How do projective representations act on the QFT vacuum?

Let $$U:\mathcal{G}\to \mathcal{U}(\mathcal{H})$$ be a unitary projective representation of a symmetry group $$\mathcal{G}$$ on a Hilbert space $$\mathcal{H}$$. It satisfies the composition rule:

$$U(g_1)U(g_2)=e^{i\phi(g_1,g_2)}U(g_1g_2).\tag{1}$$

Now suppose there is a state $$|\Omega\rangle$$ which only changes by a phase under the symmetry:

$$\tag{2} U(g)|\Omega\rangle = e^{if(g)}|\Omega\rangle,$$

for some real phases $$f(g)$$. Then acting on $$|\Omega\rangle$$ with each side of (1) we find:

$$\phi(g_1,g_2)=-f(g_1g_2)+f(g_1)+f(g_2) \mod 2\pi \tag{3}$$

So $$\phi(g_1,g_2)$$ is a trivial cocycle. By defining the "improved" symmetry operators $$\tilde{U}(g)=e^{-i f(g)}U(g)\tag{4}$$

we get a true, non-projective representation of $$\mathcal{G}$$.

To summarise, if a projective representation contains a vector which is left invariant up to phases, then it is not an "intrinsically projective" representation: rescaling the unitaries by phases makes it into a true representation.

But in any QFT without symmetry-breaking there is such an invariant state: the vacuum! So the above argument shows that unbroken symmetries can't be represented projectively in QFT.

My question: What about spinor representations of the Lorentz group - they're projective, aren't they?

• Indeed, you just proved that if $\phi$ is a non-trivial cocycle (i.e., the rep is truly projective) then neither the vacuum -- nor any other particular state -- can be invariant, not even up to a phase. This is a trivial statement, it is obvious that there are no non-trivial one-dimensional projective representations (given that a projective rep is a map to $PU(n)$ and $PU(1)\equiv\emptyset$). In conclusion, any anomalous symmetry in quantum mechanics is spontaneously broken, which is a rather well known fact. Congrats on rediscovering it! Commented Mar 2, 2023 at 0:32
• I'll choose to take your last sentence at face value. So, thank you! Commented Mar 2, 2023 at 20:54
• @AccidentalFourierTransform Do you know a grad-level (i.e. not too advanced, no nlab links pls!) reference which discusses anomalies in terms of projective representations? Commented Mar 3, 2023 at 2:58
• possibly useful? physics.stackexchange.com/q/687500/84967 Commented Mar 9, 2023 at 20:51

1. Yes, OP is right. More generally, one could consider a projective (not necessarily unitary) representation $$\rho:G\to {\rm End }(V)$$ where $$\rho(g)\rho(h)~=~c(g,h)\rho(gh),\tag{1}$$ and where $$c(g,h)\in \mathbb{C}^{\times}$$ is a 2-cocycle. Assume that there exists a 1-dimensional invariant subspace of $$V$$, $$\rho(g)|\Omega\rangle~=~b(g)|\Omega\rangle,\qquad |\Omega\rangle~\in~V\backslash\{0\}, \qquad b(g)~\in~ \mathbb{C}^{\times}~:=~\mathbb{C}\backslash\{0\}, \tag{2}$$ i.e. a 1-dimensional projective subrepresentation. So $$c(g,h)~=~b(gh)^{-1}b(g)b(h)\tag{3}$$ is a 2-coboundary, i.e $$\rho$$ can be lifted to an ordinary linear representation $$\tilde{\rho}(g)~:=~b(g)^{-1}\rho(g),\qquad \tilde{\rho}(g)\tilde{\rho}(h)~=~\tilde{\rho}(gh),\tag{4}$$ such that $$\tilde{\rho}$$ acts trivially on the vacuum $$\left.\tilde{\rho}\right|_{{\rm span}|\Omega\rangle}~=~ \left.{\bf 1}\right|_{{\rm span}|\Omega\rangle}, \tag{5}$$ cf. OP's title question.
2. OP's last subquestion about spinor representations for the restricted Lorentz group $$SO^+(1,3)$$ seems already answered in OP's other post here, i.e. one should consider its double cover $$Spin^+(1,3)~\cong~ SL(2,\mathbb{C})\tag{6}$$ instead.