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Homework Help: Group Action on a Set

  1. Apr 17, 2005 #1
    Question

    Let [itex]G=GL(2,\mathbb{R})[/itex] be the group of invertible [itex]2\times 2[/itex] martrices with real entries. Consider the action of [itex]G[/itex] on itself by conjugation. For the element


    [tex]A= \left(\begin{array}{cc}
    2 & 1 \\
    0 & 3
    \end{array}\right)[/tex]

    of [itex]G[/itex], describe (i) the orbit and (ii) the isotropy group of [itex]A[/itex]

    Sorry, I have no working out because I am completely stumped. Can anyone give me some helpful hints or pointers. Thanks
     
  2. jcsd
  3. Apr 17, 2005 #2
    The orbit is going to be of the form

    [tex]O_A = \{gA\,|\, g\in GL(2,\mathbb{R})\}[/tex]

    Here [itex]g[/itex] can be found by the following way

    [tex]gA = Ag[/tex] where [tex]g \in GL(2,\mathbb{R})[/tex]

    [tex]\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)\left(\begin{array}{cc}2 & 1 \\ 0 & 3\end{array}\right) = \left(\begin{array}{cc}2 & 1 \\ 0 & 3\end{array}\right)\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)[/tex]

    [tex]\left(\begin{array}{cc}2a & a+3b \\ 2c & c+3d\end{array}\right) = \left(\begin{array}{cc}2a+c & 2b+d \\ 3c & 3d\end{array}\right)[/tex]

    Which implies that [itex]2c = 3c = 0 \Leftrightarrow c = 0[/itex]. Hence

    [tex]\left(\begin{array}{cc}2a & a+3b \\ 0 & 3d\end{array}\right) = \left(\begin{array}{cc}2a & 2b+d \\ 0 & 3d\end{array}\right) [/tex]

    And we can write

    [tex]g = \left(\begin{array}{cc}a & d-a \\ 0 & d\end{array}\right)[/tex]

    Therefore the orbit can be described as

    [tex]O_A = \{gA\,|\, g\in GL(2,\mathbb{R})\}[/tex]

    So

    [tex]O_A = \left(\begin{array}{cc}2a & -2a + 3d \\ 0 & 3d\end{array}\right)\quad \forall a,b \in \mathbb{R}[/tex]

    how does this look?
     
    Last edited: Apr 17, 2005
  4. Apr 17, 2005 #3
    The isotropy group is the subgroup of [itex]GL(2,\mathbb{R})[/itex] consisting of the elements that do not move [itex]A[/itex]. That is

    [tex]G_A = \{gA = A\,|\,g\in GL(2,\mathbb{R})\}[/tex]

    Therefore we have

    [tex]\left(\begin{array}{cc}2a & -2a+3d \\ 0 & 3d\end{array}\right) = \left(\begin{array}{cc}2 & 1 \\ 0 & 3\end{array}\right)[/tex]

    So [itex]a = 1, \, d = 1 [/itex]. Therefore

    [tex]G_A = \left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right) = e[/tex]

    the isotropy subgroup consists of the identity element.
     
    Last edited: Apr 18, 2005
  5. Apr 19, 2005 #4
    Anyone know if I have done this correctly? Anyone?
     
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