Let $V=\{v_1,\cdots v_k\}$ a set of non zero different norm one vectors in $R^d$, $k>2$. I am trying to demonstrate that if $Q=\{P_i\in R^{k\times k},i=1,\cdots,k\}$ is a set of permutations such that: \begin{eqnarray} &&\sum_{q=1}^{k}P_q=J\\ && P_i\circ P_j =\mathbb{0}, i\neq j\\ && P_iP_j=P_l\;\;\forall\;i,j\;\;\exists\;l \end{eqnarray} ($J\in R^{k\times k}$ the all-ones matrix, $\circ$ the Hadamard product) and $$ v^T_iP_q v_j=v_{l,q}\;\;\forall\;i,j\;\;\exists\;l\\ $$ ($v_{l,q}$ the $q^{th}$ component of the $v_l$ vector) then the $V$ is a collection of vectors with one only one component different from zero equal to one. Is it true? Counterexamples? Thanks!
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$\begingroup$ Where do your conditions come from? $\endgroup$– Igor RivinCommented Sep 21, 2016 at 8:26
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$\begingroup$ Hi, originally I was wondering if there exists a set of (equivalent) Latin squares that are closed under matrix multiplication. The conditions on the permutations come from the latin square constraints. The last one from the latin squares matrix multiplication clousure condition. Cheers. $\endgroup$– FabioCommented Sep 21, 2016 at 8:43
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$\begingroup$ So, the third condition says that your permutations form a group, and the second says that they are all fixed-point-free, right? $\endgroup$– Igor RivinCommented Sep 21, 2016 at 8:45
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$\begingroup$ That would seem to imply that the basic case is when $k=d,$ in which case your permutation group is generated by a $d$-cycle. I assume that your conjecture is easily checked in that case. $\endgroup$– Igor RivinCommented Sep 21, 2016 at 8:53
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$\begingroup$ Yes, the case I am more interested in $k=d$. Sorry why is it easy to check in the case the permutation is generated by a $d-$cycle? Thanks! $\endgroup$– FabioCommented Sep 21, 2016 at 8:56
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