Maurer Cartan Form of the Heisenberg group

Question

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 $SO(3)$

Answer 1

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

$\hskip1in$

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.