Matrix Expression of the Maurer-Cartan Form

Question

I'm looking for clarification re: the 'classical' matrix expression for the Maurer-Cartan form

$$g^{-1} dg$$

(I have seen the related posts, they don't answer my specific question.)

Specifically I am confused about $dg$. I read in, e.g. Sharpe's book 'Differential Geometry', that we

Regard g in $GL_n(R)$ as 'the general point', that is, $g$ is the identity map on $GL_n(R)$. Then $dg$ is the identity map on the tangent bundle.

Again, in Frankel's 'Geometry of Physics' I read that

$dg$ takes $Y$ at $g$ into $Y$ and $g^{-1}$ takes $Y$ back to $e$.

where (I believe) Y is understood to be an element of the tangent space to the Lie group manifold $G$ at the point $g$.

My specific question is why do we bother with this identity map, i.e. why map the object Y to itself before mapping it back to the identity via $g^{-1}$? Why not just apply the left-multiplication $g^{-1}$ to the object directly?

Is it as simple as that we have to map the tangent vector to a matrix first, to put it in matrix form?

Answer 1

It's because we want to apply the exterior derivative to $dg$.

It's very reasonable to think of $dg$ as being "just a tangent vector". But after transporting it to the tangent space at the identity what would it mean to apply the exterior derivative to this object? Somehow this tangent vector must have some extra structure or be part of some more extended object to allow us to make sense of what it would mean to differentiate it. It might be possible to make some construction so that it would make sense to directly apply the exterior derivative to a tangent vector, and it would probably be a good excercise in building intuition. But at the end of the day such a construction should in some sense be equivalent to regarding $dg$ as the identity map.

From an algebraic viewpoint, note that when we extend the exterior derivative to all kinds of different object-valued forms (vector-valued, Lie algebra-valued, frame-valued, etc.), the tangent $n$-vector-valued $n$-forms are special in that the domain and codomain are the same. Out of these, the tangent vector-valued one-forms are the most fundamental and among these forms there is this unique form, the identity tangent vector form $dg: TM\to TM$. So it is perhaps expected that this seemingly trivial but unique object must be the starting point from which we are able to extract structure equations and curvature forms.