guide:9ed2cae800

[math] \newcommand{\mathds}{\mathbb}[/math]

This article was automatically generated from a tex file and may contain conversion errors. If permitted, you may login and edit this article to improve the conversion.

We have kept the best for this final chapter. Calculus, more calculus, and even more calculus, in relation with the integration over spheres. Our motivations are varied:

(1) First of all, calculus is a good thing, and calculus over spheres, using spherical coordinates, is even better. Mathematicians usually snub spherical coordinates, deemed “unconceptual”, but physicists just love them. Want to do some electrodynamics? Spherical coordinates. Want to solve the hydrogen atom? Spherical coordinates, too. So, following the physicists, we will love these spherical coordinates too, in this chapter. And let me recommend here again the delightful books of Griffiths , .

(2) Second, the spheres themselves are a very good thing too, be that in the context of the Connes noncommutative geometry , , , or in the context of our noncommutative geometry, following and related papers, and as explained so far in this book, or in the context of any other kind of noncommutative geometry theory. Also, in our setting, everything more advanced, as for instance of analysis over free manifolds type, like the work in , , , starts of course with a study in the sphere case.

(3) Finally, all this, calculus over spheres, will naturally lead us into all sorts of advanced considerations. At the core of all this will be a tough computation from , as well as a subtle twisting result from , relating the free orthogonal/unitary projective quantum group [math]PO_N^+=PU_N^+[/math] to the quantum permutation group [math]S_{N^2}^+[/math]. And with this being virtually related to pretty much everything, mathematics and physics alike, including , , , , , , , , , , , , .

As a starting point, we have the very natural question, first investigated in , of computing the laws of individual coordinates of the main 3 real spheres, namely:

[[math]] S^{N-1}_\mathbb R\subset S^{N-1}_{\mathbb R,*}\subset S^{N-1}_{\mathbb R,+} [[/math]]

We already know from chapter 5 the [math]N\to\infty[/math] behavior of these laws, called “hyperspherical”. To be more precise, for [math]S^{N-1}_\mathbb R[/math] we obtain the normal law, and for [math]S^{N-1}_{\mathbb R,+}[/math] we obtain the semicircle law. As for the sphere [math]S^{N-1}_{\mathbb R,*}[/math], this has the same projective version as [math]S^{N-1}_\mathbb C[/math], where the corresponding law becomes complex Gaussian with [math]N\to\infty[/math], as explained in chapter 5, and so we obtain a symmetrized Rayleigh variable. See .

The problem that we want to investigate is that of computing these hyperspherical laws at fixed values of [math]N\in\mathbb N[/math]. Let us begin with a discussion in the classical case. At [math]N=2[/math] the sphere is the unit circle [math]\mathbb T[/math], with [math]z=e^{it}[/math] the coordinates are [math]x=\cos t[/math], [math]y=\sin t[/math], and the integrals of the products of such coordinates can be computed as follows:

Theorem

We have the following formula,

[[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]]

where [math]\varepsilon(p)=1[/math] if [math]p[/math] is even, and [math]\varepsilon(p)=0[/math] if [math]p[/math] is odd, and where

[[math]] m!!=(m-1)(m-3)(m-5)\ldots [[/math]]

with the product ending at [math]2[/math] if [math]m[/math] is odd, and ending at [math]1[/math] if [math]m[/math] is even.

Show Proof

This is standard calculus, with particular cases of this formula being very familiar to everyone loving and teaching calculus, as we all should. Let us set:

[[math]] I_p=\int_0^{\pi/2}\cos^pt\,dt [[/math]]

We compute [math]I_p[/math] by partial integration. We have the following formula:

[[math]] \begin{eqnarray*} (\cos^pt\sin t)' &=&p\cos^{p-1}t(-\sin t)\sin t+\cos^pt\cos t\\ &=&p\cos^{p+1}t-p\cos^{p-1}t+\cos^{p+1}t\\ &=&(p+1)\cos^{p+1}t-p\cos^{p-1}t \end{eqnarray*} [[/math]]

By integrating between [math]0[/math] and [math]\pi/2[/math], we obtain the following formula:

[[math]] (p+1)I_{p+1}=pI_{p-1} [[/math]]

Thus we can compute [math]I_p[/math] by recurrence, and we obtain:

[[math]] \begin{eqnarray*} I_p &=&\frac{p-1}{p}\,I_{p-2}\\ &=&\frac{p-1}{p}\cdot\frac{p-3}{p-2}\,I_{p-4}\\ &=&\frac{p-1}{p}\cdot\frac{p-3}{p-2}\cdot\frac{p-5}{p-4}\,I_{p-6}\\ &&\vdots\\ &=&\frac{p!!}{(p+1)!!}\,I_{1-\varepsilon(p)} \end{eqnarray*} [[/math]]

Together with [math]I_0=\frac{\pi}{2}[/math] and [math]I_1=1[/math], which are both clear, we obtain:

[[math]] I_p=\left(\frac{\pi}{2}\right)^{\varepsilon(p)}\frac{p!!}{(p+1)!!} [[/math]]

Summarizing, we have proved the following formula, with one equality coming from the above computation, and with the other equality coming from this, via [math]t=\frac{\pi}{2}-s[/math]:

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

In relation with the formula in the statement, we are therefore done with the case [math]p=0[/math] or [math]q=0[/math]. Let us investigate now the general case. We must compute:

[[math]] I_{pq}=\int_0^{\pi/2}\cos^pt\sin^qt\,dt [[/math]]

In order to do the partial integration, observe that we have:

[[math]] \begin{eqnarray*} (\cos^pt\sin^qt)' &=&p\cos^{p-1}t(-\sin t)\sin^qt\\ &+&\cos^pt\cdot q\sin^{q-1}t\cos t\\ &=&-p\cos^{p-1}t\sin^{q+1}t+q\cos^{p+1}t\sin^{q-1}t \end{eqnarray*} [[/math]]

By integrating between [math]0[/math] and [math]\pi/2[/math], we obtain, for [math]p,q \gt 0[/math]:

[[math]] pI_{p-1,q+1}=qI_{p+1,q-1} [[/math]]

Thus, we can compute [math]I_{pq}[/math] by recurrence. When [math]q[/math] is even we have:

[[math]] \begin{eqnarray*} I_{pq} &=&\frac{q-1}{p+1}\,I_{p+2,q-2}\\ &=&\frac{q-1}{p+1}\cdot\frac{q-3}{p+3}\,I_{p+4,q-4}\\ &=&\frac{q-1}{p+1}\cdot\frac{q-3}{p+3}\cdot\frac{q-5}{p+5}\,I_{p+6,q-6}\\ &=&\vdots\\ &=&\frac{p!!q!!}{(p+q)!!}\,I_{p+q} \end{eqnarray*} [[/math]]

But the last term was already computed above, and we obtain the result:

[[math]] \begin{eqnarray*} I_{pq} &=&\frac{p!!q!!}{(p+q)!!}\,I_{p+q}\\ &=&\frac{p!!q!!}{(p+q)!!}\left(\frac{\pi}{2}\right)^{\varepsilon(p+q)}\frac{(p+q)!!}{(p+q+1)!!}\\ &=&\left(\frac{\pi}{2}\right)^{\varepsilon(p)\varepsilon(q)}\frac{p!!q!!}{(p+q+1)!!} \end{eqnarray*} [[/math]]

Observe that this gives the result for [math]p[/math] even as well, by symmetry. Indeed, we have [math]I_{pq}=I_{qp}[/math], by using the following change of variables:

[[math]] t=\frac{\pi}{2}-s [[/math]]

In the remaining case now, where both [math]p,q[/math] are odd, we can use once again the formula [math]pI_{p-1,q+1}=qI_{p+1,q-1}[/math] established above, and the recurrence goes as follows:

[[math]] \begin{eqnarray*} I_{pq} &=&\frac{q-1}{p+1}\,I_{p+2,q-2}\\ &=&\frac{q-1}{p+1}\cdot\frac{q-3}{p+3}\,I_{p+4,q-4}\\ &=&\frac{q-1}{p+1}\cdot\frac{q-3}{p+3}\cdot\frac{q-5}{p+5}\,I_{p+6,q-6}\\ &=&\vdots\\ &=&\frac{p!!q!!}{(p+q-1)!!}\,I_{p+q-1,1} \end{eqnarray*} [[/math]]

In order to compute the last term, observe that we have:

[[math]] \begin{eqnarray*} I_{p1} &=&\int_0^{\pi/2}\cos^pt\sin t\,dt\\ &=&-\frac{1}{p+1}\int_0^{\pi/2}(\cos^{p+1}t)'\,dt\\ &=&\frac{1}{p+1} \end{eqnarray*} [[/math]]

Thus, we can finish our computation in the case [math]p,q[/math] odd, as follows:

[[math]] \begin{eqnarray*} I_{pq} &=&\frac{p!!q!!}{(p+q-1)!!}\,I_{p+q-1,1}\\ &=&\frac{p!!q!!}{(p+q-1)!!}\cdot\frac{1}{p+q}\\ &=&\frac{p!!q!!}{(p+q+1)!!} \end{eqnarray*} [[/math]]

Thus, we obtain the formula in the statement, the exponent of [math]\pi/2[/math] appearing there being [math]\varepsilon(p)\varepsilon(q)=0\cdot 0=0[/math] in the present case, and this finishes the proof.

■

General references

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