Example: basis of left invariant vector fields and MC form

The case of E(2)

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Consider the euclidean group E(2)=R2⋊SO(2). It can be realized as a matrix group with a local chart given by

φ:(a,b,θ)↦(cos(θ)−sin(θ)asin(θ)cos(θ)b001),

for a,b,θ∈R×R×(−π,π)=:U.

I am looking for a basis of the left invariant vector fields. In the parameter space U I take a basis of TeU, {∂a,∂b,∂θ} and apply d(Lg)e to all of them, being g=(a0,b0,θ0) and

Lg:(a,b,θ)↦(cos(θ0)a−sin(θ0)b,sin(θ0)a+cos(θ0)b,θ+θ0)

the left translation by g.

This way I obtain the basis of left invariants vector fields

e1=d(Lg)e(∂a|e)=cos(θ)∂a+sin(θ)∂be2=d(Lg)e(∂b|e)=−sin(θ)∂a+cos(θ)∂be3=∂θ

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 Lg takes the shape of left multiplication by the matrix φ(g), satisfying the commutative diagram

U⟶LgUφ↓φ↓GL(3)⟶φ(g)⋅−GL(3)A⟼φ(g)⋅A

A basis for g=TeE(2)⊂TeGL(3) is B={dφe(∂a),dφe(∂b),dφe(∂θ)}, that is,

B={(001000000),(000001000),(0−10100000)}

Multiplying by the left with

φ(g)=(cos(θ)−sin(θ)asin(θ)cos(θ)b001)

we obtain the left invariant vector fields in TE(2)⊂TGL(3)

{(00cos(θ)00sin(θ)000),(00−sin(θ)00cos(θ)000),(−sin(θ)−cos(θ)0cos(θ)−sin(θ)0000)}

It can be checked that they are precisely {φ∗(ei)}.

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=v1∂a+v2∂b+v3∂θ∈TgU. It yields a vector field dφ(V)∈GL(3):

dφ(V)=(00v1000000)+(00000v2000)+(−sin(θ)v3−cos(θ)v30cos(θ)v3−sin(θ)v30000)

We can take da,db,dθ instead of v1,v2,v3 and then

dφ=(−sin(θ)dθ−cos(θ)dθdacos(θ)dθ−sin(θ)dθdb000)

The g-valued Maurer-Cartan form is, according to the formula MC form for a matrix group

[φ(−)]−1dφ=(0−dθcos(θ)da+sin(θ)dbdθ0−sin(θ)da+cos(θ)db000)

but, what are the real valued Maurer-Cartan forms? Since we are working with basis B above, we have

μ1=cos(θ)da+sin(θ)dbμ2=−sin(θ)da+cos(θ)dbμ3=dθ

Question
Coming from this question.
I find the approach of Chern, @griffiths1974cartan and Clelland very misleading. They consider maps from G=E(n) to Rn, x,e1,…,en, 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

(Av01)

with A∈O(n) and v∈Rn. And now you only have to apply the formula for MC form for a matrix group θ=g−1dg, obtaining the same 1-forms.

Is this true for every Lie group of this type? That is, whenever we have a group G≈Rn⋊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≈Rn. 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

θ=g−1dg=(0−dθcos(θ)da+sin(θ)dbdθ0−sin(θ)da+cos(θ)db000)

If we consider the basis of e(2) given by

B={(001000000),(000001000),(0−10100000)}≡≡{∂a|e,∂b|e,∂θ|e}

we have

θ=μ1⊗∂a|e+μ2⊗∂b|e+μ3⊗∂θ|e

with

μ1=cos(θ)da+sin(θ)dbμ2=−sin(θ)da+cos(θ)dbμ3=dθ

In this case the Maurer-Cartan form has "two parts": μ1,μ2 on the one hand, and μ3 on the other hand. I think that (μ1,μ2) corresponds to the canonical solder form and μ3 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.