3.2 No projective phase on S^1
Now specialize to 1+1d QFT, and consider \(M_1 = S^1\), which is the only connected closed 1-dimensional manifold. A reader might have guessed that, as a generalization of what we have learned in the case of QM, the unitary operator realizing the symmetry on \(\mathcal{H}_{S^1}\) might have projective phase in there multiplication rule. This very natural guess, however, is not correct.
For a general QFT, the Hamiltonian depends on the radius \(R\) of \(S^1\). When \(R\) is very large compared to any scale in the theory, we regard the QFT on \(S^1\times \mathbb{R}_\text{time}\) as the infrared-regulation of QFT on \(\mathbb{R}^1\times \mathbb{R}_\text{time}\).
Now, if the QFT has single vacuum, the cluster decomposition principle (which is another manifestation of locality) that the symmetry should preserve it. So \(S^1\) with large enough \(R\) has single vacuum, which contradicts with a nontrivial projective phase. More generally, if a discrete symmetry is spontaneously broken, there can be multiple vacua in the \(R\to \infty\) limit. However, if there is a domain wall connecting the two vacua , there is a exponentially suppressed amplitude between the two state with finite but large \(R\). Therefore, we expect that, without additional mechanism prohibiting domain walls16, there is a single vacuum on \(S^1\) with large but finite \(R\). As the projective phase is discrete, captured by \(H^2(G)\), it has to be constant of \(R\). Therefore we do not expect any nontrivial projective phase of \(U\) acting on \(\mathcal{H}_{S^1}\) for any radius \(R\) of \(S^1\).
Then, is there any projective-ness in the symmetry in 2d QFT? To explicitly see that, we need to think about manifolds with boundaries.17