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

http://mathoverflow.net/questions/2084/what-is-the-volume-of-a-delta-ball-in-the-orthogonal-group-on-is-there-a-sim

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.

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It seems the exact formula given in the predecessor to this question,

http://mathoverflow.net/questions/2084/what-is-the-volume-of-a-delta-ball-in-the-orthogonal-group-on-is-there-a-sim

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.