Apparent elimination of a 't Hooft anomaly in quantum spin system The simplest system with a 't Hooft anomaly is the spin $\frac{1}{2}$ system with hamiltonian $\hat{H}=0$. The 't Hooft anomaly follows from the fact that such system has a trivial $SO(3)$ symmetry, and we are dealing with a $j=1/2$ representation of $SO(3)$, that is, a projective representation, as one can easily check in http://www.damtp.cam.ac.uk/user/examples/D18S.pdf. Such projective representation prevent us from gauging the $SO(3)$, and therefore we have a 't Hooft anomaly https://www.youtube.com/watch?v=2vTvHYYl1Qk&ab_channel=BootstrapCollaboration.
However, as mentioned in the first link, the projective representation is in some sense a problem of the group $SO(3)$, which isn't simply-connected. Therefore, the same rotation symmetry of the $\hat{H}=0$ system can be implemented by $SU(2)$, instead of $SO(3)$.
By doing that, we lose the projective representation and, by consequence, the 't Hooft anomaly. Can anyone explain for me why the 't Hooft anomaly seem to disappear when we switch from $SO(3)$ to $SU(2)$.
 A: $\newcommand{\cH}{\mathcal{H}}\newcommand{\P}{\mathbb{P}}\newcommand{\U}{\mathrm{U}}\newcommand{\SU}{\mathrm{SU}}\newcommand{\SO}{\mathrm{SO}}\newcommand{\Un}{\mathrm{U}}\renewcommand{\to}{\longrightarrow}\newcommand{\H}{\mathrm{H}}$In the following I will take $G$ to be a connected Lie group. I will first give the general reasoning for a generic group and then I will specialise to the case $G=\SO(3)$.
A unitary representation of a group $G$ is a continuous map:
$$ R:G\to \Un(\cH),\tag{1}\label{rep}$$
where $\cH$ is the Hilbert space of your quantum mechanical theory, and $\Un(\cH)$ is the group of unitary operators on $\cH$. Similarly, a projective representation is a continuous map:
$$ P:G\to \Un(\P\cH),\tag{2}\label{prep}$$
$\P\cH$ being the projective Hilbert space, $\P\cH:=\cH\big/\!\!\sim$, where the equivalence relation is imposed by identifying states differing by a phase. $\Un(\cH)$ and $\Un(\P\cH)$ are related by a central extension
$$ 1\to\U(1)\to\Un(\cH)\overset{\gamma}{\to}\Un(\P\cH)\to 1.\tag{3}\label{ext}$$
A quick comment, in passing, here: In quantum mechanics, 't Hooft anomalies and projective representations are one and the same because they are both classified by the group cohomology group $\H^2(G;\U(1))$, which classifies the possible central extensions of $G$.
Coming back to the point, the question of having an 't Hooft anomaly or not, translates to the following. Given $P$ as in \eqref{prep}, and \eqref{ext}, can you always find $R$ as in \eqref{rep} such that $P = \gamma\circ R$? The answer is no, not generally. If you can, you are free of 't Hooft anomalies. If you can't, there is an 't Hooft anomaly.
However, if $\tilde{G}$ is the universal covering group of $G$, with
$$ \Pi: \tilde{G}\to G,$$
which is, up to isomorphism, uniquely defined, connected, and simply connected you can lift the anomaly. Namely, you can find a representation
$\tilde{R}:\tilde{G}\to \Un(\cH)$,
such that $\gamma\circ \tilde{R} = \tilde{P}$, where
$$\tilde{P}:=\Pi\circ P : \tilde{G}\to \Un(\P\cH).\tag{4}$$
(This you can see by building a central extension $\tilde{E}$ of $\tilde{G}$ and noticing that the diagram
$$\require{AMScd}
\begin{CD}
1@>>>\U(1) @>>> \tilde{E} @>>>\tilde{G} @>>> 1\\
{} @V{\mathrm{id}}VV @VVV @V{\tilde{P}}VV \\
1@>>>\U(1) @>>> \Un(\cH) @>{\gamma}>> \Un(\P\cH) @>>> 1,
\end{CD}$$
commutes. Then you can add the extra arrows to construct $\tilde{P}$ and $\tilde{R}$.)
Your case is precisely in the above form. Just take $G=\SO(3)$. Then $\tilde{G}=\SU(2)$ the above construction is the mechanism that lifts the 't Hooft anomaly.
