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The above are part of the articles and some background of it. And there is one claim in the articles, saying that " in order to show that no trivial ring can be formed, it is sufficient to show that no non-trivial sequence of the matrices $R_1,R_2,R_3,R_4$ is a permutation matrices". But why it only consider the permutation matrix? It may be possible that it gives another form of matrix and give the permutation of the original position vector. As we know, either one of the position vector $OA,OB,OC, OD$ can be expressed in a linear combination of another 3 vector.It seems possible to have another form matrices.The authur seem to ignore that case with no reasons provided.

Also,there are some points in the comment which i don't qutie understand. Any explaination is appreciated.

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    The answer is no, they cant.2012-05-16

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The idea is that

  • to glue regular tetrahedra forming a ring

and

  • to make a sequence of reflections on the faces such that at the end you end up in the same tetrahedron you started with

are equivalent things.

Now in the second version, you end up with the same solid, but possibly rotated or, what's the same thing, the vertices permuted.

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    @anon can you give a brief explaination to why if$O$is arbitrary they we can neglect the case.2012-04-09
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If $M$ is a matrix leaving $ABCD$ globally invariant, then if $(a\ b\ c\ d)$ is the first row of $M$ we have $a \cdot OA+b\cdot OB+c\cdot OC+d\cdot OD \in \{OA,OB,OC,OD\}$ for any reference $O$. We must have $a+b+c+d=1$ otherwise this would be true for at most 4 values of $O$. Every point in space is a unique barycentre of $A,B,C,D$, so since the coefficients of the vertices of $ABCD$ are $1,0,0,0$ up to permutation, $a,b,c,d$ is $1,0,0,0$ up to permutation.

This is true of all rows, and since $M$ is invertible no two rows have a 1 in the same column so $M$ is a permutation matrix.