Can you map a 3-vector onto a spinor?

Question

These three groups have isomorphic Lie algebras: SO(3), Spin(3), SU(2). We can think of group elements as rotating a 3-vector $\mathbf v\in R^3$ and I'll give examples for rotation by $\theta$ around the $z$ axis.

1. An ortogonal matrix $R_z(\theta)=\left(\begin{smallmatrix} \cos\theta&-\sin\theta&0 \\ \sin\theta&\cos\theta&0 \\ 0&0&1 \end{smallmatrix}\right)$ acts on a 3-vector $\mathbf v=\left(\begin{smallmatrix} v_1\\v_2\\v_3 \end{smallmatrix}\right)$: $\mathbf v\rightarrow R_z(\theta)\mathbf v$

2. A unit quaternion $u_z(\theta)=\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\,\mathbf k$ acts on a pure-imaginary quaternion $q=v_1 \mathbf i+v_2 \mathbf j+v_3 \mathbf k$: $q\rightarrow u_z(\theta)qu^{-1}_z(\theta)$

3. A special unitary matrix $S_z(\theta)=\left(\begin{smallmatrix} e^{i\theta/2}&0 \\ 0&e^{-i\theta/2} \end{smallmatrix}\right)$ acts on a spinor $\boxed{s=\left(\begin{smallmatrix} s_1=f(\mathbf v) \\ s_2=g(\mathbf v) \end{smallmatrix}\right)\in C^2}$: $s\rightarrow S_z(\theta)s$

Between the first two, we have a nice way of mapping the vector components $v_1,v_2,v_3$.

My question: Is there a mapping between 3-vectors $\mathbf v$ and spinors $s$ in the third case, so that $S_z(\theta)$ rotates $\mathbf v$ about the $z$ axis (etc. for any axis)? What would the $f(\mathbf v)$ and $g(\mathbf v)$ look like (in the box above)?

I feel this should be simple, but I'm staring myself blind...

Note: I realize that a matrix $V=\left(\begin{smallmatrix} -iv_3&-v_2-iv_1\\v_2-iv_1&v_3 \end{smallmatrix}\right)$ can represent the vector to be rotated: $V\rightarrow S_z(\theta)VS^\dagger_z(\theta)$. I was hoping for something acting directly on a spinor in $C^2$, $S_z(\theta)s(\mathbf v)$.

Answer 1

The question is poorly worded. I assume you're asking about a possible $\mathbb{R}$-linear map between $\mathbb{R}^3$ and $\mathbb{C}^2$ which intertwines the action of $G=\mathrm{Spin}(3)$ (acting as $\mathrm{SO}(3)$ and $\mathrm{SU}(3)$ respectively). In other words, if we consider the representations $\alpha:G\to\mathrm{SO}(3)$ and $\beta:G\to\mathrm{SU}(2)$, we want a map $s:\mathbb{R}^3\to\mathbb{C}^2$ satisfying the condition $s(\alpha(g)v)=\beta(g)s(v)$ for all $v\in\mathbb{R}^3$, $g\in G$ (called equivariance).

There are no nonzero maps $s$ satisfying these conditions. For instance, the unique central involution in $\mathrm{Spin}(3)$ (corresponding to $-1$ in the unit quaternions $S^3=\mathrm{Sp}(1)$ or the matrix $-I_2$ in $\mathrm{SU}(2)$), which I'll call $-1$, acts trivially on $\mathbb{R}^3$ but not on $\mathbb{C}^2$, so setting $g=-1$ in the equivariance condition yields $s(\alpha(-1)v)=\beta(-1)s(v)$, which becomes $s(v)=-s(v)$, implying $s(v)=0$ for all vectors $v$.

More generally, if we consider $\mathbb{R}^3$ and $\mathbb{C}^2$ as real representations of $G$ (3D and 4D respectively), they are both irreducible representations ($G$ acts transitively on their unit spheres, so all vectors are cyclic generators, so there are no proper nonzero simple subreps). They are not isomorphic since they have different dimensions, and Schur's lemma says all the intertwiners between irreps are either isomorphisms or vanish.