Past exam of the geometry course of the University of Oxford / 2007 / A / 3

A critical point satisfies $df=0$; it is nondegenerate when the Hessian on its two-dimensional tangent space is nonsingular. For a smooth function with only nondegenerate critical points on a compact surface, the Morse theory formula iswhere $m_i$ counts the critical points whose Hessian has $i$ negative eigenvalues.

For $f(x)=|x{|}^2$, differentiation in a tangent direction $v$ gives $d{f}_x(v)=2x\cdot v$. This vanishes for all $v\in T_{x}X$ exactly when the position vector is perpendicular to that tangent plane, equivalently the normal line at $x$ contains the origin. This includes $x=0$, if it is on the surface. On a torus $\chi =0$, while a minimum and a maximum give $m_0,m_2\ge 1$. Thus $m_1=m_0+m_2\ge 2$ and the total is at least four. Hence at least four surface normals pass through the origin under the stated nondegeneracy assumption.

Write the generating circle as center $(a,R)$ and radius $r$, with $R>r>0$. A parametrization and unit normal areTake $P=(a,R/2,0)$. If $P-X=tn$, its first coordinate gives $(r+t)\cos u=0$. The alternative $t=-r$ would force the radial distance of $P$ from the axis to be $R$, contrary to $R/2$. Thus $\cos u=0$. The remaining equations and the nonzero radial distance of $P$ force $v=0$ or $\pi$. Each of the two choices of $u$ and two of $v$ works, giving exactly four distinct points. Take $Q=(a,0,0)$ instead. All points on each of the circles $u=\pi /2$ and $u=3\pi /2$ have radial normals through $Q$, giving infinitely many.