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Show that the exponential function maps any line through the origin in $\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$ onto a circle of radius $1$ in $\mathbb{S}^3$.

I know that for any element v $\in\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$, we can write $v = \theta u$, where $u$ is a unit vector in $\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$ and $\theta$ is any real number.

Then $e^{\theta u} = cos\theta + usin\theta\in S^3$ (i.e we can map any pure imaginary quaternion onto $S^3 = SU(2)$ So the exponential function maps $\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$ onto $S^3$

What is a line through $O$ in $\mathbb{R}i +\mathbb{R}j + \mathbb{R}k$? What is a circle of radius $1$ in $\mathbb{S}^3$?

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  • $\begingroup$ A line through the origin in the purely quaternionic space $\mathbb{R}^3$ is just as you say: a line that can be parametrized by a mapping $t \mapsto t u$ where $u$ is a unit vector. $\endgroup$
    – user43208
    Commented Oct 8, 2013 at 16:26
  • $\begingroup$ spanning any unit vector is always a line through origin? so {$\theta$u | $\theta$ is any real} is always a line through origin $\endgroup$
    – sarah
    Commented Oct 8, 2013 at 16:36
  • $\begingroup$ Yes, sure; the origin is the point where $\theta = 0$. $\endgroup$
    – user43208
    Commented Oct 8, 2013 at 17:18

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If $u$ is a unit norm element in $\mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ then the elements $1$ and $u$ form an orthonormal basis for the plane spanned by them. Therefore the exponential map you wrote down by definition gives the unit circle in this plane. A line through the origin is simply the line spanned by $u$.

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