Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 1 / 3 / ii / Solution

The tangent matrices areUse the derivative of conjugation from part (i), rather than assuming antisymmetry. Since $H(s{)}^{-1}=\left(\begin{array}{cc}1& -s\\ 0& 1\end{array}\right)$, direct multiplication givesDifferentiating at zero yieldsFor the reverse order put $c=\cos t$, $s=\sin t$. The other conjugation isIts derivative at zero isBoth computations use actual curves in the special linear group; the opposite off-diagonal signs in $g(t)$ ensure its determinant is one.