= Solution
For a <Matrix Lie group> $G$, the left <Maurer-Cartan form> identifies tangent vectors with the <Lie algebra> by translating them to the identity:
$$
\boxed{\rho_g(v)=(dL_{g^{-1}})_g v=g^{-1}v,\qquad \rho=g^{-1}dg.}
$$
It is a <Lie algebra>-valued <differential form> of degree one. For a fixed $h\in G$, replacing $g$ by $hg$ gives $(hg)^{-1}d(hg)=g^{-1}dg$, so it is a <left-invariant differential form>.
Differentiate $g^{-1}g=I$ to get $d(g^{-1})=-g^{-1}(dg)g^{-1}$. The <exterior derivative> then gives
$$
d\rho=d(g^{-1})\wedge dg+g^{-1}d^2g=-g^{-1}dg\wedge g^{-1}dg=-\rho\wedge\rho.
$$
Here the <exterior product> of matrix-valued <differential forms> includes matrix multiplication; the order of the matrices matters. Hence \b[the Maurer-Cartan equation is]
$$
\boxed{d\rho+\rho\wedge\rho=0.}
$$
Write $\rho=\sigma^\alpha T_\alpha$. Antisymmetry of the <exterior product> implies
$$
\rho\wedge\rho=\frac12\sum_{\alpha,\beta}\sigma^\alpha\wedge\sigma^\beta[T_\alpha,T_\beta]
=\frac12\sum_{\alpha,\beta,\gamma}c^\gamma{}_{\alpha\beta}\sigma^\alpha\wedge\sigma^\beta T_\gamma.
$$
Comparison of the <Lie algebra> components yields the <Maurer-Cartan equation in a Lie-algebra basis>:
$$
\boxed{d\sigma^\gamma=-\frac12\sum_{\alpha,\beta}c^\gamma{}_{\alpha\beta}\sigma^\alpha\wedge\sigma^\beta.}
$$
Thus \b[the canonical antisymmetric choice is $f^\gamma{}_{\alpha\beta}=-\tfrac12c^\gamma{}_{\alpha\beta}$]. The factor one half is required because the sum includes both ordered pairs $(\alpha,\beta)$ and $(\beta,\alpha)$. If only $\alpha<\beta$ is summed, its coefficient is $-c^\gamma{}_{\alpha\beta}$. Strictly, the equality of <differential forms> determines only the antisymmetric part of $f$: one may add any tensor symmetric in $\alpha,\beta$ without changing it. This is the <symmetric-part ambiguity in Maurer-Cartan coefficients>.
A faithful <matrix representation> of the orientation-preserving <real affine group> is
$$
\boxed{g(a,b)=\begin{pmatrix}a&b\\0&1\end{pmatrix},\qquad a>0,\ b\in\mathbb R.}
$$
Its action on $(x,1)^T$ has first component $ax+b$. The <group operation> and inverse are
$$
g(a,b)g(a',b')=g(aa',b+ab'),\qquad g(a,b)^{-1}=g(a^{-1},-b/a).
$$
Different transformations have different matrix entries, proving faithfulness. With the <Lie algebra> basis
$$
D=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad T=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad [D,T]=T,
$$
the left <Maurer-Cartan form> is
$$
g^{-1}dg=\begin{pmatrix}da/a&db/a\\0&0\end{pmatrix}=\sigma^D D+\sigma^T T.
$$
The requested <left-invariant differential forms> therefore form the <Maurer-Cartan coframe of the real affine group>:
$$
\boxed{\sigma^D=\frac{da}{a},\qquad\sigma^T=\frac{db}{a},\qquad d\sigma^D=0,\quad d\sigma^T=-\sigma^D\wedge\sigma^T.}
$$
For a direct invariance check, left translation by $(a_0,b_0)$ sends $(a,b)$ to $(a_0a,b_0+a_0b)$, and the pullbacks of the two displayed <differential forms> are unchanged. Their dual <left-invariant vector fields> are $a\partial_a$ and $a\partial_b$, whose bracket is $a\partial_b$, in agreement with $[D,T]=T$.
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