xournal 192
Consider the euclidean group $E(2)=\mathbb{R}^2 \rtimes SO(2)$. It can be realized as a matrix group with a local chart given by
$$ \varphi:(a,b,\theta)\mapsto \begin{pmatrix} cos(\theta)&-sin(\theta) &a \\ sin(\theta)& cos(\theta) & b\\ 0 & 0& 1\\ \end{pmatrix}, $$for $a,b,\theta \in \mathbb{R}\times \mathbb{R}\times(-\pi,\pi)=:U$.
I am looking for a basis of the left invariant vector fields. In the parameter space $U$ I take a basis of $T_eU$, $\{\partial_a,\partial_b,\partial_{\theta}\}$ and apply $d(L_g)_e$ to all of them, being $g=(a_0,b_0,\theta_0)$ and
$$ L_g: (a,b,\theta) \mapsto (cos(\theta_0) a-sin(\theta_0)b,sin(\theta_0) a+cos(\theta_0)b,\theta+\theta_0) $$the left translation by $g$.
This way I obtain the basis of left invariants vector fields
$$ e_1=d(L_g)_e(\partial_a|_e)=cos(\theta)\partial_a+sin(\theta)\partial_b $$ $$ e_2=d(L_g)_e(\partial_b|_e)=-sin(\theta)\partial_a+cos(\theta)\partial_b $$ $$ e_3=\partial_{\theta} $$But now I want to see this basis in $GL(3)$.
Because even if this example is fairly easy to work with it in parameter space, with a group with a more complicated "product" can be difficult to deal. In $GL(3)$, left translation $L_g$ takes the shape of left multiplication by the matrix $\varphi(g)$, satisfying the commutative diagram
$$ \begin{array}{ccc} U &\stackrel{L_g}{\longrightarrow} &U\\ \varphi \Big\downarrow & &\varphi \Big\downarrow\\ GL(3)&\stackrel{\varphi(g)\cdot-}{\longrightarrow} &GL(3)\\ A&\longmapsto &\varphi(g)\cdot A \end{array} $$A basis for $\mathfrak{g}=T_e E(2)\subset T_e GL(3)$ is $B=\{d\varphi_e(\partial_a),d\varphi_e(\partial_b),d\varphi_e(\partial_\theta)\}$, that is,
$$ B=\left\{ \begin{pmatrix} 0&0&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}, \begin{pmatrix} 0&0&0&\\ 0&0&1\\ 0&0&0&\\ \end{pmatrix}, \begin{pmatrix} 0&-1&0&\\ 1&0&0\\ 0&0&0&\\ \end{pmatrix} \right\} $$Multiplying by the left with
$$ \varphi(g)=\begin{pmatrix} cos(\theta)&-sin(\theta) &a \\ sin(\theta)& cos(\theta) & b\\ 0 & 0& 1\\ \end{pmatrix} $$we obtain the left invariant vector fields in $TE(2)\subset TGL(3)$
$$ \left\{ \begin{pmatrix} 0&0&cos(\theta)\\ 0&0&sin(\theta)\\ 0&0&0 \end{pmatrix}, \begin{pmatrix} 0&0&-sin(\theta)&\\ 0&0&cos(\theta)\\ 0&0&0&\\ \end{pmatrix}, \begin{pmatrix} -sin(\theta)&-cos(\theta)&0&\\ cos(\theta)&-sin(\theta)&0\\ 0&0&0&\\ \end{pmatrix} \right\} $$It can be checked that they are precisely $\{\varphi_*(e_i)\}$.
By the way, given any left invariant vector field, its coordinates in the basis above are the Maurer-Cartan forms. Consider a vector field $V=v_1\partial_a+v_2\partial_b+v_3 \partial_{\theta}\in T_gU$. It yields a vector field $d\varphi (V)\in GL(3)$:
$$ d\varphi(V)=\begin{pmatrix} 0&0&v_1\\ 0&0&0\\ 0&0&0 \end{pmatrix}+ \begin{pmatrix} 0&0&0&\\ 0&0&v_2\\ 0&0&0&\\ \end{pmatrix}+ \begin{pmatrix} -sin(\theta)v_3&-cos(\theta)v_3&0&\\ cos(\theta)v_3&-sin(\theta)v_3&0\\ 0&0&0&\\ \end{pmatrix} $$We can take $da,db,d\theta$ instead of $v_1,v_2,v_3$ and then
$$ d\varphi=\begin{pmatrix} -sin(\theta)d\theta&-cos(\theta)d\theta&da&\\ cos(\theta)d\theta&-sin(\theta)d\theta&db\\ 0&0&0&\\ \end{pmatrix} $$The $\mathfrak{g}$-valued Maurer-Cartan form is, according to the formula MC form for a matrix group
$$ [\varphi(-)]^{-1} d\varphi=\begin{pmatrix} 0&-d\theta&cos(\theta)da+sin(\theta)db\\ d\theta&0&-sin(\theta)da+cos(\theta)db\\ 0&0&0&\\ \end{pmatrix} $$but, what are the real valued Maurer-Cartan forms? Since we are working with basis $B$ above, we have
$$ \mu_1=cos(\theta)da+sin(\theta)db $$ $$ \mu_2=-sin(\theta)da+cos(\theta)db $$ $$ \mu_3=d\theta $$Question
Coming from this question.
I find the approach of Chern, @griffiths1974cartan and Clelland very misleading. 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 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.
> 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 Levi-Civita connection.
> 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? See Euclidean plane.
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Author of the notes: Antonio J. Pan-Collantes
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