Properties of $\Lambda_{\mu \nu}=\frac{1}{2}tr(\sigma_{\mu} S \sigma_{\nu} S^{\dagger})$ For a given 2 $\times$ 2, unit determinant, complex matrix $S$, define the matrix $\Lambda$ by the following.
$$\Lambda_{\mu \nu}=\frac{1}{2}tr(\sigma_{\mu} S \sigma_{\nu} S^{\dagger})$$
Show that

*

*$\Lambda^{T} \eta \Lambda=\eta$

*$\det \Lambda=1$

*$\Lambda(S_1) \Lambda(S_2)=\Lambda(S_1 S_2)$
where $\eta=diag(-1, 1, 1, 1)$ ,  $\sigma_0=I$, and $\sigma_i$ are pauli matrices.
I'm aware of
A simple proof of this equality: $\det_{\mu\nu}\left(\frac 1 2 \text{Tr}\left[A\sigma_\mu A^\dagger \sigma_\nu \right]\right)=1$
https://physics.stackexchange.com/q/619754/209828
The answer in the second link proves property 3.
Also, I did 1 as follows.

For any $2\times 2$ matrix B, $$\sigma_{\mu}tr(\sigma_{\mu}B)=2B$$  where the summation is implied. This can be proved by direct computation. Then $$\sigma_{\mu}\Lambda_{\mu\nu}=S\sigma_{\nu}S^{\dagger}$$
Introduce $a_{\nu}$ and contract with the above and take determinant.
$$\det(S\sigma_{\nu}a_{\nu}S^{\dagger})=\det(\sigma_{\nu}a_{\nu})=-\eta_{\mu\nu}a_{\mu}a_{\nu}$$
$$\det(\sigma_{\mu}\Lambda_{\mu\nu}a_{\nu})=-\eta_{\lambda\rho}(\Lambda_{\lambda\mu}a_{\mu}
)(\Lambda_{\rho\nu}a_{\nu})$$
Since a is arbitrary, $$\eta_{\mu\nu}=\eta_{\lambda\rho}\Lambda_{\lambda\mu}\Lambda_{\rho\nu}$$
which completes the proof.

Now my question is how to prove $\det \Lambda=1$. (It suffices to prove $\det \Lambda >0$ as 1 indicates $\det \Lambda=\pm 1$) The answer seemes to be provided in the first link above but I cannot understand from the beginning.
 A: The second property can be proven by using vectorization to convert the trace into a Kronecker product, which has a simple determinant ($\tau_\alpha = \sigma_\alpha/\sqrt{2}$):
\begin{align*}
\Lambda_{\mu\nu} &= \mathrm{tr}(A\tau_\mu A^\dagger \tau_\nu),\\
&= \mathrm{vec}(\tau_\mu)^\dagger \mathrm{vec}\left(A^\dagger\tau_\nu A\right),\\
&= \mathrm{vec}(\tau_\mu)^\dagger\left(A^\mathrm{T}\otimes A^\dagger\right)\mathrm{vec}\left(\tau_\nu\right).\end{align*}
This shows that $\Lambda_{\alpha\beta}$ are the coefficients of $A^\mathrm{T}\otimes A^\dagger$ in the (orthonormal) basis given by ${\mathrm{vec}(\tau_\alpha)}$. Given that the determinant is basis independent:
\begin{align*}
\mathrm{det}(\Lambda) &= \mathrm{det}\left(A^\mathrm{T}\otimes A^\dagger\right) = \mathrm{det}(A^\mathrm{T})^2\mathrm{det}(A^\dagger)^2 = |\mathrm{det}(A)|^4.
\end{align*}
The second property then follows for $\mathrm{det}(A)=1$.
There is a little more detail in an answer I posted on the question you cite: A simple proof of this equality: $\mathrm{det}_{\mu\nu}(\mathrm{Tr}[A\sigma_\mu A^\dagger \sigma_\nu])=1$.
