# Projective representations of the Lorentz group can't occur in QFT! What's wrong with my argument?

In flat-space QFT, consider a spinor operator $$\phi_i$$, taken to lie at the origin. Given a Lorentz transformation $$g$$, we have $$\tag{1} U(g)^\dagger \phi_i U(g) = D_{ij}(g)\phi_j$$ where $$D_{ij}$$ is some spinor projective representation of $$SO^+(3,1)$$, and $$U$$ is an infinite-dimensional unitary, perhaps projective, representation of $$SO^+(3,1)$$, satisfying $$\tag{2} U(gh)= e^{i\alpha(g,h)}U(g)U(h)$$ for some real phases $$\alpha(g,h)$$. Now, using (1) and (2) we have: \begin{align} D_{ij}(gh)\phi_j &\stackrel{(1)}{=} \tag{3}U(gh)^\dagger\phi_iU(gh) \\ \tag{4} &\stackrel{(2)}{=}e^{-i\alpha(g,h)}U(h)^\dagger U(g)^\dagger\phi_iU(g)U(h)e^{i\alpha(g,h)}\\ \tag{5} &\stackrel{(1)}{=} U(h)^\dagger \big((D_{ik}(g)\phi_k\big)U(h)\\ \tag{6} &\stackrel{(1)}{=} D_{ik}(g)D_{kj}(h)\phi_j. \end{align}

Hence $$\tag{7} D_{ij}(gh) = D_{ik}(g)D_{kj}(h).$$ That is, $$D$$ is a true, non-projective, representation. Clearly this is false for spinors, so where did I go wrong?

• I think (though I'm not sure) that "projective" just means (2): you allow a phase when acting on the Hilbert space, not on the Lorentz/spinor indices. Commented Mar 1, 2023 at 18:35
• @Javier No, $D_{ij}$ is meant to be a spinor rep, so we need to allow a phase there. Commented Mar 1, 2023 at 18:40

Your observation is correct, which is why careful statements of the Wightman axiom eq. (1) (see e.g. Wiki, nLab) stipulate that $$D$$ is a representation of $$\mathrm{SL}(2,\mathbb{C})$$, the universal cover of the Lorentz group, not a representation of the Lorentz group itself.
Note that classical fields - the things on which $$D$$ acts - are not necessarily complex-valued, so the notion of a projective representation on them does not even make sense in general.