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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!

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  • $\begingroup$ Hint for the second point: in the first point, all the matrices you have written are in $SL_2$, so you can apply $\phi$ $\endgroup$
    – Exodd
    Jun 14, 2020 at 17:43
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    $\begingroup$ as for the conclusion: proving that $A(t)$ has only eigenvalues 1, says that $\phi(A(t))=I$ for every $t$, and if they generate the whole group, then you can prove that the whole group maps to the identity matrix $\endgroup$
    – Exodd
    Jun 14, 2020 at 17:57
  • $\begingroup$ @Exodd I am having difficulty in understanding the precise statement intended in the 2nd step. If you can explain the statement, I can try to prove it following your hint $\endgroup$
    – Brozovic
    Jun 14, 2020 at 17:58
  • $\begingroup$ @Exodd Thanks for explaining the conclusion point. $\endgroup$
    – Brozovic
    Jun 14, 2020 at 17:58
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    $\begingroup$ "Call $\lambda_1,\dots,\lambda_n$ the eigenvalues of $\phi(A(t))$. Prove that, for any integer $m$, the quantities $\lambda_1^m,\dots,\lambda_n^m$ coincide with $\lambda_1,\dots,\lambda_n$ in some order. Conclude that $\lambda_i=1$ for every $i$" $\endgroup$
    – Exodd
    Jun 14, 2020 at 17:59

1 Answer 1

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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 $1$.

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  • $\begingroup$ I felt that the step(3) needs separate attention. So for an extended discussion on Step(3), please visit this question $\endgroup$
    – Brozovic
    Jun 15, 2020 at 16:16

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