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Let $k$ be a field whose characteristic is zero and let $n\geq 1$. Say that a matrix $M\in {\cal M}_{n\times n}(k)$ is almost orthogonal if $M^{T}M$ is a nonzero multiple of the identity. Denote the set of those matrices by $AO_n(k)$.

When $n=2$, there is a nice parametric description of $AO_2(k)$ :

$$ AO_2(k)=\Bigg\lbrace \bigg(\begin{matrix} a & b \\ -b&a \end{matrix}\bigg) \Bigg| (a,b) \in k^2, (a,b)\neq (0,0) \Bigg\rbrace $$

Are there similar exhaustive formulas with polynomial entries for $n>2$ ?

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I think you expect more than just: $\Bigg\lbrace \bigg(\begin{matrix}1&0&0\\0& a & b \\0& -b&a \end{matrix}\bigg) \Bigg| (a,b) \in k^2 \Bigg\rbrace $. What does $k^2$ mean? Your field squared? –  draks ... Dec 6 '12 at 11:20
@draks: $k^2$ is standard notation for $k\times k$. Your set clearly doesn't contain all of $AO_3(k)$. –  joriki Dec 6 '12 at 11:26
@joriki thanks and I know. It was just meant as an example... –  draks ... Dec 6 '12 at 11:30
@Ewan: This doesn't quite work out for $n=2$. If you exclude the zero multiple, you'd have to exclude $(a,b)=(0,0)$. If you don't, then $AO_2(\mathbb F_2)$ also contains e.g. $\pmatrix{1&0\\1&0}$. –  joriki Dec 6 '12 at 11:39
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up vote 2 down vote accepted

Let's talk about $SO(n)$ instead. You do this as well implicitely, since the determinant of the matrix you have written is $a^2 + b^2 \geq 0$, so that this obviously can't parametrize even $O(2)$ which also has a component with negative determinant. The full parametrization would be two copies of $AO_2(k)$ glued along the surface of zero determinant (if you allow the multiple of the identity to be zero in your definition).

Now, this is purely a consequence of low dimension. $SO(2)$ is a circle and if you allow $M^TM$ be a multiple of identity then you also allow each matrix from $SO(2)$ to be multiplied by any number, so that you obtain a line bundle over $SO(2)$ surface and this is more or less just polar parametrization of the plane that the $SO(2)$ lies in. In higher dimensions, you similarly obtain a line bundle over $SO(n)$. But since $SO(n)$, $n>2$ is a much more complicated surface (as a subspace of the space of all matrices) than a simple circle, you can't expect anything nice here for big $n$.

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Also note the troubles regarding the joriki's comment above. So you probably want the multiple of the identity to be nonzero to exclude the zero determinant surface. –  Marek Dec 6 '12 at 12:01
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