- #1
mnb96
- 715
- 5
Hello,
let's supppose I am given a unit-quaternion q expressed as an element of [tex]\mathcal{C}\ell_{0,2}(\mathbb{R})[/tex] as follows:
[tex]\mathit{q} = a + b \mathbf{e_1} + c \mathbf{e_2} + d \mathbf{e_{12}}[/tex]
I now rearrange the terms in the following way:
[tex]\mathit{q} = (a + d \mathbf{e_{12}}) + \mathbf{e_1}(b - c \mathbf{e_{12}})[/tex]
The terms inside the brackets can be seen as ordinary complex numbers [tex]r_1e^{i\theta}[/itex] and [tex]r_2e^{i\phi}[/tex], so that q is an element of [itex]\mathbb{C}^2[/itex].
Since q is a unit quaternion we must have that [tex](r_1)^2 + (r_2)^2 = 1[/tex].
Recalling that [itex]r_1[/tex] and [itex]r_2[/tex] are non-negative, we can parametrize them with an auxiliary angle [tex]\psi \in [0,\pi / 2][/tex]
[tex]r_1 = \cos\psi[/tex]
[tex]r_2 = \sin\psi[/tex]
We have: [tex]\mathit{q} = \left( \cos\psi e^{i\theta}, \sin\psi e^{i\phi} \right)[/tex]
*** question: *** when i compute the metric on the 3-sphere I obtain the well-known form:
[tex]ds^2 = d\psi^2 + \cos^2\psi d\theta^2 + \sin^2\psi d\phi^2[/tex].
But now, [itex]\psi \in [0,\pi/2][/itex], and [itex]\theta,\phi \in [0,2\pi][/itex] which is not correct because other sources says that [itex]\psi[/itex] is in the range [itex][0,\pi][/itex].
Where is the mistake?
let's supppose I am given a unit-quaternion q expressed as an element of [tex]\mathcal{C}\ell_{0,2}(\mathbb{R})[/tex] as follows:
[tex]\mathit{q} = a + b \mathbf{e_1} + c \mathbf{e_2} + d \mathbf{e_{12}}[/tex]
I now rearrange the terms in the following way:
[tex]\mathit{q} = (a + d \mathbf{e_{12}}) + \mathbf{e_1}(b - c \mathbf{e_{12}})[/tex]
The terms inside the brackets can be seen as ordinary complex numbers [tex]r_1e^{i\theta}[/itex] and [tex]r_2e^{i\phi}[/tex], so that q is an element of [itex]\mathbb{C}^2[/itex].
Since q is a unit quaternion we must have that [tex](r_1)^2 + (r_2)^2 = 1[/tex].
Recalling that [itex]r_1[/tex] and [itex]r_2[/tex] are non-negative, we can parametrize them with an auxiliary angle [tex]\psi \in [0,\pi / 2][/tex]
[tex]r_1 = \cos\psi[/tex]
[tex]r_2 = \sin\psi[/tex]
We have: [tex]\mathit{q} = \left( \cos\psi e^{i\theta}, \sin\psi e^{i\phi} \right)[/tex]
*** question: *** when i compute the metric on the 3-sphere I obtain the well-known form:
[tex]ds^2 = d\psi^2 + \cos^2\psi d\theta^2 + \sin^2\psi d\phi^2[/tex].
But now, [itex]\psi \in [0,\pi/2][/itex], and [itex]\theta,\phi \in [0,2\pi][/itex] which is not correct because other sources says that [itex]\psi[/itex] is in the range [itex][0,\pi][/itex].
Where is the mistake?
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