Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
add comment

It seems the exact formula given by David Bar Moshe in the predecessor to this question,

What is the volume of a \delta-ball in the orthogonal group O(n)? Is there a (simple) lower bound?

is the place to start. (To cut and paste: this formula is

Vol(delta-ball) = 2^(m^2)/(pi^m * m!)* int_ phi_ 1<=phi_ 2 <= . . . <=phi_ m <=2 * arcsin(sqrt(delta/2))*

prod_ 1 < = j < k < = m (cos(phi_ k)-cos(phi_ j)^2 prod_ l sin^2(phi_ l) dphi_ 1 . . . dphi_ k.

when n=2m+1, and something very similar for n=2m (my guess is that any lower bound for the former can basically be achieved for the latter).

I am tempted to start dyadically decomposing based on the approximate size of phi_l and of phi_k-phi_j, but this is likely to lose some exponential factors of n.

An alternative is Fourier decomposition; the formula above does suggest that there is a reasonably nice formula for the integral of exp( i theta r ) where r is the distance to the identity and theta is a parameter. If one gets sufficiently good control on that integral, one can then try Fourier inversion.

share|improve this answer
add comment

The answer is affirmative. The integral can be approximately evaluated to obtain an estimate of f(n) for delta<<1 as follows:

  1. In my previous answer there is a silly mistake in the integral upper bound; there shouldn't be a square root but simply 2 * arcsin(delta/2).

  2. For delta<<1, The integral can be approximated as an integral over the Lie algebra. A simple power counting of the leading terms in the Taylor expansion gives the right exponent for delta.

  3. In the case of N-even, the leading terms in the Taylor expansion are: (phi_ k^2-phi_ j^2)^2. The total power is 4* 1/2(N/2-1)N/2 + N/2 terms from the measure = N(N-1)/2. In the odd case the count is 4 1/2((N-1)/2-1)*(N-1)/2 + 3/2 (N-1) = N(N-1)/2

  4. To obtain an estimate for f(N) in the case of even N in the delta<<1 limit we observe that the integrand's leading taylor term is proportional to the square of a Vandermonde like determinant (with x replaced by x^2). The solution of this kind of integrals is known, to be the determinant of the matrix of the "one particle" integrals, which is in our case:

M_ ij = int_ [0 1] phi^(2i)*phi^(2j) dphi = 1/(2i+2j+1)

  1. The determinant of this kind of matrices is known, see for example C. Kattenthaler's article math\9902004v3 equation (2.7). The result is:

det(M) = (prod_ 1 <=i < j<=N/2(i-j)^2))/(prod_ 1<=i,j<=N/2 (2i+2j+1))

and the final answer:

f(N) = 2^((N/2)^2)/(pi^(N/2) * (N/2)!)* det(M)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.