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Consider the gyroid surface: $$ \mathcal{G} = \{(x, y, z) \in \Bbb{R}^3 \ | \ f(x, y, z) = 0 \} $$ where $$ f(x, y, z) = \sin(x+y) + \sin(x-y) + \sin(y+z) + \sin(y-z) + \sin(x + z) + \sin(x-z). $$ Then $\mathcal{G}$ divides $\Bbb{R}^3$ in two connected regions, given by $\{f < 0\}$ and $\{f > 0\}$.

My question is:

What is a rigid motion that interchanges the two regions above?

After trying some translations by $\pi$, which did not work, I came to believe such a rigid motion is given by some rotation, but I am unable to find out what exactly.

Thanks in advance.

Danilo Gregorin Afonso
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    One has first to be familiarized with gyroid surfaces (triply periodic surfaces). Ask "images" after Gogle search with these keywords. Begin for example by the artist view here – Jean Marie Mar 21 '20 at 13:54
  • Dear Jean, thanks for the comment. This question is the last part of an exercise about triply periodic surfaces, namely the primitive, the diamond and the gyroid. I have found translations that keep these surfaces invariant and rigid motions that interchange the regions defined by the primitive and diamond surfaces. Looking at images is not helping very much :( – Danilo Gregorin Afonso Mar 21 '20 at 14:06
  • ... btw where it is said that your surface (they take the equivalent equation $\cos(x)\sin(y)+\cos(y)\sin(z)+\cos(z)\sin(x)=0$) is an approximation of a minimal surface. – Jean Marie Mar 21 '20 at 14:07
  • I did see this approximation, but at first glance it seemed to me that I would get the same problem with the formula I used. Let me take a second thought. Thanks a lot – Danilo Gregorin Afonso Mar 21 '20 at 14:09
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    My comment was directed at readers who may not have the faintest idea of these surfaces. Looking at images can be fruitful for gaining ideas. Afterwards, it's computations... – Jean Marie Mar 21 '20 at 14:11
  • In your formula, are you sure of your 2 last terms ? Isn't it $\sin(z+x) + \sin(z-x)$ instead of $\sin(x + z) + \sin(x-z)$ (which would like normal by circular permutation) : it doesn't change the first term but the second. – Jean Marie Mar 21 '20 at 14:21
  • In the book, it is as I wrote. Could be a mistake, however – Danilo Gregorin Afonso Mar 21 '20 at 14:24
  • ... in this way, your formula could be exactly the same as $\cos(x)\sin(y)+\cos(y)\sin(z)+\cos(z)\sin(x)=0$ up to a coordinates change. – Jean Marie Mar 21 '20 at 14:24

1 Answers1

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A direction of research, not a full answer.

Your issue is to find in a first step the invariance group of this surface (group of rigid motions preserving it). Those exchanging the two sides are order-2 elements ($T^2=Id$). Caution: the reciprocal isn't necessarily true.

Let us take the more compact equation :

$$\cos(x)\sin(y)+\cos(y)\sin(z)+\cos(z)\sin(x)=0\tag{1}$$

Remark : we don't take into account translations

$$x \to x+2k_x\pi, y \to y+2k_y\pi, z \to z+2k_y\pi,$$

because they shouldn't exchange interior and exterior (although it has to be verified...)

It suffices to consider symmetries.

  • central symmetry $(x,y,z)\mapsto (-x,-y,-z)$

(I don't consider here for example symmetries with respect to a line such as $(x,y,z)\mapsto (-x,-y,z)$ which can be considered as $\pi$ rotation around axis this line, knowing that, in general, rotations will be treated as compositions of symmetries with respect to planes).

  • symmetries with respect to a plane, having the following generic matrix :

$$\begin{pmatrix}x\\ y\\ z \end{pmatrix} = S_N \begin{pmatrix}X\\ Y\\ Z\end{pmatrix} \ \ \text{ with} \ \ S_N:=I_3-2N^TN = \begin{pmatrix}(1-2a^2)&-2ab&-2ac\\ -2ab&(1-2b^2)&-2bc\\ -2ac&-2bc&(1-2c^2)\end{pmatrix} \tag{2}$$

(where $N=(a,b,c)^T$ is a unit normal vector to the plane of symmetry ; we can check that trace($S_N)= 3-2(a^2+b^2+c^2)=1$ which is the sum of the eigenvalues $1,1,-1$ of $S_N$).

Then plug expressions (2) into (1) obtaining an expression that has to be identical to the initial expression, whatever the values of variables $(x,y,z)$ (or $X,Y,Z$).

  • as said upwards, rotations can be obtained by composition of plane symmetries.

Connected : this.

Jean Marie
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  • When you say that plugging (2) into (1) we obtain an expresion that has to be identical to the initial expression, in fact it will interchange the positive and negative values of $f(x, y, z) = \cos x \sin y + \cos y \sin z + \cos z \sin x$, is it correct? – Danilo Gregorin Afonso Mar 21 '20 at 16:50
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    Not necessarily because, at this step, we built "the large image" : the set of transforms preserving the boundary. Think for example to identity transform (or a $2 \pi$ translation in a certain direction). Only in a second step, we look, among them, which ones exchange interior and exterior. – Jean Marie Mar 21 '20 at 17:03
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    I just corrected the matrix of $S_N$ on its diagonal. Sorry for this error. – Jean Marie Mar 26 '20 at 12:29
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    Have you seen the astounding images here ? – Jean Marie Mar 26 '20 at 12:33
  • Incredible, indeed. Thank you very much for your help. – Danilo Gregorin Afonso Mar 26 '20 at 17:31