Wigner’s symmetry representation theorem

Uhlhorn 2 — rescaling \(\tau\)

Using \(\tau,\omega\) and arbitrary fixed vector \(x_0\ne 0\), construct bijective, additive \(\theta:H\rightarrow H'\).

We define \(\theta(y)=\tau(y)\omega(y,x_0)\). Using the properties of the components, we have additivity: \[\begin{align}\theta(y_1+y_2)&=\tau(y_1+y_2)\omega(y_1+y_2,x_0)\\&=[\tau(y_1)\omega(y_1,y_1+y_2)+\tau(y_2)\omega(y_2,y_1+y_2)]\omega(y_1+y_2,x_0)\\&=\tau(y_1)\omega(y_1,x_0)+\tau(y_2)\omega(y_2,x_0)\\&=\theta(y_1)+\theta(y_2)\end{align}\]

The function preserves the \(\perp\)-symmetry since \(T([y])=\mathbb C \tau(y)=\mathbb C \theta(y)\).

We now explore the effect of scalar multiplication by \(\alpha\ne 0\), defining \(\theta(\alpha x)=\varphi(\alpha,x)\theta(x)\), since \([\alpha x]=[x]\):\[\theta(\alpha x+\alpha y)=\theta(\alpha x)+\theta(\alpha y)=\varphi(\alpha,x)\theta(x)+\varphi(\alpha,y)\theta(y)\] and \[\theta(\alpha (x+y))=\varphi(\alpha,x+y)\theta(x+y)=\varphi(\alpha,x+y)\theta(x)+\varphi(\alpha,x+y)\theta(y).\] Hence, \(\phi(\alpha)=\varphi(\alpha,x)=\varphi(\alpha,x+y)=\varphi(\alpha,y)\) is independent of the vectors \(x,y\). We complete the definition with \(\phi(0)=0\).

We find that \(\phi\) is a homomorphism (automporphism, since non-zero) of the complex field: \(\theta((\alpha+\beta)x)=\phi(\alpha+\beta)\theta(x)\) and \(\theta(\alpha x+\beta x)=\theta(\alpha x)+\theta(\beta x)=(\phi(\alpha)+\phi(\beta))\theta(x)\) give additivity; \(\theta((\alpha\beta) x)=\phi(\alpha\beta)\theta(x)\) and \(\theta(\alpha(\beta x))=\phi(\alpha)\phi(\beta)\theta(x)\) give multiplicativity. Clearly, \(\phi(1)=1\).

Given \(x,y\) orthogonal and equal normed, \(x+\alpha y,x-\overline \alpha^{-1}y\) are also orthogonal. \(\theta\) preserve orthogonality, so: \(\Vert\theta(x)\Vert^2-\overline{\phi(\alpha)}\phi(\overline \alpha^{-1})\Vert\theta(x)\Vert^2=0\). Hence: \[\overline{\phi(\alpha)}\phi(\overline \alpha^{-1})=\frac{\Vert\theta(x)\Vert^2}{\Vert\theta(y)\Vert^2}.\] The right-hand side is independent of \(\alpha\), so \(\overline{\phi(\alpha)}\phi(\overline \alpha^{-1})=\overline{\phi(1)}\phi(1)=1\implies \overline{\phi(\alpha)}=\phi(\overline \alpha)\).