Understanding Representation theory

Question

I'm a physicist and I was reading the book of Barut "representation theory, groups an applications". I want to find the unitary irreducible representations of the group $SL(2,R)$. Barut writes that if $\delta \neq 0$, we can write $$ g = \begin{bmatrix} \alpha & \beta \\\ \gamma & \delta \end{bmatrix} = \begin{bmatrix} \delta^{-1} & \beta \\\ 0 & \delta \end{bmatrix} \begin{bmatrix} 1 & 0 \\\ \frac{\gamma}{\delta} & 1 \end{bmatrix} = \begin{bmatrix} \delta^{-1} & \beta \\\ 0 & \delta \end{bmatrix} \begin{bmatrix} 1 & 0 \\\ x & 1 \end{bmatrix}, $$ where $x \in SL(2,R)/K$ and $K$ is group of upper triangular matrix. After that he finds how transform $x$ under the entire group. Hence $$ \begin{bmatrix} 1 & 0 \\\ x & 1 \end{bmatrix} \begin{bmatrix} \alpha & \beta \\\ \gamma & \delta \end{bmatrix} = ... = \begin{bmatrix} (\beta x + \delta)^{-1} & \beta \\\ 0 & \beta x + \delta \end{bmatrix} \begin{bmatrix} 1 & 0 \\\ \frac{\alpha x + \gamma}{\beta x + \delta} & 1 \end{bmatrix} $$, so under the entire transformation we have $$ x \to xg = \frac{\alpha x + \gamma}{\beta x + \delta}. $$ What I didn't understand is that if we perform only a transformation of the "stability group" (or for the physicists "little group") and hence we take $\alpha = \delta^{-1}$ and $\gamma = 0$ we have $$ x \to xg|_{LG} = \frac{\delta^{-1} x}{\beta x + \delta}. $$ Which seems to me not invariant as it should be, because it belongs to $SL(2,R)/K$ (the right costes of $K$). What I aspected is that under the stability group it doen't change.

Answer 1

We know the group $G=\mathrm{SL}(2,\mathbb{R})$ acts on $\mathbb{R}^2$. Thus, it acts on the real projective line $\mathbb{RP}^1$. With the congruence $\big[\begin{smallmatrix} x \\ y \end{smallmatrix}\big]\sim\big[\begin{smallmatrix} x/y\\1\end{smallmatrix}\big]$, we can identify $\mathbb{RP}^1$ with the extended reals $\widehat{\mathbb{R}}=\mathbb{R}\cup\{\infty\}$. Then the action of $G$ given by $\big[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big]\big[\begin{smallmatrix} x \\ y \end{smallmatrix}\big]=\big[\begin{smallmatrix} ax+by \\ cx+dy \end{smallmatrix}\big]$ with $z=x/y$ becomes $\big[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big]z:=\frac{az+b}{cz+d}$, the Möbius aka linear fractional transformations. The stabilizer subgroup of $1/0=\infty$ is the subgroup $B$ (for "Borel") of upper triangular matrices. This means $g(\infty)=\infty$ iff $g\in B$. Thus, we can identify $G/B$ (left coset space) with $\mathbb{RP}^1$ (which is $\simeq S^1$).

Moreover, there is a subgroup $T\le G$ of lower unitriangular matrices $\big[\begin{smallmatrix}1&0\\x&1\end{smallmatrix}\big]$ which is almost a left transversal for $G/B$. (If it were, we could say $G$ is a knit product of $T$ and $B$, meaning $G=TB$ and $T\cap B=1$, or in other words every $g\in G$ is expressible as $g=tb$ for a unique choice of $t\in T,b\in B$.) But $T$ is not a transversal: it's missing left coset of $J=\big[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\big]$ corresponding to $\gamma/\delta=1/0$. Write $\widehat{T}=T\sqcup\{J\}$.

We can also think about this backwards: view $B\,\backslash\,G$ as a right coset space and have a right action of $G$, then have a sort of knit product $G=B\widehat{T}$, so $\widehat{T}$ is a right transversal. But now $B$ is the stabilizer of $0\in\widehat{\mathbb{R}}$ corresponding to the row vector $[0~1]$. This is what your source appears to be doing. It also involves a projection $G\to \widehat{T}$, which we may call $\tau(\big[\begin{smallmatrix}\alpha&\beta\\ \gamma&\delta\end{smallmatrix}\big])=\gamma/\delta$, with the inclusion $T\hookrightarrow G$ a distinguished section. Note $\tau$ is left $B$-invariant, i.e. $\tau(bg)=\tau(g)$ for all $g\in G$, because $\tau(\cdot)$ extracts the representative $g$ of all elements of a right coset $Bg$. Acting on the right, $B$ is only defined as the stabilizer of a particular point (namely, the identity $I$), so we have $\tau(I\cdot b)=\tau(I)$ for all $b\in B$, but generally $\tau(gb)\ne\tau(g)$ for other $g\in G$.

So in summary, $B$ is the stabilizer of a particular point/value, not the kernel of the action (the matrices which have no effect on anything). For a simpler example, consider $\mathrm{SO}(3)$ acting on the sphere $S^2$ by rotations. If you pick the $z$-axis, then the stabilizer subgroup is a copy of $\mathrm{SO}(2)$ (block-diagonally in $\mathrm{SO}(3)$) which is made up of the rotations around the $z$-axis; this fixes the North/South poles of the sphere, but doesn't fix any other points or axes.