Representing $\mathbb{R}/\mathbb{Z}$ as a matrix group.

Question

It was told to me that $G = \mathbb{R}/\mathbb{Z}$ is a real matrix group.

Can someone help me understand how to represent $G$ in $Gl_n(\mathbb{R})$ for some $n$?

(Supposedly, $n = 1$? But that's confusing because then G is [0,1) with a modular addition group operation, which doesn't seem like a matrix group to me.)

Any help would be much appreciated!

Thanks! -Dan

Answer 1

$$\theta \to \begin{pmatrix} \cos(2 \pi \theta) & \sin(2 \pi \theta) \\ -\sin(2 \pi \theta) & \cos(2 \pi \theta) \end{pmatrix}$$

Answer 2

Maybe she or he meant $ℝ/ℤ \cong \cong \mathrm{SO}_2(ℝ)$?

Answer 3

Actually, there is a good reason for your confusion. The group map $\mathbb R \to \mathbb R/\mathbb Z \to \mathrm{GL}(2,\mathbb R)$ given by $t \mapsto \bigl( \begin{smallmatrix} \cos 2\pi t& \sin 2\pi t\\ -\sin 2\pi t& \cos 2\pi t\end{smallmatrix}\bigr)$ is not algebraic, in the sense that it is not given by polynomial functions. Often in the theory of matrix groups, it is interesting to ask just for representaitons that are algebraic. The group $(\mathbb R,+)$ can be represented algebraically, by $t \mapsto \bigl( \begin{smallmatrix} 1 & t \\ 0 & 1 \end{smallmatrix}\bigr)$, but no algebraic representation of $\mathbb R$ can have kernel $\mathbb Z$ (since $\mathbb Z$ is not the vanishing locus of a polynomial). Thus, from the point of view of algebraic groups, there isn't a group "$\mathbb R/\mathbb Z$". What there is is a group called $\mathrm{U}(1) = \mathrm{SO}(2,\mathbb R)$, which is equal to $\mathbb R/\mathbb Z$ as a (smooth) group, but is algebraic. It is defined as the group of real matrices of the form $\bigl(\begin{smallmatrix} a & b \\ -b & a \end{smallmatrix}\bigr)$ satisfying $a^2+b^2 = 1$, with the understanding that "polynomial" means "polynomial in $a$ and $b$" rather than "polynomial in $\arccos(a) = \arcsin(b)$".

Of course, a second good reason for your confusion is that $\mathrm{GL}(1,\mathbb R) = \mathbb R \smallsetminus\{0\}$ is one-dimensional, and cannot receive a nontrivial group map from the circle. (You can think of $\mathbb R \smallsetminus \{0\}$ as a group of pairs $(t,s)$ satisfying $ts = 1$, and then a "polynomial" means any polynomial function in $t$ and $s$.) What you may have heard was that the circle $\mathbb R / \mathbb Z$ has a one-dimensional complex matrix representation: $\mathrm{GL}(1,\mathbb C) = \mathbb C \smallsetminus \{0\}$ and you can use the (nonalgebraic) representation $t \mapsto \exp(2\pi i t)$ from $\mathbb R \to \mathrm{GL}(1,\mathbb C)$ with kernel $\mathbb Z$.