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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

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

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

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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 $\mathbb Z$.

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    $\begingroup$ Hi @DanDouglas: An algebraic group is a subspace cut out by polynomial equations, with a group structure for which the group multiplication is polynomial. It's not that hard to show that every algebraic group is a matrix group, so I might as well use the definition "an algebraic group is a subgroup of $\mathrm{GL}(n)$ defined by polynomial equations", although this isn't really the primary definition. Given an algebraic group, it makes sense to ask whether a given representation is algebraic, which is to say the matrix coefficients are polynomial functions. $\endgroup$ – Theo Johnson-Freyd Jul 17 '13 at 1:46
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    $\begingroup$ In the second paragraph, you can see immediately that there cannot be a faithful homomorphism from the circle to $\mathbb R^\times$, since the former has torsion elements (i.e. elements with finite order), whereas the latter does not. By thinking about torsion elements, you can see moreover that every continuous homomorphism is trivial. When I wrote that paragraph, I was thinking about Lie (i.e. smooth) homomorphisms; a Lie homomorphism $G \to H$, when $G$ is connected, is determined by its first derivative at $e\in G$. (In fact, every continuous homomorphism is smooth.) $\endgroup$ – Theo Johnson-Freyd Jul 17 '13 at 1:50
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    $\begingroup$ So, the point is that "algebraic" is not a property, but extra structure on a group. The abstract definition is that "algebraic group" usually means "group object in affine schemes" (and occasionally "affine" is dropped). The better definition is that "algebraic group" means "group equipped with a distinguished class of functions called regular or polynomial functions, for which multiplication is polynomial". Then by definition a morphism of algebraic groups should by polynomial. As an example, $\mathbb R$ has a notion of polynomials (the usual one), and the multiplication is ... $\endgroup$ – Theo Johnson-Freyd Jul 19 '13 at 2:43
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    $\begingroup$ ... clearly polynomial. (When I say that a map $\phi : X \to Y$ is polynomial, I mean that for any polynomial $f$ on $Y$, its pullback $\phi^*f = f\circ\phi$ is a polynomial on $X$.) A final definition of algebraic group is "commutative Hopf algebra" (with the understanding that homomorphisms of algebraic groups correspond to but go in the opposite direction of homomorphisms of Hopf algebras). From this perspective, $\mathbb R$ corresponds to the Hopf algebra $\mathbb R[x]$, where the Hopf comultiplication is $x \mapsto x\otimes x$. $\endgroup$ – Theo Johnson-Freyd Jul 19 '13 at 2:48
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    $\begingroup$ Maybe the confusion is what "cut out" means. Given any vector space (e.g. $n\times n$ matrices), you can talk about polynomials on that vector space (i.e. polynomial functions in the coordinates). Given any set of polynomials, you can talk about the set of vectors for which all those polynomials vanish. Such a set is "cut out by polynomials". Most sets are not cut out by polynomials; for example, no infinite discrete subset of the vector space is cut out by polynomials. $\endgroup$ – Theo Johnson-Freyd Jul 19 '13 at 2:52

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