Just a quick question, I'm hoping someone can clarify how this probably small issue can be resolved.
It is said that a Lorentz transformation $\Lambda$ is a linear tranformation of $\mathbb{R}^3_1$ into itself such that $\vert\vert\Lambda\mathbf{x}\vert\vert^2=\vert\vert\mathbf{x}\vert\vert^2$ for all $\mathbf{x}\in\mathbb{R}^3_1,$ where $\mathbb{R}^3_1$ denotes Minkowski space. If we identify a vector $\mathbf{x}=\left(x_0,x_1,x_2,x_3\right)\in\mathbb{R}^3_1$ with the Hermitian matrix $$x:=\sum\limits_{j=0}^3x_j\sigma_j=\left(\begin{array}{cc}x_0+x_3 & x_1-ix_2\\ x_1+ix_2 & x_0-x_3\end{array}\right),$$ then we let $A$ be any $2\times 2$ matrix and $A:x\to AxA^*,$ and $\mathbf{x}\to\phi(A)\mathbf{x},$ then we have $\det\left(AxA^*\right)=|\det A|^2\det x$ which is just $\det x$ if $A\in SL(2,\mathbb{C}).$
Here is the small issue. If we choose $\vert\vert\mathbf{x}\vert\vert^2=x_0^2-x_1^2-x_2^2-x_3^2,$ then we have $\det x=\vert\vert\mathbf{x}\vert\vert^2$ and $\vert\vert\phi(A)\mathbf{x}\vert\vert^2=\vert\vert\mathbf{x}\vert\vert^2,$ and we have that $\phi(A)$ is a Lorentz transformation, with $\phi(AB)=\phi(A)\phi(B),$ and we therefore have a homomorphism $\phi:SL(2,\mathbb{C})\to \mathcal{L},$ where $\mathcal{L}$ denotes the Lorentz group (I don't know if there is a proper notation for this). But if we choose $(-,+,+,+),$ then $\det x=-\vert\vert\mathbf{x}\vert\vert^2$ and $\vert\vert\phi(A)\mathbf{x}\vert\vert^2\neq\vert\vert\mathbf{x}\vert\vert^2$ in general.
Can someone either point out to me my mistake in thinking about the $(-,+,+,+)$ part (I feel like I've made a dumb error that I just can't see here, like dropping a sign or something, that would explain this), or show a modification that would give the correct homomorphism if we change the sign of the metric?