# Maurer-Cartan form involved in Lie bracket

1. Jul 31, 2014

### center o bass

The Maurer-Cartan one-form $\Theta = g^{-1} dg$ is though of as a lie algebra valued form.
It arises in connection with Yang-Mill's theory where the gauge potential transforms as
$$A \mapsto g Ag^{-1} - g^{-1} dg.$$

However, one also defines for lie-algebra valued differential forms $\alpha, \beta \in \Omega_p(M,\mathfrak g)$, the Lie bracket
$$[\alpha, \beta] = [\xi_k, \xi_l] \alpha^k \wedge \beta^l.$$

The question then arise, what does one mean by the lie-bracket when $g^{-1} dg$ is involved?
For example, how would one compute
$$[g^{-1} dg, g\alpha g^{-1}],$$
for a lie algebra valued form $\alpha$?

2. Aug 12, 2014

### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Sep 9, 2014

### TamTamTam

Hi,

An exterior product exists on the exterior algebra of vector space valued differential forms only if the vectorspace carries in addition the structure of an algebra.

Let G be a lie group and $$\varOmega^{*}:=\bigoplus_{r=0}^{dim(G)}\Omega^{r}(G,\mathfrak{g})$$
the exterior algebra of the associated lie algebra valued differential forms. If so, the exterior product is defined by

$$[\alpha,\beta](v_{1},\ldots,v_{p},v_{p+1},\ldots,v_{p+q}):=\frac{1}{p!q!}\sum_{\sigma\in S_{p+q}}sign(\sigma)[\alpha(v_{\sigma(1)},\ldots,v_{\sigma(p)}),\beta(v_{\sigma(p+1)},\ldots,v_{\sigma(p+q)})],$$
So the bracket is not a lie bracket, the bracket denotes the product of the algebra.

Last edited: Sep 9, 2014
4. Sep 15, 2014

### lavinia

Correct. So what is your question?

5. Sep 16, 2014

### Blazejr

The question was asked in the first post.