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Discrete abelian subgroups

  1. Aug 29, 2010 #1
    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!
     
  2. jcsd
  3. Aug 29, 2010 #2

    Office_Shredder

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    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 Zk for some k
     
  4. Aug 29, 2010 #3
    Could you elaborate on that? Dummy here. So you are saying the isotropy group of R^n cannot be discrete?
     
  5. Aug 29, 2010 #4
    I believe that Office_Shredder is incorrect. He seems to be thinking about subgroups of the translations group of [itex]R^n[/itex]. 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 [itex]Z_k[/itex]) is finite and all its elements have finite order.

    I also have to question whether you're using the term isotropy group correctly.
     
  6. Aug 29, 2010 #5
    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.
     
  7. Aug 29, 2010 #6
  8. Aug 29, 2010 #7
    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.
     
  9. Aug 30, 2010 #8

    Office_Shredder

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