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If we define the Pauli matrices as

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

and define the vector space isomorphism $\phi:\mathbb R^{1,3}\ni(x^0,x^1,x^2,x^3)\mapsto\sum_{\mu=0}^3 x^\mu\sigma_\mu\in M_2(\mathbb C)$, then it is straightforward to compute that $\det\phi(x)=(x^0)^2-(x^1)^2-(x^2)^2-(x^3)^2=:Q(x),$ where $Q$ is the standard quadratic form on $\mathbb R^{1,3}$. I'm looking for a coordinate-free, geometric understanding of this fact.

To illustrate what I'm looking for, consider the real algebra generated by the Pauli matrices, which is isomorphic to the Clifford algebra over $\mathbb R^3$ and also the biquaternions $\mathbb C\otimes\mathbb H$. Multiplying the matrices $\{\sigma_1,\sigma_2,\sigma_3\}$ by $i$, along with the same $\sigma_0$, gives a representation of $\mathbb H$, whose elements act naturally on $\mathbb H$ (from the left, say) before choosing a matrix representation. I discuss in the comments to this answer how we can understand $\det (q)=|q|^2$ as a geometric consequence of quaternion multiplication.

But unlike $\mathbb H$, the Pauli matrices do not form a subalgebra of $\text{Cl}_{3,0}(\mathbb R)\simeq\mathbb C\otimes\mathbb H$, so it is harder to find a "natural" two-dimensional complex vector space on which these matrices act, let alone finding a geometric picture of the action of these matrices on such a space.

WillG
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  • One way or another this is going to boil down to the exceptional isomorphism $SL_2(\mathbb{C}) \cong Spin(3, 1)$ but I'm not sure what the details look like exactly. – Qiaochu Yuan Jan 31 '23 at 20:32
  • This is almost certainly related to the fact that your $\psi$ embeds a vector as a paravector of $\mathrm{Cl}{3,0}(\mathbb R)$, and these paravectors descend naturally from vectors in $\mathrm{Cl}{1,3}(\mathbb R)$. See my answer here for more on paravectors. – Nicholas Todoroff Feb 01 '23 at 00:06
  • Just replace $\sigma_k$ by $i*\sigma_k$ for $k\neq 0$, and use these "phase-shifted" pauli matrices as your new terms. That changes your norm to the usual: $\det(\phi(x))=(x^0)^2+(x^1)^2 + (x^2)^2 +(x^3)^2$ and the algebra between these new terms behaves like quaternions.

    Edit: the geometric space on which Pauli matrices act can be considered the same as $2$ dimensional Complex space on which quaternions act. Just multiply the Pauli matrices by scalar $i$ to get the unit quaternions (or vice-versa).

     – user3257842 Feb 13 '25 at 18:28

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