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I suppose that there are formulas which parametrize all the orthogonal matrices with rational coefficients. Does anyone know anything about it? And what are some publications that discuss this? Thanks.

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The formula in @Carlo Beenakker answer does not give all rational orthogonal matrices, but only those for which $-1$ is not an eigenvalue. Here is a rational parametrization which gives all orthogonal matrices for $n=3$:

$$\frac{1}{a^2+b^2+c^2+d^2}\left( \begin{array}{ccc}a^2+b^2-c^2-d^2 & 2bc-2ad & 2ac+2bd\\ 2ad+2bc & a^2-b^2+c^2-d^2 & 2cd-2ab\\ 2bd-2ac & 2ab+2cd & a^2-b^2-c^2+d^2\end{array}\right),$$ where $a,b,c,d$ are integers with no common factor, and not all equal to $0$. This is due to Euler, and can be proved using quaternions. There is a similar formula for $n=4$.

The answer for $n=4$ is somewhat inconvenient to write here. It was obtained by Euler, Problema algebraicum ob affectiones prorsus singulares memorabile, Novi Commentarii academiae Scientiarum Petropolitanae, XV, (1771) 75-106.

The formula was proved using quaternions in 1937 by D. Grave, and published in an obscure Ukrainian journal (Zhurnal Instituta Matematiki AN UkrSSR, 3, 73-74.) and in his book Treatease on Algebraic Analysis, vol. I Kiev, 1938.

In English, the formula for $n=4$ is reproduced in the book S. Khrushchev, Orthogonal polynomials and continued fractions from Euler's point of view, Cambridge UP, 2008. (Encyclopedia of Math and Appl., vol. 122), on page 297.

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  • $\begingroup$ Where can I find a formula for n=4? $\endgroup$ Jun 14, 2022 at 9:45
  • $\begingroup$ @Bogdan Grechuk: I added references. $\endgroup$ Jun 15, 2022 at 3:31
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you can use the Cayley transform: $O=(I+H)(I-H)^{-1}$, with $I$ the $n\times n$ identity matrix and $H$ an $n\times n$ skew-symmetric matrix; choose rational values for the matrix elements of $H$ and $O$ will be rational orthogonal; see The Generation of All Rational Orthogonal Matrices (1991)

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    $\begingroup$ Notice, though, that this doesn't give everything: If $A$ is orthogonal and $A+I$ is not invertible, then $A$ cannot be written in the form $(I+H)(I-H)^{-1}$ where $H$ is skew-symmetric. Thus, you are missing the orthogonal matrices for which $-1$ is an eigenvalue. $\endgroup$ Mar 22, 2014 at 14:46
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    $\begingroup$ I don't think one can rationally parametrize literally all orthogonal matrices. For one, some have determinant $1$ and some have determinant $-1$, so you'd have to do that separately. For another, even for two-by-two orthogonal matrices you can't rationally parametrize all of them, because they form a circle and any birational isomorphism between the circle and the line always uses the point at $\infty$ of the line for one point of the circle, so you always miss that point. Presumably something similar is true in higher dimensions. $\endgroup$
    – Will Sawin
    Mar 22, 2014 at 16:39
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    $\begingroup$ Once you've parametrized an open set on an algebraic group, cover the group with finitely many left translates, to get everything. Which is what's done in the reference above, where the translates are by diagonal matrices with entries $\pm 1$. Also: Kostant and Michor have a Cayley transform for other semisimple groups. (In the split case you can use the Bruhat decomposition.) $\endgroup$ Mar 22, 2014 at 16:45

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