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:


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?

  • 1
    $\begingroup$ 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! $\endgroup$ Mar 2, 2023 at 0:32
  • $\begingroup$ I'll choose to take your last sentence at face value. So, thank you! $\endgroup$ Mar 2, 2023 at 20:54
  • $\begingroup$ @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? $\endgroup$ Mar 3, 2023 at 2:58
  • $\begingroup$ possibly useful? physics.stackexchange.com/q/687500/84967 $\endgroup$ Mar 9, 2023 at 20:51

1 Answer 1

  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.


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