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Group Action on a Set

  • Thread starter Oxymoron
  • Start date
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
 
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:
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:
Anyone know if I have done this correctly? Anyone?
 

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