For any $n \in \Bbb N$, for any representation $\phi:SL_2(\Bbb R) \to U(n)$ we must have $\phi \begin{pmatrix} 0 &1 \\ -1 &0 \end{pmatrix}= I_{n}$

Question

Actually this is a continuation of this question, but am asking it separately as it deserves separate discussion as an independent problem. Thanks to comments by Exodd and an answer by Tsemo Aristide, we could solve upto step(2) and the only thing left to be proven was :

Let $A(t)=\begin{pmatrix}1 &t\\0 &1\end{pmatrix}, \forall t \in \Bbb R$ . Then show that the normal subgroup of $G=SL_2(\Bbb R)$, generated by $\{A(t):t \in \Bbb R\}$ is the whole group.

As I mentioned in the previous discussion, I found out this question, which implies that all I need to show that the normal subgroup of $G$ generated by $\{A(t):t \in \Bbb R\}$ contains $\begin{pmatrix} 0 &1 \\ -1 &0 \end{pmatrix}$, which I was unable to prove!

NOW I THINK THAT THIS CLAIM IS ACTUALLY FALSE!

Since, for any $\begin{pmatrix} a &b \\ c &d\end{pmatrix} \in G$ we have , $\begin{pmatrix} a &b \\ c &d\end{pmatrix} \begin{pmatrix} 0 &1 \\ -1 &0 \end{pmatrix} \begin{pmatrix} d &-b \\ -c &a\end{pmatrix}=\begin{pmatrix} -bd-ac &b^2+a^2 \\ -d^2-c^2 &bd+ac\end{pmatrix}$ .

Now if the claim in step(3) has to be true, then for certain $a,b,c,d \in \Bbb R$ with $ad-bc=1$ one must have that $\begin{pmatrix} -bd-ac &b^2+a^2 \\ -d^2-c^2 &bd+ac\end{pmatrix}=\begin{pmatrix} 1 &t \\ 0 &1\end{pmatrix}$ . But then $d^2+c^2=0 \implies d=c=0 \implies 1= -bd-ac=0$ , A contradiction!

So rather it's enough to just show that

For any $n \in \Bbb N$, for any representation $\phi:SL_2(\Bbb R) \to U(n)$ we must have $\phi \begin{pmatrix} 0 &1 \\ -1 &0 \end{pmatrix}= I_{n}$

Note that $\begin{pmatrix} 0 &1 \\ -1 &0 \end{pmatrix}^4=I_2$, then since $\phi$ is a group homomorphism $\phi \begin{pmatrix} 0 &1 \\ -1 &0 \end{pmatrix}$ is a Unitary matrix of order $1 \text{ or }2 \text{ or } 4$ .

Now that's all I could come up with, unable to see how to proceed from here. Thanks in advance for help!

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. Please don't use them, as I don't have them at my disposal!

Answer 1

I don't know if it is the fastest way, but

$$ \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1& 1\\ 0& 1 \end{pmatrix} \begin{pmatrix} 0&1 \\ -1&0 \end{pmatrix} = \begin{pmatrix} 1& 0\\ -1&1 \end{pmatrix} $$ $$ \begin{pmatrix} 1 & 0\\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1& 1\\ 0& 1 \end{pmatrix} = \begin{pmatrix} 1& 1\\ -1&0 \end{pmatrix} $$ $$ \begin{pmatrix} 1& 1\\ -1&0 \end{pmatrix} \begin{pmatrix} 1& 0\\ -1& 1 \end{pmatrix} = \begin{pmatrix} 0& 1\\ -1&0 \end{pmatrix} $$