MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does anyone have a reference to learn more about the Cartan $3$-form on a group manifold $G$? I have read that the WZW term is nothing more than the integral of the pullback of the Cartan $3$-form via $g:W\rightarrow G$

$WZW = -\frac{1}{6}\int_W \langle \phi_g\wedge[\phi_g\wedge\phi_g]\rangle$,

where $\phi_g=g^\ast(\phi)$ is the pullback of the Maurer-Cartan form, and would like to learn more about the math behind WZW terms. For eg., why is it the generator of $H^3(G,\mathbb{R})$ when $G$ is a connected, simply connected, compact Lie group?

share|cite|improve this question
What you have written is only the Wess-Zumino term. The WZW lagrangian also has a standard sigma model term. Why not read Witten's original paper Non-abelian bosonization in two dimensions (CMP 1984)? – José Figueroa-O'Farrill Apr 26 '11 at 4:13
Yes, of course, sorry about that. I will edit it now. – Kevin Wray Apr 26 '11 at 4:26
Also, I have tried to go through Witten's paper, but he really doesn't discuss the math behind the $3$-form, $\langle \phi_g[\phi_g\wedge\phi_g]\rangle$, but will go back and check again. – Kevin Wray Apr 26 '11 at 4:29
If $G$ is compact and simple, then $H^3(G;\mathbb{R}) \cong \mathbb{R}$, with generator the 3-form you have written down. This follows from the isomorphism between $H^3(G;\mathbb{R})$ and $H^3(\mathfrak{g})$, with $\mathfrak{g}$ the Lie algebra of $G$. – José Figueroa-O'Farrill Apr 26 '11 at 5:02
(And the calculation of $H^3(\mathfrak{g})$, of course!) – José Figueroa-O'Farrill Apr 26 '11 at 5:03

You should check out Theodore Frankel's Geometry of Physics. There is a section on the gauge group . The cartan three form is the three form from the Chern Simons term at a flat connection. A flat connection is when the potential A is equal to the $g^{-1}dg$. In the Chern Simons 3 form you have A^A^A. At pure gauge this becomes $Tr(g^{-1}dg\wedge g^{-1}dg\wedge g^{-1}dg)$. This term is proportional to the cartan 3 form. I hope this helped.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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