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Let $\alpha : (so(3), [\cdot, \cdot]) \to (\mathbb{R}^3, \times)$ be the Lie algebra isomorphism that maps the usual $l_x, l_y, l_z$ basis to the $e_1, e_2, e_3$ basis. Recall that, for any $R \in SO(3)$, one has

$$\alpha Ad_R \alpha^{-1} = R$$

where $Ad_R : so(3) \to so(3)$ is the adjoint action.

Let $\beta : sp(1) \to \mathbb{R}^3$ be the vector space isomorphism that maps the $i, j, k$ basis to the $e_1, e_2, e_3$ basis. We define the covering morphism $\Phi : Sp(1) \to SO(3)$ by

$$\Phi(q) = \beta Ad_q \beta^{-1}$$

where $Ad_q : sp(1) \to sp(1)$ is the adjoint action, i.e. conjugation of a pure quaternion in $sp(1)$ by the unit quaternion $q$. Let $\phi : sp(1) \to so(3)$ be the induced Lie algebra morphism (i.e. the derivative of $\Phi$ at the identity), which is an isomorphism since the kernel of $\Phi$ is discrete (namely, $\ker \Phi = \{\pm 1\}$)

Recall that, if $F : G \to H$ is a morphism of Lie groups with induced Lie algebra map $f : \mathfrak{g} \to \mathfrak{h}$, then, for any $g \in G$ and $X \in \mathfrak{g}$, one has
$$f(Ad_g X) = Ad_{F(g)} f(X)$$

In the case at hand, this means that, for any $q \in Sp(1)$ and $u \in sp(1)$, one has
$$\phi(Ad_q u) = Ad_{\Phi(q)} \phi(u) = (\alpha^{-1} \Phi(q) \alpha)(\phi(u))$$
Since this is valid for any $u \in sp(1)$ we conclude that $\phi Ad_q = \alpha^{-1} \Phi(q) \alpha \phi$, so that
$$\Phi(q) = \alpha \phi Ad_q \phi^{-1} \alpha^{-1}$$
and hence
$$\beta Ad_q \beta^{-1} = \alpha \phi Ad_q \phi^{-1} \alpha^{-1}$$
which is equivalent to
$$Ad_q \beta^{-1} \alpha \phi = \beta^{-1} \alpha \phi Ad_q$$
i.e. the (vector space) map $\beta^{-1} \alpha \phi : sp(1) \to sp(1)$ commutes with the adjoint representation of $Sp(1)$.

_Proposition_: If $f : sp(1) \to sp(1)$ is a linear map which commutes with the adjoint representation of $Sp(1)$ then $f$ is a multiple of the identity.

_Proof_: By differentiating the condition that $f$ commutes with the adjoint action of $Sp(1)$ we obtain the fact that $f$ commutes with the adjoint action of $sp(1)$ on itself, i.e.
$$f([u, v]) = [u, f(v)]$$
for all $u, v \in sp(1)$. The map $sp(1) \to (\mathbb{R}^3, \times)$ which sends $i/2, j/2, k/2$ to the $e_1, e_2, e_3$ basis is a Lie algebra isomorphism, so it suffices to prove that if $g : \mathbb{R}^3 \to \mathbb{R}^3$ is a linear map which satisfies $g(X \times Y) = X \times g(Y)$ for all $X, Y \in \mathbb{R}^3$ then $g$ is a multiple of the identity.

Note $g(e_3) = g(e_1 \times e_2) = e_1 \times g(e_2)$, so that $g(e_3)$ is perpendicular to $e_1$. Also $g(e_3) = -g(e_2 \times e_1) = -e_2 \times g(e_1)$, so $g(e_3)$ is also perpendicular to $e_2$. Hence $g(e_3) = \lambda_3 e_3$. Similarly $g(e_1) = \lambda_1 e_1$ and $g(e_2) = \lambda_2 e_2$.

We have $\lambda_3 e_3 = g(e_3) = e_1 \times g(e_2) = \lambda_2 e_3$, and so $\lambda_3 = \lambda_2$. Similarly we also have $\lambda_1 = \lambda_2$, and so setting $\lambda = \lambda_1 = \lambda_2 = \lambda_3$ we have $f = \lambda I$. QED.

Hence we conclude $\beta^{-1} \alpha \phi = \lambda I$, i.e.
$$ \phi = \lambda \alpha^{-1} \beta $$
Since $\alpha^{-1} \beta : sp(1) \to so(3)$ is the map which sends $i, j, k$ to $l_x, l_y, l_z$, we see $\phi$ is a Lie algebra isomorphism which maps $i, j, k$ to $\lambda l_x, \lambda l_y, \lambda l_z$. To preserve brackets we must have $\lambda = 2$, and that is why quaternions double rotation angles.

Fri, 08 Jan 2021 14:34 GMT