Explicit construction of $SO(3)$ elements from $SU(2)$ and $\mathfrak{su}(2)$

Question

Let $R \in SU(2)$ and let $\tau_a$ , $a=1,2,3$ be Pauli-Matrices (generators of $SU(2)$). I would like to show that the following matrix actually belongs to $SO(3)$. It is defined in components by: $$O_{ab}:=\frac{1}{2}\mathrm{Tr}(\tau_aR\tau_bR^{\dagger})$$ So I have to show that $OO^{T}=Id$ and that $\det(O)=1$. I started computing $OO^{T}$ in index notation, but ultimately failed to show it. Is there a nice way to show that $O \in SO(3)$?

Answer 1

To show that $O O^T=Id$, one can use that (Einstein summation convention used) $$ \tau^a_{ij}\tau^a_{kl}=2\delta_{il}\delta_{jk}-\delta_{ij}\delta_{kl}.$$ Here $a,b,c$ represents the $SO(3)$ indices, while $i,j,k,l$ represents the "internal" $SU(2)$ indices (i.e. $\tau^3_{ij}=\delta_{ij}(\delta_{i1}-\delta_{i2})$ and so on). After some calculation, one gets $$O_{ab}O_{ac}=\frac12 {\rm Tr}( \tau^a\tau^c)-\frac14 {\rm Tr}( \tau^a) {\rm Tr}(\tau^c)=\delta_{bc},$$ which is what we wanted to show.