I am coming from this question, which has not being completely answered but I think is very interesting.

In several works ([Chern], [Griffiths] and [Clelland]) the Maurer-Cartan form for $E(n)$ is worked out in the following manner. They consider maps from $G=E(n)$ to $\mathbb{R}^n$, $x, e_1,\ldots, e_n$, and express their differentials in terms of the frame in which we are. But for me that doesn't seem natural because is something very particular of this example: the frame itself can be described in terms of the objects it describe. I consider more natural the general approach: the group $E(n)$ can be seen like a matrix group of a special type, that one with elements of the form $$ \begin{pmatrix} A & v\\ 0 & 1\\ \end{pmatrix} $$ with $A\in O(n)$ and $v\in \mathbb{R}^n$. And now you only have to apply the formula for MC form for a matrix group, $\theta=g^{-1}dg$, obtaining the same 1-forms.

Question 1 (solved)
Is this true for every Lie group of this type? That is, whenever we have a group $G\approx \mathbb{R}^n \rtimes H$ it can be seen as a subgroup of $GL(n+1)$ as above (see this QA in MSE) and we can interpret the columns as vectors in the homogeneous space $G/H\approx \mathbb{R}^n$. Then, does the Maurer-Cartan form tell us the variation of these vectors expressed in the current frame?

Back to the case of $E(2)$, for simplicity. The MC form is $$ \theta=g^{-1}dg=\begin{pmatrix} 0&-d\theta&cos(\theta)da+sin(\theta)db\\ d\theta&0&-sin(\theta)da+cos(\theta)db\\ 0&0&0&\\ \end{pmatrix} $$ If we consider the basis of $\mathfrak{e}(2)$ given by $$ B=\left\{ \begin{pmatrix} 0&0&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}, \begin{pmatrix} 0&0&0&\\ 0&0&1\\ 0&0&0&\\ \end{pmatrix}, \begin{pmatrix} 0&-1&0&\\ 1&0&0\\ 0&0&0&\\ \end{pmatrix} \right\}\equiv $$ $$ \equiv\{\partial_a|_e,\partial_b|_e,\partial_{\theta}|_e\} $$ we have $$ \theta=\mu_1 \otimes\partial_a|_e+\mu_2 \otimes\partial_b|_e +\mu_3\otimes \partial{\theta}|_e $$ with $$ \mu_1=cos(\theta)da+sin(\theta)db $$ $$ \mu_2=-sin(\theta)da+cos(\theta)db $$ $$ \mu_3=d\theta $$ In this case the Maurer-Cartan form has "two parts": $\mu_1, \mu_2$ on the one hand, and $\mu_3$ on the other hand. I think that $(\mu_1, \mu_2)$ corresponds to the canonical solder form and $\mu_3$ is the connection form of the Levi-Civita connection.

Question 2
Why is this the Levi-Civita connection? What relationship does it have (if any) with the group reduction of $GL(2)$ to $O(2)$ by means of the standard metric?

I have an intuition about some relation but I cannot grasp what it is... I know that the frame bundle for $\mathbb{R}^2$ is $\mathbb{R} \rtimes GL(2)$ and that the standard metric let us reduce the structure group of this principal bundle to $O(2)$...

[Chern]: Chapter 6 of S.S. Chern's book "Lectures on differential geometry"

[Griffiths]: On Cartan's method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry

[Clelland]: From Frenet to Cartan: The Method of Moving Frames


2 Answers 2


In question 1, I think you are just asking, if we have elements of $G$ written as $$g=\begin{pmatrix}h&v\\ 0&1\end{pmatrix}$$ then $$g^{-1}=\begin{pmatrix}h^{-1}&-h^{-1}v\\ 0&1\end{pmatrix}$$ hence $$g^{-1}dg=\begin{pmatrix}h^{-1}dh&-h^{-1}dv\\ 0&1\end{pmatrix}.$$ This is clearly since the Lie group operation is matrix multiplication.

Question 2: In books which discuss the moving frame, one learns that the soldering forms $\omega_1,\omega_2$ are precisely those for which the projection of orthonormal frame bundle to surface pulls back the dual of the orthonormal frame to $\omega_1,\omega_2$ and the Levi-Civita connection form $\omega_{12}$ is precisely the one for which $d\omega_1=-\omega_{12}\wedge\omega_2$ and $d\omega_2=\omega_{12}\wedge\omega_1$, so you can check that $\mu_3$ is the Levi-Civita. For an example of such a book, my Introduction to Exterior Differential Systems discusses all of this in its many appendices, in particular in appendix G the structure equations are derived for Riemannian metrics on surfaces.

  • $\begingroup$ Yes, you are right. I asked yesterday, but this morning I realized that question 1 was a triviality. I had no time to edit. Thanks for the confirmación. Now I am struggling with question 2. $\endgroup$ Oct 17, 2022 at 14:45
  • $\begingroup$ Thank you for the reference, I'll take a look $\endgroup$ Oct 21, 2022 at 14:34
  • $\begingroup$ I think LibGen links on MO should be discouraged (though opinion on that is far from universal), but I guess no-one can argue about you giving a link to your own book on LibGen! $\endgroup$
    – LSpice
    Oct 31, 2022 at 19:56

I have been working in question 2 and I think I have a good explanation. At the end is a triviality, but that's what (almost) always happens in math when you understand something.

I am going to discuss here the Euclidean plane from two different perspectives.

Euclidean plane

From the point of view of classical differential geometry

The Euclidean plane is the manifold $M=\mathbb{R}^2$, with coordinates $(x_1,x_2)$, together with the natural Riemannian metric $$ g=dx_1\otimes dx_1+dx_2\otimes dx_2. $$ It is, therefore, a Riemannian manifold.

Seen as a plain manifold we can consider that $M$ is endowed with a natural linear connection $\nabla$, such that $\nabla_{\partial_{x_i}}\partial_{x_j}=0\partial_{x_1}+0\partial_{x_2}$. The induced principal connection on the frame bundle is given by the 1-form $$ \omega= \begin{pmatrix} c_{11}&c_{12}\\ c_{21}&c_{22}\\ \end{pmatrix}^{-1}\cdot\begin{pmatrix} dc_{11}&dc_{12}\\ dc_{21}&dc_{22}\\ \end{pmatrix}\in \Omega(\mathbb R^2,\mathfrak{gl}(2)) $$ at $$ f= \begin{pmatrix} c_{11} & c_{12}& x_1\\ c_{21} & c_{22}& x_2\\ 0 & 0& 1\\ \end{pmatrix}\in FM $$ Here it is shown how to construct $\omega$ from an arbitrary $\nabla$.

On the other hand, if we think of $M$ as a Riemannian manifold we can consider the Levi-Civita connection $\nabla_{LC}$. Since the metric is constant, the covariant derivative $\nabla_{LC}$ coincides with the natural covariant derivative $\nabla$, so it induces the same connection $\omega$ on $FM$. But the metric $g$ also specifies a orthonormal frame bundle $OM$ (see here why). The elements of this principal bundle are $$ f= \begin{pmatrix} c & -\sqrt{1-c^2}& x_1\\ \sqrt{1-c^2} & c& x_2\\ 0 & 0& 1\\ \end{pmatrix}\in OM $$ with $c\in [-1,1]$. Since $$ \omega|_{OM}=\begin{pmatrix} c&\sqrt{1-c^2}\\ -\sqrt{1-c^2}&c\\ \end{pmatrix}\cdot\begin{pmatrix} dc&\frac{c}{\sqrt{1-c^2}}dc\\ \frac{-c}{\sqrt{1-c^2}}dc&dc\\ \end{pmatrix}= $$ $$ =\begin{pmatrix} 0&\frac{dc}{\sqrt{1-c^2}}\\ -\frac{dc}{\sqrt{1-c^2}}&0\\ \end{pmatrix} \in \Omega(\mathbb R^2,\mathfrak{o}(2)), $$ according to Proposition 4.7. in vicenteBundles, this connection can be reduced to a connection on the orthonormal frame bundle determined by the metric $g$.

If we parameterize this principal bundle $OM$ with $$ (x_1,x_2,\theta)\mapsto \begin{pmatrix} \cos(\theta) & -\sin(\theta)& x_1\\ \sin(\theta) & \cos(\theta)& x_2\\ 0 & 0& 1\\ \end{pmatrix} $$ we obtain the more famous expression for $\omega$: $$ \begin{pmatrix} 0&d\theta\\ -d\theta&0\\ \end{pmatrix} $$ Remember: this 1-form tells us how much the bases at $f$ and $f'$ "fail to be constant" when we pass from the frame $f$ to a nearby frame $f'$, but expressing this mistake with respect to the frame $f$ itself.

From the point of view of Cartan geometry

The Euclidean plane is a Cartan geometry modeled over $(E(2),O(2))$ , indeed is the Klein geometry $(E(2),O(2))$. Moreover, it is a reductive Klein geometry since $$ \mathfrak e(2)=\left\{\begin{pmatrix} C & v\\ 0 & 0\\ \end{pmatrix} :C\in \mathfrak{o}(2), v\in \mathbb R^2\right\}=\mathfrak o(2)\oplus \mathfrak p $$ We have a natural choice for $\mathfrak p$ $$ \mathfrak{p}=\left\{\begin{pmatrix} 0 & p\\ 0 & 0\\ \end{pmatrix} :p\in \mathbb R^2\right\}. $$

With this in mind, remember that the Maurer-Cartan form describes all possible "infinitesimal displacements" of the frame $f$, but from the point of view of the frame $f$ itself. That is, if we pass from the frame $f$ to another frame $f'$, the Maurer-Cartan form at $f$ applied to the "vector" $\vec{ff'}=(dx_1,dx_2,d\theta)$ is a packet of information $$ A=\begin{pmatrix} 0&-d\theta&cos(\theta)dx_1+sin(\theta)dx_2\\ d\theta&0&-sin(\theta)dx_1+cos(\theta)dx_2\\ 0&0&0&\\ \end{pmatrix}\in \mathfrak e(2) $$ Here is encoded, on the one hand, how much have we moved the base point of $f$ to the base point of $f'$ and, on the other, how much have we changed the basis itself. The natural choice of $\mathfrak p$ let us think that the information about the change of base point is in the $v$ part (the projection of the Maurer-Cartan form on $\mathfrak p$), and so the projection of the Maurer-Cartan form on $\mathfrak o(2)$, $\begin{pmatrix}0&d\theta\\-d\theta&0\\\end{pmatrix}$, tell us how much has the basis changed. That is, the same as the connection 1-form of the connection on the orthonormal bundle induced by the metric $g$ (which is the Levi-Civita connection).

To summarize:
In the orthonormal frame bundle induced by the metric $g$ we consider a displacement from a frame $f=\begin{pmatrix}C & p\\0 & 1\\\end{pmatrix}$ to a frame $f'=\begin{pmatrix}C' & p'\\0 & 1\\\end{pmatrix}$.

The principal connection $\omega$ induced by the Levi-Civita connection measures the change from $C$ to $C'$ as an element of $\mathfrak o(2)$.

The Cartan connection (Maurer-Cartan form) measures the change from $f$ to $f'$ as an element of $\mathfrak e(2)$. This change can be decomposed like the union of a change from $C$ to $C'$ and a change from $p$ to $p'$. This is reflected in the fact that $\mathfrak e(2)=\mathfrak o(2)\oplus \mathfrak p$. If we focus on the change from $C$ to $C'$ we have the principal connection $\omega$.


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