0
$\begingroup$

I'm not really a math person, and apologize if the question here is too simple. I've ended up with the following type of question for a few Lie groups, but state it for SO(n).

I start with a subgroup G of SO(n) generated by rotations in $k$ fixed 2-dimensional planes $P_{1}, \ldots, P_{k}$. If I fix a permutation $\sigma \in S_{k}$, can I generate any element of $G$ that is 'sufficiently close' to the identity by multiplying rotations in the planes $P_{\sigma(1)}, \ldots, P_{\sigma(k)}$ in that order?

I know that this is nonsense for discrete groups. I don't care too much about the metric as long as it is 'reasonable' (i.e. any of the $L^{p}$ norms from being in Euclidean space, or Hilbert-Schmidt, are fine; 0-1 metric not so much). If the above factorization is possible, I'd also be interested in knowing if it ever requires going 'very far' from the origin in order to hit elements that are 'very close' to the origin.

$\endgroup$
1
  • $\begingroup$ By rotation in a plane, do you mean you are thinking of a copy of SO(2) in SO(n)? $\endgroup$ Commented Apr 11, 2011 at 16:12

1 Answer 1

2
$\begingroup$

This is false for G=SO(3). This can be generated by rotations in the two planes orthogonal to the x axis and the y axis, but not every element close to the identity can be written as a product of a rotation in the first plane and a rotation in the second plane.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .