Quaternions and metric of the 3-sphere

[mnb96]

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 [tex]r_1e^{i\theta}[/itex] and $$r_2e^{i\phi}$$, so that q is an element of $\mathbb{C}^2$.

Since q is a unit quaternion we must have that $$(r_1)^2 + (r_2)^2 = 1$$.
Recalling that [itex]r_1[/tex] and [itex]r_2[/tex] are non-negative, we can parametrize them with an auxiliary angle $$\psi \in [0,\pi / 2]$$

$$r_1 = \cos\psi$$
$$r_2 = \sin\psi$$

We have: $$\mathit{q} = \left( \cos\psi e^{i\theta}, \sin\psi e^{i\phi} \right)$$

*** question: *** when i compute the metric on the 3-sphere I obtain the well-known form:

$$ds^2 = d\psi^2 + \cos^2\psi d\theta^2 + \sin^2\psi d\phi^2$$.

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 $[0,\pi]$.

Where is the mistake?

 

[mnb96]

ok...it doesn't matter.
Actually I think there might no mistake, in fact I've just noticed that Hopf coordinates are defined with the same ranges as I have used.