Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 24H (Differential Geometry)

For $v\in T_{p}S$ for which the geodesic exists up to time one, the exponential map is ${\exp }_p(v)={\gamma }_v(1)$, where ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The initial velocity determines the geodesic uniquely. Since $d({\exp }_p{)}_0$ is the identity, this map is a local diffeomorphism near zero.

Choose an orthonormal basis of $T_{p}S$ and set $e(\theta )=(\cos \theta ,\sin \theta )$. The geodesic polar coordinates arein a normal neighbourhood, away from its centre. For sufficiently small positive $r$, the geodesic circle iswithin that neighbourhood. Outside the injectivity range the distance sphere need not have this smooth coordinate description.

The Gauss lemma says that the radial and angular directions of the exponential map are orthogonal, with radial lengths preserved. On a surface it givesso the metric in geodesic polar coordinates is $d{r}^2+G(r,\theta )d{\theta }^2$.

To prove it, consider a smooth variation $\gamma (r,s)={\exp }_p(rv(s))$ of geodesics with unit initial vectors $v(s)$. Put $T={\gamma }_r$ and $J={\gamma }_s$. Each radial geodesic has constant unit speed, and the Levi-Civita connection is torsion free. HenceAt $r=0$, $J=0$ because all geodesics start at $p$, so the constant inner product is zero. Unit radial length follows from constant speed. General initial-vector variations split into a radial component, whose image has the same length, and a component tangent to the initial sphere, whose image is orthogonal to it. This proves the full Gauss lemma.

The literal assertion needs $q\ne p$. If $q=p$, the two geodesic circles coincide, have equal tangent lines at every point, and do not intersect transversally; their intersection is not finite. We prove that close geodesic circles intersect twice for nearby distinct centres, and in fact find exactly two intersections.

Choose a strongly convex normal neighbourhood about $p$ and a sufficiently small $r>0$ so that every circle considered and all geodesics joining its points to nearby centres remain there. Squared distance $d(q,x{)}^2$ is then smooth in both arguments, and the exponential map is nonsingular on the relevant radius-$r$ tangent circles. Consequently all these distance spheres are smooth embedded one-dimensional manifolds.

Write $q={\exp }_p(\epsilon v)$ with $|v|=1$ and $\epsilon >0$, and parameterize $S_r(p)$ by $x(\theta )={\exp }_p(re(\theta ))$. DefineThe first variation of geodesic energy givesIndeed differentiating the energy of the joining geodesic cancels the interior term by the geodesic equation, leaving the initial endpoint term $-2⟨re(\theta ),v⟩$. This is also the endpoint form of the Gauss lemma.

Since $F(0,v,\theta )=0$, the quotient $F/\epsilon$ extends smoothly to $\epsilon =0$ by integrating ${\partial }_{\epsilon }F$ from zero to $\epsilon$. At zero it is $-2r\cos (\theta -\arg v)$, with exactly two simple zeros. The implicit function theorem gives two nearby simple zeros for small positive $\epsilon$. Compactness of the unit vector circle makes the choice of small $\epsilon$ uniform in $v$; away from fixed neighbourhoods of those two zeros the cosine is bounded away from zero, so there are no additional zeros.

A nonzero angular derivative at either zero says that the defining function of $S_r(q)$ has a nonzero derivative along $S_r(p)$. Thus their tangent lines differ: the intersections are transverse. Shrinking the neighbourhood of centres accordingly proves