# Wigner proof of the non-existence of finite unitary representation of the Lorentz group

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.

• Related: the "modern" proof this posts alludes to. Dec 17 '18 at 22:00

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.