$SU(2)$ as an algebraic group

Question

The $\mathbb R$-valued points of the algebraic group $SU(2)$ can be identified with the real 3-sphere. But how does one define $SU(2)$ over the base field $\mathbb R$ as an algebraic group? What are its defining equations? What is the Hopf algebra structure on its coordinate ring?

Answer 1

You can equate $SU(2)$ with the group of unit quaternions. If you do it that way, then the coordinate ring becomes $$ \Gamma=\Bbb{R}[x_0,x_1,x_2,x_3]/\langle x_0^2+x_1^2+x_2^2+x_3^2-1\rangle. $$ The neutral element is $1$, so the counit is given by $$ \varepsilon: \Gamma\to\Bbb{R}, x_0\mapsto1, x_\ell\mapsto0,\ell=1,2,3. $$ The inverse of the quaternion $q=x_0+x_1i+x_2j+x_3k$ is its conjugate, so the antipode is determined by $$ S:\Gamma\to\Gamma, x_0\mapsto x_0,\ x_\ell\mapsto -x_\ell, \ell=1,2,3. $$ The product of $q$ and $q'=x_0'+x_1'i+x_2'j+x_3'k$ is $$ \begin{aligned} qq'&=(x_0x_0'-x_1x_1'-x_2x_2'-x_3x_3')+\\ &+(x_0x_1'+x_1x_0'+x_2x_3'-x_3x_2')i+\\ &+(x_0x_2'+x_2x_0'-x_1x_3'+x_3x_1')j+\\ &+(x_0x_3'+x_3x_0'+x_1x_2'-x_2x_1')k. \end{aligned} $$ This implies that the coproduct $\Delta:\Gamma\to\Gamma\otimes_{\Bbb{R}}\Gamma$ is determined by $$ \begin{aligned} \Delta(x_0)&=x_0\otimes x_0-x_1\otimes x_1-x_2\otimes x_2-x_3\otimes x_3,\\ \Delta(x_1)&=x_0\otimes x_1+x_1\otimes x_0+x_2\otimes x_3-x_3\otimes x_2,\\ \Delta(x_2)&=x_0\otimes x_2+x_2\otimes x_0-x_1\otimes x_3+x_3\otimes x_1,\\ \Delta(x_3)&=x_0\otimes x_3+x_3\otimes x_0+x_1\otimes x_2-x_2\otimes x_1.\\ \end{aligned} $$