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By Givens' rotation $R(1,2;\theta)$ I mean a matrix which has the $$\begin{pmatrix} \hphantom{-}\cos \theta &\sin \theta \cr -\sin \theta & \cos \theta \end{pmatrix}$$ $2 \times 2$ block at the first and second rows and columns and a diagonal matrix for the remaining dimension. Similarly define $R(i,j;\theta)$ to be the matrix which is a rotation along the $i \wedge j$ 2-plane, $i < j$. Define the following map $$ \phi_{I,J}: T^t \to SO(n); \phi(\theta_1,\ldots, \theta_t) = R(i_1,j_1;\theta_1) \ldots R(i_t,j_t;\theta_t)$$.

My question is given a sequence of pairs $(I,J)=(i_1,j_1),(i_2,j_2),\ldots, (i_t,j_t)$, is there a way to determine whether every element in $SO(n)$ can be expressed as a product of the form $$ \phi_{I,J}(\theta_1,\ldots,\theta_t)$$ for some $\theta_1, \ldots, \theta_t$?

In fact one sufficient condition for a product of such sequence to contain an open set is obtained simply by looking at a neighborhood of the identity and taking derivatives. But does the local surjectivity imply global surjectivity? If not, then how big of a neighborhood of the identity does the image of $\phi$ cover?

For people who are comfortable with probability: imagine the $\theta_j$'s to be iid uniform $[0,2\pi)$ random variables. If the image of $\phi_{I,J}$ does contain an open set of $SO(n)$, what does the singular set of the density of the resulting measure on $SO(n)$ look like?

Answer to any of the above question will be considered a correct answer to this entry. Helps are deeply appreciated!

edit note: thanks to Robert Bryant for clarifying about the local condition.

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I edited the question so that the matrix would render as intended. –  José Figueroa-O'Farrill Oct 14 '11 at 21:14
You write "In fact it's not too hard to determine whether a product of such sequence contains an open set, simply by looking at a neighborhood of the identity and taking derivatives." I assume that you are aware that it's not enough to check that the map is open at $\theta_i=0$. For example, when $n=3$ and $(I,J) = (1,2),(1,3),(1,2)$, the map $T$ covers an open set, but its differential at $\theta_i=0$ is not surjective. –  Robert Bryant Oct 14 '11 at 22:14
Oh no I was not aware of that. But it makes sense now that you bring it up. Thanks! –  John Jiang Oct 14 '11 at 22:30
I suspected that you might have been laboring under this misunderstanding. In fact, in the example I gave, the map $T$ is surjective, but the local behavior of the map is kind of strange at some places. –  Robert Bryant Oct 14 '11 at 22:48
You were indeed right. Can you elaborate on the local strangeness? Even better if you can characterize the strangeness for general such maps, or give a plausible reference. –  John Jiang Oct 14 '11 at 23:14

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