# Question

Let $$A\in \text{SL}(2,\mathbb{C})$$, so $$\det(A)=1$$. Define the following (Pauli) matrices:

\begin{align} \sigma_0=\begin{pmatrix}-1 & 0 \\ 0 & -1 \end{pmatrix} & &\sigma_1=\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} & \\ &\\ \sigma_2=\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} & &\sigma_3=\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix} & \end{align}

Now define the following $$4\times 4$$ matrix.

$$L_{\mu\nu}\equiv \frac{1}{2}\text{Tr}\left[A\sigma_\mu A^\dagger \sigma_\nu\right]$$

What I am trying to prove is $$\det (L)=1$$, that's it. But I am having so much trouble. I have verified that it is true via Mathematica. The following is know exactly:

$$\det (L)=\left|\det(A)\right|^4=1$$

Actually, it would be sufficient for my current purposes to show $$\det(L)\geq 0$$, but even that is very hard for me to show.

# Mathematica Code.

ClearAll[s0, s1, s2, s3, s, A]; (* Define the Pauli-Matrices and A \
matrix *)
s0 = {{-1, 0}, {0, -1}};
s1 = {{0, 1}, {1, 0}};
s2 = {{0, -I}, {I, 0}};
s3 = {{1, 0}, {0, -1}};
s = {s0, s1, s2, s3};
A = {{a, b}, {c, d}};

L00 = 1/2*
Tr[A.s0.A\[ConjugateTranspose].s0];  (* Define L(A) through the \
equation LSubscript[(A)^\[Mu], \[Nu]] = 1/2Tr[Subscript[A\[Sigma], \
\[Mu]]A\[ConjugateTranspose]Subscript[\[Sigma], \[Nu]]] *)
L01 =
1/2*Tr[A.s0.A\[ConjugateTranspose].s1];
L02 = 1/2*Tr[A.s0.A\[ConjugateTranspose].s2];
L03 = 1/2*Tr[A.s0.A\[ConjugateTranspose].s3];
L10 = 1/2*Tr[A.s1.A\[ConjugateTranspose].s0];
L11 = 1/2*Tr[A.s1.A\[ConjugateTranspose].s1];
L12 = 1/2*Tr[A.s1.A\[ConjugateTranspose].s2];
L13 = 1/2*Tr[A.s1.A\[ConjugateTranspose].s3];
L20 = 1/2*Tr[A.s2.A\[ConjugateTranspose].s0];
L21 = 1/2*Tr[A.s2.A\[ConjugateTranspose].s1];
L22 = 1/2*Tr[A.s2.A\[ConjugateTranspose].s2];
L23 = 1/2*Tr[A.s2.A\[ConjugateTranspose].s3];
L30 = 1/2*Tr[A.s3.A\[ConjugateTranspose].s0];
L31 = 1/2*Tr[A.s3.A\[ConjugateTranspose].s1];
L32 = 1/2*Tr[A.s3.A\[ConjugateTranspose].s2];
L33 = 1/2*Tr[A.s3.A\[ConjugateTranspose].s3];
L = {
{L00, L01, L02, L03},
{L10, L11, L12, L13},
{L20, L21, L22, L23},
{L30, L31, L32, L33}
};


$$\renewcommand{\vec}[1]{\boldsymbol{#1}}$$ $$\DeclareMathOperator{\Tr}{Tr}$$
Let $$\vec{\tau}=\mathrm{i}\vec{\sigma}$$, $$A=\mathrm{e}^{\vec{a}\cdot{\vec{\tau}}}$$. Then the goal is to find $$\det\mathsf{A}_1$$, where $$\mathsf{A}_t$$ is the linear operator $$\mathsf{A}_t(x)=\mathrm{e}^{t\vec{a}\cdot\vec{\tau}}x\mathrm{e}^{-t\vec{a}^*\cdot\vec{\tau}}\text{.}$$ (Here $$x=x^0+\vec{x}\cdot\vec{\tau}$$—slightly different from your convention). By differentiating with respect to $$t$$ we find $$\frac{\mathrm{d}}{\mathrm{d}t}\ln \det \mathsf{A}_t=\Tr \mathsf{a}$$ where $$\mathsf{a}$$ is the linear operator $$\mathsf{a}(x)=\vec{a}\cdot\vec{\tau}x - x\vec{a}^*\cdot\vec{\tau}\text{.}$$ Written out in components, $$\mathsf{a}(x)$$ is given by $$[\mathsf{a}][x]= \begin{bmatrix}0 & (\vec{a}-\vec{a}^*)\cdot\\ \vec{a}-\vec{a}^* &-(\vec{a}+\vec{a}^*)\times \end{bmatrix} \begin{bmatrix}x^0 \\ \vec{x} \end{bmatrix}\text{.}$$ The diagonal components of this $$4\times 4$$ matrix vanish, so $$\Tr \mathsf{a}=0$$, so $$\det \mathsf{A}_t$$ is constant in $$t$$, so $$\det \mathsf{A}_t=1$$, whence $$\det\mathsf{A}_1=1$$ as required.