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Math Help - Discrete subgroups of Isom(R^n)

  1. #1
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    Discrete subgroups of Isom(R^n)

    If there exists a number d(X)>0, for every point X in Rn, such that, for every isometry f in H which satisfies f(X)!=X, d(X,f(X))>=d(X), we say that a subgroup H of Isom(Rn) is discrete.
    I don't understand the definition.
    Also, I have to give some examples of discrete subgroups acting on R2.

    Can somebody help me?

    Thanks
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  2. #2
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    Re: Discrete subgroups of Isom(R^n)

    consider the following group of isometries:

    G = {Tn: n in Z} where Tn(x,y) = (x+n,y).

    how close can (x+n,y) and (x,y) be?

    (G can be thought of as modelling an "asymmetric (infinitely) repeating pattern", which is 1 unit long).

    the "discreteness" of G essentially derives from the discreteness of Z in R (under the usual metric topology for the real line).
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    Re: Discrete subgroups of Isom(R^n)

    Deveno, do I understand the definition in the OP right?

    ∀X ∀f ∈ H. f(X) ≠ X => d(X, f(X)) ≥ d(X)?

    That is, every point is equipped with a positive number so that every isometry either leaves this point in place or moves it farther than this number.
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    Re: Discrete subgroups of Isom(R^n)

    Thank you so much! I think I finally understand.
    One more thing, how can I prove the result: "If H is a discrete group of Isom(Rn) and P is a given point, the intersection of any ball and the set {f(P): f∈H} is a finite set"?
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    Re: Discrete subgroups of Isom(R^n)

    Quote Originally Posted by emakarov View Post
    Deveno, do I understand the definition in the OP right?

    ∀X ∀f ∈ H. f(X) ≠ X => d(X, f(X)) ≥ d(X)?

    That is, every point is equipped with a positive number so that every isometry either leaves this point in place or moves it farther than this number.
    i think there is an unfortunate double-use of "d", here. one refers to a real number, which may depend on the vector x. the other refers to the metric function on Rn.

    in the example i gave, any real number < 1 will do. your interpretation is essentially correct: there is an ε-ball around every x such that f(x) ≠ x implies f(x) is outside that ε-ball.
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