I'm trying to understand meaning and application of the Maurer Cartan Form, but I'm still not quite there. I'm then trying to do some examples and trying understand how it works.

I begun with the Heisenberg group: I took an element $$ A=\left(\begin{array}{ccc} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{array}\right).$$ found the inverse, i.e.$$A^{-1}=\left(\begin{array}{ccc} 1 & -x & -y+xz\\ 0 & 1 & -z\\ 0 & 0 & 1 \end{array}\right),$$ Than the differential $$dA=\left(\begin{array}{ccc} 0 & dx & dy\\ 0 & 0 & dz\\ 0 & 0 & 0 \end{array}\right).$$ Then the Maurer Cartan form should be $$A^{-1}dA=\left(\begin{array}{ccc} 0 & dx & dy-xdz\\ 0 & 0 & dz\\ 0 & 0 & 0 \end{array}\right)$$ Which I understand as the forms should be $$ \omega^{1}=dx , \omega^{2}=dy-xdz ,\omega^{3}=dz$$ Now the questions:

  1. Is this right? If not what am I doing wrong?

  2. What are the main applications/proprieties with this form?

  3. How could this procedure workout in $SO(3)$


1) It's correct. (assuming you've calculated the inverse matrix right)

2) Maurer-Cartan is related to curvature of a Cartan geometry through the strutural equation and is essential to the characterization of a Cartan Geometry and Klein Geometry. The pullback of the section of a Maurer-Cartan form defines a $\mathfrak g$-valued $1$-form called gauge which is a really important object in physics.

The Maurer-Cartan $\omega_G: T(G) \to \mathfrak g$ form satisfies

(i) $\omega_G$ is a linear isomorphism on each fiber;

(ii) $R^*_h\omega_G = Ad (h^{-1})\omega_G$ for all $h \in H$, where $H$ is a closed subgroup of the Lie group $G$ and $\mathfrak g$ the Lie algebra of $G$;

(iii) $\omega_g (X^{\dagger}) =X$ for all $X \in \mathfrak h$.

3) Maybe it would be more ilustrative to start off with $SO(2) \simeq S^1 = G = \{z \in \mathbb C; |z| = 1\}$. Here is the example

We have $T(S^1) = \{(e^{i\theta}, ire^{i\theta}) ; r,\theta \in \mathbb R\}$ the left action of $S^1$ on $T(S^1)$ is given by

$$\begin{align}S^1 \times T(S^1) &\to T(S^1)\\(e^{i\varphi}, (e^{i\theta},rie^{i\theta})) &\mapsto (e^{i(\theta + \varphi)}, rie^{i(\theta + \varphi)})\end{align}$$

The Maurer-Cartan form is then given by

$$\omega_G (e^{i\theta}, ir e^{i \theta}) = L_{e^{-i\theta}*}(e^{i \theta}, ire^{i \theta}) = (1, ir)$$

in a picture

$\hskip1in$enter image description here

in terms of matrix of $SO(2)$ the Maurer-Cartan form looks like

$$\omega_G = g^{-1}dg= \begin{pmatrix}0 & -d\theta \\ d\theta & 0\end{pmatrix}$$

using the parametrization of $SO(2)$ $$g (\theta) =\begin{pmatrix}\cos \theta & -\sin\theta \\\sin\theta & \cos \theta\end{pmatrix} \,\,\, , \,\,\theta \in \mathbb R$$

Edit: Regarding $SO(3)$, we have that its elements are matrices, relative to orthonormal basis for $\mathbb R^3$, of the linear transformation that rotate $\mathbb R^3$. To give a general element in $SO(3)$ let $\phi$ denote a real number and $\vec n = (n_1, n_2, n_3)$ a normal vector in $\mathbb R^3$. Define a $3 \times 3$ matrix $R(\phi, \vec n) $ by $$g := R(\phi, \vec n) = id + (\sin \phi)N + (1 - \cos \phi)N^2$$

where $id$ is the $3 \times 3$ idendity matrix $$N = \begin{pmatrix}0 &-n_3& n_2 \\n_3 & 0 & -n_1\\-n_2 & n_1 & 0\end{pmatrix}$$


$$N^2 = \begin{pmatrix}-((n_2)^2 + (n_3)^2) & n_1 n_2 & n_1n_3\\n_1n_2 &-((n_1)^2 + (n_3)^2) & n_2n_3\\n_1n_3 & n_2n_3 & -((n_1)^2 + (n_2)^2)\end{pmatrix}$$

verifying that $R(\phi,\vec n)$ is the matrix of the rotation through angle $\phi$ about an axis along $\vec n$ and the computation of the Maurer-Cartan form is up to you.

  • 1
    $\begingroup$ The structure constants of the Lie algebra also come from the structure equation $$d\omega_G = -\omega_G\wedge\omega_G.$$ In this case the only interesting entry in this equation is $d\omega^2 = -\omega^1\wedge\omega^3$. $\endgroup$ – Ted Shifrin Jul 11 '16 at 21:16
  • $\begingroup$ Thank you for your answer in point 1 and 2 and for your efforts in point 3. I really appreciated your answer and upvoted. I worked out SO(2) too but I was wondering how to proceed in the case of SO(3) do you have any idea? $\endgroup$ – Dac0 Jul 12 '16 at 6:29
  • $\begingroup$ In $SO(2)$ you can "see" what is happening in order to better understand the Maurer-Cartan form, which was the core of your question. And yes, the idea is rather simple, as $SO(n)$ is a Lie subgroup of $Gl(n)$ then a Maurer-Cartan form on $SO(n)$ is just the restriction of the Maurer-Cartan form on $Gl(n)$. Thus, it may be written as $g^{-1}dg$. In fact similar statements hold for all Lie subgroups of $Gl(n)$. $\endgroup$ – Aaron Maroja Jul 12 '16 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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