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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-313/3/a/solution