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We know that in $\mathbb{R}^3$, any line normal to a sphere passes through the origin. Is the converse true?

Let $F(x,y,z) = 0$ be such a surface. Then, we have for some $t(x,y,z) \in \mathbb{R}$ $$ \nabla F = \left(\begin{matrix} F_x \\ F_y \\ F_z\end{matrix}\right) = t(x,y,z)\left(\begin{matrix} x \\ y \\ z\end{matrix}\right).$$

If we take $t \equiv g(x^2 + y^2 + z^2)$, the surface turns out to be a sphere. For other choices of $t$ that I've used ($t(x,y,z) = x^n, xyz,$ etc), I seem to get a contradiction. Is there some way we can definitively prove that the surface must be a sphere?

The Ginger Prince
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1 Answers1

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We can show that if all normals to a connected surface pass through a fixed point, the surface is contained in a sphere. This is an exercise I have done in my differential geometry course.

We can use the following lemma to prove the statement.

Let $f:S\to \mathbb{R}$ be a differentiable function on a connected regular surface $S$. Assume that $df_p=0$ for all $p\in S$, then $f$ is constant on $S$.

And then suppose $\forall p \in S$, the normal line passes through fixed $p_0$, that is, $pp_0$ is the normal at $p$, we take $f(x)=|x-p_0|^2$. Then the differential $df_p(v)=2 (p-p_0)\cdot v=0 \quad \forall v \in T_p(S)$, since $p_0$ lies on the normal line by assumption. Hence we get $f(x)$ by constant.

John
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  • @ZHANG at ur solution how we use the sentence 'all normals to a connected surface pass through a fixed point ' i cant see it. – bytrz Dec 26 '14 at 14:20
  • @rmznyzgyr thanks for your reply. I have modified more answer. – John Dec 28 '14 at 05:15