2
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ 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
    Feb 7, 2014 at 18:06
  • $\begingroup$ Related by OP: math.stackexchange.com/q/667502 $\endgroup$
    – Kyle Kanos
    Feb 7, 2014 at 18:12
  • 3
    $\begingroup$ @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). $\endgroup$ 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
    Feb 7, 2014 at 18:28
  • 1
    $\begingroup$ @Dilaton As someone who answered that other question; I probably agree that it should have stayed here. $\endgroup$ Feb 7, 2014 at 18:30

1 Answer 1

2
$\begingroup$

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 $\mathbb R^2$.)

$\endgroup$
10
  • $\begingroup$ 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
    Feb 8, 2014 at 14:33
  • $\begingroup$ Also I am interested in understanding how to find the compactness of a Group $\endgroup$
    – user35952
    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. $\endgroup$ 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
    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 $|U_{rs}|^2 \leq \sum_{i} ( \sum_{j}U_{ij}U_{ij}^*) = \sum_{i} \delta_{ii} = n$. $\endgroup$ Feb 8, 2014 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.