I am reading a proof that $SU(2)$ is a double cover of $SO(3)$. My source is this set of notes: http://www.damtp.cam.ac.uk/user/examples/D18S.pdf. The proof begins near the bottom of page 4.
I have had no trouble with most of the proof, but I am getting stuck at an assertion toward the very end. Between equations $(7)$ and $(8)$ the author claims that $R\in SO(3)$ because (presumably using Einstein summation notation)
$$
\epsilon_{ijk}R_{i1}R_{j2}R_{k3} = \det R = 1.
$$
I don't see how this follows. The best I have tried is (with $e_\ell$ the standard basis vectors for $\mathbb{R}^3$)
$$
\begin{align*}
\det R
&= \epsilon_{ijk}R_{i1}R_{j2}R_{k3} \\
&= \epsilon_{ijk}(e_i\cdot R e_1)(e_j\cdot R e_2)(e_k\cdot R e_3)\\
&= \frac{1}{2^3}\epsilon_{ijk}~\text{trace}(e_i\cdot\sigma~R e_1 \cdot\sigma)~\text{trace}(e_j\cdot\sigma~R e_2\cdot\sigma)~\text{trace}(e_k\cdot\sigma~R e_3\cdot\sigma)\\
&= \frac{1}{8}\epsilon_{ijk}~\text{trace}~(\sigma_i U \sigma_1 U^*)~\text{trace}(\sigma_j U \sigma_2 U^*)~\text{trace}(\sigma_k U \sigma_3 U^*).
\end{align*}
$$
I don't see how to conclude that this is equal to $+1$. I am hoping to conclude something like
$$
\text{trace}(\sigma_i U \sigma_j U^*) = \text{trace}(\sigma_i\sigma_j) = 2\delta_{ij}
$$
in which case I would have the desired result, but I am not sure if this is true nor how to go about proving it.
I would appreciate it if someone would explain this line of the proof in greater detail.
Note: I am reading this proof (and nothing else from these notes) because I need the description of the homomorphism $SU(2) \to SO(3)$ specifically to construct the binary icosahedral group. I am not looking for a shorter proof of the double cover using theorems of Lie groups or Lie algebras; I am using this proof in particular because I do not have prior exposure to these topics, and I do not want to acquaint myself with the general theory of Lie groups at the moment.
If you know of another exposition of a proof that does not assume any knowledge of Lie groups, I would gladly accept it as an answer as well.
Best Answer
First, note that
$$\text{Tr}(\sigma_i\sigma_j\sigma_k)=\text{Tr}\{(\delta_{ij}+i\epsilon_{ijk}\sigma_k)\sigma_k\}=\text{Tr}(i\epsilon_{ijk}\sigma_k^2)=2i\epsilon_{ijk}$$
Then, on the one hand
$$\text{Tr}\{(Re_1)\cdot\sigma(Re_2)\cdot\sigma(Re_3)\cdot\sigma\}=\text{Tr}\{(Re_1)_i\sigma_i(Re_2)_j\sigma_j(Re_3)_k\sigma_k\}=(Re_1)_i(Re_2)_j(Re_3)_k\text{Tr}\{\sigma_i\sigma_j\sigma_k\}= 2i\epsilon_{ijk}(Re_1)_i(Re_2)_j(Re_3)_k = 2i\epsilon_{ijk}R_{i1}R_{j2}R_{k3}$$
on the other hand
$$\text{Tr}(\sigma_1\sigma_2\sigma_3)=2i \; .$$
Therefore, combining the two hands, you clap... I mean you obtain the desired result.