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.

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    $\begingroup$ This is a perfectly legitimate technical physics question question. What is the point of the close vote? I have to say that it is rather worrysome that more and more such and similar legitimate questions of people who are seriously interested in studying physics at a technical level seem to be no longer welcome here ... $\endgroup$
    – Dilaton
    Commented Feb 7, 2014 at 18:06
  • $\begingroup$ Related by OP: math.stackexchange.com/q/667502 $\endgroup$
    – Kyle Kanos
    Commented Feb 7, 2014 at 18:12
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    $\begingroup$ @Dilaton The close vote was a migrate vote (to math.SE); I re-read the question and found it also asks about angular momentum towards the end. Close vote retracted (even though I'm not entirely sure if it should be considered primarily a math question or not). $\endgroup$ Commented Feb 7, 2014 at 18:19
  • $\begingroup$ @joshphysics the other question Kyle Kanos linked too should not have been migrated either. 3 community members indeed took it out of the close queue by saying leave open, but then it got migrated by a mod anyway. As I said to Kyle Kanos, imho it is, as in this and the other case not always possible to draw a clear line because physics is written in the language of math, explaining consevation laws needs group theory for example, and the more advanced/theoretical the topic the more math is needed to technically talk about it. I would really like to see some more tolerance towards rather $\endgroup$
    – Dilaton
    Commented Feb 7, 2014 at 18:28
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    $\begingroup$ @Dilaton As someone who answered that other question; I probably agree that it should have stayed here. $\endgroup$ Commented Feb 7, 2014 at 18:30

1 Answer 1


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 $\mathbb R^2$.)

  • $\begingroup$ I really appreciate your answer. But I am just beginning to learn Lie Algebra (and I am student of Physics). Although I understand the crux of your answer I am at loss to understand the mathematical details. $\endgroup$
    – user35952
    Commented Feb 8, 2014 at 14:33
  • $\begingroup$ Also I am interested in understanding how to find the compactness of a Group $\endgroup$
    – user35952
    Commented Feb 8, 2014 at 14:34
  • $\begingroup$ Hi, if you are just beginning to learn these things, maybe your question is too complicated, as it needs a technical answer as you saw. however I am a physicist too (including my PhD). Regarding compactness the story is easy. Almost all interesting groups in theoretical physics are groups of matrices, so they are subsets of $\mathbb R^{n^2}$ with $n$ large enough. The topology and the differentiable structure are those induced by $\mathbb R^{n^2}$. As in $\mathbb R^N$ compacts sets are all of closed bounded sets, you should only check these two conditions. $\endgroup$ Commented Feb 8, 2014 at 17:25
  • $\begingroup$ Oh ok !! Thanks, but I just made up the question in curiosity. Thinking further, shouldn't there be a similar connection to $SO(3)$ and $U(2)$ (so is that what gives us these spin-half particles. $\endgroup$
    – user35952
    Commented Feb 8, 2014 at 17:32
  • $\begingroup$ So, for instance $U(n)$ can be seen as a subset of $\mathbb R^{(2n)^2}$. It is a closed subset of that space because it includes its limit points (if $A_kA_k^\dagger =I$ and $A_k \to A$ in $\mathbb R^{(2n)^2}$ then $AA^\dagger =I$). It is also compact because is bounded: If $U \in U(n)$ then $|U_{rs}|^2 \leq \sum_{i} ( \sum_{j}U_{ij}U_{ij}^*) = \sum_{i} \delta_{ii} = n$. $\endgroup$ Commented Feb 8, 2014 at 17:37

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