3
$\begingroup$

There is an attractive theorem that says that if two plane figures are directly similar, then so is any convex combination of them. Below, $P_1$ and $P_2$ are directly similar polygons: they have the same angles in the same sequence; They are (positively) scaled, rotated, translated versions of one another. The convex combination illustrated $P_{12} = \frac{1}{3} P_1 + \frac{2}{3} P_2$ is also directly similar.
   DirectlySimilar
(Let me henceforth abbreviate "direcly similar" with "similar.") My question is:

Q1. Is there a natural extension to $\mathbb{R}^3$ and to higher dimensions?

Any extension cannot be an exact extension—some aspect has to give. For example, here are two cubes $P_1$ and $P_2$, the latter a $\frac{3}{4}$-scale rotated version of $P_1$. Shown is $P_{12} = \frac{1}{2} P_1 + \frac{1}{2} P_2$, and it is clearly not a cube. I would define similarity in $\mathbb{R}^3$ to require all faces to be similar with the same scale factor as well preserving all dihedral angles—in other words, $B$ is similar to $A$ if $B$ is a rotated, scaled, translated version of $A$.
   SimilarNotCube
But perhaps this is true?

Q2. Is the convex combination of two cubes (scaled, rotated, translated) always a parallelopiped? A parallelotope in $\mathbb{R}^d$?

Another approach would be to restrict the transformations:

Q3. Is there some condition on the transformations applied to the shapes that permits the similarity conclusion? Or: What is the widest class of transformations that leads to the similarity conclusion?

Certainly if $P_2$ is just a translated, scaled copy of $P_1$, then any convex combination is similar, in any dimension. Rotations are the culprit. But perhaps some rotations still lead to similarity.

It seems likely this has all been well-explored. If so, thanks for pointers!

$\endgroup$

1 Answer 1

6
$\begingroup$

First of all, we may forget about translations, since they just shift the convex combinations. Thus we may regard the direct similarity as a linear transform given by an orthogonal matrix $A$ multiplied by some constant $\alpha$. Then the transform mapping the otiginal polytope to the convex combination is given by $$ B=\lambda I+(1-\lambda)\alpha A. $$ This is also a linear transform, so the image of a parallelotope is also a parallelotope (if the transform is non-degenerate; notice that it may be degenerate only if $\lambda=\pm(1-\lambda)\alpha$ since the complex eigenvalues of $A$ have unital absolute values; in this case $A$ should have eigenvalue $\pm 1$, and this may happen even in the planar case). This answers Q2.

As for Q1 (and Q3), we need to check whether $B$ is an orthogonal matrix multiplied by a real number $c$ (provided that the polytope is solid). This is the case iff all eigenvalues of $B$ have the same absolute value, which reduces to the following two options:

1) All eigenvalues of $A$ coincide (thus $A$ is scalar, and the transform is a scaling); or

2) All the eigenvalues of $A$ are $e^{\pm i\theta}$ for some fixed $\theta$ (surely these two eigenvalues coinciding multiplicities). Thus $A$ acts as a composition of rotations in $d/2$ mutually orthogonal planes. This case cannot happen in odd-dimensional space (which answers Q3), but it sometimes happens in all even-dimensional spaces.

The phenomenon of the planar case is that for the plane there are no more options.

$\endgroup$
2
  • $\begingroup$ If object $P_1$ lies in a $k$-flat, and $P_2$ is obtained by rotation about an axis orthogonal to that $k$-flat (and then scaled), then I believe the similarity holds for convex combinations. For example, polygons in parallel planes in $\mathbb{R}^3$. $\endgroup$ Jan 9, 2014 at 1:59
  • 1
    $\begingroup$ @Joseph: that's why I speak on the case when the polytope is solid (i.e. contains an interior point). But it seems also that you need not the rotation in this axis, but again some $k/2$ rotations at the same angle in orthogonal 2-flats. $\endgroup$ Jan 9, 2014 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.