= Solution
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 is
$$
\boxed{\chi(X)=m_0-m_1+m_2,}
$$
where $m_i$ counts the <critical points> whose <Hessian> has $i$ negative eigenvalues.
For $f(x)=|x|^2$, differentiation in a tangent direction $v$ gives $df_x(v)=2x\cdot v$. This vanishes for all $v\in T_xX$ 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 $m_0,m_2\geq1$. Thus $m_1=m_0+m_2\geq2$ and the total is at least four. Hence \b[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> are
$$
X(u,v)=(a+r\cos u,(R+r\sin u)\cos v,(R+r\sin u)\sin v),\quad n(u,v)=(\cos u,\sin u\cos v,\sin u\sin v).
$$
Take \b[$P=(a,R/2,0)$]. 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 \b[$Q=(a,0,0)$] instead. All points on each of the circles $u=\pi/2$ and $u=3\pi/2$ have radial normals through $Q$, giving infinitely many.
Back to article page
