I know that similiar answers have already been given but i could not find an explaination of this exact passage.
I want to build the usual homomorphism between the lorentz group $SO(1,3)$ and $SL(2,\mathbb{C})$ and obtain an explicit form for the map but i do not understand few passages.
My goal il to prove: $\Lambda^\mu \ _\nu = \frac{1}{2} \text{Tr}{(\sigma^\mu A \sigma_\nu A^\dagger)}$. $\ \ \ \Lambda \in SO(1,3)$, $\ A \in SL(2,\mathbb{C})$ and the sigmas are Pauli matrices.
I start with the transformations laws of $X=\sigma_\mu x^\mu$ that is $X'=AXA^\dagger$ and $x'^\mu = \Lambda^\mu\ _\nu x^\nu$.
$$X'=AXA^\dagger \rightarrow \sigma_\mu x'^\mu = \sigma_\mu \Lambda^\mu\ _\nu x^\nu = A\sigma_\mu x^\mu A^\dagger$$
Here I am not 100% sure but i think that i can say that on the right the $\mu$ is a free index so i can swap it with a $\nu$, then $x$ commutes with $A^\dagger$ and i can move it to the right, obtaining: $$\sigma_\mu \Lambda^\mu\ _\nu x^\nu = A\sigma_\mu A^\dagger x^\mu \rightarrow \sigma_\mu \Lambda^\mu\ _\nu x^\nu = A\sigma_\nu A^\dagger x^\nu \rightarrow \sigma_\mu \Lambda^\mu\ _\nu = A\sigma_\nu A^\dagger$$
And finally i multiply both sides by $\sigma_\lambda$ to exploit the fact that $\text{Tr}(\sigma_\mu \sigma_\nu) = 2g_{\mu \nu}$
$$\sigma_\lambda \sigma_\mu \Lambda^\mu\ _\nu = \sigma_\lambda A\sigma_\nu A^\dagger$$
Here i have the biggest problem: I think i need to take the trace on both sides and use some property of the sigmas to obtain that $1/2$ factor but i dont know how to exactly procede further:
$$\text{Tr}(\sigma_\lambda \sigma_\mu \Lambda^\mu\ _\nu) = \text{Tr}(\sigma_\lambda A\sigma_\nu A^\dagger)$$
Sorry if the question is a duplicate.
