Wess–Zumino–Witten model

Example

Let $G$ be a compact, connected, simply connected simple Lie group with Lie algebra $\mathfrak{g}$. Let $\ip{\cdot}{\cdot}$ be the invariant inner product on $\mathfrak{g}$ normalized so that long roots have squared length $2$. Let $\Sigma$ be a closed Riemann surface and $k$ a positive integer, the level. The fields of the Wess–Zumino–Witten model are smooth maps $g \colon \Sigma \to G$, and its action is $$S_k(g) = \frac{k}{8\pi} \int_\Sigma \ip{g^{-1}\partial_\mu g}{g^{-1}\partial^\mu g} \, d^2x + k \, \Gamma(g).$$ The first term is the action of the non-linear sigma model with target $G$. The second, the Wess–Zumino term, is $\Gamma(g) = \int_B \tilde g^* \chi$. Here $B$ is a $3$-manifold with boundary $\Sigma$, and $\tilde g \colon B \to G$ extends $g$. The form $\chi$ is the bi-invariant $3$-form $\chi(X, Y, Z) = c \ip{X}{[Y, Z]}$ on $G$. The constant $c$ is chosen so that the integral of $\chi$ over every closed $3$-cycle in $G$ lies in $2\pi\Z$.

Check that the model is well defined: two extensions $\tilde g$ and $\tilde g'$ glue to a map from a closed $3$-manifold to $G$. So $\Gamma(g)$ changes by an element of $2\pi\Z$, and $e^{iS_k(g)}$ does not depend on the extension when $k$ is an integer. An extension exists because $\pi_1(G) = \pi_2(G) = 0$.