Every finite subgroup of SL(2,R) is isomorphic to a finite subgroup of the group of all rotations of R^2. Since this is just S^1, it is easy to prove that such a group is cyclic.
On Nov 3, 6:22 am, José Carlos Santos <jcsan...@fc.up.pt> wrote:
> On 03-11-2009 6:46, leiko wrote:
> > Why is a finite subgroup of SL (2,R) cyclic?
> Every finite subgroup of SL(2,R) is isomorphic to a finite subgroup of > the group of all rotations of R^2. Since this is just S^1, it is easy to > prove that such a group is cyclic.
> Best regards,
> Jose Carlos Santos
Thanks a lot for your replies. The Sl(3,r) example is neat.
About Sl(2,r) - the group of rotations is SO(2,r), and this is just a subgroup of Sl(2,r), am I right? I think I can show that So(2,r) is isomorphic to S^1, but how can i show that a finite subgroup of Sl(2,r) is in fact a subgroup of So (2,r)?
A finite subgroup of SL_n is conjugate to a subgroup of SO_n by an averaging argument. In case n=2, SO_2 is rotations of a circle so the finite subgroups are cyclic. In case of n=3 the finite subgroups of SO_3 (rotations of a sphere) are cyclic, dihedral, or one of the groups of symmetries of a regular polyhedron: the tetrahedron of order 12, octahedron/cube of order 24 or icosohedron/dodecahedron of order 60. The finite subgroups of SO_4 are classified too but it is more complicated of course. I don't know what happens for n>=5.
>>> Why is a finite subgroup of SL (2,R) cyclic? >> Every finite subgroup of SL(2,R) is isomorphic to a finite subgroup of >> the group of all rotations of R^2. Since this is just S^1, it is easy to >> prove that such a group is cyclic.
> Thanks a lot for your replies. The Sl(3,r) example is neat.
> About Sl(2,r) - the group of rotations is SO(2,r), and this is just a > subgroup of Sl(2,r), am I right?
Yes.
> I think I can show that So(2,r) is isomorphic to S^1, but how can i > show that a finite subgroup of Sl(2,r) is in fact a subgroup of So > (2,r)?
You don't; it is not true. For instance, the group generated by
-1 -2 A = 1 1
has four elements, but neither A nor A^3 belong to SO(2,R).
What I wrote was that a finite subgroup of SL(2,R) is _isomorphic_ to a finite subgroup of S^1 (or SO(2,R), if you prefer). As I.M. Soloveichik wrote as a reply to your original post, you prove this "by an averaging argument". If you still don't see how, I will tell you (or someone else will).
> >>> Why is a finite subgroup of SL (2,R) cyclic? > >> Every finite subgroup of SL(2,R) is isomorphic to a finite subgroup of > >> the group of all rotations of R^2. Since this is just S^1, it is easy to > >> prove that such a group is cyclic.
> > Thanks a lot for your replies. The Sl(3,r) example is neat.
> > About Sl(2,r) - the group of rotations is SO(2,r), and this is just a > > subgroup of Sl(2,r), am I right?
> Yes.
> > I think I can show that So(2,r) is isomorphic to S^1, but how can i > > show that a finite subgroup of Sl(2,r) is in fact a subgroup of So > > (2,r)?
> You don't; it is not true. For instance, the group generated by
> -1 -2 > A = > 1 1
> has four elements, but neither A nor A^3 belong to SO(2,R).
> What I wrote was that a finite subgroup of SL(2,R) is _isomorphic_ to a > finite subgroup of S^1 (or SO(2,R), if you prefer). As I.M. Soloveichik > wrote as a reply to your original post, you prove this "by an averaging > argument". If you still don't see how, I will tell you (or someone else > will).
> Best regards,
> Jose Carlos Santos
I am not quite sure what "averaging argument" means, so i don't really see it :( I am sorry.
>>>>> Why is a finite subgroup of SL (2,R) cyclic? >>>> Every finite subgroup of SL(2,R) is isomorphic to a finite subgroup of >>>> the group of all rotations of R^2. Since this is just S^1, it is easy to >>>> prove that such a group is cyclic. >>> Thanks a lot for your replies. The Sl(3,r) example is neat. >>> About Sl(2,r) - the group of rotations is SO(2,r), and this is just a >>> subgroup of Sl(2,r), am I right? >> Yes.
>>> I think I can show that So(2,r) is isomorphic to S^1, but how can i >>> show that a finite subgroup of Sl(2,r) is in fact a subgroup of So >>> (2,r)? >> You don't; it is not true. For instance, the group generated by
>> -1 -2 >> A = >> 1 1
>> has four elements, but neither A nor A^3 belong to SO(2,R).
>> What I wrote was that a finite subgroup of SL(2,R) is _isomorphic_ to a >> finite subgroup of S^1 (or SO(2,R), if you prefer). As I.M. Soloveichik >> wrote as a reply to your original post, you prove this "by an averaging >> argument". If you still don't see how, I will tell you (or someone else >> will).
> I am not quite sure what "averaging argument" means, so i don't really > see it :( I am sorry.
> Thanks a lot for your patience.
Let ( , ) be the standard scalar product in R^2. Given a finite subgroup G of SL(2,R), define a new scalar product by
<v,w> = sum_{g in G}(g.v,g.w)/#G
(here's the averaging). Then, for each g in G and each v,w in R^2, <g.v,g.w> = <v,w>. So, G becomes a group of isometries with respect to this new scalar product. Furthermore, since each _g_ in G has determinant 1, G becomes a subgroup of SO(2,R) (again, with respect to this new scalar product).
> >>>>> Why is a finite subgroup of SL (2,R) cyclic? > >>>> Every finite subgroup of SL(2,R) is isomorphic to a finite subgroup of > >>>> the group of all rotations of R^2. Since this is just S^1, it is easy to > >>>> prove that such a group is cyclic. > >>> Thanks a lot for your replies. The Sl(3,r) example is neat. > >>> About Sl(2,r) - the group of rotations is SO(2,r), and this is just a > >>> subgroup of Sl(2,r), am I right? > >> Yes.
> >>> I think I can show that So(2,r) is isomorphic to S^1, but how can i > >>> show that a finite subgroup of Sl(2,r) is in fact a subgroup of So > >>> (2,r)? > >> You don't; it is not true. For instance, the group generated by
> >> -1 -2 > >> A = > >> 1 1
> >> has four elements, but neither A nor A^3 belong to SO(2,R).
> >> What I wrote was that a finite subgroup of SL(2,R) is _isomorphic_ to a > >> finite subgroup of S^1 (or SO(2,R), if you prefer). As I.M. Soloveichik > >> wrote as a reply to your original post, you prove this "by an averaging > >> argument". If you still don't see how, I will tell you (or someone else > >> will).
> > I am not quite sure what "averaging argument" means, so i don't really > > see it :( I am sorry.
> > Thanks a lot for your patience.
> Let ( , ) be the standard scalar product in R^2. Given a finite subgroup > G of SL(2,R), define a new scalar product by
> <v,w> = sum_{g in G}(g.v,g.w)/#G
> (here's the averaging). Then, for each g in G and each v,w in R^2, > <g.v,g.w> = <v,w>. So, G becomes a group of isometries with respect to > this new scalar product. Furthermore, since each _g_ in G has > determinant 1, G becomes a subgroup of SO(2,R) (again, with respect to > this new scalar product).
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