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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. – TROLLKILLER May 16 '12 at 7:32

The idea is that

  • to glue regular tetrahedra forming a ring


  • 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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If $v$ is $w$ but with the components permuted, there may exist a nonpermutation matrix $A$ such that $v=Aw$. This is the crux of OP's problem in my opinion. I think the key fact is that $O$ is arbitrary... – anon Apr 6 '12 at 5:57
But not simultaneously for all four vertices! – Mariano Suárez-Alvarez Apr 6 '12 at 6:00
How do we know that? – anon Apr 6 '12 at 6:15
Why $O$ is arbitary would give any information about there doesn't exist such matrice? It seems that no matter which o we choose, there wouls also be a chance to exist such matrice – Mathematics Apr 6 '12 at 6:29
The matrix tracks the position of the vertices after successive reflections - it is a different concept from a matrix which acts on basis vectors. If the transformed tetrahedron ends up in the same place as the original one, the vertices will have been permuted, so the matrix which tracks their positions will be a permutation matrix. – Mark Bennet Apr 6 '12 at 10:17

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.

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