Vtyjkyuk

$textbfNotation:$

$bullet$ For $N$ vectors $textbfu_1, ldots, textbfu_N in S^n-1 subset mathbbR^n$, let $alpha_ij$ ($i < j$) denote the angle between $textbfu_i$ and $textbfu_j$

$bullet$ Let $omega_n$ be the surface area (ie. $(n-1)$-dimensional volume) of the sphere $S^n-1$.

$bullet$ Let $sigma$ be the surface measure on $S^n-1$.

$bullet$ Let $I( cdots)$ be the indicator function.

I'm reading an article in which the author is considering the integral
beginalign
frac1omega_n^N cdot int_S^n-1 cdots int_S^n-1 I(d_ij < alpha_ij leqslant d_ij' : 1 leqslant i < j leqslant N ) hspace0.1cm d sigma(textbfu_1) cdots d sigma(textbfu_N).
endalign
By using the spherical symmetry of the integrand, he then proceeds by rewriting (1) as
beginalign
omega_n^N-1 prod_k = 1^N-1 omega_n-k int_phi_12=0^pi cdots int_phi_(N-1)N=0^pi I(d_ij < alpha_ij leqslant d_ij' : 1 leqslant i < j leqslant N) \ nonumber times prod_1 leqslant i < j leqslant N sin^n-i-1 phi_ij hspace0.1cm dphi_(N-1)N cdots d phi_13 d phi_12
endalign
($tfrac12N(N-1)$ integrals in total), where now he only considers unit vectors of the form
beginalign*
textbfu_1 &= (1, 0, ldots, 0) \
textbfu_2 &= (cos phi_12, sin phi_12, 0, ldots, 0) \
textbfu_3 &= (cos phi_13, sin phi_13 cos phi_23, sin phi_13 sin phi_23, 0 ldots, 0) \
&hspace0.2cmvdots \
textbfu_N &= (cos phi_1N, sin phi_1N cos phi_2N, ldots, sin phi_1N cdots sin phi_(N-1)N, 0, ldots, 0).
endalign*
The idea behind the transformation of (1) into (2) is that, by restricting the unit vector $textbfu_1$ to a $1$-dimensional subspace, one has to compensate by adding a factor of $omega_n$ in front of the integral; by restricting the unit vector $textbfu_2$ to a $2$-dimensional subspace, one has to compensate by adding a factor of $omega_n-1$ in front of the integral; and generally, by restricting $textbfu_t$ to a $t$-dimensional subspace, one has to compensate by adding a factor of $omega_n+1-t$ in front of the integral. This makes sense intuitively, but I haven't been able to actually prove it.

Any thoughts on this are very welcome!