1-dimensional Ring geometry - Group of Translations I considered a Ring-like one dimensional geometry. In this, if we fix an origin (at some point on the circumference), we can think of set of all displacements along the circumference to form a vector space. Now one vector can be denoted by (for some reasons that will become clear),
$$ \left( \begin{array}{ccc}
x  \\
1  \end{array} \right) $$
Further one can obtain any other vector in the space by translating the vector, say $ x_0 \rightarrow x_0+a $. We can use the linear transformation : 
$$ T(a) =
\left( \begin{array}{cc}
0 & a \\
0 & 0\end{array} \right) $$
such that 
$$ \left( \begin{array}{ccc}
x + a  \\
1  \end{array} \right) 
= \left( \begin{array}{ccc}
x  \\
1  \end{array} \right)+
T(a)\left( \begin{array}{ccc}
x  \\
1  \end{array} \right)
$$
Now the set of all such linear transformations will form a group. 
Most important part of this transformation is that, if the circumference of the ring is some $L$, then the transformation $T(nL)$ where $ n \in \mathbb Z $ should not change the vector. Mathematically, 
$$ T(nL) \left( \begin{array}{ccc}
x_0  \\
1  \end{array} \right)
=
\left( \begin{array}{ccc}
x_0  \\
1  \end{array} \right) $$
Now my question is, with these definitions is the group of Translations a Compact one ? And if it is the generator of the translations will have some properties like angular momenta (although this is a generator of translations) ?
PS : I hope I am not talking about rotations. I am just talking translations along the circumference of the circle.
 A: First of all I try to restate your question into a more clear form.
Consider $\mathbb R$ equipped with the equivalence relation:
$x \sim y$ if and only if $x-y= 2k\pi$ with $k \in \mathbb Z$. 
The space ${\mathbb R}/ \sim$  of equivalence classes $[x]$ is $\mathbb S^1$ also as a topological space using the quotient topology. 
Next consider the standard actions of the Lie group of translations $\mathbb R$  on the real line $\mathbb R$: $$T(a)x:=x+a\quad \forall x,a \in \mathbb R\:,$$
and define the representation of the translation group on $\mathbb S^1$  as $$T′(a)[x]:=[T(a)x]\:\forall x,a \in \mathbb R\:. \quad (1)$$ 
The map $\mathbb R\ni a \mapsto T′(a)$ is in fact a representation of the translation group on $\mathbb S^1$ in terms of isometries of the circle (when equipped with the standard metric). In particular, one has $T'(0)= id$ and
$T'(a)T'(b)= T'(a+b)$.
However all that has nothing to do with compactness (false!) of the translation group, even if the outlined procedure gives rise to a representation of that (non-compact) Lie group on a compact manifold, in terms of isometries of that manifold.
Let us eventually come to the relation with the rotations group of $\mathbb R^2$: $SO(2) \equiv U(1)$. 
As $\mathbb R$ is the universal covering of $U(1)$, with covering (surjective Lie group) homomorphism:
$$\pi : \mathbb R^1 \ni a \mapsto e^{ia} \in U(1)\:,\qquad (2) $$
every representation of the group of $\mathbb R^2$ rotations $U(1)$ is also a representation of the group of  translations $\mathbb R$. 
Identifying $\mathbb S^1$ with $U(1)$ in the standard way, the natural action (representation) of $U(1)$ on the circle is trivially
$$R(e^{ia}) e^{ix} = e^{i(a+x)} \qquad (3)$$ 
where the first $e^{ia}$ is viewed as an element of the group $U(1)\equiv SO(2)$ and the other two are viewed as  elements of the circle $U(1) \equiv \mathbb S^1$.
The interplay of $T', R$ and $\pi$, as  one easily proves is:
 $$R(\pi(a))= T'(a)\quad \forall a \in \mathbb R\:.\qquad (4)$$
This is in agreement with the remark above that reps of $SO(2)$
 are also reps of $\mathbb R$.
Thus, as a matter of fact, it is not possible to distinguish between the action of $\mathbb R$ and that of $SO(2)$ on the circle $\mathbb S^1$, though they are different groups and only the latter is compact (and in a certain way related with the component of angular momentum orthogonal to $\mathbb R^2$.)
