Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 309 / 1

Put $D=\mathrm{diag}(a_1,\dots ,a_{n+1})$. The invertible linear map $x\mapsto u=Dx$ takes the ellipsoid to the unit sphere $S^n$. Write $u=(v,s)$ with $v\in {\mathbb{R}}^n$, and let $N=(0,1)$ and $S=(0,-1)$ be its two poles. The following two stereographic projections define manifold charts on the ellipsoid:Their ranges are ${\mathbb{R}}^n$. With $r^2=|w{|}^2$, their inverses, followed by $D^{-1}$, areThese are continuous and smooth functions, and directly satisfy the defining constraint. On the overlap, the smooth transition map isIt is its own inverse and is smooth away from zero. The two domains are open in the subspace topology and cover the ellipsoid. Its subspace topology is inherited from Euclidean space, so the ellipsoid is Hausdorff and has a second-countable space topology. Thus these charts form a smooth atlas and make it an $n$-dimensional smooth manifold, with an explicit diffeomorphism to $S^n$. For the degenerate dimension $n=0$, each chart consists of a single point with range ${\mathbb{R}}^0$, and their overlap is empty; the same conclusion holds.

At $p\in \mathcal{Q}$, the differential of a smooth map defines the pushforward of a vector field byThis is a tangent vector at $\psi (p)$, and varies smoothly as a section of the pullback tangent bundle. In local coordinates its components are $X^i(p){\partial }_i{\psi }^a(p)$.

For a general smooth map between manifolds, this is a vector field along a map, rather than an intrinsic vector field on all of $N$. A vector field $Y$ on the image can be defined only if these vectors agree whenever two points have the same image, and the resulting field must also be smooth; a projectable vector field must meet that requirement. For a diffeomorphism, there is no ambiguity and $Y_q=d{\psi }_{{\psi }^{-1}(q)}{X}_{{\psi }^{-1}(q)}$. For example, $\psi (x)=x^2$ on $\mathbb{R}$ and $X={\partial }_x$ give $2x{\partial }_y$, with opposite nonzero values at the two preimages of $y>0$. Thus the source's notation must be read with this qualification.

On a branch where $(\rho ,\varphi )$ are valid local coordinates, the inclusion map is $j(\rho ,\varphi )=(x(\rho ),\rho \cos \varphi ,\rho \sin \varphi )$, with $x^2=1-a^2{\rho }^2$. Its differential givesStrictly this is first a vector field along a map on the surface. The displayed formula extends it canonically to a smooth vector field on all of ${\mathbb{R}}^3$. Its integral curves of a vector field through $(x_0,y_0,z_0)$ obey $\dot{x}=0$, $\dot{y}=-z$, $\dot{z}=y$, henceThese are circles around the $x$-axis, traversed with angular speed one; points on the axis are stationary. Since $K(x^2+a^2(y^2+z^2))=0$, curves starting on the ellipsoid stay there. The local angular chart can expire while the ambient flow map and the geometric curve remain well defined.

The induced metric is obtained from the ambient Euclidean metric by the pullback of a Riemannian metric operation. Differentiating $x^2+a^2{\rho }^2=1$ on a branch with $x\ne 0$ gives $dx=-a^2\rho d\rho /x$. Also $d{y}^2+d{z}^2=d{\rho }^2+{\rho }^{2}d{\varphi }^2$. ConsequentlyThe assumption that $(\rho ,\varphi )$ are coordinates already excludes the equator $x=0$ as well as the poles $\rho =0$. On such a chart, $0<\rho <1/|a|$ and $E>0$.

The conformal flattening of a surface of revolution supplies the particularly simple conformal factorSince ${\sigma }^{\prime }>0$, these are isothermal coordinates with a flat rescaled metric. An angular coordinate is understood on a local branch; the flat metric may equally be viewed locally as a cylinder metric. The apparent divergence of $E$ at the equator is a coordinate failure, not a singularity of the induced metric. For example, $(x,\varphi )$ give $g=[1+x^2/(a^2(1-x^2))]d{x}^2+(1-x^2)d{\varphi }^2/a^2$, regular at $x=0$.