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Error in proof for symmetry groups?

  1. Aug 1, 2013 #1
    [itex]\renewcommand{\vec}[1]{\mathbf{#1}}[/itex]
    Here is an excerpt from the text:

    "[...]Theorem 12.5 The only finite symmetry groups in [itex]ℝ^2[/itex] are [itex] \mathbb{Z}_n[/itex] and [itex]D_n.[/itex]

    PROOF. Any finite symmetry group [itex]G[/itex] in [itex]\mathbb{R}^2[/itex] must be a finite subgroup of [itex]O(2)[/itex]; otherwise, [itex]G[/itex] would have an element in [itex]E(2)[/itex] of the form [itex](A,\vec{x})[/itex], where [itex]\vec{x} ≠ \vec{0}[/itex]. Such an element must have infinite order. [...]"

    But if I understand this correctly, the last sentence is false. As a counterexample, I can give the element [itex](-I, \vec{x})[/itex] which will have order 2 for any [itex]\vec{x}[/itex].

    So is this an error, or did I misunderstand something?
     
  2. jcsd
  3. Aug 7, 2013 #2
    Anyone have an idea?
    I think the rest of the proof still works even if I ignored this statement, but this is bothering me because I obviously did not fully understand how elements in the Euclidean Group behave. Sorry for bumping this thread, I will only do it once.
     
  4. Aug 7, 2013 #3

    Stephen Tashi

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    What does the "[...]" say?


    I don't really know anything about this theorem, but I notice that the PDF http://www.google.com/url?sa=t&rct=...sg=AFQjCNFmCggqAcvxgPjA2FIBNgWFYK7Cug&cad=rja Theorem 5.2.2 gives a proof where reflections are explicitly mentioned as one case.
     
  5. Aug 7, 2013 #4
    Yes, the proof continues (hence the [...] ) but it wouldn't really fit in the page. In any case, the sentence I had a problem with is the last one.

    The original proof can be found in this link:
    http://abstract.ups.edu/download.html

    in the 2012 edition. Chapter 12: Matrix Groups and Symmetry.
    page 190 (in document) or 200 (in the pdf file)


    Previously in the section, they defined the Euclidean group E(n) as the group of isometries in R^n fixing a subset of points in R^2. Every element in E(n) is a pair (A,x) where A is an nxn orthogonal matrix and x is a vector in R^n. Multiplication is defined as (A,x)(B,y) = (AB, Ay + x ). (See page 187, or p.197 in the pdf document). I guess your proof was based on a different definition for E(n).
     
    Last edited: Aug 7, 2013
  6. Aug 8, 2013 #5

    Stephen Tashi

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    Isn't [itex] -I [/itex] an element of [itex] O(2) [/itex] ? - a rotation matrix with [itex] \theta = \pi [/itex].
     
  7. Aug 8, 2013 #6
    Yes, this is why I used it as a counter-example. Using the multiplication definition,
    (-I,x)(-I,x) = (-I * -I, -Ix + x ) = ( I, -x + x ) = ( I, 0 )

    So (-I,x) has order 2 even though x is nonzero. But their sentence seems to imply that if x is nonzero, the order must be infinite.
     
  8. Aug 8, 2013 #7

    Stephen Tashi

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    To be a counterexample to what is stated in the proof [itex] G [/itex] must not be a subgroup of [itex] O(2) [/itex]. You are giving a counterexample only to the assertion that an arbitrary [itex] (A,x) [/itex] with [itex] x \ne 0 [/itex] must be of infinite order.

    Perhaps the proof means than an element "of the form" [itex] (A,x) [/itex] with [itex] x \neq 0 [/itex] is one that cannot also be expressed as [itex] (A,y) [/itex] with [itex] y = 0 [/itex] because the proof is dealing with an [itex] (A,x) [/itex] that is not in the rotation group.
     
  9. Aug 8, 2013 #8
    I agree with the theorem's statement itself. My counterexample is not against the theorem, but against the statement that (A,x) has infinite order.

    This may be correct. But then you mean that the pair (-I, x) could be written as a pair of the form (A, 0)? I know what they were trying to say: a translation must have an infinite order. But it seems that not every (A,x) for nonzero x is like a translation.
     
  10. Aug 8, 2013 #9

    micromass

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    For clarification, the element ##(A,\mathbf{x})## is just shorthand for the map

    [tex]\mathbf{v} \rightarrow A\mathbf{v} + \mathbf{x}[/tex]

    ?
     
  11. Aug 8, 2013 #10

    micromass

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    I checked the pdf. So yes, it seems to be the case.

    In this case, the proof is wrong. A correct proof is given in Artin's Algebra.
     
  12. Aug 13, 2013 #11
    I think that I have a proof, and I'll do it it for general E(n).

    Its symmetry groups have elements (A,x) with A in O(n) and x in Rn.
    Its product: (A1,x1).(A2.x2) = (A1.A2, x1+A1.x2)
    Its identity: (I,0)
    Its inverse: (A,x)-1 = (A-1, -A-1.x)

    Let us consider what possible finite ones there can be.

    Every E(n) symmetry group has a quotient group in O(n) whose elements are all the values of A. If the group is finite, then all its quotient groups must also be finite, and the A's must form a finite subgroup of O(n).

    Let us consider what sets of elements map on to each quotient-group element. Consider (A,x1) and (A,x2), which both map onto A. Consider
    (A,x1).(A,x2)-1 = (I,x1-x2)

    If x1 != x2, then we get (I,x) where x != 0. It's easy to show that (I,x)k = (I,k*x), thus giving an infinite subgroup of our group. To keep that from happening, the only permissible x in (I,x) must be 0.

    This means that every other element has a unique value of x associated with it, though that value may be nonzero.

    Thus, the only finite subgroups of E(n) are isomorphic to finite subgroups of O(n). For O(2), those are Z(n) and D(n).
     
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