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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 $SO(3)$

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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 $$N = \begin{pmatrix}0 &-n_3& n_2 \\n_3 & 0 & -n_1\\-n_2 & n_1 & 0\end{pmatrix}$$

and

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

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.

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    $\begingroup$ 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

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