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 $$q 2 \langle V,W \rangle \overline{q} = 2 \langle V,W \rangle q \overline{q} = 2 \langle V,W \rangle .$$

Added on request:

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 $\text{Re}(V\overline{W}) = \frac{1}{2}(V\overline{W} + \overline{V\overline{W}}) = \frac{1}{2}(V\overline{W} + W\overline{V})$.

  • $\begingroup$ 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 $2 \langle V, W \rangle \in \mathbb R$? $\endgroup$ – Elliot Herrington Apr 1 at 9:22
  • $\begingroup$ This is well-known. I add a proof in my answer. $\endgroup$ – Paul Frost Apr 1 at 9:38
  • $\begingroup$ Very nice! Thank you. $\endgroup$ – Elliot Herrington Apr 1 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.