Trying to understand better the construction of the $n$-sphere

Question

I try to understand better the parametrisation of the $n-$sphere (suppose $R=1$). I know that it is given in $\mathbb R^{n+1}$ by $$\begin{cases}\cos\theta _n\cos\theta _{n-1}...\cos\theta _1\\ \cos\theta _n...\cos\theta _2\sin\theta _1\\ \cos\theta _n...\cos\theta _3\sin\theta _2\\ \vdots \\\cos \theta _n\sin\theta _{n-1}\\ \sin\theta _n\end{cases},\quad \theta _i\in [0,2\pi[, i=1,...,n-1, \theta _n\in [-\pi/2,\pi/2]$$

So, for $n=1$, it is just $(\cos\theta ,\sin\theta )$ with $\theta \in [0,2\pi]$ and in cartesian is $x^2+y^2=1$. For $n=2$, we have have in cartesian $$x^2+y^2+z^2=1,$$ So by setting $u^2=x^2+y^2$, we have half circle in $Ouz$ parametrized by $(u,z)=(\cos\theta ,\sin\theta )$ with $\theta \in [-\pi/2,\pi/2]$ and a circle in $Oxy$ parmetrized by $(u\cos\varphi,u\sin\varphi)$ with $\varphi\in [0,2\pi[$. At the end, in $Oxyz$, we get $$(x,y,z)=(\cos\theta \cos\varphi,\cos\theta \sin \varphi,\sin \theta )$$ with $\varphi\in [0,2\pi]$ and $\theta \in [-\pi/2,\pi/2]$.

I know that the cartesian equation of an hypersphere is given by $$x_1^2+\cdots+x_n^2=1.$$ Let take $n=4$, so $$x_1^2+\cdots+x_4^2=1.$$ If I follow the previous construction, we should have a sphere on $Ox_1x_2x_3$ and a half circle on $Oux_4$, where $u^2=x_1^2+x_2^2+x_3^2$. Therefore, $(u,x_4)=(\cos \theta _1,\sin\theta _1)$ with $\theta _1\in [-\pi,\pi]$ and $$(x_1,x_2,x_3)=(u\cos\theta \cos\varphi,u\cos\theta \sin\varphi,u\sin\theta ),\quad \theta \in [-\pi/2,\pi/2],\varphi\in [0,2\pi],$$ and thus $$(x_1,x_2,x_3,x_4)=(\cos\theta _1\cos\theta \cos\varphi,\cos\theta _1\cos\theta \sin\varphi,\cos\theta _1\sin\theta,\sin\theta _1 ),\quad \theta _1,\theta \in [-\pi/2,\pi/2],\varphi\in [0,2\pi].$$

We can then continue like this, and thus, we should obtain

$$\begin{cases}\cos\theta _n\cos\theta _{n-1}\cdots\cos\theta _1\\ \cos\theta _n\cdots\cos\theta _2\sin\theta _1\\ \cos\theta _n\cdots\cos\theta _3\sin\theta _2\\ \vdots \\\cos \theta _n\sin\theta _{n-1}\\ \sin\theta _n\end{cases},\quad \theta _1\in [0,2\pi[, \theta _i\in [-\pi/2,\pi/2], i=2,\cdots,n.\tag{E}$$

Question 1 : Why such a construction does not work ?

Question 2 : What would be the surface obtained by $(E)$?

Question 3 : Finally, how is constructed the $n$-sphere?

Answer 1

(1) We try to parametrize the $n$-sphere, taking as pattern the story known for the circle of radius $r$ in the two dimensional plane.

Here, the natural parametrization is given by $$ \begin{aligned} x &= r\cos t\ ,\\ y &= r\sin t\ ,\\ &\qquad\text{where the roles are clear, $r>0$ is the radius, $t\in(0,2\pi)$ the angle to $Ox$.} \end{aligned} $$ Now let us go in the $n$-dimensional space, use the coordinate variables $x,y,z,\dots$ for $\Bbb R^n$, so the equation for the corresponding sphere is $$ x^2+y^2+z^2+\dots =1\ . $$ To use the polar pattern, we write it in the shape $$ x^2 +\underbrace{(y^2+z^2+\dots)}_{x_1^2}=1\ . $$ $x_1$ is a substitution for the above function in $y,z,\dots$ .

And we would like to parametrize so that $x^2 +x_1^2=1$ uses the polar pattern. There are two chances,

• $x=\cos t_x$, $x_1=\sin t_x$, and

• $x=\sin t_x$, $x_1=\cos t_x$.

And of course, we have to make a choice of some interval where $t_x$ canonically lives in. A good choice would be correspondingly:

• $x=\cos t_x$, $x_1=\sin t_x$, and $0<t_x<\pi$, so $x\in(-1,1)$, and $x_1>0$, and $x_1$ can be used further as a "radius", and

• $x=\sin t_x$, $x_1=\cos t_x$, and $-pi/2<t_x<\pi$, so $x\in(-1,1)$, and $x_1>0$, and $x_1$ can be used further as a "radius".

In both cases we had to use the corresponding half-circle making $x$ brush the interval $(-1,1)$.

From here, we rewrite $y=x_1y_1$, $z=x_1z_1$, $\dots$ and pass inductively till the end is imminent.

Then, the last variable will use the full circle.

Now let us go back to the relations (E). The stand alone variable is the last one, and it corresponds in my notation to $x$, $$ x = \sin \theta_n\ ,\qquad \theta_n \in(-\pi/2,\pi/2)\ ,$$ (well closed interval in the post) this corresponds to my second chance above, all other variables have the expected factor $\cos \theta_n>0$, let us declare we feel comfortable, we tacitly "factor $\cos\theta_n$ in the bucket of all other variables", and let us pass to the next stand alone variable in the "factored bucket". It is $\sin\theta_{n-1}$, well done, same game. We inductively pass through all "refactored buckets", till we reach the last step, and yes, it corresponds to $(\cos\theta_1,\sin\theta_1)$, $\theta_1\in(0,2\pi)$, parametrizing $S^1$ in a standard way.

Note: The initial parametrization, inserted after "I know that..." cannot work, there are too many $(0,2\pi)$ intervals. To see immediately why, let us replace all of them with only the quarter $(0,\pi/2)$ (of the full interval $(0,2\pi)$ used there - there is also a "special interval" of half length... ). Then using $(0,\pi/2)^{\text{sphere dimension}}$ we parametrize in the plane only the part of the sphere in the first quadrant, in the 3D space only the part in the first octant, and so on. All other parts (obtained by chosing for the one or the other $\theta_i$ "the other half") make a difference only in $\pm$ sign on some components, so that we cover all quadrants, octants, ... and in affine dimension $n+1$ there are $2^{n+1}$ of them. So in the corresponding "sphere dimension" $n$ we expect $2^{n+1}$ of them. For the $n$ dimensional sphere we need $n$ parameters, so it is natural that up to one "special" parameter, all other parameters will live in correctly chosen intervals of length $\pi$. The special one in an interval of length $2\pi$.

(2) It is the sphere.

(3) As in (E) for instance.

Answer 2

In fact, for $n \geqslant 3$, the range of parameters given at the beginning is incorrect whereas that in (E) is correct. To see this, consider an arbitrary point $(x_1, \cdots, x_{n + 1}) \in S^n$ such that $x_1, \cdots, x_{n + 1} ≠ 0$. Suppose$$ \begin{cases} x_1 = \cos θ_1 \cos θ_2 \cdots \cos θ_n\\ x_2 = \sin θ_1 \cos θ_2 \cdots \cos θ_n\\ x_3 = \sin θ_2 \cos θ_3 \cdots \cos θ_n\\ \vdots\\ x_{n + 1} = \sin θ_n \end{cases}, $$ where $θ_1, \cdots, θ_{n - 1} \in [0, 2π)$, $θ_n \in \left[\dfrac{π}{2}, \dfrac{π}{2}\right]$. Now take$$ (θ_1', θ_2', θ_3', \cdots, θ_n') = \begin{cases} (θ_1 + π, π - θ_2, θ_3, \cdots, θ_n), & 0 < θ_1 < π,\ 0 < θ_2 < π\\ (θ_1 - π, π - θ_2, θ_3, \cdots, θ_n), & π < θ_1 < 2π,\ 0 < θ_2 < π\\ (θ_1 + π, 3π - θ_2, θ_3, \cdots, θ_n), & 0 < θ_1 < π,\ π < θ_2 < 2π\\ (θ_1 - π, 3π - θ_2, θ_3, \cdots, θ_n), & π < θ_1 < 2π,\ π < θ_2 < 2π \end{cases}, $$ then $(θ_1', \cdots, θ_n') \mapsto (x_1, \cdots, x_{n + 1})$.