Wigner’s symmetry representation theorem

Uhlhorn 1 — from rays to \(H\)

Given \(\perp\)-symmetric \(T:[H]\rightarrow [H']\) choose function \(\tau: H\rightarrow H'\) and explore structure.

We axiom-of-choice choose a function \(\tau:H\rightarrow H'\), such that \(\forall (x (\ne 0) \in H),\tau(x) (\ne 0) \in T([x])\), and complete the definition with \(\tau(0)=0\). The function \(\tau\) preserves \(\perp\)-symmetry, and thus linear independence/dependence, due to Lemma 7.

By Lemma 7.2, if \(x,y\) linearly independent, \(x,y,x+y\) are linearly dependent, and thus \(\tau(x),\tau(y),\tau(x+y)\) are linearly dependent. We define \(\omega\):\[\tau(x+y)=\tau(x)\omega(x,x+y)+\tau(y)\omega(y,x+y)\] [We have altered Chevalier’s definition to mimic the form of Dirac bra-ket projections.]

Considering three linearly independent vectors \(x,y,z\) (\(\mathrm{dim}H\ge 3\)), and associativity we derive from \(\tau(x+(y+z))=\tau((x+y)+z)\): \[\begin{aligned} \omega(x,x+y+z)&=\omega(x,x+y)\omega(x+y,x+y+z) \\\omega(y,x+y)\omega(x+y,x+y+z)&=\omega(y,y+z)\omega(y+z,x+y+z)\\ \omega(z,x+y+z)&=\omega(z,y+z)\omega(y+z,x+y+z)\end{aligned}\] Since \(x,x+y,x+y+z,\) are linearly independent, we can rewrite, making appropriate substitutions, the first and last relations for general three linearly independent vectors \(x,y,z\) as \(\omega(x,z)=\omega(x,y)\omega(y,z)\).

From \(x,y\) independent, we derive \[\begin{align}\tau(x)&=\tau(x +y-y)\\&=\tau(x +y)\omega(x +y,x)+\tau(-y)\omega(-y,x)\\&=\tau(x)\omega(x,x +y)\omega(x +y,x)+[\tau(y)\omega(y,x +y)\omega(x +y,x)+\tau(-y)\omega(-y,x)]\end{align}\] Since \(y,-y\) are in the same ray, so are their \(\tau\)-transforms. [However, we have made no assumption as to (anti)linearity of the mapping.] The part in square brackets is therefore the zero vector, and equating the \(\tau(x)\) parts, we derive \(\omega(x,x +y)\omega(x +y,x)=1\), or relabelling, \(\omega(x,y)\omega(y,x)=1\) for independent \(x,y\).

This relation enables us to show for \(x,x'\) dependent and \(x,y,z\) independent that \(\omega(x,y)\omega(y,x')=\omega(x,y)\omega(y,z)\omega(z,y)\omega(y,x')=\omega(x,z)\omega(z,x')\), so one can define \(\omega(x,x')=\omega(x,y)\omega(y,x')\) from any \(y\) independent of \(x,x'\), so long as \(\mathrm{dim}H\ge 3\).