1

I am a robotics student who has very poor knowledge of topology, thus I hope my question is not ill-posed.

Studying the classical textbook [1], I found an interesting diffeomorphism from stars* to spheres. Both the book [1] and the original paper [2], report a formula to transform points form the star world to the sphere world. Since it is a diffeomorphism, I looked for the inverse smooth transformation but I cannot find it.

Does anyone know it?

I tried to compute the inverse algebraically but the map from stars to open balls is similar to $f(x) = r \frac{x-x_0}{\|x-x_0\|}+c$ where $r,c$ are the radius and the center of the ball, respectively, and $x_0$ is the vantage point of the star.

Thank you very much for your attention!

* a star-shaped set $S$ is a set where there exists at least one point (also called vantage point) that is within line of sight of all other points in the set.

REFERENCES

[1] Choset, Howie, et al. "Principles of robot motion: theory, algorithms, and implementations". MIT press, 2005.

[2] Koditschek, Daniel E, et al. "Robot Navigation Functions on Manifolds with Boundary". Advances in applied Mathematics 11, 412-442 (1990)

Antonio Bono
  • 21
  • The exact expression of the inverse map will depend on the shape of the star* set (those sets are often called star domains). The map from the star domain to the sphere probably maps the "vantage point" of the star to the center of the sphere and then linearly scales all segments starting from this point to the same length (radius of the sphere). To construct a reverse map you need to the scaling factor for each radius of the sphere to scale it back to its original length (basically you need the shape of the star domain) – david_sap May 30 '22 at 20:40
  • *you need to know (last sentence) – david_sap May 30 '22 at 20:46
  • 1
    But you should not call it a "sphere," the right terminology is an "open ball." For a proof, see here. – Moishe Kohan May 30 '22 at 22:01
  • @david_sap Thank you for your answer. The map does what you have written: it scales the ray starting at the "center" point of the star set (the book and paper do not mention the vantage point but I think it is the same) through its unique intersection with the boundary in order to make it of the same lenght of the sphere radius. I also think that you need to know such a scaling factor to compute the inverse transformation. But then, since it is prooved to be a diffeomorphism, should we not be able to have such analytical inverse transformation? Thanks! – Antonio Bono May 31 '22 at 09:51
  • Yes, the analytical inverse transformation will always exist, but its exact expression will depend on the shape of the boundary of the star domain. If you have an expression for the direct map, finding the inverse should be doable: what you need is the distance between the "center" of the star set and the "unique intersection of every radius with the boundary". And, as @MoisheKohan correctly pointed out, we should talk of a ball instead of a sphere (a sphere is the external surface/boundary of a ball) – david_sap May 31 '22 at 17:50
  • As you can see in the provided link the proof of the diffeomorphism does not explicitly construct an inverse mapping, it relies on some form of inverse function theorem to prove its existence. You can use the same approach for a diffeomorphism $\Omega \to \mathring{D^2}$, where $\Omega \subset \mathbb{R}^2$ is the star domain and $\mathring{D^2} \subset \mathbb{R}^2$ is the open ball. – david_sap May 31 '22 at 17:58
  • I thank you both david_sap and @MoisheKohan. Unfortunately my very poor knowledge of topology and math in general makes the proof beyond comprehension for me. My very basic question could be reformulated in this way: If $f(x)$ is an analytical diffeomorphism, should I be able to complute $f(x)^{-1}$ like we do with a simple invertible function $f(x)= ax+b$? Or should I build a new function based on the information about the center of the star and rays intersections with the boundary? Thanks again! – Antonio Bono Jun 01 '22 at 08:27
  • You would have to be much more specific about the word "compute." It's easy to say what this means in the case of inverting a linear function. But what if I gave you a general polynomial of degree 5. What would it mean to "compute" its roots? Note also that the answer in the link I gave yields only $C^\infty$-smooth diffeomorphism, not an analytic diffeomorphism. – Moishe Kohan Jun 01 '22 at 10:50
  • @MoisheKohan It is even more complicated. The authors consider configuration spaces (containing all possible positions of a robot arm). A sphere-space is a special configuration space; it is defined as a ball $B$ minus the union of finitely many disjoint balls $B_i \subset B$. I am not sure whether these balls are assumed to be open or closed. A star-space is defined as a star-shaped set $S$ minus the union of finitely many disjoint star-shaped sets $S_i \subset S$. Again I am not sure whether these objects are open or closed (and have at least interior points). – Paul Frost Jun 04 '22 at 10:38
  • @PaulFrost: Oh, I see. Then the question needs a major revision since in the present form it is very misleading. Without reading their papers, it's had to tell if the spaces are diffeomorphic. On general grounds, configuration spaces should be compact though. – Moishe Kohan Jun 04 '22 at 13:05
  • @MoisheKohan The authors allow compact and non-compact configuration spaces. – Paul Frost Jun 04 '22 at 17:10

0 Answers0