Canonical representation of $\operatorname{SL}(2,\mathbb{R})$ on $L^2(\mathbb{R}^2)$

Question

As a unimodular subgroup of the group of automorphisms of $\mathbb{R}^2$, $\operatorname{SL}(2,\mathbb{R})$ can be represented as a subgroup of $\mathcal{U}(L^2(\mathbb{R}^2))$ (the group of unitary operators on $L^2(\mathbb{R}^2)$), where each element $A \in \operatorname{SL}(2,\mathbb{R})$ is mapped to a unitary operator $R_A$ defined by $$ R_A \xi(X):= \xi(A^{-1}X) $$ for all $\xi \in L^2(\mathbb{R}^2)$ and $X \in \mathbb{R}^2$. Applying the Fourier transform on $L^2(\mathbb{R}^2)$, it seems to me that this representation has to be irreducible. Is this true?

If that is so, to which irreducible unitary representation of $\operatorname{SL}(2,\mathbb{R})$, is it equivalent? (I found the list of irreducible unitary representations of $\operatorname{SL}(2,\mathbb{R})$ in Lang's book.)

Answer 1

The representation is highly reducible, and it is a direct integral of principal unitary series representations of $\operatorname{SL}_2({\mathbb R})$. You may look at the book by Gelfand-Pjatetski Shapiro where a full decomposition of this rep is given. As Yves Cornulier has pointed out, the rep is the space of square summable functions on $G/U={\mathbb R}^2\setminus \{0\}$ where $U$ is the upper triangular unipotent matrices. Hence the group ${\mathbb R}^*$ acts on this space $f\mapsto (x\mapsto \frac{1}{{\sqrt \mid \lambda \mid }}f(\lambda x))$ . This action commutes with the $\operatorname{SL}_2({\mathbb R})$ action on the left. Decomposing this action of scalars (an abelian group) you get Mellin transforms of $L^2$ functions at a parameter $s$ (corresp to a representation of diagonals) which yields the unitary principal series paratemtrised by $s$.