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

The identity and inverse axioms give

Setting either argument to zero in the expansion and comparing coefficients givesThus, through quadratic order,

Write ${\bar{x}}^r=-x^r+q^r{}_{st}{x}^s{x}^t+O(x^3)$. Substitution into $\phi (\bar{x},x)=0$ givesand henceOnly the symmetric part of $C^r{}_{st}$ contributes to this expression.

Insert the expansions from parts ii and iii successively into
$g(w)=g(x{)}^{-1}g(y{)}^{-1}g(x)g(y)$. All constant and linear terms cancel, leavingThe antisymmetric part of the local multiplication law is therefore the infinitesimal group commutator.

Since $z+dz=\phi (z,\theta )$,Thus

The infinitesimal change $dy$ can be writtenRight multiplication of $g(y)$ by $g(\theta )$ is also right multiplication of $g(z)=g(x)g(y)$ by the same element. Part v therefore givesso

Applying $T_a(y)={\mu }_a{}^s(y)\partial /\partial y^s$ to a function of $z$ and using part vi,HenceThese are the same left-invariant vector fields in different coordinates.

Differentiate the identity in part vi with respect to $y^t$, multiply by ${\mu }_a{}^s(y){\mu }_b{}^t(y)$, and use $\lambda \mu =I$. The product rule gives

Differentiate ${\lambda }_s{}^c{\mu }_a{}^s={\delta }_a{}^c$ to obtainMultiplying part viii by ${\lambda }_r{}^c(z)$ consequently gives

Antisymmetrize part ix in $a,b$. The left side vanishes because the second derivative is symmetric in $s,t$. ThereforeSince multiplication by an arbitrary $g(x)$ moves $y$ to any $z$ in the connected coordinate neighborhood, the $f^c{}_{ab}$ are constants. They are the structure constants.

Directly commuting the vector fields givesUsing the definition in part x and $\lambda \mu =I$ yields

In exponential coordinates, the Baker--Campbell--Hausdorff formula givesComparison with ${\phi }^c(x,y)=x^c+y^c+C^c{}_{ab}{x}^a{y}^b+O(3)$ gives