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Group action

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Let G = (R,+) and define az = eiaz for all z in C and a in G. Show that this definition makes C into a G-set, describe the action geometrically, and find the orbits and the stabilizers.


    3. The attempt at a solution

    A mapping G x C -> C, denoted (a,z) -> az = eiaz for z in C.
    Let eiaz = (cos(a) + isin(a))(x+iy)
    Thus az = a(x+iy) = (cos(a) + isin(a))(x+iy)
    We can be called this an action of G as satisfied the follwing:
    Since 0 is identity of G, 0z = (cos(0) + isin(0))(x+iy) = (x+iy) = z
    Also this satisfies such as a(bz) = (ab)z for all z in C and for all a,b in G
    Thus G acts on C, called a G-set.
    Correct??
    I have no idea how to show and find orbits and stabilizers.
    Thanks
     
  2. jcsd
  3. May 5, 2009 #2

    Dick

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    Looks ok so far. You might want to elaborate a little on why a(bz)=e^(ia)(e^(ib)z)=(ab)z. If you think of the complex plane C as R^2, with coordinates (x,y), what kind of geometric operation does e^(ia)z correspond to?
     
  4. May 5, 2009 #3
    Thanks very much,
    I think I can modify to more correct..
    But I still have no idea about orbits and stabilizers.
    Could you give me any hints?
    Are the orbits just circle?
    like, r is distance of z and denoted |z| and
    e^ia is the circle which is 1 of radius??
    Thanks!
     
    Last edited: May 5, 2009
  5. May 5, 2009 #4

    Dick

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    Yes, the orbits are circles. e^(ia)z is z rotated around the origin by an angle a. Write z in polar form z=r*e^(it) to see this.
     
  6. May 5, 2009 #5
    and the stabilizers are when a = 0?
     
  7. May 5, 2009 #6

    Dick

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    Depends on what value of z you are talking about. And the stabilizer of a given z is a subgroup. If you are saying it's the trivial subgroup {0} I don't agree with that.
     
  8. May 5, 2009 #7
    I do not know what you mean.
    Yeah, I know the stabilizers are subgroup of G but
    I don't know how I can apply to this problem..
    I have to find in case of az = z, right?
     
  9. May 5, 2009 #8

    Dick

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    Right. 1) suppose z=0, what's the stabilizer group? 2) suppose z is not zero. What about a=2pi?
     
  10. May 5, 2009 #9
    hmm,

    I found something in my textbook,
    If X is a G-set and x in X, the orbit of x is Gx = {ax | a in G}. combining this with "|Gx| = |G:S(x)| for each x in X" gives equivalent conditions that the orbit is a singleton:
    Gx = {x} <=> ax=x for all a in G <=> S(x)=G.

    And, when z = 0 and z is not zero, both cases seem just G = (R,+)..
    Right??
     
  11. May 5, 2009 #10

    Dick

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    If z=0 then a0=0 for all a. So, yes, the stabilizer is G. If z is not equal to zero that's definitely not true. Which rotation angles take a point to itself? It's not just zero. What does a rotation by 2pi do?!!
     
  12. May 5, 2009 #11
    Thank you very much!
     
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