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We have two points $(x_1, y_1), (x_2, y_2) \in \mathbb{R}^2$ with polar parameterizations $(r_1, \theta_1), (r_2, \theta_2)$ (i.e., $x_1=r_1 cos(\theta)$ and so on.) How would the point $(x_1, y_1, x_2, y_2) \in \mathbb{R}^4$ be represented with four-dimensional spherical coordinates $(r, \phi_1, \phi_2, \phi_3)$?

Please correct my proposed answer if I have made a mistake or upvote it if you concur.

Finally, how would one inductively generalize equivalency to $\mathbb{R}^{2^n}$ or $\mathbb{R}^{2n}$?

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3 Answers 3

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We begin by equating the components of the two numbers in polar coordinates and the single number in spherical coordinates. Here, $r_3 = \sqrt{r_1^2 + r_2^2}$.

$$r_1 \cos \left( \theta_1 \right) = r_3 \cos \left( \phi_1 \right)$$ $$r_1 \sin \left( \theta_1 \right) = r_3 \sin \left( \phi_1 \right) \cos \left( \phi_2 \right)$$ $$r_2 \cos \left( \theta_2 \right) = r_3 \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \cos \left( \phi_3 \right)$$ $$r_2 \sin \left( \theta_2 \right) = r_3 \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \sin \left( \phi_3 \right)$$

Next, the standard definition of $arccos$ does not include the full input plane, so we form a continuous extension to $\mathbb{R}^2 \setminus \mathbb{R}^+$. (We exclude the point intersecting $\mathbb{R}^+$ with $\theta = 0$ or $\theta = 2 \pi$). With an abuse of notation, we will still refer to this continuous extension as $arccos$. An interesting application of this definition of $arccos$ is that $\mathbb{R}^3 \setminus 0$ is not homologically trivial, which is an exercise in Munkre's Analysis on Manifolds.

$$\phi_1 = \arccos \left( r_1 r_3^{-1} \cos \left( \theta_1 \right) \right)$$ $$\phi_2 = \arccos \left( r_1 r_3^{-1} \frac{ sin (\theta_1) }{sin (\phi_1)} \right)$$ $$\phi_3 = \arccos \left( r_2 r_3^{-1} \frac{ \cos \left( \theta \right) }{ \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) } \right)$$

Fortunately, the other direction is simpler to compute.

$$\theta_{1} = \arccos(r_3 r_1^{-1} \cos \left( \phi_1 \right))=\arcsin \left( r_3 r_1^{-1} \sin \left( \phi_1 \right) \cos \left( \phi_2 \right) \right)$$

$$\theta_2 = \arccos \left( r_3 r_2^{-1} \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \cos \left( \phi_3 \right) \right)=...$$

$$\frac{ r_1 \cos \left( \theta_1 \right) }{ \cos \left( \phi_1 \right) } = \frac{ r_1 \sin \left( \theta_1 \right) }{ \sin \left( \phi_1 \right) \cos \left( \phi_2 \right) }$$

$r_1 = \left|\frac{ \cos \left( \theta_1 \right) }{ \cos \left( \phi_1 \right) } - \frac{ \sin \left( \theta_1 \right) }{ \sin \left( \phi_1 \right) \cos \left( \phi_2 \right) }\right|^{-1}$ provided that $cos(\phi_1)\neq0$, $cos(\phi_2)\neq 0$, and $sin(\phi_1)\neq 0$.

$r_2 = \left|\frac{1}{sin(\phi_1) sin(\phi_2)} \left( \frac{cos (\theta_2) }{cos(\phi_3)} - \frac{sin(\theta_2)}{sin(\phi_3)} \right)\right| ^{-1}$ provided that $sin(\phi_1)\neq 0$, $sin (\phi_2) \neq 0$, $cos(\phi_3)\neq 0$, and $sin(\phi_3) \neq 0$.

I wonder if there would be an easier formula for $r_1$ and $r_2$ by taking the components of $r_3$ along two axes. (In this case, $r_1$, $r_2$, and $r_3$ would all be tangent vectors on the same plane, though finding a parameterization or normal vector for the plane is not trivial; one cannot simply compute $\vec{r_1} \times \vec{r_2}$.)

By stipulating $r_3>0$ and $\theta_1, \theta_2, \theta_3 \in [0, 2 \pi)$, proof of injection (being a one-to-one map) should follow immediately. Likewise, every pair of $(x_1, y_1)=(r_1, \theta_1)$ and $(x_2, y_2) = (r_2, \theta_2)$ should produce a value for $(r, \phi_1, \phi_2, \phi_3)$, allowing surjection to follow as well. $f(r_1, \theta_1, r_2, \theta_2)= (r, \phi_1, \phi_2, \phi_3)$ should then satisfy the criteria of an isomorphic map. If feel that I have missed something. Please correct me in the comments or my answer directly if you have sufficient reputation.

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While the other answer would take you to the standard spherical coordinates, the formula gets very unwieldy especially for higher dimensions. An alternative will get you the same radial coordinate but different angles

$$\phi_1=\theta_1$$ $$\phi_2=\theta_2$$ $$r_3\cos\phi = r_1$$ $$r_3\sin\phi = r_2$$

The interesting thing about these coordinates is instead of having the angles be maps $[0,2\pi]\times[0,\pi]^{2n-2}\to S^{2n-1}$, these are maps $[0,2\pi]^{n}\times\left[0,\frac{\pi}{2}\right]^{n-1}\to S^{2n-1}$ (embedded in $\Bbb{R}^{2n}$). The generalization to $\Bbb{R}^{2n}$ is simply setting all of the radial coordinates to the standard spherical coordinates in $\Bbb{R}^n$ but restricted to the first quadrant/octant/etc..

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  • $\begingroup$ Thank you for the excellent answer. Do you agree with my belief that the mapping may be made isomorphic? How are the angles necessarily different? I'm sorry, I don't see an immediate consequence of the proposed mapping save for restricting the domains, specifically for $\phi$. $\endgroup$
    – Talmsmen
    Commented Mar 6, 2021 at 6:13
  • $\begingroup$ @JPwin I don't understand. What is not isomorphic about it that has to be "made" isomorphic? $\endgroup$ Commented Mar 6, 2021 at 6:15
  • $\begingroup$ My apologies. I had already thought the mapping was isomorphic but hadn't made a rigorous proof for it. You do agree that it is one-to-one and onto, right? $\endgroup$
    – Talmsmen
    Commented Mar 6, 2021 at 6:17
  • $\begingroup$ @JPwin sure. You don't need go prove the total mapping is both of those things from scratch. You have that it is a chain of two mappings, both of which you already have are bijections or can prove more easily. Chaining bijections is still a bijection. $\endgroup$ Commented Mar 6, 2021 at 6:19
  • $\begingroup$ Thank you. I'll remember that the composition of bijections is sufficient for a formal proof. I haven't had much experience with maths outside of the classroom. If one was to, say, take the output of a biholomorphic map $f_1:\mathbb{R}^4 \to \mathbb{R}^4$, your mapping $f_2: \mathbb{R}^{+} \times \mathbb{R}^{+} \times [0,2\pi] \times [0,\frac{\pi}{2}]^{2n-2}\to \mathbb{R}^4_{\geq}$ where $\mathbb{R}^4_{\geq}$ is $\mathbb{R}^4$ with one of the planes restricted to the upper (first) quadrant would have an inverse as it is isomorphic. I could work in the simplified space, and apply the inverse... $\endgroup$
    – Talmsmen
    Commented Mar 6, 2021 at 6:28
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$$r_1 \sin \left( \theta_1 \right) = r_3 \sin \left( \phi_1 \right) \cos \left( \phi_3 \right)$$ $$r_1 \cos \left( \theta_1 \right) = r_3 \cos \left( \phi_{1} \right)$$

We divide the two equations by each other to develop $\theta_1=\arctan \left( \frac{ \sin \left( \phi_1 \right) \cos \left( \phi_3 \right) }{ \cos \left( \phi_1 \right)} \right) + \vartheta \left( \phi_1, \phi_2, \phi_3 \right)$. Intuitively, $\phi_1 \to \pm \frac{ \pi }{ 2 }$ yields $\theta_1 \to \pm \frac{ \pi }{ 2 }$. $\vartheta$ is a correction term to account for the quadrant of $\theta$.

$$r_2 \sin \left( \theta_2 \right) = r_3 \sin \left( \phi_1 \right) \sin \left( \phi_3 \right) \sin \left( \phi_2 \right)$$ $$r_2 \cos \left( \theta_2 \right) = r_3 \sin \left( \phi_1 \right) \sin \left( \phi_3 \right) \cos \left( \phi_2 \right)$$

Likewise, $\tan \left( \theta_2 \right) = \frac{ r_3 \sin \left( \phi_1 \right) \sin \left( \phi_3 \right) \cos \left( \phi_2 \right) }{ r_3 \sin \left( \phi_1 \right) \sin \left( \phi_3 \right) \sin \left( \phi_2 \right)}$ can be continuously extended to $\theta_2 = \phi_2$ If we restrict ourselves to $\mathbb{R}^2$, $\phi_3 = 0$, so $\theta_1 = \phi_1$. I am still somewhat unsure of a formula for the radii $r_1$ and $r_2$. $$r_1 = \begin{cases} \frac{ r_3 \cos \left( \phi_1 \right) }{ \cos \left( \theta_1 \right) } & \text{ if }\cos \left( \theta_1 \right) \neq 0 \\ \frac{ r_3 \sin \left( \phi_1 \right) \cos \left( \phi_2 \right)}{ \sin \left( \theta_1 \right)} & \text{ if } \cos \left( \theta_1 \right) = 0 \end{cases}$$ $$r_2 = \sqrt{r_3^2 - r_1^2}$$ Unfortunately, I don't believe that my formula for $r_1$ is properly defined as continuity around $\theta_1=0$ may fail. Once more, $\mathbb{R}^2$ could present a simplification $r_1=r_3$.

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