16b. More calculus

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In order to discuss the higher spheres, we will use spherical coordinates:

Theorem

We have spherical coordinates in [math]N[/math] dimensions,

[[math]] \begin{cases} x_1\!\!\!&=\ r\cos t_1\\ x_2\!\!\!&=\ r\sin t_1\cos t_2\\ \vdots\\ x_{N-1}\!\!\!&=\ r\sin t_1\sin t_2\ldots\sin t_{N-2}\cos t_{N-1}\\ x_N\!\!\!&=\ r\sin t_1\sin t_2\ldots\sin t_{N-2}\sin t_{N-1} \end{cases} [[/math]]

the corresponding Jacobian being given by the following formula:

[[math]] J(r,t)=r^{N-1}\sin^{N-2}t_1\sin^{N-3}t_2\,\ldots\,\sin^2t_{N-3}\sin t_{N-2} [[/math]]

Show Proof

The fact that we have spherical coordinates as above is clear. Regarding now the Jacobian, by developing the determinant over the last column, we have:

[[math]] \begin{eqnarray*} J_N &=&r\sin t_1\ldots\sin t_{N-2}\sin t_{N-1}\times \sin t_{N-1}J_{N-1}\\ &+&r\sin t_1\ldots \sin t_{N-2}\cos t_{N-1}\times\cos t_{N-1}J_{N-1}\\ &=&r\sin t_1\ldots\sin t_{N-2}(\sin^2 t_{N-1}+\cos^2 t_{N-1})J_{N-1}\\ &=&r\sin t_1\ldots\sin t_{N-2}J_{N-1} \end{eqnarray*} [[/math]]

Thus, we obtain the formula in the statement, by recurrence.

■

With the above results in hand, we can now compute arbitrary polynomial integrals, over the spheres of arbitrary dimension, the result being is as follows:

Theorem

The spherical integral of [math]x_{i_1}\ldots x_{i_k}[/math] vanishes, unless each index [math]a\in\{1,\ldots,N\}[/math] appears an even number of times in the sequence [math]i_1,\ldots,i_k[/math]. We have

[[math]] \int_{S^{N-1}_\mathbb R}x_{i_1}\ldots x_{i_k}\,dx=\frac{(N-1)!!l_1!!\ldots l_N!!}{(N+\Sigma l_i-1)!!} [[/math]]

with [math]l_a[/math] being this number of occurrences.

Show Proof

First, the result holds indeed at [math]N=2[/math], due to the following formula proved above, where [math]\varepsilon(p)=1[/math] when [math]p\in\mathbb N[/math] is even, and [math]\varepsilon(p)=0[/math] when [math]p[/math] is odd:

[[math]] \int_0^{\pi/2}\cos^pt\sin^qt\,dt=\left(\frac{\pi}{2}\right)^{\varepsilon(p)\varepsilon(q)}\frac{p!!q!!}{(p+q+1)!!} [[/math]]

In general, we can assume [math]l_a\in 2\mathbb N[/math], since the other integrals vanish. The integral in the statement can be written in spherical coordinates, as follows:

[[math]] I=\frac{2^N}{V}\int_0^{\pi/2}\ldots\int_0^{\pi/2}x_1^{l_1}\ldots x_N^{l_N}J\,dt_1\ldots dt_{N-1} [[/math]]

In this formula [math]V[/math] is the volume of the sphere, [math]J[/math] is the Jacobian, and the [math]2^N[/math] factor comes from the restriction to the [math]1/2^N[/math] part of the sphere where all the coordinates are positive. The normalization constant in front of the integral is:

[[math]] \frac{2^N}{V} =\frac{2^N}{N\pi^{N/2}}\cdot\Gamma\left(\frac{N}{2}+1\right) =\left(\frac{2}{\pi}\right)^{[N/2]}(N-1)!! [[/math]]

As for the unnormalized integral, this is given by:

[[math]] \begin{eqnarray*} I'=\int_0^{\pi/2}\ldots\int_0^{\pi/2} &&(\cos t_1)^{l_1} (\sin t_1\cos t_2)^{l_2}\\ &&\vdots\\ &&(\sin t_1\sin t_2\ldots\sin t_{N-2}\cos t_{N-1})^{l_{N-1}}\\ &&(\sin t_1\sin t_2\ldots\sin t_{N-2}\sin t_{N-1})^{l_N}\\ &&\sin^{N-2}t_1\sin^{N-3}t_2\ldots\sin^2t_{N-3}\sin t_{N-2}\\ &&dt_1\ldots dt_{N-1} \end{eqnarray*} [[/math]]

By rearranging the terms, we obtain:

[[math]] \begin{eqnarray*} I' &=&\int_0^{\pi/2}\cos^{l_1}t_1\sin^{l_2+\ldots+l_N+N-2}t_1\,dt_1\\ &&\int_0^{\pi/2}\cos^{l_2}t_2\sin^{l_3+\ldots+l_N+N-3}t_2\,dt_2\\ &&\vdots\\ &&\int_0^{\pi/2}\cos^{l_{N-2}}t_{N-2}\sin^{l_{N-1}+l_N+1}t_{N-2}\,dt_{N-2}\\ &&\int_0^{\pi/2}\cos^{l_{N-1}}t_{N-1}\sin^{l_N}t_{N-1}\,dt_{N-1} \end{eqnarray*} [[/math]]

Now by using the above-mentioned formula at [math]N=2[/math], this gives:

[[math]] \begin{eqnarray*} I' &=&\frac{l_1!!(l_2+\ldots+l_N+N-2)!!}{(l_1+\ldots+l_N+N-1)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(N-2)}\\ &&\frac{l_2!!(l_3+\ldots+l_N+N-3)!!}{(l_2+\ldots+l_N+N-2)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(N-3)}\\ &&\vdots\\ &&\frac{l_{N-2}!!(l_{N-1}+l_N+1)!!}{(l_{N-2}+l_{N-1}+l_N+2)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(1)}\\ &&\frac{l_{N-1}!!l_N!!}{(l_{N-1}+l_N+1)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(0)} \end{eqnarray*} [[/math]]

Now observe that the various double factorials multiply up to quantity in the statement, modulo a [math](N-1)!![/math] factor, and that the [math]\frac{\pi}{2}[/math] factors multiply up to:

[[math]] F=\left(\frac{\pi}{2}\right)^{[N/2]} [[/math]]

Thus by multiplying with the normalization constant, we obtain the result.

■

In connection now with our probabilistic questions, we have:

Theorem

The even moments of the hyperspherical variables are

[[math]] \int_{S^{N-1}_\mathbb R}x_i^kdx=\frac{(N-1)!!k!!}{(N+k-1)!!} [[/math]]

and the variables [math]y_i=x_i/\sqrt{N}[/math] become normal and independent with [math]N\to\infty[/math].

Show Proof

The moment formula in the statement follows from Theorem 16.3. Now observe that with [math]N\to\infty[/math] we have the following estimate:

[[math]] \begin{eqnarray*} \int_{S^{N-1}_\mathbb R}x_i^kdx &=&\frac{(N-1)!!}{(N+k-1)!!}\times k!!\\ &\simeq&N^{k/2}\times k!!\\ &=&N^{k/2}M_k(g_1) \end{eqnarray*} [[/math]]

Thus we have, as claimed, the following asymptotic formula:

[[math]] x_i/\sqrt{N}\sim g_1 [[/math]]

Finally, the independence assertion follows as well from the formula in Theorem 16.3, via some standard probability theory.

■

In the case of the half-classical sphere, we have the following result:

Theorem

The half-classical integral of [math]x_{i_1}\ldots x_{i_k}[/math] vanishes, unless each index [math]a[/math] appears the same number of times at odd and even positions in [math]i_1,\ldots,i_k[/math]. We have

[[math]] \int_{S^{N-1}_{\mathbb R,*}}x_{i_1}\ldots x_{i_k}\,dx=4^{\sum l_i}\frac{(2N-1)!l_1!\ldots l_n!}{(2N+\sum l_i-1)!} [[/math]]

where [math]l_a[/math] denotes this number of common occurrences.

Show Proof

As before, we can assume that [math]k[/math] is even, [math]k=2l[/math]. The corresponding integral can be viewed as an integral over [math]S^{N-1}_\mathbb C[/math], as follows:

[[math]] I=\int_{S^{N-1}_\mathbb C}z_{i_1}\bar{z}_{i_2}\ldots z_{i_{2l-1}}\bar{z}_{i_{2l}}\,dz [[/math]]

In order to get started, and prove the first assertion, let us apply to this integral transformations of the following type, with [math]|\lambda|=1[/math]:

[[math]] p\to\lambda p [[/math]]

We conclude from this that the above integral [math]I[/math] vanishes, unless each [math]z_a[/math] appears as many times as [math]\bar{z}_a[/math] does, and this gives the first assertion.

Assume now that we are in the non-vanishing case. Then the [math]l_a[/math] copies of [math]z_a[/math] and the [math]l_a[/math] copies of [math]\bar{z}_a[/math] produce by multiplication a factor [math]|z_a|^{2l_a}[/math], so we have:

[[math]] I=\int_{S^{N-1}_\mathbb C}|z_1|^{2l_1}\ldots|z_N|^{2l_N}\,dz [[/math]]

Now by using the standard identification [math]S^{N-1}_\mathbb C\simeq S^{2N-1}_\mathbb R[/math], we obtain:

[[math]] \begin{eqnarray*} I&=&\int_{S^{2N-1}_\mathbb R}(x_1^2+y_1^2)^{l_1}\ldots(x_N^2+y_N^2)^{l_N}\,d(x,y)\\ &=&\sum_{r_1\ldots r_N}\binom{l_1}{r_1}\ldots\binom{l_N}{r_N}\int_{S^{2N-1}_\mathbb R}x_1^{2l_1-2r_1}y_1^{2r_1}\ldots x_N^{2l_N-2r_N}y_N^{2r_N}\,d(x,y) \end{eqnarray*} [[/math]]

By using the formula in Theorem 16.3, we obtain:

[[math]] \begin{eqnarray*} &&I\\ &=&\sum_{r_1\ldots r_N}\binom{l_1}{r_1}\ldots\binom{l_N}{r_N}\frac{(2N-1)!!(2r_1)!!\ldots(2r_N)!!(2l_1-2r_1)!!\ldots (2l_N-2r_N)!!}{(2N+2\sum l_i-1)!!}\\ &=&\sum_{r_1\ldots r_N}\binom{l_1}{r_1}\ldots\binom{l_N}{r_N} \frac{(2N-1)!(2r_1)!\ldots (2r_N)!(2l_1-2r_1)!\ldots (2l_N-2r_N)!}{(2N+\sum l_i-1)!r_1!\ldots r_N!(l_1-r_1)!\ldots (l_N-r_N)!} \end{eqnarray*} [[/math]]

We can rewrite the sum on the right in the following way:

[[math]] \begin{eqnarray*} &&I\\ &=&\sum_{r_1\ldots r_N}\frac{l_1!\ldots l_N!(2N-1)!(2r_1)!\ldots (2r_N)!(2l_1-2r_1)!\ldots (2l_N-2r_N)!}{(2N+\sum l_i-1)!(r_1!\ldots r_N!(l_1-r_1)!\ldots (l_N-r_N)!)^2}\\ &=&\sum_{r_1}\binom{2r_1}{r_1}\binom{2l_1-2r_1}{l_1-r_1}\ldots\sum_{r_N}\binom{2r_N}{r_N}\binom{2l_N-2r_N}{l_N-r_N}\frac{(2N-1)!l_1!\ldots l_N!}{(2N+\sum l_i-1)!} \end{eqnarray*} [[/math]]

The point now is that the sums on the right can be computed, by using the following well-known formula, whose proof is elementary:

[[math]] \sum_r\binom{2r}{r}\binom{2l-2r}{l-r}=4^l [[/math]]

Thus the sums on the right in the last formula of [math]I[/math] equal respectively [math]4^{l_1},\ldots,4^{l_N}[/math], and this gives the formula in the statement.

■

As before, we can deduce from this a probabilistic result, as follows:

Theorem

The even moments of the half-classical hyperspherical variables are

[[math]] \int_{S^{N-1}_{\mathbb R,*}}x_i^kdx=4^k\frac{(2N-1)!k!}{(2N+k-1)!} [[/math]]

and the variables [math]y_i=x_i/(4N)[/math] become symmetrized Rayleigh with [math]N\to\infty[/math].

Show Proof

The moment formula in the statement follows from Theorem 16.5. Now observe that with [math]N\to\infty[/math] we have the following estimate:

[[math]] \begin{eqnarray*} \int_{S^{N-1}_{\mathbb R,*}}x_i^kdx &=&4^k\times\frac{(N-1)!}{(N+k-1)!}\times k!\\ &\simeq&4^k\times N^k\times k!\\ &=&(4N)^kM_k(|c|) \end{eqnarray*} [[/math]]

Here [math]c[/math] is a standard complex Gaussian variable, and this gives the result.

■

As a comment here, it is possible to prove, based once again on the integration formula from Theorem 16.5, that the rescaled variables [math]y_i=x_i/(4N)[/math] become “half-independent” with [math]N\to\infty[/math]. For a discussion about half-independence, we refer to .

General references

Banica, Teo (2024). "Affine noncommutative geometry". arXiv:2012.10973 [math.QA].