In exponential coordinates, the <Baker--Campbell--Hausdorff formula> gives
$$
g(x)g(y)
=\exp\left\{(x^c+y^c)T_c
+\frac12x^ay^b[T_a,T_b]+O(3)\right\}.
$$
Comparison with $\varphi^c(x,y)=x^c+y^c+C^c{}_{ab}x^ay^b+O(3)$ gives
$$
\boxed{f^c{}_{ab}=C^c{}_{ab}-C^c{}_{ba}
=2C^c{}_{[ab]}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 302 / 2 / xii / Solution

For the <Strang splitting>
$$
F(t)=e^{tA/2}e^{tB}e^{tA/2},
$$
reversing $t$ reverses all three factors, so
$$
F(-t)=e^{-tA/2}e^{-tB}e^{-tA/2}=F(t)^{-1}.
$$
It is therefore a <time-symmetric numerical method>. Multiplication of the three <matrix exponential>[exponential] series shows agreement with $e^{t(A+B)}$ through degree two. Equivalently, the <Baker--Campbell--Hausdorff formula> gives an odd modified generator
$$
\log F(t)=t(A+B)+t^3D+O(t^5)
$$
for a matrix $D$ made from nested <commutator>[commutators]. Exponentiating gives
$$
\boxed{F(t)=e^{t(A+B)}+Ct^3+O(t^4)}
$$
for a matrix $C$ depending on $A$ and $B$.

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 5 / b / Solution

{c}
{wiki=Baker–Campbell–Hausdorff_formula}

The Baker--Campbell--Hausdorff formula expresses the local <matrix logarithm> of a product of exponentials as
$$
\log(e^Xe^Y)=X+Y+\frac12[X,Y]+\frac1{12}[X,[X,Y]]+\frac1{12}[Y,[Y,X]]+\cdots.
$$

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