# Quaternion conjugation map is orthogonal linear transformation

The following concerns an exercise from an undergraduate level textbook on Lie groups, which provides an elementary proof of $$SU(2)/\mathbb{Z}_2 \cong SO(3)$$.

Think of $$Sp(1)$$ as the group of unit-length quaternions; that is, $$Sp(1) = \left\{ q \in \mathbb H : |q| = 1 \right\}.$$ For every $$q \in Sp(1)$$, show that the conjugation map $$C_q: \mathbb H \to \mathbb H$$, defined as $$C_q(V) = q V \overline{q}$$, is an orthogonal linear transformation. Thus, with respect to the natural basis $$\left\{ 1, i, j, k \right\}$$ of $$\mathbb H$$, $$C_q$$ can be regarded as an element of $$O(4)$$.

My attempt so far has been as follows: To show it is an orthogonal linear transformation, we must show $$\langle C_q(V), C_q(W) \rangle = \langle V, W \rangle$$ for $$V, W \in \mathbb H$$. So we have

$$LHS = (qV \overline q) \cdot (\overline{qW \overline q}) = qV \overline q \cdot q \overline W \overline q = q V \overline W \overline q,$$

and I just can't seem to figure out what to do from here. We know $$q$$, $$\overline q$$ are unit-length, but that does not mean they commute with $$V, \overline W.$$ I have a feeling that perhaps $$V$$ and $$W$$ must be pure imaginary quaternions, but even making that assumption doesn't seem to help. If anyone has any insight, I'd be very grateful. It's possible I'm using a bad/incorrect definition of orthogonal linear transformation, but it definitely seems like the error in question is something simple/obvious. Thanks for your time.

You have $$\langle V,W \rangle = \frac{1}{2}(V \overline{W} + W \overline{V})$$. Therefore $$2\langle C_q(V),C_q(W) \rangle = q V\overline{q} q \overline{W} \overline{q} + q W \overline{q} q \overline{V} \overline{q} = q V\overline{W} \overline{q} + q W\overline{V} \overline{q} = q (V \overline{W} + W \overline{V}) \overline{q} \\ = q 2 \langle V,W \rangle \overline{q} .$$ But $$2\langle V,W \rangle \in \mathbb{R}$$, hence $$q 2 \langle V,W \rangle \overline{q} = 2 \langle V,W \rangle q \overline{q} = 2 \langle V,W \rangle .$$
Write $$V = a + bi + cj + dk, W = a' + b'i + c'j + d'k$$. Then by definition $$\langle V, W \rangle = aa' +bb' + cc' + dd' \in \mathbb{R}$$. Moreover $$V\overline{W} = (a + bi + cj + dk)(a' - b'i - c'j - d'k) = aa' +bb' + cc' + dd' + r$$ with a suitable $$r = ui + vj + wk$$. Hence $$\langle V, W \rangle = \text{Re}(V\overline{W})$$. But $$\text{Re}(V\overline{W}) = \frac{1}{2}(V\overline{W} + \overline{V\overline{W}}) = \frac{1}{2}(V\overline{W} + W\overline{V})$$.
• Looks good. I'm just wondering where you got the first equality from, i.e., the statement $\langle V, W\rangle = \frac{1}{2} (V \overline W + W \overline V).$ Also, why is $2 \langle V, W \rangle \in \mathbb R$? – Elliot Herrington Apr 1 at 9:22