# What's up with $\mathrm{U}(1)$ regarding the spin homomorphism?

Let $$\mathcal H(2)$$ be the space of hermitian matrices of size $$2\times 2$$, and let $$\sigma:\mathbb R^{4}\rightarrow\mathcal H(2)$$, $$\sigma(x)=x^\mu\sigma_\mu=\left(\begin{matrix} x^0+x^3 & x^1-ix^2 \\ x^1+ix^2 & x^0-x^3\end{matrix}\right)$$ be the isomorphism between Minkowski spacetime and $$\mathcal H(2)$$. We have $$\det\sigma(x)=\eta_{\mu\nu}x^\mu x^\nu$$ (if the $$(+---)$$ signature is used).

If $$A\in\mathrm{GL}(2,\mathbb C)$$, then $$A\sigma(x)A^\dagger$$ is also hermitian, so this realizes a linear transformation on $$\mathcal H(2)$$ and thus on $$\mathbb R^4$$. if we take determinants, then $$\det(A\sigma(x)A^\dagger)=\left|\det A\right|^2\det\sigma(x),$$ so $$A$$ preserves the Minkowski norm if $$\det A\in\mathrm U(1)$$. Therefore, $$A$$ represents a Lorentz tranformation if its determinant is a unit-length complex number.

However the usual spin homomorphism is between $$\mathrm{SL}(2,\mathbb C)$$ and $$\mathrm O(3,1)$$, and the matrices in $$\mathrm {SL}(2,\mathbb C)$$ have $$\det A=1$$, rather than $$\det A=e^{i\varphi}$$.

Why are those matrices with unit-length, but not $$1$$ determinant excluded, when they also determine Lorentz transformations?

An element of $${\rm GL}(2,{\mathbb C})$$ of the form $$A= e^{i\phi}{\mathbb I}$$ has no effect on $$x^\mu \sigma_\mu$$ beause such matrices commute with all the Pauli's, so such group elements can be quotiented out and the effective group is $${\rm SL}(2, {\mathbb C})$$.