How projective representations can lead to 't Hooft anomalies in quantum mechanics?

Question

In Shao's talk https://youtu.be/2vTvHYYl1Qk?t=1554, he argues that in quantum mechanics "if a symmetry acts projectively on states, then we have a t' Hooft anomaly". But I'm having trouble to understand his arguments that we can't have any gauge invariant state, and this should be viewed as an obstruction to gauging.

I think there is something to do with the phase that appears because of the projective representation. If the symmetry acts like $$ U_{g_2}U_{g_1}\psi=e^{i\alpha(g_1,g_2)}U_{g_1g_2}\psi $$ on states. Then, when we try to gauge this symmetry, we would be able to introduce a gauge field to cancel the terms like $\partial_{\mu}\Lambda(x)$, $\Lambda(x)$ being the gauge parameter of the transformation $U_g$, but we will fail to get rid of the term $\partial_{\mu}\alpha(g_1,g_2)$. Of course, my argument is more related to the arguments given in QFTs when we're dealing with 't Hooft anomalies. Where we try to gauge a symmetry and the partition function gain a non-trivial phase $$ Z[A^\prime]=e^{i\int \alpha(A, \Lambda)}Z[A]. $$ But Shao's arguments seems to be easier and cleaner than that.

Can anyone clarify this for me?

Answer 1

Shu-Heng Shao's argument is essentially the statement that gauging means projecting to $G$-invariant subspace in (0+1)d. In a way this is the definition of gauging in (0+1)d in the Hamiltonian formalism. I will provide an Euclidean QFT "derivation". Here derivation in quote because one has to start from somewhere, so it is kind of a matter of which definition is most comfortable for you. I will also assume $G$ is finite to keep things simple. In Euclidean QFT, one natural definition of gauging is to sum over all $G$-bundles. Let us put the theory on $S^1$, i.e. say we are considering the Euclidean partition function $\mathrm{Tr} e^{-\beta H}$ where $H$ is the Hamiltonian. Then the $G$ bundle is just the holonomy around $S^1$, labeled by $g\in G$. Such a bundle can be created by inserting a $g$ symmetry operator somewhere along $S^1$, meaning $\mathrm{Tr} U_g e^{-\beta H}$. After gauging the new partition function is

$$ Z_\mathrm{gauged}=\frac{1}{|G|}\sum_{g\in G} \mathrm{Tr} (U_g e^{-\beta H})=\mathrm{Tr}\left(\frac{1}{|G|}\sum_{g\in G}U_g\cdot e^{-\beta H}\right)=\mathrm{Tr}\, P_G e^{-\beta H} $$ Here $P_G$ is the projector to $G$-invariant subspace. If there is no $G$-invariant subspace, then the result is $0$, which means that we can not gauge the system.

Since I'm at this, let me give a slightly different argument but along the same line. If the theory can be gauged, then all gauge-equivalent bundles must produce the same partition function. In particular, consider the following two bundles: one with $g_2$ inserted first followed by $g_1$, the other with $g_1g_2$ inserted. These are equivalent bundles, so they should give the same partition function. One is $\mathrm{Tr} U_{g_1}U_{g_2}e^{-\beta H}$, the other is $\mathrm{Tr} U_{g_1g_2} e^{-\beta H}$, and they differ by $e^{i\alpha(g_1,g_2)}$. Unless this phase can be absorbed into the definition of $U_g$, the result is not consistent -- in other words, there is no gauge invariance. This is essentially the same as your argument, as to go from one configuration to another one should apply a gauge transformation.