On homomorphisms of the compact symplectic group Each $q$ in the symplectic group $Sp(1)$, defines an operator on the imaginary quaternions:
$\rho(q):\operatorname{Im}\mathbb{H} \rightarrow \operatorname{Im}\mathbb{H}$ by $\rho(q)(x)=qx\bar{q}$
We can see here that $\rho:Sp(1) \rightarrow O(3)$ is a group homomorphism, and the image of the symplectic group $\rho[Sp(1)]$ is a subset of the special orthogonal group $SO(3)$. Here we see that that $\rho$ maps $Sp(1)$ onto $SO(3)$
Is there an isomorphism of Lie groups from $Sp(1)$ onto the orthogonal group $O(3)$? Since $\ker(\rho)=\mathbb Z_2$, can we compose $\rho$ with a suitable reflection or a change of orientation for half of the unit quaternions to achieve such an isomorphism?
Are $Sp(1)$ and $O(3)$ isomorphic as topological groups? Are they at least isomorphic as groups or homeomorphic as topological spaces?
Thank you for your help.
 A: Just for clarity, the group called $\operatorname{Sp}(1)$ here is just the group $U(1,\Bbb H)$ of unit quaternions (I'm not personally used to see Sp followed by an odd index).
The morphism $\rho:U(1,\Bbb H)\to O(3,\Bbb R)$ has kernel $\{1,-1\}$ of order $2$, and image $SO(3,\Bbb R)$ of index$~2$, but that does not mean that $ U(1,\Bbb H)$ is isomorphic to $O(3,\Bbb R)$. Indeed it cannot be since $SO(3,\Bbb R)$, being a connected index $2$ subgroup, is necessarily a connected component of $O(3,\Bbb R)$, while $U(1,\Bbb H)$ is itself connected. And the same argument shows that the cannot even be a section (right inverse) $f:SO(3,\Bbb R)\to U(1,\Bbb H)$ of $\rho$, since its image having index $2$ would have to be a connected component of $U(1,\Bbb H)$, but cannot.
A simple purely algebraic argument that $U(1,\Bbb H)$ is not isomorphic to $O(3,\Bbb R)$ is that $-1\in U(1,\Bbb H)$ can be written as the square of an other element (in infinitely many many ways) but $-I\in O(3,\Bbb R)$ (the unique non identity central element) cannot. Or somewhat more generally, in $U(1,\Bbb H)$ every element is a square (as it lies in an $\Bbb R$-subalgebra of$~\Bbb H$ isomorphic to$~\Bbb C$) so there are no nontrivial group morphisms $U(1,\Bbb H)\to\{1,-1\}$, but one has $\det:O(3,\Bbb R)\to\{1,-1\}$ which is surjective.
