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

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?

• What is not clear about Shao's argument? Is it the statement that gauging is the same as projecting to $G$-invariant subspace in (0+1)d? Jan 7, 2022 at 21:39
• Yes, I don't understand when he says to project on the symmetric space Jan 7, 2022 at 21:46

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.