Lie bracket of $\mathfrak{so}(3)$

Question

I know that for $\mathfrak{so}(3)=\mathcal{L}(SO(3))$, the set of $3\times 3$ real antisymmetric matrices, we can define a basis $$T^1=\begin{pmatrix}0&0&0\\ 0&0&-1\\ 0&1&0\end{pmatrix}\quad T^2=\begin{pmatrix}0&0&1\\ 0&0&0\\ -1&0&0\end{pmatrix}\quad T^3=\begin{pmatrix}0&-1&0\\ 1&0&0\\ 0&0&0\end{pmatrix}$$ such that $$\left(T^a\right)_{bc}=-\epsilon_{abc}$$ and the Lie bracket is then $$\left[T^a,T^b\right]=\epsilon_{abc}T^c$$ I can see that it works by explicitly putting in the matrices $T^a$ defined above, but how would one manipulate the $\epsilon$ symbols to show it formally?

Answer 1

Is this are you looking for? (using Einstein convention) $$\left[T^a,T^b\right]^i_j=(T^aT^b )_{ij}-(T^bT^a )_{ij}$$ $$=(T^a)^i_p(T^b)^p_j-(T^b)^i_q(T^a)^q_j=\epsilon_{aip}\epsilon_{bpj}-\epsilon_{biq}\epsilon_{aqj}.$$ And then whatever you need to do you probably will need to use the following identity: $$\epsilon_{ijk}\epsilon_{mnk}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$$

EDIT: Applying the formula to the previous one (if I'm not mistaking) you get $$\left[T^a,T^b\right]^i_j=\delta_{aj}\delta_{bi}-\delta_{bj}\delta_{ai}$$ Which indeed is $$[T^a,T^b]=\epsilon_{abc}T^c$$