# Reducing an additive group (simple example)

1. Jan 1, 2010

### 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!

1. The problem statement, all variables and given/known data

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.

3. 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. :)