Show that $G=SL_2(\Bbb R)$ has no finite dimensional unitary representations except the trivial one. Let $A(t)=\begin{pmatrix}1 &t\\0 &1\end{pmatrix}, \forall t \in \Bbb R$ .
Steps to follow : (1) For $m \in \Bbb N$ show $$\begin{pmatrix}m &0\\0 &m^{-1}\end{pmatrix} A(t){\begin{pmatrix}m&0\\0 &m^{-1}\end{pmatrix}}^{-1}=A(m^2t)={A(t)}^{m^2}$$ (2) Let $\phi : G \to U(n)$ be a representation. Show that the eigenvalues of $\phi(A(t))$ are a permutation of their $m$-th powers for every $m \in \Bbb N$ . Conclude that they all must be equal to 1.
(3) Show that the normal subgroup of $G$ generated by $\{A(t):t \in \Bbb R\}$ is the whole group.
I've verified the computation in Step(1)
But am a bit confused about the statement made in Step (2). What does the Authors actually intend to say by "the eigenvalues of $\phi(A(t))$ are a permutation of their $m$-th powers for every $m \in \Bbb N$" ?
EDIT : And for Step(3), According to Derek Holt's comment in a linked question : The group $PSL_2(𝐾)$ is simple for any field $𝐾$ with $|𝐾|>3$, so in particular $PSL_2(ℝ)$ is simple. So the only normal subgroups of $SL_2(ℝ)$ are the trivial group, the whole group, and its centre $\{\pm I_2\}$. So the normal subgroup generated by $𝐴(𝑡)$ is indeed the whole group.
And for the conclusion, Exodd's comments resolve it completely.
Thanks everyone for discussing and helping me solve this question 🙂
Just a short comment: There are statements like "An irreducible finite-dimensional representation of a noncompact simple Lie group of dimension greater than 1 is never unitary" which would give the result immediately, but I want to prove the statement in the Question ONLY as in the instruction/hints given in the question!
Best Answer
You need to use $2$ to show that the eigenvalues of $\phi(A(t))$ are $1$, we know that $\phi(A(t))$ and $\phi(A(t))^{m^2}$ have the same eigenvalues. Let $c$ be an eigenvalue of $\phi(A(t))$, for every positive integer $c^{m^2}$ is an eigenvalue of $\phi(A(t))$ since the number of the eigenvalues of $\phi(A(t))$ is finite, there exists $n\neq m$ such that $c^{n^2}=c^{m^2}$ we deduce that the order of $c$ is finite. Let $N$ be a multiple of the order of the eigenvalues of $\phi(A(t))$. The eigenvalues of $\phi(A(t))^{N^2}$ are the eigenvalues of $\phi(A(t))$ and are $1$ since they are $N^2$ power of eigenvalues of $A(t)$, we deduce that the eigenvalues of $A(t)$ are $1$.