I define the $2\times2$ matrix sphere as the quotient set $$\boxed{\mathbb{S}_{2}:=\{(X,Y)\in M_2(\mathbb{R})^2:X^2+Y^2=I\}/\sim}$$ where $M_2(\mathbb{R})^2=\mathbb{R}^{2\times2}\times\mathbb{R}^{2\times2}$ is the set of pairs of $2\times2$ real matrices and $\sim$ is the equivalence relation $$(A,B)\sim(C,D)\iff\exists P\in GL_2(\mathbb{R}):(C,D)=(PAP^{-1},PBP^{-1})$$ The motivation comes from generalizing the $1$-sphere $S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}$ to $2\times2$ matrices identifying those pairs that are essentially the same since every matrix solution $(A,B)$ of $X^2+Y^2=I$ leads to an infinite amount of solutions as $$\forall P\in GL_2(\mathbb{R}):(PAP^{-1})^2+(PBP^{-1})^2 = P^2(A^2+B^2)P^{-2}=P^2IP^{-2}=I$$
Q: My question then is, analogous to the group structure of $(S^1,\cdot)$ considering it as a subgroup of $\mathbb{C}^\times$, is there any natural operation $\odot:\mathbb{S}_{2}\times\mathbb{S}_2\to\mathbb{S}_2$ that $(\mathbb{S}_2,\odot)$ is a group?
The thing with $\mathbb{S}_2$ is that it is not abelian. So, if we considered the operation $$\odot:[M_2(\mathbb{R})^2/\sim]\times[M_2(\mathbb{R})^2/\sim]\to M_2(\mathbb{R})^2/\sim$$ $$(A,B)\odot(C,D):=(AC-BD,AD+BC)$$ which is an intuitive generalization of the complex product, $\mathbb{S}_2$ is not closed under $\odot$ so $(\mathbb{S}_2,\odot)$ is not a group. In fact, if we define the norm $N:M_2(\mathbb{R})^2\to M_2(\mathbb{R})$ as $N(A,B)=A^2+B^2$, then $\forall (A,B),(C,D)\in\mathbb{S}_2$ $$N((A,B)\odot(C,D))=N(A,B)N(C,D)\iff ACBD+BDAC=ADBC+BCAD$$ So $N$ is only multiplicative given that condition.
I don't know if this helps but it is easy to see that the orthogonal group $\mathcal{O}_2=\{Q\in\mathbb{R}^{2\times2}:Q^TQ=I\}$ acts on $\mathbb{S}_2$ as $$\begin{pmatrix}\alpha& \beta\\ \gamma& \delta\end{pmatrix}(A,B)=(\alpha A+\beta B,\gamma A+\delta B)$$