Reducing an additive group (simple example)

[Rory9]

I've found what looks like a really easy problem. If I can be clear about this one, I'll probably start to get somewhere with some of the more difficult ones. So I'd really appreciate it if someone could point out any mistakes/missunderstandings in my attempt at the answer; I'd like to nail this!

Homework Statement

Show that the 2x2 set of matrices

$$\Gamma(x) = \left( \begin{array}{ccc} 1 & 0 \\ x & 1 \\ \end{array} \right)$$

provides a representation of the additive group of all real numbers. Determine whether this representation is irreducible, is reducible but not completely reducible, or it completely reducible.

The Attempt at a Solution

I find that $$\Gamma(x_{1})\Gamma(x_{2}) = \Gamma(x_{1}x_{2})$$, given that the group 'multiplication' operation here is addition (so $$x_{1} + x_{2} = x_{1}.x_{2}$$.

I think if a non-finite group is irreducible, it is supposed to have - this is a textbook result -

$$\int |\chi(T)|^{2} dT = 1$$

So am I correct to compute $$\chi(T) = 1 + 1$$ and then conclude

$$\int |\chi(T)|^{2} dT$$

is not 1 (does T get replaced by x here, in the integral?).

I suppose if it is completely reducible, there would be a non-singular matrix that would diagonalise $$\Gamma$$, via

$$S^{-1}\Gamma(T)S$$

but there isn't. So is it correct to conclude that it's reducible, but not completely reducible?

How exactly should I use that textbook result in this case?

I'd really appreciate some insight here. Thanks in advance. :)

 

[Jhero]

Hi there,

First of all, great job on attempting to solve this problem! It looks like you have a good understanding of group representations and how to determine if they are reducible or not.

To answer your question about using the textbook result, you are correct in computing \chi(T) = 1+1, which gives you the character of the representation. However, the integral \int|\chi(T)|^2 dT is not necessary for this problem. This result is typically used to determine the dimension of the representation, but in this case, the dimension of the representation is already known (2x2 matrices).

To show that this representation is irreducible, you can use the fact that the representation is faithful, meaning that the map from the group to the matrices is injective. In other words, there is no non-trivial subgroup of the real numbers that is mapped to the identity matrix. Therefore, the representation is irreducible.

To determine if the representation is completely reducible, you can try to find a non-singular matrix S that diagonalizes \Gamma(T). However, as you have correctly pointed out, there is no such matrix in this case. Therefore, the representation is reducible but not completely reducible.

I hope this helps and clarifies any misunderstandings you may have had. Keep up the good work!