Talk:The Seven Limit Symmetrical Lattices: Difference between revisions

From Xenharmonic Wiki
Jump to navigation Jump to search
Mike Battaglia (talk | contribs)
m 1 revision imported: Moving archived Wikispaces discussion to subpage
Sintel (talk | contribs)
No edit summary
 
Line 1: Line 1:
{{WSArchiveLink}}
{{WSArchiveLink}}
== Some notes ==
Interesting page, actually. I'm trying to work through it bit by bit to make sure I understand it.
The page doesn't really explain what is "symmetric" about these lattices. The original lattice is of course already full of symmetries since it's just the usual cubical lattice.
For the first part, there is some bit about coming up with various norms and then comes to the conclusion that you can make a cuboctahedron out of the 7-odd limit.
I feel like you can explain this quite simply by considering the transformation:
:<math>
A = \begin{bmatrix}
0 & -1 & -1 \\
-1 &  0 & -1 \\
-1 & -1 &  0 \\
\end{bmatrix}
</math>
Computing explicitly:
:<math>
\begin{aligned}
A[1,0,0]^\mathsf{T} &= [0, -1, -1]^\mathsf{T} \\
A[0,1,0]^\mathsf{T} &= [-1, 0, -1]^\mathsf{T} \\
A[0,0,1]^\mathsf{T} &= [-1, -1, 0]^\mathsf{T} \\
A[1,-1,0]^\mathsf{T} &= [1,-1,0]^\mathsf{T} \\
A[1,0,-1]^\mathsf{T} &= [1,0,-1]^\mathsf{T} \\
A[0,1,-1]^\mathsf{T} &= [0,1,-1]^\mathsf{T} \\
\end{aligned}
</math>
Which are the vertices of a cuboctahedron, as we wanted.
– [[User:Sintel|Sintel🎏]] ([[User_talk:Sintel|talk]]) 22:00, 26 April 2025 (UTC)

Latest revision as of 22:00, 26 April 2025

This page also contains archived Wikispaces discussion.

Some notes

Interesting page, actually. I'm trying to work through it bit by bit to make sure I understand it. The page doesn't really explain what is "symmetric" about these lattices. The original lattice is of course already full of symmetries since it's just the usual cubical lattice.

For the first part, there is some bit about coming up with various norms and then comes to the conclusion that you can make a cuboctahedron out of the 7-odd limit. I feel like you can explain this quite simply by considering the transformation:

[math]\displaystyle{ A = \begin{bmatrix} 0 & -1 & -1 \\ -1 & 0 & -1 \\ -1 & -1 & 0 \\ \end{bmatrix} }[/math]

Computing explicitly:

[math]\displaystyle{ \begin{aligned} A[1,0,0]^\mathsf{T} &= [0, -1, -1]^\mathsf{T} \\ A[0,1,0]^\mathsf{T} &= [-1, 0, -1]^\mathsf{T} \\ A[0,0,1]^\mathsf{T} &= [-1, -1, 0]^\mathsf{T} \\ A[1,-1,0]^\mathsf{T} &= [1,-1,0]^\mathsf{T} \\ A[1,0,-1]^\mathsf{T} &= [1,0,-1]^\mathsf{T} \\ A[0,1,-1]^\mathsf{T} &= [0,1,-1]^\mathsf{T} \\ \end{aligned} }[/math]

Which are the vertices of a cuboctahedron, as we wanted.

Sintel🎏 (talk) 22:00, 26 April 2025 (UTC)