Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 24H / d / Solution

The assertion is false without further global hypotheses. For a concrete non-spherical surface of constant Gaussian curvature one, fix $a=1/2$ and setHere $f>0$, $g^{\prime }>0$ and $f^{\prime 2}+g^{\prime 2}=1$, so this is a connected regular embedded surface of revolution with unit-speed meridian. Its metric is $d{u}^2+f(u{)}^{2}d{v}^2$, whose Gaussian curvature is $-f^{\prime \prime }/f=1$. This formula also follows by multiplying its principal curvatures $-f^{\prime \prime }/g^{\prime }$ and $g^{\prime }/f$.

At $u=0$ these principal curvatures are $a=1/2$ and $1/a=2$. Every point of a radius-one sphere is umbilical with both principal curvature magnitudes one. The displayed surface therefore cannot lie in any such sphere, despite having $K=1$ everywhere.