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Proving a set is connected

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that the group G where [itex]G = \left(\begin{array}{ccc} \cos\theta & -
    sin\theta & u \\ \sin\theta & \cos\theta & v \\ 0 & 0 & 1 \end{array} \right) u,v \in \Re, \theta \in \Re/2\pi Z [/itex]


    2. Relevant equations

    I know that a set is connected if it is not the disjoint union of two non-empty sets, if any two elements in G can be joined by a [itex]C^k[/itex] path in G, and if G is generated by a neighbourhood of 1.

    3. The attempt at a solution

    I thought that it would be easiest to show that G is not the disjoint union of two non-empty sets, and I was trying to do it by contradiction, but got nowhere.

    Any hints/nudges in the right direction would be appreciated.
     
  2. jcsd
  3. Nov 10, 2011 #2

    Office_Shredder

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    Path connectivity is often the easiest direction to go in, especially when your set is parametrized by real numbers
     
  4. Nov 10, 2011 #3
    Would it then work to say that you can take the function f: [0,1] -> G, where f = tg + (1-t)g' with g, g' members of G. f(0) =g and f(1) = g' so G is indeed connected?

    Or am I missing something here?
     
  5. Nov 10, 2011 #4

    Office_Shredder

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    That's not quite right. Imagine
    [tex]g = \left(\begin{array}{ccc} 1 &
    0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) [/tex]

    [tex]g' = \left(\begin{array}{ccc} 0 &
    -1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) [/tex]

    then the halfway point is

    [tex]g = \left(\begin{array}{ccc} 1/2 & -
    1/2 & 0\\ 1/2 & 1/2 & 0 \\ 0 & 0 & 1 \end{array} \right) [/tex]

    which isn't actually of the desired form. You need to be a little more careful about how you deal with the sin and cos terms



    2. Relevant equations

    I know that a set is connected if it is not the disjoint union of two non-empty sets, if any two elements in G can be joined by a [itex]C^k[/itex] path in G, and if G is generated by a neighbourhood of 1.

    3. The attempt at a solution

    I thought that it would be easiest to show that G is not the disjoint union of two non-empty sets, and I was trying to do it by contradiction, but got nowhere.

    Any hints/nudges in the right direction would be appreciated.[/QUOTE]
     
  6. Nov 11, 2011 #5
    Yeah, I realised that last night when I was showering.

    However, I can't seem to find a curve that will actually suit all of the conditions, namely that:
    f(t) = a(t)g + b(t)g' with b(t) + a(t) = 1 from the bottom right corner and that a(t)cos(x) + b(t)cos(y) = cos(z) and a(t)sin(x) + b(t)sin(y) = sin(z).

    I think it may just be easier to show that any g in G can be written as an exponential, and so then all points are joined together.
     
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