April 24

MathWorld gives formulae for the volumes of these polyhedra (and annoyingly don't define what they mean by the volume of a self-intersecting polyhedra). I tried to rederive them, using the density method (the one where the centre of a regular pentagram counts twice towards its total area). The small stellated dodecahedron seems rather easy (the central dodecahedron has density 3, because the outside of it is the centre of a pentagram and hence has density 2: the 12 pyramidal peaks have density 1), but it's not quite clear to me how to divide the other three polyhedra up into regions by density. Double sharp (talk) 15:45, 24 April 2014 (UTC)[reply]

I was helping a friend's son to go through some question sheets, revising their first year mathematics-for-physics course. One of the questions to get them to practice basic calculation with matrices involved the matrix

${\displaystyle \mathbf {M} =\left({\begin{matrix}0&1&\cos \theta \\-1&0&\sin \theta \\\cos \theta &\sin \theta &0\\\end{matrix}}\right)}$

They were asked to confirm that M³ = 0, and then that if

${\displaystyle \mathbf {M} _{x}\;{\overset {\underset {\mathrm {def} }{}}{=}}\;\mathbf {I} +x\mathbf {M} +{\tfrac {1}{2}}x^{2}\mathbf {M} ^{2}}$

that

${\displaystyle \mathbf {M} _{x+y}=\mathbf {M} _{x}\mathbf {M} _{y}}$

All straightforward enough to confirm. But I wondered if this was a system I ought to be recognising?

The second half, although without using the name, is essentially working through the exponential-of-sums property of matrix exponentials for matrices that commute, in the special case that the power series terminates at the second term.

The first half is setting this up by introducing a matrix that successively maps the orthogonal vectors

${\displaystyle {\tfrac {1}{2}}\left({\begin{matrix}-\sin \theta \\\cos \theta \\1\end{matrix}}\right)\to \left({\begin{matrix}\cos \theta \\\sin \theta \\0\end{matrix}}\right)\to \left({\begin{matrix}\sin \theta \\-\cos \theta \\1\end{matrix}}\right)\to \left({\begin{matrix}0\\0\\0\end{matrix}}\right)}$

-- so the matrix is essentially annihilating one direction at each application, while rotating the rest of the space into the firing line. A little more involved than ${\displaystyle \mathbf {M} ^{\prime }=\left({\begin{smallmatrix}0&1&0\\0&0&1\\0&0&0\\\end{smallmatrix}}\right)}$, but not much.

So the question(s) that occurred to me were: is this a matrix whose properties I should recognise; and/or associate with anything? Has it come from some particular physical model? Was there any original significance to the θ parameter?

The matrix exponentiation suggests that M might represent a commutator underlying a Lie algebra. In fact the set-up looks not too dissimilar to that discussed in the article Heisenberg group if in that article one sets b=a and c=a²/2 (give or take some rotation, and the non-zero singular values being ${\displaystyle {\sqrt {2}}}$ rather than 1).

But I wondered if anyone had come across this matrix and its exponentials, and whether their form or parametrisation is a standard one to address any particular topic? (And if so then what is the significance of it?) Jheald (talk) 22:49, 24 April 2014 (UTC)[reply]

This is a nilpotent matrix. Nilpotent matrices are similar to block diagonal matrices with blocks as shift matrices. So this may be a similarity transform of such a matrix. Matrix exponentials form a finite sum for such matrices. If every matrix in a Lie algebra is nilpotent, then that forms a nilpotent Lie algebra. Nilpotent matrices are also related to nilpotent orbits in Lie groups and algebras. --Mark viking (talk) 23:35, 24 April 2014 (UTC)[reply]

Thanks, useful links. It is indeed a similarity transformation of a shift matrix,

${\displaystyle \mathbf {M} =\mathbf {P} \mathbf {S} \mathbf {P} ^{-1};\mathbf {S} =\left({\begin{matrix}0&{\sqrt {2}}&0\\0&0&{\sqrt {2}}\\0&0&0\\\end{matrix}}\right);\mathbf {P} =\left({\begin{matrix}{\frac {1}{\sqrt {2}}}\sin \theta &\cos \theta &-{\frac {1}{\sqrt {2}}}\sin \theta \\-{\frac {1}{\sqrt {2}}}\cos \theta &\sin \theta &{\frac {1}{\sqrt {2}}}\cos \theta \\{\frac {1}{\sqrt {2}}}&0&{\frac {1}{\sqrt {2}}}\end{matrix}}\right)}$

But I'm still curious as to whether this particular matrix comes from anywhere. Jheald (talk) 08:24, 25 April 2014 (UTC)[reply]