I would like to know if it is possible to show that :

Given $S$ a compact surface of $R^3$, for any vector $a \in R^3$, there is $p \in S$ such that $T_pS$ is orthogonal to $a$

without using a classification theorem on the compact connected surfaces (or else you "just" have to show it for $S^2$, the tori and the projective spaces after taking a connected part of $S$).


1 Answer 1


Since $S$ is compact there exists a maximum (or minimum) $x=p$ of the continuous function $f(x)=x\cdot a$ on $S$, where dot is the Euclidean inner product in $\mathbb {R}^3$. The tangent plane at $p$ will be orthogonal to $a$.

To clarify, the surface $S$ lies entirely on one side of the plane $(x-p)\cdot a=0$ which is therefore its tangent plane at $p\in S$.

  • $\begingroup$ Thank you ! Is it possible to take the minimum instead ? Thus you apply anyway Lagrange. $\endgroup$
    – Labo
    Commented Oct 25, 2016 at 11:29
  • $\begingroup$ I am sorry by I don't understand how to fill the gap between the maxi/minimality and the orhogonality property. $\endgroup$
    – Jean Marie
    Commented Oct 25, 2016 at 13:41

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