Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 24H (Differential Geometry)

For $v\in T_{p}S$ sufficiently small, let ${\gamma }_v$ be the geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. Define ${\exp }_p(v)={\gamma }_v(1)$. Geodesic uniqueness and rescaling give ${\gamma }_v(t)={\exp }_p(tv)$, soThus its derivative at zero is the identity, an isomorphism. Assuming the stated smoothness, the inverse function theorem proves ${\exp }_p$ is a diffeomorphism between neighbourhoods of zero and $p$.

For a parametrization $X(u^1,u^2)$, put $g_{ij}=X_i\cdot X_j$. Decompose the second derivatives into tangent and normal components:These tangent coefficients are the Christoffel symbols. Differentiate $g_{ij}$ and combine the resulting three identities to obtainHere $(g^{kl})$ is the inverse matrix. The formula proves that these symbols depend only on the first fundamental form and its first derivatives, not on the second fundamental form.

In normal coordinates, each radial line $u(t)=tv$ is an affinely parametrized geodesic. Its coordinate geodesic equation gives ${\Gamma }_{ij}^k(tv)v^i{v}^j=0$. At $t=0$ this holds for every tangent vector $v$. Since ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$, polarization of this quadratic form gives every coefficient zero. Thus all Christoffel symbols vanish at the centre $p$; they need not vanish elsewhere.