Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 26F (Differential Geometry)

For each $p\in S$, the first fundamental form is defined without coordinates bythe restriction of the Euclidean inner product to the tangent space of $S$.

If $\varphi (u,v)$ is a local parametrization andthenTaking their dot product giveswhere

Since $z>0$, the cone has the parametrizationIts coordinate derivatives satisfyso its first fundamental form is

On a sufficiently narrow angular patch, defineand map the cone to the plane point with polar coordinates $(R,\Theta )$. The Euclidean metric pulls back asThe map is therefore a local isometry from a circular cone to the plane, proving that $\Sigma$ is locally isometric to the Euclidean plane.

For a chosen unit normal field $N$, the second fundamental form isIn a local parametrization $\varphi$, its coefficients, up to the equivalent simultaneous sign convention, areThe Gaussian curvature is the determinant of the shape operator, henceTheorema Egregium states that $K$ is determined entirely by the first fundamental form. In particular, it is intrinsic and is preserved by every local isometry.

LetIts gradient isIf this vanishes, the first and third coordinates give $x=y=z$, and the middle coordinate then gives $y=0$. Thus the origin is the only critical point of $F$. It has been removed from $V$, so $0$ is a regular value of the restriction under consideration. The regular level set theorem shows thatis a smooth surface.

Completing the square givesThus $V$ is a two-napped cone. On the nappe where $\epsilon y>0$, with $\epsilon \in \{1,-1\}$, useThis has the conical formIts second derivatives obeyThe latter vector is tangent, so for either choice of unit normal,Since the parametrization is regular, $EG-F^2>0$, and thereforeat every point of $V$. This is the Gaussian curvature of a cone away from its vertex.