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This question is more in recreational mathematics area

By accident I came across the concept of "hyperbolic rotation" where we use a matrix containing $\cosh$ and $\sinh$ instead of the trigonometric $\cos$ and $\sin$ for the rotation by matrix-multiplication such that we have: $$ \tag{trigonometric} \qquad T_t(\varphi) = \begin{bmatrix} \cos \varphi & \sin \varphi \\ -\sin \varphi & \cos \varphi \end{bmatrix} $$
$$ \tag{hyperbolic} \qquad T_h(\varphi) = \begin{bmatrix} \cosh \varphi & \sinh \varphi \\ \sinh \varphi & \cosh \varphi \end{bmatrix} $$
A key-feature is surely, that both rotation-matrices have a determinant of $1$ ( because $$\small \cos^2(\varphi) + \sin^2(\varphi) = 1 \tag{trigonometric} $$ and $$\small \cosh^2(\varphi) - \sinh^2(\varphi) = 1 \tag{hyperbolic} $$ ) .

Now I toyed a bit to extend this to the case of 3-multisection series of the exponential; an example for what I mean is my older question; let's call that three functions just $f(x),g(x),h(x)$ such that $f(x)+g(x)+h(x) = \exp(x)$ and the analogon to the square-formulae which equal $1$ is: $$ f(x)^3 + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1 \tag{3-multisection} $$

At least, such a "rotation"-matrix must have size of $3 \times 3$ but possibly even more - if it is constructable at all.

Qu1: Is such a generalization to higher multisections (here order 3) possible?
Qu2: and if, how could such a "rotation"- matrix be contructed?

[Update]: I've just found, that $$ \tag{3-multisection} \qquad T_{3m}(\varphi) = \begin{bmatrix} f(\varphi) & h(\varphi) & g(\varphi)\\ g(\varphi)&f(\varphi)&h(\varphi)\\ h(\varphi) & g(\varphi) &f(\varphi) \end{bmatrix}$$ has determinant $1$ and could be a candidate model. But I didn't find nice properties so far. Perhaps it is even better to not to stick to the determinant 1-condition, but allow determinant $-1$ here; the "rotation"-matrix could then be a simple circulant one, like in the hyperbolic case.

The log of that matrix looks like $$ \log(T_{3m}(\varphi))=\small \begin{bmatrix} 0 & 0 & \varphi \\ \varphi & 0 & 0 \\ 0 & \varphi & 0 \end{bmatrix}$$ and I think it is a good hint, that this is equivalent to the rotation-matrices in the trigonometric/hyperbolic-cases, where the form of the matrix-log comes out to be much similar.

Community
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Gottfried Helms
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2 Answers2

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The analogy I would suggest comes from quadratic forms. Your first two examples are are linear changes of variable that preserve a quadratic form. That is, a matrix $P$ such that $P^T A P = A.$ In the first one, $A= I,$ in the second $A$ is the diagonal matrix with diagonal entries $(1,-1.)$ This is very convenient for homogeneous degree two because of matrices; but we could write $$ u = a x + b y, v = c x + d y, u^2 + v^2 = x^2 + y^2 ?? $$ and ask about possible $a,b,c,d.$ Same for $u^2 - v^2 = x^2 - y^2.$

So that suggests $$ u = ax+by+cz, v=dx+ey+fz, w=gx+hy+iz, $$ what about $$ u^3 + v^3 + w^3 - 3 u v w = x^3 + y^3 + z^3 - 3 x y z ?? $$ and ask about possible $a,b,c,d,e,f,g,h,i.$ These linear variable changes make a group over the reals. Hmmm. They should, but I need to think about why the determinant of the evident 3 by 3 matrix is nonzero; obvious for the quadratic form case.

EDIT: got it, in the version I am pushing. Given cubic form $$ f(x,y,z) = x^3 + y^3 + z^3 - 3 x yz. $$ $$ f(x,y,z) = \det \left( \begin{array}{ccc} x & y & z \\ z & x & y \\ y & z & x \end{array} \right) . $$

$f(x,y,z) = \det(xI + y A + z A^2),$ where $$ A = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) $$ Then $A^3 = I$ and $A^4 = A .$

Now, since both the characteristic and minimal polynomials for $A$ are $x^3 - 1,$ it follows that all matrices that commute with $A$ are of the form $rI + s A + t A^2.$

Will Jagy
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  • Ah, wow! I'll try this just now... – Gottfried Helms Oct 30 '13 at 18:23
  • I've played with a version giving $\det()=-1$ and the second version giving $ \det()=1$. The second version has a nice matrix-logarithm which has a comparable form to the matrix-logarithms of the trigonometric and the hyperbolic rotations... (the matrix-logarithm is possible via diagonalization) – Gottfried Helms Oct 30 '13 at 18:39
  • @GottfriedHelms, sounds good. – Will Jagy Oct 30 '13 at 18:53
  • Now if we define a matrix in that circulant manner $T_{3m}(x) = \begin{bmatrix} f(x) & h(x) & g(x) \ g(x) & f(x) & h(x) \ h(x) & g(x) & f(x) \end{bmatrix}$ then we have the nice property $T_{3m}^k(x) = T_{3m}(kx )$ like in the trigonometric/hyperbolic case – Gottfried Helms Oct 30 '13 at 19:15
  • @GottfriedHelms, seems it is working for you. I got some information about this type of thing at http://math.stackexchange.com/questions/329936/primes-represented-integrally-by-a-homogeneous-cubic-form and then http://math.stackexchange.com/questions/336191/numbers-represented-by-a-cubic-form where I was mostly interested in integer values for everything. – Will Jagy Oct 30 '13 at 19:32
  • Yes, very well. Thanks Will! – Gottfried Helms Oct 30 '13 at 21:39
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Isn't this just an element of the Lorenz group from Special Relativity? For Example: $$Q(\beta) = \left[ \begin{matrix} \cosh(\beta) & 0 & 0 & \sinh(\beta) \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \sinh(\beta) & 0 & 0 & \cosh(\beta) \end{matrix} \right]$$

Alan
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  • I'm sorry, I didn't see your multisection question. – Alan Oct 30 '13 at 17:46
  • Well, for the hyperbolic question I've already found much interesting and explanative pages which relate this to Lorentz-transformation and special relativity. That triggered me to think, whether I could introduce some of my old lego-toys $f(),g(),h()$ ... – Gottfried Helms Oct 30 '13 at 17:54