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An $n\times n$ orthogonal real matrix $A$ is a set ${A_{ij}}$ of $n^2$ real numbers that satisfy the constraints:

  1. $$\sum_k A_{ik} A_{kj} = \delta_{ij} $$

for all $1\leq i,j\leq n$. The equations (1.) represent

$$ n + \binom{n}{2} = \frac{n(n+1)}{2} $$

constraints on the $n^2$ components of $A$. Therefore an orthogonal matrix $A$ has

$$ n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2} $$

degrees of freedom. Is there a simple representation for these $n(n-1)/2$ parameters? In other words, is there a coordinate system where the constraints (1.) take a simple form? For example, in two dimensions, one can use polar coordinates:

$$ x = r \cos \phi $$ $$ y = r \sin \phi $$

Then the constraint of orthonormality of two vectors $(x_1,y_1)$, $(x_2,y_2)$ takes the simple form: $r_1=r_2=1$, $\phi_1 = \phi_2 \pm \pi/2 $, and a real orthogonal $2\times 2$ matrix has the general parametric form:

$$A = \pm\left(\begin{array}{cc} \cos\phi & \sin\phi\\ -\sin\phi & \cos\phi \end{array}\right)$$

Is there a relatively simple parametric representation like this for higher dimensions?

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up vote 3 down vote accepted

The cleanest representation is this: let $S$ be an $n$ by $n$ skew symmetric matrix, that is $S^T = -S.$ Then $$ e^S $$ is an orthogonal matrix with positive determinant, which is therefore $1.$ In order to get an orthogonal matrix with negative determinant, you may negate the first row.

There is a standard proof for the necessary properties, which can sometimes more elaborate. For this one, you just need to know that $$ e^{\left( A^T \right)} = \left( e^A \right)^T $$ and $$ e^{\left( -A \right)} = \left( e^A \right)^{-1} $$

Surjectivity: there are elementary treatments for $n=2$ and for $n=3,$ showing that all special orthogonal matrices can be written in such a way. The discussion for all $n$ begins to use material that is roughly a first graduate course in Lie groups, Lie algebras, and the exponential map. Since the orthogonal group is one of the earliest compact Lie groups found, I imagine there is an elementary treatment somewhere, maybe someone's online notes. The theorem is that the exponential map is surjective on a compact connected Lie group, in this case the special orthogonal group. See Proposition 1 at TAO BLOG Note that these are sometimes called matrix groups, also this particular example is one of the classical groups. Tao's discussion uses material from a second graduate differential geometry course, early Riemannian manifolds. Again, if all you want is this one fact, surjectivity, there ought to be treatments that prove everything or almost everything using the one parameter groups $$ \gamma(t) = e^{tS} $$

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+1 That is a really nice representation. But how can I prove that $e^S$ (with $S = -S^T$) goes through all orthogonal matrices with determinant $=1$? – becko May 8 '13 at 13:38
@becko, Proposition 1 at… – Will Jagy May 8 '13 at 20:12

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