Is there a lower bound on the volume of a delta-ball in the orthogonal group O(n) of the type
f(n) \* delta^{n(n-1)/2}? For which f(n)? How can it be proven?


n(n-1)/2 is the number of degrees of freedom in the orthogonal group.


The volume in the orthogonal group is measured by the Haar measure, which is the up to scaling unique measure that is invariant under the group operation. I consider the usual metric that is induced by the spectral norm |M| = max |Mx| where x ranges over all vectors of length 1 and the vector norm is the Euclidean one. A delta-ball is the set of all orthogonal matrices that have distance less or equal delta to a fixed matrix M. Because of the invariance of the Haar measure, for a fixed delta, all delta-balls have the same volume.



Lower bound on the volume of a delta-ball in the orthogonal group O(n) of the type f(n)*delta^{n(n-1)/2}
Well, there is certainly a bound of that form. Here is a crude proof:


There is an exponential map from skew-symmetric matrices to O(n). This is differentiable with nonsingular Jacobian near the origin. The preimage of the delta ball, if I understand your notations, is the log(1+delta) ball. So the volume of this preimage is c(n)\*log(1+delta)^{n(n-1)/2} where c(n) is the volume of the unit ball. Now just multiply by some lower bound for the Jacobian near 0.


By the way, do you want to insist on using the sup norm? A lot of these formulas are nicer for the L^2 norm. For example, I think the preimage in the skew symmetric matrices is an ordinary ball, so its volume would be well known.


