I have seen similar post asking for interpretation of the Maurer-Cartan form, but I am still struggling to understand it, so let me try to work a specific example and pose a specific question.
Let $G$ be the group containing all matrices of the form
\begin{bmatrix}
x &y \\
0 &1/x
\end{bmatrix}
where $x \neq 0$.
I read that the Maurer-Cartan form is defined as $\Omega = g^{-1} dg$ so I computed
$\Omega=
\begin{bmatrix}
1/x &-y \\
0 &x
\end{bmatrix}
\begin{bmatrix}
dx &dy \\
0 &-dx/x^2
\end{bmatrix}=
\begin{bmatrix}
dx/x &ydx/x^2+dy/x \\
0 &-dx/x
\end{bmatrix}$
So, for example, at the point
$g=
\begin{bmatrix}
5 &3 \\
0 &1/5
\end{bmatrix}
$
we have
$\Omega_g =
\begin{bmatrix}
dx/5 &3dx/25+dy/5 \\
0 &-dx/5
\end{bmatrix}
$
But I read that at a specific point $g\in G$, the entity $\Omega_g$ is supposed to be a linear map from $T_gG$ to $T_eG$. Looking at what I computed above, I don't see how $\Omega_g$ is a linear map on $T_gG$. For example, what is $\Omega_g (\frac{\partial}{\partial x})$?
I'm also interested in the claim that $\Omega$ is left-invariant. I am able to show that the 1-forms appearing as entries of $\Omega$ are each left-invariant via explicit computation.
–Proof:
However, I have seen the following "direct" proof, applicable in general, which appears similar in spirit:
–Proof:
However, because $dh$ is not a 1-form in the usual sense, and we are not really working in some coordinates, I am not sure how I am supposed to interpret this "proof" besides recognizing the analogy. Why is the above proof true?
As for background, you may assume I understand Lee's Smooth Manifold text (2nd edition) chapters 1-14. Lee's treatment of 1-form is always real-valued, so this different kind of 1-form is throwing me off.
Best Answer
The matrix $dg$ is an element of $T_gG$ and is mapped to $T_eG$ by simple multiplication from the left by $g^{-1}$ which is of course a linear operation.
The Lie algebra of your group are the matrices of the form $$\begin{bmatrix}a & b\\ 0 & -a \end{bmatrix}$$ and in your example the resulting matrix $$\begin{bmatrix}dx/5 & 3dx/25+dy/5\\ 0 & -dx/5 \end{bmatrix}$$ is indeed in this form and belongs to $T_eG$.
To compute $\Omega_g(\partial_x)$, simply put $dx=1$ and $dy=0$ since these are the coordinates for the basis vector $\partial_x$.