Past exam of the geometry course of the University of Oxford / 2010 / A / 2

For a regular parametric surface $r(u,v)$, setThe first fundamental form is the induced Riemannian metricThe second fundamental form, for this choice of unit normal vector, isThe Gaussian curvature is the product of the principal curvatures, equivalently the determinant of the shape operator:Reversing the normal reverses the second fundamental form and both principal curvatures, while leaving $K$ unchanged.

Work on a regular patch with $f(u)>0$. Differentiating the surface of revolution givesThe arc-length condition gives $E=1$, $F=0$, $G=f^2$. Also $\Vert r_u\times r_v\Vert =f$, so takeThe second derivatives yieldThus the first fundamental form and second fundamental form areThe opposite normal reverses the second displayed form. At a smooth pole the revolution coordinates degenerate, so curvature is understood by a regular local chart or by continuity.

Differentiating the arc-length identity gives $f^{\prime }{f}^{\prime \prime }+h^{\prime }{h}^{\prime \prime }=0$. Multiply by $f^{\prime }$ and use $(f^{\prime }{)}^2+(h^{\prime }{)}^2=1$ to obtainThis derivation does not divide by $h^{\prime }$, so it remains valid where $h^{\prime }=0$. Inserting the first fundamental form and second fundamental form computed above gives the Gaussian curvature of a surface of revolution:

Choose $f(u)=\sin u$ and $h(u)=\cos u$ for $0\le u\le \pi$. Adding the smooth poles at the endpoints gives the unit sphere. Its meridian is parametrized by arc length, and $-f^{\prime \prime }/f=1$ away from the poles, hence everywhere by continuity. Thus the unit sphere has constant Gaussian curvature one.

There is no compact smooth surface of revolution without boundary of constant Gaussian curvature minus one in this setting. On a compact such surface, the distance $f$ from the rotation axis attains a positive maximum. The maximum lies on a regular revolution patch, where the meridian arc length coordinate gives $f^{\prime \prime }\le 0$. Hence $K=-f^{\prime \prime }/f\ge 0$ there, contradicting $K=-1$. Equivalently, constant negative curvature would require $f^{\prime \prime }=f>0$, excluding any positive radial maximum. Compact patches with boundary would not be excluded by this argument, but they are not closed surfaces.