Suppose we have $x\_1^2 + y\_1^2 + x\_2^2 + y\_2^2 + x\_3^2 + y\_3^2 + x\_4^2 + y\_4^2 = 1$ and we define $z\_j = x\_j + iy\_j$, where $j = 1,\,2,\,3,\,4$.


The problem is finding or approximating the following integral (The actual problem is more complex than this !)
$$I(k) = \int\limits\_{\mathbb{S}^7}\lvert(z\_1z\_4 - z\_2z\_3)^k\rvert,$$ where $\mathbb{S}^7$ is the $7$-sphere, and the integral is over its surface measure. The ultimate aim is to find or get an estimate of the ratios of the type $\displaystyle \frac{I(k)}{I(k+1)}$.


I have never done an integration over such a surface, the hypersphere and don't know where to begin. Any hint or any pointers would really help me to get started.


(I have been suggested to make use of the fact that the points $x\_i$'s and $y\_i$'s lie on a sphere; hence we can use it to reduce one variable and make use of Fubini's Theorem with proper limits for integration to proceed. But I can't see how to make use of this.)




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What have I done till now?


As the problem say, it would be acceptable if we can approximate the ratios or the integral, so
$$\lvert z\_1z\_4 - z\_2z\_3\rvert \, \leq \, \lvert z\_1z\_4\rvert + \lvert z\_2z\_3\rvert \, = \, \lvert z\_1\rvert\lvert z\_4\rvert + \lvert z\_2\lvert\rvert z\_3\rvert.$$
This converts everything into real variables, and then I have used Mathematica to calculate this. But the problem is Mathematica gives the solution for small values of $k$, but as I have to find it for the general case, it is not of much use.


Also, I have tried using the spherical coordinates to help with the integral, but they make it more complicated.


Any pointers on this front are also welcome.




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Here are the values of $I(k)$ for $k=1,\,\cdots,\,10$.


$z\_1 = x\_1 + ix\_2, \ \ z\_2 = x\_3+ix\_4, \ \ z\_3 = x\_5+ix\_6, \ \ z\_4 = x\_7+ix\_8$ then


$$\mid z\_1z\_4 - z\_2z\_3| \ \ = \ \ \sqrt{\left(x\_4 x\_5+x\_3 x\_6-x\_2 x\_7-x\_1 x\_8\right){}^2+\left(x\_3 x\_5-x\_4 x\_6-x\_1 x\_7+x\_2 x\_8\right)^2} $$


To integrate over $\mathbb{S}^7$, I have used two methods, one is using `NIntegrate` and the second using `integrateSphere` function from [HFT](https://www.axler.net/HFT_Math.html) Software package.


(**Note** : HFT package doesn't work for odd values of $k$, but for even values of $k$, the results match perfectly with that of `INtegrate`)


$$\begin{align}
I(1) \ \ &\approx \ \ 9.54903\\
I(2) \ \ &= \ \ \frac{\pi^4}{3\times 10} \ \ \approx \ \ 3.24697\\\\
I(3) \ \ &\approx \ \ 1.19691\\
I(4) \ \ &= \ \ \frac{\pi^4}{3\times70} \ \ \approx \ \ 0.463853\\\\
I(5) \ \ &\approx \ \ 0.188085\\
I(6) \ \ &= \ \ \frac{\pi^4}{3\times420} \ \ \approx \ \ 0.0773088\\\\
I(7) \ \ &\approx \ \ 0.0329181\\
I(8) \ \ &= \ \ \frac{\pi^4}{3\times2310} \ \ \approx \ \ 0.0140561\\\\
I(9) \ \ &\approx \ \ 0.00620447\\
I(10) \ \ &= \ \ \frac{\pi^4}{3\times12012} \ \ \approx \ \ 0.0027031\\\\
\end{align}$$



Integration of a function over 7-sphere
This is not a complete answer but shows how to simplify your integral:
As pointed out by mlk in the comments the Hopf fibration plays a crucial role: Consider $\mathbb S^7\subset\mathbb C^4.$ On $\mathbb C^4=\mathbb C^2\oplus\mathbb C^2$ the special unitary group $\mathrm{SU}(2)$ acts by the direct sum of the standard representation. Note that $\mathrm{SU}(2)=\mathbb S^3.$ The Hopf fibration $$\mathrm{SU}(2)\to\mathbb S^7\to\mathbb S^4$$ is given by taking the quotient of this action restricted to $\mathbb S^7\subset\mathbb C^4.$ Then, one directly computes that $$z\_1z\_4-z\_2z\_3$$ is invariant under the action of $\mathrm{SU}(2).$ Hence, by Fubini using the fact that we consider a Riemannian submersion, the integral $I(k)$ is given by a constant factor $C\neq0$ (comming from the volume of $\mathrm{SU}(2)$ ) times an integral over the quotient $\mathbb S^4=\mathbb S^7/\mathrm{SU}(2).$ For your final question concernig $I(k)/I(k+1)$, the constant $C$ does not play any role.


Let us identify the 4-sphere with the quaterninic projective line $\mathbb H P^1.$ Using affine coordinates (being basically equivalent to coordinates using a suitable stereographic projection, your integrand becomes $$|z\_1 z\_2-z\_2 z\_3|^k( [x\_1+x\_2i+(x\_3+x\_4 i) j,1])=|x\_3+x\_4 i|^k.$$ The volume form (what you call the surface measure) on the quotient is the standard volume form and can be written as $$vol= \frac{4}{(1+x\_1^2+x\_2^2+x\_3^2+x\_4^2)^4} dx\_1\wedge dx\_2\wedge dx\_3\wedge dx\_4.$$ So you should integrate
$$\int\_{\mathbb R^4}\frac{4 |x\_3+x\_4 i|^k}{(1+x\_1^2+x\_2^2+x\_3^2+x\_4^2)^4} dx\_1\wedge dx\_2\wedge dx\_3\wedge dx\_4.$$ I do not have time to finish the computation, but I am sure the last integral can be simplified further.


