# Isometry of surfaces in $\mathbb{R}^3$

Let $F$ be an isometry of the Euclidean space $\mathbb{R}^3$.

Hence $F$ is orthogonoal transform followed by translation by a constant vector.

Let M be a surface of $\mathbb{R}^3$ that is connected, not in any plane.

How to prove that if $F(m)=m$ for all $m \in M$, then $F$ is the identity map?

• Could you tell us please what you tried so far? – Sergio Parreiras Dec 15 '13 at 9:16
• Well, I tried to find out the role of M being nonplanar, but cannot do it. – noot Dec 15 '13 at 9:28
• If $M$ were contained in a plane, the mapping could be reflection through that plane. – Harald Hanche-Olsen Dec 15 '13 at 9:30

You know that $F(x)=Ux+b$, where $U$ is a $3\times 3$ orthogonal matrix and $b\in\mathbb R^3$.

What is the set $S$ of fixed points of such an $f$? ($S=\{x:f(x)=x\}$). We have $$Ux+b=x \quad\Longleftrightarrow\quad (U-I)x=-b.$$

Case I. $U-I$ invertible. Then $S$ has a single element.

Case II. $\mathrm{rank}(U-I)=2$, then $S$ is a straight line or the empty set.

Case III. $\mathrm{rank}(U-I)=1$, then $S$ is a plane or a subset of it.

Case IV. $\mathrm{rank}(U-I)=0$, then $S=\mathrm R^3$, if $b=0$ or $S=\varnothing$ if $b\ne 0$.

Therefore, only Case IV works, and for $b=0$.

You know that $F$ and $I$ are both affine maps. So, to prove they are equal it is enough to prove that they coincide in $n+1=4$ points which are affinely-independent: i.e. such that the smallest affine space containing them is the whole space.

addendum You known that the smallest affine space containing $M$ is the whole space (otherwise it would be contained in an affine $2$-dimensional space), so it must contain at least $4$ affinely independent points.

• Which makes one wonder where the connectedness comes into the picture. Perhaps the problem proposer threw it in just to confuse the issue. – Harald Hanche-Olsen Dec 15 '13 at 9:32
• Could anyone elaborate on the proof and the use of the conditions such as nonplanar? – noot Dec 15 '13 at 9:35