# How do you show $f(x,y)=\left( \frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right)$ is a diffeomorphism?

Given this mapping:$$f(x,y)=\left( \frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right), \quad (x,y) \in \mathbb{R}^2$$

$$(\mathbb{R}^2= \{ (x,y):x,y \text{ are real numbers}, \text{ excluding } (x,y)=(0,0) \})$$

How do you check if it is a diffeomorphism?

I would say $f$ is continuous on the given domain... now to how would you check for differentiability? Is it true to say that if the determinant of the Jacobian matrix is non-zero for all $(x,y)$ in the given domain then f is differentiable and therefore a diffeomorphism?

This is a specific function that is its own inverse but considering a function that wasn't its own inverse does the fact that the determinant of the Jacobian matrix being non-zero (and existing) for the whole domain show that the inverse function exists without restricting the domain?

• Either you ask a question for this specific function or you ask in a general setting – Guy Fsone Jan 24 '18 at 20:58
• By viewing $(x,y)$ as $z=x+iy$, you can realize your map as sending $z$ to $f(z) = z/|z|^2$. Then $f(f(z)) = (z/|z|^2)/ 1/|z|^2 = z$, so $f$ is a bijection. It suffices you now check $f$ is smooth. – Pedro Tamaroff Jan 24 '18 at 20:58

Solving $$u=\frac{x}{x^2+y^2}~~~~v=\frac{y}{x^2+y^2}$$

gives $$x=\frac{u}{u^2+v^2}~~~~y=\frac{v}{u^2+v^2}$$

Namely you can check that $$\color{red}{f\circ f(x,y) = (x,y)}$$

Hence $f$ is a bijection and since $f$ is differentiable on $\Bbb R^2\setminus \{(0,0)\}$ you conclude that it is a diffeomorphism from $\Bbb R^2\setminus \{(0,0)\}$ to $\Bbb R^2\setminus \{(0,0)\}$

• Why a down vote here? – Guy Fsone Jan 24 '18 at 20:53
• Must be the same reason as my answer's downvote :-/ – Fly by Night Jan 24 '18 at 20:54
• +1 Sometimes they downvote for no reason... Anyway thanks to both of you for your answers – LostInSpace Jan 24 '18 at 20:58
• This transformation $(x,y) \to f(x,y)$ is classical (and very important) in geometry : it is called inversion (encyclopediaofmath.org/index.php/Inversion) – Jean Marie Jan 25 '18 at 9:21
• @Guy Fsone I have been working on inversion. And your question has triggered in me the desire to give my own answer to the question, and I think you will be interested when you see it (in some hours). Indeed, the Jacobian matrix is akin to a symmetry matrix, and there is a geometrical reason for this... Besides the reference you just gave is for nD conformal mappings, underlines the importance of nD inversions (a famous 3D inversion being stereographic projection (cut-the-knot.org/pythagoras/StereoProAndInversion.shtml). – Jean Marie Jan 25 '18 at 18:32

The Jacobian matrix being non-singular only shows that the mapping is a local diffeomorphism. For the inverse to exist, you need the mapping to be one-to-one.

• Thank you, I will go google local diffeomorphism, does that mean if the mapping is one-to-one and the Jacobian matrix is non-singular then it is a diffeomorphism? – Desmoz Jan 24 '18 at 21:02
• @Enigma It'll be a diffeomorphism onto the image. As an example to contrast local and global diffeomorphisms: think of the real line, and the unit circle. The mapping $x \mapsto (\cos x,\sin x)$ is a local diffeomorphism in a neighbourhood of every point $x \in \mathbb R$, but the line and the circle are not diffeomorphic manifolds. (They're not even homeomorphic!) The definition of a global diffeo is a differentiable bijection with differentiable inverse. – Fly by Night Jan 24 '18 at 21:10

This is not properly an answer to the question of diffeomorphism, but aremark about geometrical aspects of this transformation $f$ conveyed by the Jacobian matrix.

As I already mentionned it, $f$ is a classical transform called "inversion" (https://www.encyclopediaofmath.org/index.php/Inversion).

Written under a vectorial form, $M'(x',y')$ is the image of $M(x,y)$ if and only if:

$$\tag{1}\vec{CM'}=\frac{1}{\|\vec{CM}\|^2} \ \vec{CM}.$$

Intuitively, consider a group of nearby points $M_k$, say exterior to unit circle but close to it. Their images will be nearby points $M'_k$ interior to the unit circle, with the farthest points exchanged with the closest points.

The Jacobian matrix of $f$ has a remarkable structure :

$$\tag{2}J \ \ = \ \ k \ \ \underbrace{\begin{pmatrix}-a&b\\ \ \ b&a \end{pmatrix}}_S \ \ \ \text{with} \ \ \ \begin{cases}a=\dfrac{x^2 - y^2}{x^2 + y^2}\\b=\dfrac{-2xy \ \ }{x^2 + y^2}\end{cases} \ \ \text{and} \ \ k=\dfrac{1}{x^2 + y^2}$$

reflecting in the "language" of linear algebra the local behaviour of inversion in the vicinity of a certain point $(x_0,y_0)$ (first order approximation) :

$$\tag{3}f(x,y)-f(x_0,y_0)\approx \underbrace{k \begin{pmatrix}-a&b\\ \ \ b&a \end{pmatrix}}_J \binom{x-x_0}{y-y_0}.$$

(by definition of a Jacobian matrix). $J$ is the composition of

• a symmetry operation due to matrix $S$ ($a^2+b^2=1$ and $\det(S)=-1$), and

• a central homothety (brought by factor $k$).

The homothety ratio $k=\frac{1}{\|\vec{CM}\|^2}$ doesn't come as a surprise (see (1)). The appearance of a clearcut symmetry matrix is, at first look, more surprizing, but can be easily understood when looking at some shapes and their images like (see picture) the way the image of the fox is, in first approximation a kind of reflection of it : its legs in particular can be considered as almost perfectly mirrored with respect to certain tangent to the unit circle ; whereas, for a point like the fox muzzle, the homothety ratio $k$ cannot be ignored : first order approximation (2) is mirroring and shrinking/enlarging by a shrinking/enlargment factor in inverse ratio with the square of the distance of the considered point with respect to the origin $C$.

The closer to the unit circle is a group of points $(x_k,y_k)$, the more faithfully their image is "reflected" with respect to the unit cercle, because the dilation/contraction factor $k \approx 1$.

Remark: The image of the little planet with a fox and a star is a tribute to "The Little Prince" (http://www.yoanaj.co.il/uploadimages/The_Little_Prince.pdf))