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## nt.quantity concept – Diophantine equation in regards to the automorphism group of lattice by constraints

Fixed $ sigma_x = left ( commence {array} {cc} 0 & 1 1 & 0 aim {array} privilege) $ and $ Ok = left ( commence {array} {ccc} 3 & 32 & -64 1 & 32 & -32 -2 & -32 & 64 aim {array} privilege), $

I wish to discover $ n $ in order that, $ K_n = sigma_x oplus sigma_x oplus .. sigma_x oplus Ok $. and the corresponding $ W in GL (2n + 3, mathbb {Z}) $ in order that $ W ^ {T} K_n ^ {- 1} W = K_n ^ {- 1}, $ the one restrictions on $ W $ the privilege nook ought to breathe $ left ( commence {array} {cc} 8 & -5 5 & 13 aim {array} privilege). $ For instance when $ n = 1, $ I cerebrate $ W $ appears to be like love $ W = commence {bmatrix} a_1 & a_2 & a_3 & a_4 & a_5 b_1 & b_2 & b_3 & b_4 & b_5 c_1 & c_2 & c_3 & c_4 & c_5 d_1 & d_2 & d_3 & 8 & – 5 e_1 & e_2 & e_3 & 5 & 13 aim {bmatrix} $ in order that:

commence {equation} W ^ {T} K_1 ^ {- 1} W = K_1 ^ {- 1}. aim {equation}

I’m making an attempt to employ Mathematica and may’t discover any options. Then I strive $ n = 2 $ and I quiet cannot discover a resolution. But I do not know methods to show that there isn’t a resolution for $ W ^ {T} K_n ^ {- 1} W = K_n ^ {- 1}. $ for every $ n $. I cerebrate there ought to breathe $ 1 $.

My observation on this downside:

1 I can not discover mighty literature on fixing the Diophantine equation in integer quadratic figure (at the least for $ 4 occasions $ 4). But my fantasy is that when $ n $ grows, there ought to breathe extra liberty and extra variables. So you need to breathe capable of finding an answer, or at the least show that the answer exists.

2 pages $ K_n oplus sigma_x $ is an indeterminate matrix, so there isn’t a computational system to seek out this automorphism group of $ K_n $. See associated query right here. An indeterminate matrix results in an innumerable automorphism group, so the capability of the automorphism group ought to breathe sizable sufficient to accommodate such $ W $.

3 Notice that Smith’s regular figure for $ = left ( commence {array} {ccc} 1 & 0 & 0 0 & 32 & 0 0 & 0 & 32 aim {array} privilege). $ So for everybody $ K_n $ it all the time has two nontrivial elements $ 32 $. So there’s a limitation for $ 2 occasions $ 2 Corner for $ W $.

My analysis has an issue associated to this kind of given query $ Ok $ and $ W $ (limit $ W $ for the scowl privilege nook and pretense that yow will discover sufficiently big $ Ok $ (extend $ Ok $ and $ W $) with the identical and nontrivial Smith regular ($ neq 1 $) train and answer $ W ^ {T} K_1 ^ {- 1} W = K_1 ^ {- 1}. $ )

Comments and outcomes are very welcome, thanks very mighty.

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