According to wikipedia, a Wess-Zumino term is well-defined when the Lie group (target space) $G$ is compact and simply connected, because that implies that $\pi_2(G)$ is trivial. But there are Lie groups with trivial $\pi_2(G)$ that are not simply connected (e.g., $S^1$). Is $S_\mathrm{WZ}(G)$ well-defined in such a case? Is a trivial $\pi_2(G)$ necessary and sufficient, or only necessary? If it is not sufficient, then what is the sufficient condition on $G$ for it to admit a WZW model?
On a second thought, it appears that the Wikipedia page contains a few misleading statements. For example, it says that $\pi_2(G)$ is trivial because $G$ is simply-connected; but, as mentioned by user gj255 in the comment section, $\pi_2(G)$ is trivial for any $G$, simply-connected or not. Moreover, in the sub-section Topological obstructions, Wikipedia claims that $kS_\mathrm{WZ}(G)$ is well-defined if $k\in\pi_3(G)$, so it seems that the relevant homotopy group is the third one instead of the second one. If this is correct, then I would guess $kS_\mathrm{WZ}(S^1)$ is only well-defined at level zero (because $\pi_3(S^1)=\{0\}$), and therefore there is no Wess-Zumino term for such a target space. Is this correct? More generally, is $kS_\mathrm{WZ}(G)$ well-defined if and only if $k\in\pi_3(G)\neq\{0\}$?