Is Every Discrete Isotropy Group of an R^n Action Isomorphic to Z/kZ?

[FreHam]

Quick question:

Suppose I have a (transitive) R^n action on a manifold M. If the isotropy group of R^n is discrete, does that mean that it is automatically isomorphic to Z/kZ, with 0<=k<=n?

Basically, my discrete subgroup is a lattice then, right?

Thanks!

 

[Office_Shredder]

The isotropy group is a subgroup of Rⁿ so certainly can't have elements of finite order.

It is true that discrete subgroups of Rⁿ have to look like Z^k for some k

 

[FreHam]

The isotropy group is a subgroup of Rⁿ so certainly can't have elements of finite order.

Could you elaborate on that? Dummy here. So you are saying the isotropy group of R^n cannot be discrete?

 

[hamster143]

I believe that Office_Shredder is incorrect. He seems to be thinking about subgroups of the translations group of $R^n$. Since that group is itself isomorphic to R^n, its subgroups can be discrete, but they can't have elements of finite order (elements such that a^n=1 for some n), aside from the identity. Z/kZ (better known as $Z_k$) is finite and all its elements have finite order.

I also have to question whether you're using the term isotropy group correctly.

 

[FreHam]

I also have to question whether you're using the term isotropy group correctly.

Why? By isotropy group at some point m\in M, I mean the {g\in G | g(m)=m}. In my result I find that dim(iso-group)=0, so it is discrete. As G=G^n, that means the isotropy group is a discrete (abelian) subgroup of G, which I kind of hoped to be isomorphic to Z^k.

 

[hamster143]

Is your group basically a subgroup of the orthogonal group O(n)? Discrete subgroups of O(n) are called point groups and they aren't always abelian, e.g.:

http://en.wikipedia.org/wiki/Point_groups_in_three_dimensions

 

[FreHam]

My group is isomorphic to R^n (group operation is addition). The left R-action acts on n factors, which I write as R^n.

 

[Office_Shredder]

To re-iterate, discrete subgroups of Rⁿ are isomorphic to Z^k for some k no larger than n, just like you would expect