The action of $SL(2,\mathbb C)$ on $\mathbb CP^n$

Question

Let $G$ be a semisimple algebraic group acting linearly on $\mathbb CP^n$. Let $S\cong SL(2,\mathbb C)$ be a closed subgroup of $G$.

How one can see that the orbit of $S$ is a copy of $\mathbb CP^1$ and the isotropy is a Borel subgroup?

Answer 1

By the representation theory of $SL(2,\mathbb{C})$, we know that $V = \mathbb{C}^{n+1}$ decomposes into summands of the form $S^k(W)$, where $W = \mathbb{C}^2$ is the standard representation of $SL(2,\mathbb{C})$. $S^k(W)$ can be thought of as the space of homogeneous polynomials of degree $k$ in $2$ complex variables (or rather its complex dual space).

I have some doubts about the claim in general. Just consider $S^3(W)$, and its projectivization. If we take an element in $S^3(W)$ corresponding to a cubic polynomial with 3 different roots, then the isotropy subgroup fixing this polynomial (up to scalar) is a finite subgroup, by property of Moebius transformations. Thus in this case, the orbit is biholomorphic to a quotient of $SL(2,\mathbb{C})$ by a finite subgroup.