Isometry of affine space that conserves the area of triangles

Question

Could you give me some suggestions to prove that an affine transformation of $R^3$ that conserves the area of triangles is an affine isometry? I do not expect you to give me the solution, only a piece of advice. Thanks

Answer 1

This isn't true. The matrix $$\left[\begin{matrix} 2 & 0\\ 0 & 1/2\end{matrix}\right]$$ preserves the area of triangles, but isn't an isometry of $\mathbb{R}^2$.

Answer 2

Suppose $T:\mathbb{R}^3\to \mathbb{R}^3$ is an affine transformation which preserves the areas of triangles.

Our goal is to show that $T$ is an isometry.

By an appropriate translation we can assume $T(0)=0$, hence $T$ is linear.

Since $T$ preserves the areas of triangles, it follows that $T$ is injective, hence $T$ is nonsingular.

Let $S$ denote the standard unit sphere in $R^3$.

Since $T$ is linear and nonsingular, $T(S)$ is an ellipsoid containing the origin in its interior.

To show that $T$ is an isometry, it suffices to show that $|T(v)|=|v|$, for all $v\in\mathbb{R}^3$.

Reducing the task even further, it suffices to show $T(S)=S$.

We'll use the term "planar region" to mean a subset with positive area of a hyperplane in $R^3$.

Since $T$ is linear, $T$ preserves convexity and coplanarity.

It follows that $T$ maps convex planar regions to convex planar regions.

By considering limits of sums of areas of pairwise disjoint small triangular regions, it follows that if $D$ is a convex planar region, then $T(D)$ is a convex planar region with the same area as $D$.

In particular, $T$ maps elliptical regions to elliptical regions with same area.

Choosing an appropriate orthonormal basis, we can assume $T(S)$ is the ellipsoid with equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ where $a,b,c > 0$.

Then the cross-sectional ellipses cut out by the coordinate plane have equations $$ \frac{y^2}{b^2}+\frac{z^2}{c^2}=1,\;x=0\\ \frac{z^2}{c^2}+\frac{x^2}{a^2}=1,\;y=0\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\;z=0 $$ with interior areas $$\pi bc,\;\;\;\pi ca,\;\;\;\pi ab$$ respectively.

These $3$ areas must be equal to the areas of their inverse images under $T$, which are circular cross-sections of $S$, all of which are unit disks, hence
$$bc=ca=ab=1$$ which yields $$a=b=c=1$$ It follows that $T(S)=S$.

This completes the proof.