SO(2)(K) isomorphic to a $K$-split Torus and its rank Let $K$ be a field and $SO(2)(K):= \{Q \in GL_2(K) \ \vert \ Q^T Q= I \text{ and } \det(Q)=1 \}$ the special orthogonal group.
Is there any characterisation known to decide for which fields $K$ the special orthogonal group $SO(2)(K)$ is toric, ie $SO(2)(K) \cong (K^{\times})^n $ and if that's the case the rank $n$ is always $1$?
My naive approach to define a map $K^{\times} \to SO(2)(K)$ via $$ x \mapsto \begin{pmatrix} x & -\sqrt{1-x^2} \\ \sqrt{1-x^2} & x \end{pmatrix}$$
suggest that the rank of $SO(2)(K)$ is always at least one (rank = maximal $n$ such that $SO(2)(K)$ contains a torus $(K^{\times})^n $ as subgroup).
Can $n$ be bigger than $1$? Under which conditions on $K$ the map above is an isomorphism?

First as an algebraic variety, $SO_2(K)\simeq V(x^2+y^2=1)$. Indeed, let $\begin{pmatrix} x & a \\ y & b\end{pmatrix}\in SO_2(K)$, we have the two conditions $xa+yb=0$ and $xb-ya=1$ have completely determined $a,b$ since $\det\begin{pmatrix} x & y \\ -y & x\end{pmatrix}=1\not=0$.
On the other hand, $K^{\times}\simeq V(xy=1)$. If $\sqrt{-1}\in K$, then we have the splitting $x^2+y^2=(x+\sqrt{-1}y)(x-\sqrt{-1}y)=1$, which connects the two varieties by solving the equations $x=a+b\sqrt{-1}y, y=a-b\sqrt{-1}y$ assuming $\text{char}(K)\not=2$ to get $$a=\frac{x+y}{2}=\frac{x+\frac{1}{x}}{2}, b=\frac{x-y}{2\sqrt{-1}}=\frac{x-\frac{1}{x}}{2\sqrt{-1}}$$
In other words, $x\in K^{\times}\mapsto \begin{pmatrix} \frac{x+\frac{1}{x}}{2} & -\frac{x-\frac{1}{x}}{2\sqrt{-1}} \\ \frac{x-\frac{1}{x}}{2\sqrt{-1}} & \frac{x+\frac{1}{x}}{2} \end{pmatrix}$ is the desired isomorphism $K^{\times}\simeq SO_2(K)$.
So $\text{char}(K)\not=2$ and $\sqrt{-1}\in K$ are sufficient. Now we show they are also necessary.
As $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^2=-I_2$, if there is a group isomorphism $K^{\times}\simeq SO_2(K)$, $K^{\times}$ must contain $\sqrt{-1}$.
If $\text{char}(K)=2$, then $x^2+y^2=(x+y)^2=1$, $x+y=1$. And from here it's not hard to show $\begin{pmatrix} x & a \\ y & b \end{pmatrix}=\begin{pmatrix} x & 1-x \\ 1-x & x\end{pmatrix}$ must be symmetric, hence its square is just $1$. But in $K^{\times}$, the only element whose square is $1$ is $1$ itself, and $SO_2(K)$ contains at least two elements $I_2, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, hence $SO_2(K)\not\simeq K^{\times}$ for any $K$ with characteristic $2$. Indeed, $x\mapsto\begin{pmatrix} 1-x & x \\ x & 1-x\end{pmatrix}$ defines an isomorphism from $(K, +)$ to $SO_2(K)$.
Note that while we have used a little algebraic geometry to figure out the exact isomorphism, our arguments are purely group theoretical. In any case, $n$ cannot be strictly greater than $1$, since the varieties $V(x^2+y^2=1)$ and $V(xy=1)$ have the same dimension.
