Isotropy subgroup of $SL(2,\mathbb{C})$ action.

Question

I am trying to compute explicitly a particular subgroup of $(P)SL(2,\mathbb{C})$ arising from the following action.

Let $X=\mathbb{P}^1\odot \mathbb{P}^1$ be the symmetric product of the complex projective line. Assume that $SL(2,\mathbb{C})$ acts on $X$ in the natural way, namely $$ g\cdot[p,q] = [g\cdot p, g\cdot q] $$ for every $[p,q]=[q,p]\in X$. Here $g\cdot p$ means the natural action of $SL(2,\mathbb{C})$ on the complex projective line:

$$ \begin{pmatrix} a & b\\c & d \end{pmatrix} \cdot (x:y) = (ax+by: cx+dy) $$ in homogeneous coordinates.

Problem

Calculate the isotropy subgroup of $[(1:0),(0:1)]$ for the above action.

Progress so far

If we want to fix $[(1:0),(0:1)]$ (I will denote $N=(1:0), S=(0:1)$ ) we can either fix both points or interchange them as we are in the symmetric product space:

$$ g \in \text{Iso}[N,S] \iff (g\cdot N = N \land g\cdot S = S)\lor (g\cdot N = S \land g\cdot S = N)$$

In the first case, the matrix $g$ is forced to be of the form $$ g=\begin{pmatrix}a & 0 \\0 & a^{-1}\end{pmatrix} $$ while for the second case we get $$ g'=\begin{pmatrix}0 & b \\b^{-1} & 0\end{pmatrix} $$ If we denote the above by $g(a)$ and $g'(b)$ we get that these elements form a group whenever we vary $a,b\in\mathbb{C}^*$. What I do not know is what kind of Lie group is this? I suspect there is a semidirect product lurking somewhere, since we have

1. $g(a_1)\cdot g(a_2) = g(a_1a_2)$
2. $g(a)\cdot g'(b) = g(ab)$
3. $g'(b)\cdot g(a) = g'(ba^{-1})$

I fail to grasp the structure of this group. Is it "just" $\mathbb{C}^*$ with an extra component with which it interacts weirdly? It is still complex 1-dimensional, that is for sure, so it cannot be a semidirect product.

Answer 1

We can also do this by using the extended complex plane for $\mathbb{P}^1$ and Mobius transformations $(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix})z=\frac{az+b}{cz+d}$. We are looking for the setwise stabilizer of $\{0,\infty\}$. If it fixes both points, then $g(\infty)=\infty$ implies $c=0$ so $g$ is affine, then $g(0)=0$ implies further that $g$ is linear, represented by a diagonal matrix. If $g$ swaps both points, then $g(0)=\infty$ says $d=0$ and then $g(\infty)=0$ implies $a=0$ which implies $g$ is a scalar multiple of the reciprocal function, i.e. $g(z)=\beta/z$, represented by $(\begin{smallmatrix}0 & i\sqrt{\beta} \\ i/\sqrt{\beta} & 0 \end{smallmatrix})$ for some choice of square root $\sqrt{\beta}$ of $\beta$.

The group comprised of homotheties $g(z)=\beta z$ and scaled inversions $g(z)=\alpha/z$ is in particular generated by homotheties and the one inversion $g(z)=1/z$, corresponding to $\sigma=(\begin{smallmatrix} 0 & i \\ i & 0 \end{smallmatrix})$, or equivalently the inversion $g(z)=-1/z$, corresponding to $\tau=(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix})$. Then the diagonal matrices form a copy of $\mathbb{C}^\times$, the subgroups generated by $\sigma$ or $\tau$ give copies of $\mathbb{Z}_2$ (in $\mathrm{PSL}_2\mathbb{C}$ where $-I$ is trivial), and we can see that conjugating a homothety of $\beta$ by either $\sigma$ or $\tau$ results in a homothety of $1/\beta$, so either copy of $\mathbb{Z}_2$ acts on $\mathbb{C}^\times$ by inversion and indeed the stabilizer is a semidirect product $\mathbb{C}^\times\rtimes\mathbb{Z}_2$.