SL(n,R) Lie group as submanifold of GL(n,R)

[cianfa72]

How? $\operatorname{SL}(1,\mathbb{Q})=\{A\in \mathbb{M}(1,\mathbb{Q})=\mathbb{Q}\,|\,\det A=1\}=\{1\}$ and $\operatorname{SU}(2,\mathbb{C})$ is real three-dimensional.

In his lectures, he claims $SL(1,\mathbb Q)$ is simply connected like $SU(2)$ and it is (group) isomorphic and (topologically) homeomorphic to $SU(2)$. Note that $\mathbb Q$ is the field of quaternions.

 

[fresh_42]

In his lectures, he claims $SL(1,\mathbb Q)$ is simply connected like $SU(2)$ and it is (group) isomorphic and (topologically) homeomorphic to $SU(2)$. Note that $\mathbb Q$ is the field of quaternions.

That was confusing. $\mathbb{Q}$ is usually the rational numbers. The quaternions are usually represented by $\mathbb{H},$ the Hamilton quaternions. Your isomorphism would then be
$$
\operatorname{U}(1,\mathbb{H})\cong \operatorname{SU}(2,\mathbb{C})\,.
$$