For a matrix <Lie group>, the left-invariant <Maurer-Cartan form> is
$$
\boxed{\rho=g^{-1}dg}.
$$
Differentiating $g^{-1}g=I$ gives $d(g^{-1})=-g^{-1}(dg)g^{-1}$. Hence
$$
d\rho=d(g^{-1})\wedge dg
=-g^{-1}dg\wedge g^{-1}dg
=-\rho\wedge\rho,
$$
and therefore
$$
\boxed{d\rho+\rho\wedge\rho=0}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 3 / a / Solution

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A Lie group is a <group> that is also a <smooth manifold>, with smooth multiplication and inversion. Its tangent space at the identity carries a natural <Lie algebra> structure.

= Lie groups
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