The natural representation of the real group $G=SO(2)$ on $V=\Bbb R^2$ is irreducible.

Question

Prove that the natural representation of the real group $G=SO(2)$ on $V=\Bbb R^2$ is irreducible.

This is the followed up from this question
The natural representation of $G=SO(2)$ on $V=\Bbb C^2$ is not irreducible
I proved the following fact which may be useful:

Proposition: Let $G$ a commutative Lie group and $(\pi, V)$ a finite-dim unitary representation of $G$.

1. $\pi$ is irreducible iff ${\rm dim}\, V = 1$

2. There exist mutually orthogonal onedimensional invariant linear subspaces $V_1,..., V_n$ of $V $ such that $V = V_1 \oplus · · · \oplus V_n$.

3. The natural representation of $SO(2)$ on $\Bbb C^2$ is not irreducible.

My attempt:
The natural representation of $ G=SO(2)$ consisting of matrices with real entries on $V=\Bbb R^2$ is given by:

$$\pi(g):V\to V: v\mapsto gv $$

Question 1:
I am not sure why the proof in 3. breaks down? It seems like I can state pretty much the same things.

I can verify the hypotheses of the propositon above:

• $G=SO(2)$ is a commutative Lie group.

• The representation $(V,\pi)$ is finite-dimensional because $\Bbb R^2$ is finite-dimensional

• The representation $(V,\pi)$ is unitary because rotations preserve the scalar product on $\Bbb R^2$

Then I can use 1. in the proposition to say that because the dim $V=2 \neq 1$ as a $\Bbb R$-vector space, $\pi$ is not irreducible, which is giving the opposite of what I want to prove. So there must be some step that is not valid anymore

Question 2: How do I prove it then?

Assume that the natural representation is not irreducible, then 1. implies that the ${\rm dim}\, V\neq 1$. This is obvious, so it does not seem to help

Answer 1

A unitary representation requires that the representation space is a vector space over the complex numbers, which $\Bbb{R}^2$ is not. Thus, you cannot use $(a)$. This is not a minor annoyance, it is in fact the crux of the problem.

The fact that you are not guaranteed to be able to solve every polynomial means that you are not ensured enough eigenvalues. In fact, you will find that almost no element of $\text{SO}(2)$ has real eigenvalues, and so the representation has no invariant one-dimensional subspace. Since the proper subspaces are one-dim. subspaces, the representation is irreducible.