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Sometimes a point set in Euclidean space may have a shadow with an unexpected symmetry. The purpose here is to ask when this happens or when it doesn't happen (in some generality). This requires a few definitions.

First, suppose $V$ is a vector space and $\Omega\subset V$ is a subset. Define a "shadow" of $\Omega$ as the image $\pi(\Omega)$, where $\pi:V\rightarrow W$ is a linear transformation. In this context, I am thinking of $\pi$ as an orthogonal projection, where $W$ is a subspace of $V$, although it is not required to make this identification.

Next, suppose $V$ is a vector space and $\Omega\subset V$ is a subset. Define a "permutation symmetry" of $\Omega$ as a permutation of $\Omega$ which extends to a linear operator on $V$. For example, suppose $V=\mathbb{R}^d$ and $\Omega\subset V$ has $n$ points. Then one may regard $\Omega$ as a $d\times n$ matrix and a permutation $\sigma\in\mathrm{Perm}(\Omega)$ is a permutation symmetry if there is a $d\times d$ matrix $M$ such that $M\Omega=\Omega P_\sigma$, where $P_\sigma$ is the permutation matrix corresponding to $\sigma$.

Next, suppose $V$ is a vector space and $\Omega\subset V$ is a subset. Call a linear transformation $\pi:V\rightarrow W$ "generic" with respect to $\Omega$ if the restriction of $\pi$ to $\Omega$ is a bijection. For simplicity, regard $\Omega$ as finite so that a projection $\pi$ is generic if $|\pi(\Omega)| = | \Omega|$.

Write $\Omega_1\mapsto\Omega_2$ if $\Omega_1\subset V_1$, $\Omega_2\subset V_2$ and there is a linear transformation $\pi:V_1\rightarrow V_2$ which is generic with respect to $\Omega_1$ and such that $\pi(\Omega_1)=\Omega_2$. If $\Omega_1\mapsto\Omega_2$, then one has a natural way to identify permutation symmetries of $\Omega_2$ with permutations of $\Omega_1$. However, such a permutation may not be a permutation symmetry of $\Omega_1$. Thus, if $\Omega_1\mapsto\Omega_2$, then call a permutation symmetry $\sigma$ of $\Omega_2$ "inherited" if it is also a permutation symmetry of $\Omega_1$.

Here is an example of a non-inherited permutation symmetry of a shadow. Define a point configuration by $$\Omega_1=\left[\begin{matrix} 1 & 1 & -1 & -1 & 0 & 0 & 0 & 0 \\ 1 & -1 & 1 & -1 & 0 & 0 & 0 & 0 \\ 1 & -1 & -1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 & -1 & 1 \end{matrix}\right].$$ Then $\Omega_1$ has a shadow $$\Omega_2=\left[\begin{matrix} 1 & 1 & -1 & -1 & 0 & 0 & 0 & 0 \\ 1 & -1 & 1 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 1 & -1 & 1 & -1 \\ \end{matrix}\right].$$ The latter is obtained from the former by deleting rows 3 and 6, and one may see that this projection is generic. Notice that $\Omega_1$ consists of the vertices of two regular tetrahedra in complementary subspaces in $\mathbb{R}^6$ and $\Omega_2$ is the set of vertices of a 4-dimensional cross polytope. One may check that there are permutation symmetries of $\Omega_2$ which do not extend to permutation symmetries of $\Omega_1$.

Here are some general questions: Which objects have generic shadows with non-inherited permutation symmetries? The set $\Omega_1$ above provides an example of this. Which objects have the property that every permutation symmetry of every generic shadow is inherited? The vertex set of the simplex, for example, has this property since the permutation symmetry group of this set is the entire symmetric group. What other families of point configurations have this property? I am most interested in point sets possessing a transitive permutation symmetry group.

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Related (but related only) is this MO question asking whether the most symmetric 3D polyhedra can be obtained as projections of the most symmetric 4D polytopes (definitely not "unexpected"): mathoverflow.net/questions/45503/… –  Joseph O'Rourke Dec 5 '10 at 22:16
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Your current definition of $\Omega_1 \mapsto \Omega_2$ just as a binary relation (“there exists a linear map such that…”) isn’t quite right, I think: there could be multiple linear maps sending $\Omega_1$ to $\Omega_2$, and your later definitions (inherited symmetries etc.) depend heavily on which map is under consideration. So surely the definitions need to involve choosing a specific linear map? (Mistakes of this sort have a long pedigree: there is one in Hartshorne, for instance :-P The good news is that they’re generally easily fixed…) –  Peter LeFanu Lumsdaine Dec 6 '10 at 0:43
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Peter: After you prompted me to look at it more carefully, I made a correction to my explanation of the notation $\Omega_1\mapsto\Omega_2$. Thanks! (I still don't understand the mistake you are talking about, however. Can you tell me how to correct it?) –  David Richter Dec 6 '10 at 1:38
    
@David: You say “If $\Omega_1 \mapsto \Omega_2$ then one has a natural way to identify [their] permutation symmetries…” I don’t think you have, without specifying which linear map you have in mind witnessing $\Omega_1 \mapsto \Omega_2$. For instance, take $\Omega_1$ to be for example the standard basis $\{e_1,e_2,e_3\}$ in $\mathbb{R}^3$; take $\Omega_2$ to be the cube roots of unity $\omega,\omega^2,1$ in $\mathbb{C}$. Certainly we have $\Omega_1 \mapsto \Omega_2$. Take the symmetry of $\Omega_1$ switching $e_1$, $e_2$. What symmetry does it correspond to on $\Omega_2$? –  Peter LeFanu Lumsdaine Dec 6 '10 at 4:34
    
(cont’d) The answer has to depend on which of $e_1,e_2,e_3$ you were sending to which roots of unity! So I would correct the definition by: instead of just defining a relation “$\Omega_1 \mapsto \Omega_2$”, define something like “$f : \Omega_1 \mapsto \Omega_2$”, and then give the later definitions in terms of $f$ as well as $\Omega_i$. (If I’m not misunderstanding something, there is also a second issue with the current “inheritance” definition, which Gerhard brings up in the comments to his answer.) –  Peter LeFanu Lumsdaine Dec 6 '10 at 4:40
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If you consider ways of going backwards, i.e. finding a point set that realizes a given shadow, I expect you will find many ways of getting non-inherited symmetries.

Fascinated as I am by constellations, I sometimes remember that Orion's belt is not three stars almost in a row, but three stars that look to me like they are in a row. Similarly, take any 2-D symmetric configuration of points, and throw some of them straight up into 3-D at different heights, you will get a configuration of points which has few if any symmetries. About the only thing that you can depend on is that each such collection of symmetries for a point set will be a group. I expect many other possible things to be realized.

If you place certain restrictions on the original point set, you may get a theorem. The only thing I see is that the full symmetric group contains the group of the shadow which in turn contains the group of the original point set.

Gerhard "Ask Me About System Design" Paseman, 2010.12.05

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My brain has suddenly kicked into (maybe out of) gear. What if we project an equilateral triangle onto a line? Do we not lose some symmetries that way? I am left wondering if the group inclusions written above are correct. Gerhard "Interruption Due to Technical Difficulties" Paseman, 2010.12.05 –  Gerhard Paseman Dec 5 '10 at 22:45
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@Gerhard: I would agree: I don’t think one gets a map of group in either direction. I guess what one can say is: given a map $f:V \to W$ sending $\Omega_1$ to $\Omega_2$, we can say a symmetry $\sigma$ of $(V,\Omega_1)$ (“upstairs”) is $f$-related to a symmetry $\tau$ downstairs if $f \cdot \sigma = \tau \cdot f$. Then a symmetry $\tau$ downstairs is inherited if there is a symmetry $\sigma$ upstairs that is $f$-related to it? I guess something like this fits the intent of the original definition? –  Peter LeFanu Lumsdaine Dec 6 '10 at 0:47
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