The group $SO(N)$ has $m=N(N-1)/2$ generators and one thus needs $m$ angles to parametrize it.

In the case of $SO(2)$ and $SO(3)$, one can also parametrize using unit vectors $\vec q$ in $\mathbb R^{m+1}$, i.e. $\vec q=(q_1,...q_{m+1})$ with $\vec q.\vec q=1$. Then the $m$ angles parametrizes the corresponding coordinates of $\vec q$, and one gets the corresponding orthogonal matrix using some specific mapping $O=\Phi(\vec q)$.

Furthermore, still for $N=2$ and $N=3$, one can show by direct calculation that ${\rm Tr}(O(\vec q_1)O^T(\vec q_2))$ is invariant under rotations of $\vec q_1$ and $\vec q_2$, that is ${\rm Tr}(O(\vec q_1)O^T(\vec q_2))={\rm Tr}(O(\vec q_1')O^T(\vec q_2'))$, with $\vec q_i'=Q \vec q_i$ and $Q \in O(m+1)$. For $SO(2)$ ($SO(3)$) matrices, with thus have an invariance of this trace under $O(2)$ ($O(4)$) rotations of the unit vectors parametrizing them.

My questions are : Is this parametrization with unit vectors true for all $N$ ? And if so, and more importantly to me, is this invariance under $O(m+1)$ of the trace also always true ?

Finally, is there a deeper meaning to this invariance ?

EDIT: It seems that my conjecture is not true. For $SO(4)$, the parametrization is in term of two unit 4-vectors $\vec q\in \mathbb R^4$ and $\vec p\in \mathbb R^4$, $\vec q^2=\vec p^2=1$ (see Wikipedia).

Furthermore, it is not too hard using this parametrization to show that for $O_1$ and $O_2$ in SO(4), $$ \rm{ Tr}(O_1^T O_2)=\rm{ Tr}(O(\vec q_1,\vec p_1)O(\vec q_2,\vec p_2))=f(\vec q_1.\vec q_2,\vec p_1.\vec p_2), $$ is invariant under $O(4)\times O(4)$ corresponding to independent rotations of the $\vec q$'s and $\vec p$'s.

Even though it is not as simple as I thought, there is nevertheless a symmetry of these traces, associated to the parametrization in terms of unit vector.

Does any one knows more about this ?


Ingeneral, $SO(n)$ has topology of a "twisted product" of spheres $S^{n-1} \ltimes S^{n-2} \ltimes \ldots \ltimes S^1$, obtained inductively by observing that for the defining action of $SO(n)$ on $\mathbb{R}^n$ orbit of any vector is $S^{n-1}$ and the stabilizer is $SO(n-1)$. The case $n=2$ gives a sphere trivially, and in the case $n=3$ we get a fibration which is double covered by the Hopf fibration. This means that $SO(3)$ is $RP^2$, and the "parametrization" by $S^3$ is the universal cover map $ S^3 \mapsto RP^2$. When $n=4$ we get $SO(4)=S^{3} \ltimes RP^2$, from which the double cover of $SO(4)$ by $S^{3} \times S^{3}$ comes. In general, you have a "tautological" parametrization by the orthonormal frame to which the standard frame is rotated - that is, by the columns of the rotation matrix. Of course then $Tr(O_1^T O_2)$ is the sum of dot products of the corresponding columns, $\sum_j q_{j,1} \cdot q_{j,2}$.

Apart from these ''small n coincidences", $SO(n)$ can not be covered by a sphere, since it has "wrong topology" (for example, homotopy groups of $SO(n)$ are subject to Bott periodicity, while homotopy groups of spheres are famously complicated. I think that also means that $SO(n)$ can not be covered by a fixed number of spheres, but I have not thought about the details of this.

Of course one can parametrize almost all of any compact m-manifold by $ \mathbb{R}^m$, and hence by part of $S^m$ or some such, but that will not have the type of symmetry you seem to be looking for.

  • $\begingroup$ Thanks for the answer. I am not sure how for example the double cover of $SO(4)$ insures that $Tr(O^T_1 O_2)$ will be invariant under $O(4)\times O(4)$ (although of course it has the smell of it...). Note that this is not equivalent to your tautological parametrization (which would involve 4 vectors, whereas the symmetry I mention involves only 2). $\endgroup$ – Adam Jun 9 '18 at 9:22
  • $\begingroup$ The last vector in tautological parametrization is redundant, so 3, not 4. And the $SO(3)=\mathbb{R}P^2$ makes the last 2 into one, so you get 2 instead of 3. $\endgroup$ – Max Jun 9 '18 at 13:39
  • $\begingroup$ I'm not sure I understand what you mean... For sure because I don't know enough about projective spaces and their relation to the orthogonal groups. Also, I don't understand why this way to go from 4 to 2 vectors insure that the symmetry associated with the trace is O(4)xO(4). $\endgroup$ – Adam Jun 9 '18 at 16:16
  • $\begingroup$ My comment was a bit off: what it says is that you could get a "two vector" parametrization from the tautological one, but the problem is that I can't see how to make that very natural (and that's why the symmetry is not visible). $\endgroup$ – Max Jun 11 '18 at 10:12

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