I posted this is mathematics SE and it didn't get much attention, so I'm trying here. I apologize if it is not relevant.

Take a set $X \in \mathbb{R}^2$ of nonzero measure $\mu(X) \neq 0$. I am attempting to design a set that has the following symmetries (continuous or discrete)

$1.$ Scale symmetry

$2.$ Rotation symmetry

$3.$ Translation symmetry

It's intuitively clear that only $X = \mathbb{R}^n$ itself simultaneously satisfies those 3 symmetries (which I call the trivial solution).

It seems that when you get one symmetry, you are forced to make the second discrete, and you cannot take a non-trivial third symmetry.

Is there a reference that would help me in my task? Is it impossible in some sense that I should be aware of?

The application I have in mind is recognizing an 'Engineered pattern', that is, the set $X$ within a 2D image.

To search a pattern we need to search the space of rotations (parametrized by $\theta$), translations (parametrized by $\Delta_x, \Delta_y$) and scaling (parametrized by $S$), that is, a cube like $[0,360]\times[0,720]\times[0,1280]\times[1/5,5]$, which is quite extensive to search into due to the high dimensionality.

By making the pattern symmetric I hope I can instead restrict my search to a cube of smaller volume: if the pattern is symmetric w.r.t. parameters $P_1, ..., P_k$, it is enough to search the space of parameters $P_{k+1}, ..., P_n$, which has reduced volume.

In practice the pattern will have some noise superposed and the rest a picture alongside. The method I was aiming for searching is finding maxima of the cross correlation of the pattern with the image.

Here are some sketches of patterns that illustrate the issue of conflicting symmetries:


$(1)\ (2)$

$(3)\ (4)$

$(1)$ Single axis translation and discrete scale

$(2)$ Rotation and discrete scale

$(3)$ Scale and discrete rotation

$(4)$ Two axis discrete translation and discrete rotation

  • 1
    $\begingroup$ The set of symmetries of $X$ (up to null sets) is a closed subgroup (and hence Lie subgroup) of the Lie group $({\bf R}^+ \times O(n)) \ltimes {\bf R}^n$ of conformal affine transformations. Conversely, given such a subgroup, one can look at a partial set of orbits of the group action on ${\bf R}^n$ to obtain a symmetric set with these symmetries. So the question is basically one of classifying the Lie subgroups of the conformal affine group; as far as the continuous symmetries are concerned, this is equivalent to classifying the Lie subalgebras of the conformal affine Lie algebra. $\endgroup$
    – Terry Tao
    May 9 '15 at 22:10
  • $\begingroup$ If you are not already familiar with the basic theory of Lie groups and Lie algebras, I would recommend starting with a textbook on these topics. $\endgroup$
    – Terry Tao
    May 9 '15 at 22:11

It is ok if your set $X$ is closed, and your rotation is different from a half-turn.

The first example in your picture is invariant by the half-turn around the origin, so you need more than just an half-turn.

A non-closed counterexample would be the complement of ${\bf Q}^2$. This set is invariant by a translation by a rational vector, a homothety with rational scaling factor, and some rotation such that the cosine and sine of the angle are rational (e.g. $\arccos(3/5)$), just because ${\bf Q}^2$ is invariant by these transformations.

If you don't want to assume $X$ to be closed, it is still possible to show that $X$ has a complement of zero measure.

Here is a sketch of proof. The set $X$ is invariant by a translation, and also by another translation in different direction, obtained by conjugating the first translation by the rotation. Choose a frame of ${\bf R}^2$ along these two directions, such that the origin belongs to $X$ and is a Lebesgue density point. Restricting our attention to one of the axis of our frame and using the scaling symmetry, it is not difficult to show that there is a dense subset of the axis that belongs to $X$. Same for the other axis. Because the origin is a density point, there is a small rectangle $R$ around the origin in $X$ such that $R\cap X$ spawns a large fraction of $R$. Now scaling and translating back close to the origin, we get an arbitrary large rectangle $R'$ near the origin such that $R'\cap X$ spawns a large fraction of $R'$, which is sufficient to conclude.

  • $\begingroup$ Thanks for your answer. The need for linearly independent directions arises because otherwise we might not be able to pick a Lebesgue density point at the origin, is that right? This seems to settle the question for 3 simultaneous (non-trivial) symmetries! I'll have to look for approximate symmetries then. $\endgroup$
    – Real
    May 9 '15 at 23:44
  • $\begingroup$ Just leaving as a bonus an interesting set I've found with $\mu(x) = 0$ en.wikipedia.org/wiki/Rep-tile#/media/… Thanks again for the fantastic proof! $\endgroup$
    – Real
    May 9 '15 at 23:47

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