- #1
jdstokes
- 523
- 1
Let [itex]\rho : \mathbb{H} \to \mathbb{H}; q \mapsto u^{-1}q u[/itex]
where u is any unit quaternion. Then [itex]\rho[/itex] is a continuous automorphism of H.
I'm asked to show that [itex]\rho[/itex] preserves the inner product and cross product on the subspace [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] consisting of purely imaginary quaternions.
The only thing I can think of is that [itex]\rho[/itex] acts on [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] by rotating that subspace (for which I know a proof), and rotations preserve angles and orientation.
Is there a more direct method which avoids using the fact that [itex]\rho[/itex] rotates [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] ?
where u is any unit quaternion. Then [itex]\rho[/itex] is a continuous automorphism of H.
I'm asked to show that [itex]\rho[/itex] preserves the inner product and cross product on the subspace [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] consisting of purely imaginary quaternions.
The only thing I can think of is that [itex]\rho[/itex] acts on [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] by rotating that subspace (for which I know a proof), and rotations preserve angles and orientation.
Is there a more direct method which avoids using the fact that [itex]\rho[/itex] rotates [itex]\mathbf{i}\mathbb{R} + \mathbf{j}\mathbb{R} + \mathbf{k}\mathbb{R}[/itex] ?