1
$\begingroup$

I am trying to understand the action of the nonlinear sigma model in the context of understanding WZW-models. On Wikipedia, its action is given as

$S_k\left(\gamma\right)=-\frac{k}{8\pi}\int_{S^2}\textrm{d}^2x\,\mathcal{K}\left(\gamma^{-1}\partial^\mu\gamma,\gamma^{-1}\partial_\mu\gamma\right)+2\pi k S^{\text{WZ}}\left(\gamma\right)$.

Here, $G$ is a compact simply-connected Lie group with $g$ its Lie algebra, $\gamma$ is a $G$-valued field and $\mathcal{K}$ is the Killing form on $g$. My difficulty is in understand how the products of the $\gamma$'s in the Killing form are supposed to work.

To my understanding, for the arguments in the Killing form to make sense, we require $\gamma^{-1}\partial^\mu\gamma$ and $\gamma^{-1}\partial_\mu\gamma$ to be elements of the Lie algebra $g$. I do not see, however, how this is supposed to be true.

To make sense of $\gamma^{-1}$, I would expect it to be the inverse of the element $\gamma$ maps to, i.e. for $z\in S^2$, we would have $\gamma\left(z\right)^{-1}\in G$. Furthermore, the derivative of $\gamma$ as seen from a manifold point of view would be a map from $TS^2$ to $TG$, if I am not mistaken. How then, would this be 'multiplied' by $\gamma^{-1}$? And how is it the resulting product to take arguments from $S^2$ into an element of $g$?

I have been following the book on Conformal Field Theory by Philippe Di Francesco et al. (http://www.springer.com/jp/book/9780387947853) and have been trying to find answers in other sources, but to no avail as of yet.

$\endgroup$

1 Answer 1

4
$\begingroup$

That $g^{-1}\mathrm{d} g$ is Liealgebra-valued for a Lie group-valued function $g$ has nothing to do with the particular model or with physics, it is true for all matrix groups. Write $g(x) = \exp(k(x))$, where $k(x)$ is now Lie algebra-valued and $\exp$ is the usual power series in the case of a matrix group. Then $\partial_\mu g = \partial_\mu k\exp(k)$ by the chain rule, and multiplying this with $g^{-1} = \exp(-k)$ gives $g^{-1}\partial_\mu g = \partial_\mu k$, which is Lie algebra-valued.

For Lie groups which are not matrix groups, writing $g^{-1}\mathrm{d}g$ is non-sensical, and you actually need to use the pullback of the Cartan-Maurer form by the function into the group.

$\endgroup$
1
  • 1
    $\begingroup$ Why do you assume that $\partial_{\mu}k$ commutes with $exp(\pm k)$ ? $\endgroup$
    – Nogueira
    Jul 5, 2018 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.