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 $k\in\pi_3(G)\neq\{0\}$?

  • $\begingroup$ It's my understanding that every Lie group has trivial second homotopy group. See mathoverflow.net/q/8957. $\endgroup$
    – gj255
    Apr 26, 2018 at 16:56
  • $\begingroup$ @gj255 huh, that was unexpected. If $\pi_2(G)$ is always trivial, then it's weird that Wikipedia mentions that it follows from $G$ being simply connected. I'm not sure what to think, but the question is still the same: "for what $G$ is WZ well-defined?". In any case, thank you for the comment! $\endgroup$ Apr 26, 2018 at 17:00

1 Answer 1


It's wrong to study the homotopy groups to classify WZW terms because we're not just interested in spherical spacetimes.

Let $X$ be the target space and $M$ an arbitrary closed spacetime $n$-manifold equipped with a map $\sigma:M \to X$. Let $\omega$ be a closed $(n+1)$-form on $X$ with integer periods.

If $\sigma$ is null-homotopic, we can extend it to the cone $\hat\sigma:CM \to X$, which is defined by forming the prism $M \times [0,1]$ and collapsing one end to a point. Then we may define $$WZW(M,\sigma) = \exp 2\pi i \int_{CM} \hat \sigma^*\omega.$$ And because $\omega$ is closed and has integer periods, this can be shown to be independent of the extension $\hat \sigma$.

It's very important for this definition that we can extend $\sigma$ to the cone, or at least to some $n+1$-chain whose boundary is $M$ (we can use a homology theory whose chains are singular manifolds with corners so it still makes sense to form the pullback $\hat \sigma^*\omega$ everywhere except a measure zero set). The condition for this more general situation is a homological one: we need $\sigma_* [M] = 0 \in H_n(X,\mathbb{Z})$, where $[M]$ is the fundamental class of $M$. If $H_n(X,\mathbb{Z}) = 0$, then we're always in business.

Note that there was recently some study of sigma model topological terms coming from cobordism that was pretty interesting: https://arxiv.org/abs/1707.05448 . It explains what happens to the theta angle one expects in a 2+1D theory with $X = S^2$ from $\pi_3 S^2 = \mathbb{Z}$.

  • $\begingroup$ Nice, thank you! It's going to take me some time to digest this answer; if there is something I don't understand I'll report back. Cheers! $\endgroup$ May 3, 2018 at 14:34
  • $\begingroup$ By the way, I think it's possible to subvert the topological obstruction by choosing a differential refinement of $\omega$. I could tell you more if you sent me an email, but it's something I'm currently working on, so I prefer not to share it here. $\endgroup$ May 3, 2018 at 14:38
  • $\begingroup$ @I'm rather new to all this topology stuff, so I'm quite sure the details would go over my head. I'll stick to basic stuff for now, but thank you very much anyway :-) $\endgroup$ May 3, 2018 at 14:44

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