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?


1 Answer 1


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 ${\rm SL}(2, {\mathbb C})$.


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