives of $G/H$.

The induced representation of $C_n$ to $D_{2n}$ for fixed $k$ is not hard to describe then since $D_{2n}/C_n$ has representatives $\{1, \tau\}$. So we just need to describe the $D_{2n}$-vector space $\mathbf{C}\oplus\tau\cdot\mathbf{C}$ where $\mathbf{C}$ has basis consisting only of $w_1$. The neat part is seeing how the $D_{2n}$ action turns into an actual matrix representation.

Specifically, we can find out how $\sigma$ acts on each summand using our representation of $C_n$: $$\sigma\cdot w_1 = e^{2\pi ki/n}w_1, \text{ and}$$ $$\sigma\cdot(\tau\cdot w_1) = \sigma\tau\cdot w_1 = \tau\sigma^{-1}\cdot w_1 = e^{-2\pi ki/n}\tau\cdot w_1$$ which tells us that $\sigma$ acts by the matrix $\pmatrix{e^{2\pi ki/n}&0\\0&e^{-2\pi ki/n}}$!

We can also figure out how $\tau$ acts. $\tau$ obviously takes $w_1$ to $\tau\cdot w_1$, and $\tau$ takes $\tau\cdot w_1$ to $\tau^2 w_1=w_1$, so $\tau$ simply interchanges the two summands. This tells us that $\tau$ acts by the matrix $\pmatrix{0&1\\1&0}$.

So that's another way to get 2-dimensional representations of $D_{2n}$. Note that the $\sigma$ matrix I described with $k=1$ is the diagonalized form of the rotation matrix you mention.

  • $\begingroup$ How does $D_{2n}/C_n$ has representatives $\left\{1, \tau\right\}$? $\endgroup$ – Omar Shehab May 1 '16 at 21:12

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