Group Action on GL(2,R) by Conjugation: Orbit and Isotropy Group of a Matrix

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Homework Help Overview

The discussion revolves around the group action of GL(2, R) on itself by conjugation, specifically analyzing the orbit and isotropy group of a given matrix A. The matrix A is defined as a specific 2x2 invertible matrix with real entries.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster seeks hints or pointers to understand the orbit and isotropy group of the matrix A. Some participants attempt to describe the orbit in terms of the action of the group on the matrix, while others explore the conditions under which the isotropy group remains unchanged.

Discussion Status

Participants are actively discussing their interpretations of the orbit and isotropy group. Some have provided calculations and reasoning, while others are questioning the correctness of these approaches. There is no explicit consensus yet, as some participants are still seeking validation of their methods.

Contextual Notes

The original poster expresses uncertainty and indicates a lack of working out, which suggests that they may be constrained by their current understanding of group actions and matrix properties.

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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}<br /> 2 & 1 \\ <br /> 0 & 3<br /> \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
 
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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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