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I am reading Wigner's paper ”On unitary representations of the inhomogenous Lorentz group” (Annals of Mathematics, Vol. 40, No.1, p. 149) found here: https://www.maths.ed.ac.uk/~jmf/Teaching/Projects/Poincare/Wigner.pdf, or officially here https://www.jstor.org/stable/1968551 (DOI 10.2307/1968551) on the unitary representations of the Poincaré group but I got stuck on something.

At the end on the proof (p. 18 of the pdf), he states that $$ \mathbf{M}(\alpha) \mathbf{\Lambda}_e(\gamma) \mathbf{M}(\alpha)^{-1} = \mathbf{\Lambda}_e(\alpha \gamma) $$ is impossible for finite unitary matrices but I don't really see why and it is a key point of the demonstration.

By the way, I know that nowadays we prove it using the fact that the group is non-compact but I just want to understand the original proof.

DanielC
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user65854
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1 Answers1

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Let, as in Wigner's article, $D$ be a finite unitary representation of the Lorentz group. We prove that $D$ is trivial. As $D$ is a representation, your formula above gives $$ D(M(\alpha))\,D(\Lambda(\gamma))\,D(M(\alpha))^{-1}=D(\Lambda(\alpha\gamma)). $$ In particular, the unitary matrices $$ D\Lambda(\gamma)\quad \mathrm{and}\quad D\Lambda(\alpha\gamma) $$ have the same finite set of eigenvalues, for all real numbers $\alpha$ and $\gamma$.

Wigner constructs the Lorentz transformations $\Lambda(\gamma)$, for a real parameter $\gamma$, in such a way that $$ \Lambda(\gamma)\Lambda(\gamma')= \Lambda(\gamma+\gamma'). $$ In particular, substituting $\frac12\gamma$ for $\gamma$ and $\gamma'$, one has $$ \Lambda(\tfrac12\gamma)^2=\Lambda(\tfrac12\gamma+\tfrac12\gamma)=\Lambda(\gamma), $$ i.e., $\Lambda(\tfrac12\gamma)$ is a square root Lorentz transformation of $\Lambda(\gamma)$. Hence, the set of eigenvalues of $D\Lambda(\tfrac12\gamma)$ is a set of square roots of the set of eigenvalues of $D\Lambda(\gamma)$, as $D\Lambda(\frac12\gamma)$ is diagonalizable. However, by what we have seen above, the set of eigenvalues of $D\Lambda(\tfrac12\gamma)$ also is equal to the set of eigenvalues of $D\Lambda(\gamma)$. It follows that the finite set of eigenvalues of $D\Lambda(\gamma)$ contains a square root of each of its elements. Therefore, the only eigenvalue of $D\Lambda(\gamma)$ is $1$, and $D\Lambda(\gamma)$ is the identity. Since the generic elements of the Lorentz group are of the form $\Lambda(\gamma)$, the representation $D$ is trivial.

Johannes Huisman
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