# Finding a representation of $D_n$ on $\mathbb{R}^2$?

I am trying to write down a representation of $D_n = \langle \sigma, \tau \mid \sigma^n = \tau^2 =e, \tau \sigma = \sigma^{-1} \tau \rangle$ over $\mathbb{R}^2 \cong \mathbb{C}$ (as an $\mathbb{R}$-vector space).

What I Know:

My representation has to be a group homomorphism $\rho: D_n \rightarrow \text{GL}(2, \mathbb{R})$. I know that the homomorphism will be fully specified by specifying where I send the generators of $D_n$. Intuitively I know that I should be sending $\sigma$ to the $2 \times 2$ rotation matrix for an angle of $2 \pi /n$ and $\tau$ to a horizontal reflection.

This makes me think I should be choosing \begin{array}{c c c} \sigma^k \rightarrow \begin{pmatrix} \cos{(2\pi k /n)} & -\sin{(2\pi k /n)} \\ \sin{(2\pi k /n)} & \cos{(2\pi k /n)} \end{pmatrix} & \text{and }& \tau \rightarrow \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ \end{array}

My Problem:

I am having issues checking that this map respects the group operation. I know I want to check that: $$\rho((\sigma^a \tau^b)(\sigma^c\tau^d)) = \rho(\sigma^a \tau^b) \rho(\sigma^c\tau^d)$$ since an arbitrary member of $D_n$ can be written as $\sigma^a \tau^b$. I think it would suffice to show that $\rho(\sigma \tau)= \rho(\sigma) \rho(\tau)$. I can easily write down the matrices on the RHS as I have just defined them above. However, I don't know how to check to see if this product matrix equals the matrix for $\rho(\sigma \tau)$, since I don't know what that is! How should I go about verifying my homomorphism is in fact a homomorphism?

• I think this answer should help. The gist is that, to verify your map is a homomorphism, you simply need to make sure that the relations defining $D_n$ hold for $\rho(D_n)$; e.g., that $\rho(\sigma)^n = 1$, that $\rho(\sigma\tau\sigma\tau) = 1$, etc. I know this sort of thing was (is) a gap in my group theory knowledge... – pjs36 Jan 25 '18 at 4:19

You can avoid a lot of work by just using the presentation of $D_n$. The presentation of $D_n$ means that given a group $G$ and elements $s,t\in G$ such that $s^n=t^2=e$ and $ts=s^{-1}t$, there is a unique homomorphism $\rho:D_n\to G$ such that $\rho(\sigma)=s$ and $\rho(\tau)=t$. In this case, $G=GL(2,\mathbb{R})$, so you just have to check that the matrices $s$ and $t$ which you want to be $\rho(\sigma)$ and $\rho(\tau)$ satisfy the relations.

Alternatively, if you really want to compute everything explicitly, here's what you can do. The trick is to choose some specific way of writing each element of $D_n$ and use that to define $\rho$. For instance, every element of $D_n$ can be written uniquely in the form $\sigma^a\tau^b$ where $a\in\{0,1,\dots,n-1\}$ and $b\in \{0,1\}$. So use this as your definition: define $\rho(\sigma^a\tau^b)=\rho(\sigma)^a\rho(\tau)^b$ when $a\in\{0,1,\dots,n-1\}$ and $b\in \{0,1\}$, where $\rho(\sigma)$ and $\rho(\tau)$ are the two matrices you chose. This uniquely defines $\rho$ on every element of $D_n$. Now you just have to check that this is a homomorphism, in other words that $$\rho((\sigma^a \tau^b)(\sigma^c\tau^d)) = \rho(\sigma^a \tau^b) \rho(\sigma^c\tau^d)$$ whenever $a,c\in\{0,1,\dots,n-1\}$ and $b,d\in \{0,1\}$. To check this, you first need to write $(\sigma^a \tau^b)(\sigma^c\tau^d)$ in the form $\sigma^e\tau^f$ so that you can evaluate $\rho$ on it using the definition above. Once you have that, you just have to compute the matrices and check they are equal.

• Perhaps this is trivial, but why should $\rho$ in your first paragraph be unique? – gabe Jan 25 '18 at 6:02
• That is part of the definition of a group presentation. If you like, it is because $\sigma$ and $\tau$ generate $D_n$. – Eric Wofsey Jan 25 '18 at 6:13
• Thanks, this is very helpful. – gabe Jan 25 '18 at 6:43

All you have to do is set $$\rho(\sigma)=\pmatrix{\cos (2\pi/n)&-\sin(2\pi/n)\\\sin(2\pi/n)&\cos(2\pi/n)}$$ and $$\rho(\tau)=\pmatrix{1&0\\0&-1}$$ and then verify these preserve the relations, that is $\rho(\sigma)^n=I$ (standard property of rotation matrices), $\rho(\tau)^2=I$ (obvious) and $\rho(\tau)\rho(\sigma)=\rho(\sigma)^{-1}\rho(\tau)$ (a couple of lines of calculation).

Anyway these are just the matrices representing the usual action of $D_n$ as the symmetry group of the regular $n$-gon with centre at the origin.

• Thanks, I didn't realize showing that if the imagine of sigma (tau) obeys the same relations as sigma (tau), I have a homomorphism. – gabe Jan 25 '18 at 6:44