Let $\mathbb{P}^2 = \mathbb{P}(Sym^2\mathbb{C}^2)$$\mathbb{P}^2 = \mathbb{P}(\operatorname{Sym}^2\mathbb{C}^2)$ be the projective space of $2\times 2$ symmetric matrices over $\mathbb{C}$ modulo scalar.
Define an $SL(2)$$\mathrm{SL}(2)$-action on $\mathbb{P}^2$ by $(A,Z)\mapsto AZA^t$, where $A\in SL(2)$$A\in \mathrm{SL}(2)$ and $Z\in \mathbb{P}^2$. The stabilizer of the identity is then $$S = \{A\in SL(2)\: | \: AA^t = \lambda I\}$$$$S = \{A\in \mathrm{SL}(2)\: | \: AA^t = \lambda I\}$$ for some $\lambda\in\mathbb{C}^{*}$. However, some papers refer to $\mathbb{P}^2$ as the wonderful compactification of $SL(2)/SO(2)$$\mathrm{SL}(2)/\mathrm{SO}(2)$ even though it seems to me that it would be more correct to say that $\mathbb{P}^2$ is the wonderful compactification of $SL(2)/S$$\mathrm{SL}(2)/S$.
For instance, via the map $SL(2)\rightarrow\mathbb{P}^2,\: A\mapsto AA^t$$\mathrm{SL}(2)\rightarrow\mathbb{P}^2,\: A\mapsto AA^t$ the matrix $$ \left( \begin{array}{cc} -i & 0 \\ 0 & i \end{array} \right) $$ is mapped to $$ -\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) $$ Note that $$ \left( \begin{array}{cc} -i & 0 \\ 0 & i \end{array} \right)\in S\setminus SO(2) $$$$ \left( \begin{array}{cc} -i & 0 \\ 0 & i \end{array} \right)\in S\setminus \mathrm{SO}(2) $$ so the morphism $SL(2)/SO(2)\rightarrow\mathbb{P}^2$$\mathrm{SL}(2)/\mathrm{SO}(2)\rightarrow\mathbb{P}^2$ is not of degree one.
What am I misunderstanding here?
