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I'm learning about rotations in 3D space. I've come across Pauli vectors and quaternions. Now as I understand, one can associate to a regular vector $\vec{v} = (v_x, v_y, v_z)$ a Pauli vector and a quaternion as follows:

$$V = \begin{bmatrix} v_z & v_x - i v_y\\ v_x + i v_y & -v_z \end{bmatrix} = \sum_i v_i \sigma_i$$

$$ q = 0 + iv_x + jv_y + kv_z$$

which are unique. However, when applying a rotation:

$$ \vec{v} \to R\vec{v}, \ \ \vec{v} \in \mathbf{R}^3\left(+,\cdot, \mathbb{R}^3\right),R \in SO(3) $$ $$ V \to UVU^\dagger, \ \ V \in Span\{\sigma_i\}, U\in SU(2) $$ $$ v \to qvq^*, \ \ v \in Span\{v \in \mathbb{H} : v = ai + bj + ck, \ \ a,b,c \in \mathbb{R}\}, q \in Sp(1) $$

Intuitively I was thinking that if the three vector spaces are isomorphic, the mapping between them is bijective, whereas the fact that the groups $SU(2)$ and $Sp(1)$ are the double cover of $SO(3)$ implies that the same transformed vector $\vec{v}'$ is associated to two different transformations:

$$ \vec{v}' = R\vec{v} \longleftrightarrow \left( \pm U \right) V \left( \pm U \right)^\dagger \longleftrightarrow \left( \pm q \right) v \left( \pm q \right)^*$$

Is it possible to say that the vector space spanned by Pauli matrices, the one of all quaternions whose real part is zero and the one of regular vectors are isomorphic as vector spaces over $\mathbf{R}$?

Thank you!

Anton Odina
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Luke__
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  • What is $\sigma_i$ in the Pauli vector? – Anton Odina Aug 12 '23 at 10:10
  • I used that notation for the basis Pauli matrices: $\sigma_x = \sigma_1 = \begin{bmatrix} 0 & 1\ 1 & 0 \end{bmatrix} $, $\sigma_y = \sigma_2 = \begin{bmatrix} 0 & -i\ i & 0 \end{bmatrix}$, $\sigma_z = \sigma_3 = \begin{bmatrix} 1 & 0\ 0 & -1 \end{bmatrix} $ – Luke__ Aug 12 '23 at 10:13
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    In your notation the three vector spaces $\mathbf{R}^3$ and $V$ and the set of all quaternions $\mathbb H$ with zero real part (socalled pure quaternions) are isomorphic. – Kurt G. Aug 12 '23 at 10:21
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    Do you want to prove that those three vector spaces over $\mathbf{R}$ are isomorphic? Because you can send generators to generators and that will give you a well-defined linear map which will be bijective. – Anton Odina Aug 12 '23 at 10:21
  • see here – Jean Marie Aug 12 '23 at 10:32
  • Thanks! My perplexity was this: the fact that the vector spaces are isomorphic and the fact that $SU(2)$ and $Sp(1)$ double cover $SO(3)$ are not related, else there would be a contradiction, right? – Luke__ Aug 12 '23 at 10:38
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    No contradiction whatsoever. $SU(2)$ can be used to rotate vectors in $\mathbb R^3$ (aka $V$ aka pure quaternions). $SO(3)$ does that as well but differently. Continue reading. You will come accross further differences between those groups. – Kurt G. Aug 12 '23 at 11:07
  • Ok, thanks again! – Luke__ Aug 12 '23 at 13:18

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