- #1
mnb96
- 711
- 5
Hello,
let's supppose I am given a unit-quaternion q expressed as an element of \mathcal{C}\ell_{0,2}(\mathbb{R}) as follows:
\mathit{q} = a + b \mathbf{e_1} + c \mathbf{e_2} + d \mathbf{e_{12}}
I now rearrange the terms in the following way:
\mathit{q} = (a + d \mathbf{e_{12}}) + \mathbf{e_1}(b - c \mathbf{e_{12}})
The terms inside the brackets can be seen as ordinary complex numbers r_1e^{i\theta}[/itex] and r_2e^{i\phi}, so that <i>q</i> is an element of \mathbb{C}^2.<br /> <br /> Since <i>q</i> is a unit quaternion we must have that (r_1)^2 + (r_2)^2 = 1.<br /> Recalling that r_1and r_2[/tex] are non-negative, we can parametrize them with an auxiliary angle \psi \in [0,\pi / 2]<br /> <br /> r_1 = \cos\psi<br /> r_2 = \sin\psi<br /> <br /> We have: \mathit{q} = \left( \cos\psi e^{i\theta}, \sin\psi e^{i\phi} \right)<br /> <br /> <b>*** question: ***</b> when i compute the metric on the 3-sphere I obtain the well-known form:<br /> <br /> ds^2 = d\psi^2 + \cos^2\psi d\theta^2 + \sin^2\psi d\phi^2.<br /> <br /> But now, \psi \in [0,\pi/2], and \theta,\phi \in [0,2\pi] which is not correct because other sources says that \psi is in the range [0,\pi].<br /> <br /> Where is the mistake?
let's supppose I am given a unit-quaternion q expressed as an element of \mathcal{C}\ell_{0,2}(\mathbb{R}) as follows:
\mathit{q} = a + b \mathbf{e_1} + c \mathbf{e_2} + d \mathbf{e_{12}}
I now rearrange the terms in the following way:
\mathit{q} = (a + d \mathbf{e_{12}}) + \mathbf{e_1}(b - c \mathbf{e_{12}})
The terms inside the brackets can be seen as ordinary complex numbers r_1e^{i\theta}[/itex] and r_2e^{i\phi}, so that <i>q</i> is an element of \mathbb{C}^2.<br /> <br /> Since <i>q</i> is a unit quaternion we must have that (r_1)^2 + (r_2)^2 = 1.<br /> Recalling that r_1and r_2[/tex] are non-negative, we can parametrize them with an auxiliary angle \psi \in [0,\pi / 2]<br /> <br /> r_1 = \cos\psi<br /> r_2 = \sin\psi<br /> <br /> We have: \mathit{q} = \left( \cos\psi e^{i\theta}, \sin\psi e^{i\phi} \right)<br /> <br /> <b>*** question: ***</b> when i compute the metric on the 3-sphere I obtain the well-known form:<br /> <br /> ds^2 = d\psi^2 + \cos^2\psi d\theta^2 + \sin^2\psi d\phi^2.<br /> <br /> But now, \psi \in [0,\pi/2], and \theta,\phi \in [0,2\pi] which is not correct because other sources says that \psi is in the range [0,\pi].<br /> <br /> Where is the mistake?
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