Inverse of function of projective space $\mathbb{R}P^1$ in $\mathbb{S}^1$?

Question

Is it possible to explicitly define the inverse function of this function of the projective space $\mathbb{R}P^1$ in $\mathbb{S}^1$?

$$f([x,y]) = \left( \dfrac{2xy}{x^2+y^2}, \dfrac{x^2-y^2}{x^2+y^2}\right).$$

Answer 1

EDIT: The map with a correct geometrical intuition should be $$ f([x,y]) = \left( \frac{2xy}{x^2+y^2}, \frac{y^2-x^2}{x^2+y^2} \right). $$ Note that this map differs by the map in the question by an automorphism of $\mathbb RP^1$, namely the map $[x,y] \mapsto [y,x]$. To avoid confusing the geometrical intuition behind the question, I will use my map $f$, not the one from the OP. Keep that in mind if you need the map in coordinates.

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Yes, it is very easy. You just have to take $$ g : \mathbb{S}^1 \to \mathbb{R}P^1, \qquad (x,y) \mapsto [1-y,x] $$ when seeing $\mathbb{S}^1$ as the unit circle in $\mathbb R^2$ centered at the origin. Interpret this over affine space for the moment, i.e. take $x=t$ and $y=1$. The map $f$ becomes the map $$ f(t) = \left( \frac {2t}{t^2+1} , \frac {1-t^2}{t^2+1} \right). $$ This is exactly the map which takes a real number $t$ (seen on the image as $(t,0)$) and associates the corresponding point on the unit circle as in the image below (and this map is also used to characterize all Pythagorean triples when $t = a/b \in \mathbb Q$, so that $[t,1] = [a,b] \in \mathbb QP^1$ is a $\mathbb Z$-point):

You need to interpret $f$ geometrically to understand this. The map $f$ basically sends the point on the real line to the intersection of that other line through $(0,1)$ with the unit circle. Inverting it just means running this bijection backwards, i.e. take the point $q$ on the unit circle, trace the line $L$ through $p = (0,1)$ and $q$, and the inversion map sends $q$ to the intersection of $L$ with the embedding of $\mathbb R^1$ in $\mathbb R^2$ depicted above ($t \mapsto (t,0)$).

In geometric terms, consider the unit circle centered at $(0,0)$. Given $(x,y)$ such that $x^2 + y^2 = 1$, the line $L$ passing through $(0,1)$ and $(x,y)$ intersects this circle in precisely $2$ points; the point $(x,y)$ and the point $(0,1)$. Its slope is $\Delta y / \Delta x = (y-1)/(x-0) = (y-1)/x$, so the map $g$ sends the point $(x,y)$ to $[1-y,x]$ if $(x,y) \neq (0,1)$ (the slope argument stops working for $(x,y) = (0,-1)$ because $\Delta x = 0$, but the result in $\mathbb RP^1$ is still correct). The vector $(1-y,x)$ is the one orthogonal to $L$. If $(x,y) = (0,1)$, the tangent line to the circle is the one with two points of intersection at $(0,1)$ (counting the point of tangency as a double point), so we associate the line $[1,0]$ to the point $(0,1)$.

(A hint of why we use the vector orthogonal to $L$ to represent $L$: it was a fundamental observation of Grothendieck that the duality between lines and points in the plane generalizes better when we see it as the duality of hyperplanes and points in a vector space, where a hyperplane is seen as the locus of an affine map. Therefore, in the plane, a line should be interpreted as the affine map whose set of solutions corresponds to the $1$-dimensional affine subspace we are familiar with, and that line is represented by a point in its locus and a vector orthogonal to that line. This representation works for lines as well as for hyperplanes, and when we compute things this way, the coordinate systems fit nicely with the abstract theory.)

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Pick the coordinates of $[x,y]$ in such a way so that $x^2 + y^2 = 1$. It follows that $f([x,y]) = (2xy, y^2-x^2)$. This point lies on the unit circle since $$ (y^2-x^2)^2 + (2xy)^2 = x^4 + y^4 - 2x^2 y^2 + 4x^2y^2 \\ = x^4 + y^4 + 2x^2y^2 \\ = (x^2+y^2)^2 \\ = 1 \\ $$ Now consider a point $(t,0)$ on the real line (see it as in the above image). The line passing through $(0,1)$ and $(t,0)$ intersects the unit circle in a second point $(0,1) + \lambda((t,0)-(0,1)) = (\lambda t,1-\lambda)$ precisely when $$ (\lambda t)^2 + (1-\lambda)^2 = 1 \\ \lambda^2(t^2+1) - 2\lambda + 1 = 1 \\ \lambda = 0 \qquad \text{ or } \qquad \lambda(t^2+1) - 2 = 0 $$ so that the $\lambda$ of interest is $\frac 2{t^2+1}$. This gives the coordinates of the point on the unit circle: $$ (x', y') = \left( \frac{2t}{t^2+1}, 1 - \frac 2{t^2+1} \right) = \left( \frac{2t}{t^2+1}, \frac{t^2 - 1}{t^2+1} \right). $$ Another way to interpret the real number $t$ is as follows: consider the point $p=(0,1)$ and the line $L_p$ passing through $(0,1)$ and $(t,0)$. A vector orthogonal to $(t,0) - (0,1) = (t,-1)$ is $(1,t)$, so $[1,t]$ is the correct representation for $L_p$. Replacing $[1,t]$ by $[1,x/y] = [y,x]$ and re-writing our expression, we recover the original $f$ (note that the flip $[x,y] \mapsto [y,x]$ needs to occur here, so this is probably why the OP formulated $f$ the way it was). So now, to invert $f$, given a point $(x,y)$ on the unit circle, we need to consider the line through $(x,y)$ and $(0,1)$. A vector orthogonal to $(x,y)-(0,1) = (x,y-1)$ is $(1-y,x)$, so the point $(x,y)$ maps to $[1-y,x]$. This gives $$ g(x,y) = \left\{ \begin{matrix} [1-y,x] & \text{ if } (x,y) \neq (0,1) \\ [0,1] & \text{ if } (x,y) = (0,1) \end{matrix} \right.. $$

Just as a confirmation, we evaluate $$ g(f([x,y])) = g\left( \frac{2xy}{x^2+y^2}, \frac{y^2-x^2}{x^2+y^2} \right) \\ = \left[1-\frac{y^2-x^2}{x^2+y^2}, \frac{2xy}{x^2+y^2} \right] \\ = \left[(x^2+y^2) - (y^2-x^2), 2xy \right] \\ = \left[2x^2, 2xy\right] \\ = [x,y] $$ This solution assumes $x\neq 0$, so one also computes $g(f([0,1])) = g(0,1) = [0,1]$. Conversely, when $x \neq 0$ and $x^2+y^2 = 1$, we have $$ x^2+(1-y)^2 = x^2 + (y^2 -2y+1) = 2-2y = 2(1-y), \\ x^2 - (1-y)^2 = (1-y^2)-(1-y)^2 = (1-y^2)-(1-2y+y^2) = 2y(1-y), $$ therefore $$ f(g(x,y)) = f([1-y,x]) = \left( \frac{2(1-y)x}{(1-y)^2 + x^2}, \frac{ x^2 - (1-y)^2}{(1-y)^2 + x^2} \right) = (x,y). $$

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A third interpretation of $f$ goes as follows: look at the unit circle $S^1$ and consider its blow-up at the point $p=(0,1)$. Given two distinct points $p,q \in \mathbb R^2$, there is a unique affine map $L : \mathbb R^2 \to \mathbb R$ satisfying $L(p) = L(q) = 0$. We can write it by letting $x = (x_1,x_2)$, $p = (p_1, p_2)$, $q = (q_1,q_2)$ and $$ L(x) = L(x) - L(p) + L(p) \overset{(!)}= \langle (w,z), x-p \rangle = w(x_1-p_1) + z(x_2-p_2). $$ because if $L(x) = Ax + b$ for some matrix $A$ and vector $b$, then $L(x) - L(y) = A(x-y)$, so $L(x)-L(y)$ is linear in $x-y$, hence (!). The condition $L(q) = 0$ gives the equation $$ w(q_1-p_1) + z(q_2-p_2) = 0, $$ which determines $(w,z)$ up to a scalar multiple; when $L(p) = 0$ (meaning that the line $L$ passes through the point $p$), we write $L_p = [w,z]$ and we write $q \in L_p$ when $L(q) = 0$. When $q = (x,y)$, we also write $L_p(x,y)$ instead of $L_p((x,y))$. In this interpretation, the vector $(w,z)$ is orthogonal to $q-p$, and the geometric line $L_p$ corresponds to the set of points $q$ such that $q-p$ is orthogonal to $(w,z)$.

We wish to have a parametrization of this blow-up (especially for the lines), which in general looks like $$ \mathrm{Bl}_p(\mathbb RP^2) = \{ ((x,y),L_p) \in \mathbb RP^2 \times \mathbb RP^1 \, | \, L_p(x,y) = 0 \}. $$

Restricting our attention to $S^1$, we have $$ \mathrm{Bl}_p(S^1) = \{ ((x,y),[w,z]) \in \mathbb RP^2 \times \mathbb RP^1 \, | \, x^2 + y^2 = 1, wx + z(y-1) = 0 \}. $$ Now since $S^1$ is smooth and we blew up at a smooth point of a curve, the correspondence $$ S^1 \leftarrow \mathrm{Bl}_p(S^1) \rightarrow \mathbb RP^1, \quad (x,y) \leftarrow ((x,y),[w,z]) \mapsto [w,z] $$ actually consists of two isomorphisms of varieties. Going from right to left gives the map $f$, and from left to right gives the map $g$.

To see this more in detail, the line $[w,z] \in \mathbb RP^1$ maps to $((x,y), [w,z]) \in \mathrm{Bl}_p(S^1)$, then $wx + z(y-1) = 0$, so $wx = z(1-y)$. Solving the system of equations $$ wx + z(y-1) = 0, \qquad x^2 + y^2 = 1 $$ by assuming $w \neq 0$ gives $$ z^2(y^2-2y+1) = (z(1-y))^2 = (wx)^2 = w^2x^2 = w^2(1-y^2) \quad \Rightarrow \quad (w^2+z^2)y^2 + (-2z^2)y + (z^2-w^2) = 0 $$ which means $$ y = \frac{2z^2 \pm \sqrt{4z^4 - 4(z^2-w^2)(z^2+w^2)}}{2(w^2+z^2)} = \frac{z^2 \pm w^2}{w^2+z^2}. $$ One solution is not of interest (it corresponds to $(x,y) = p$) and the other solution is the point $\left(\frac{2zw}{w^2+z^2}, \frac{z^2-w^2}{w^2+z^2}\right)$. Note that $f([w,z]) = (x,y)$ corresponds to our map of interest. When $w=0$, a similar process occurs except the quadratic will have a double root, which will lead to $f([0,1]) = (0,1)$.

In the other direction, suppose $x^2 + y^2 = 1$. The line $L_p$ through $p = (0,1)$ and $(x,y)$ is given by coordinates $[w,z]$ satisfying $w(x-0) + z(y-1) = 0$. An easy choice of $w$ and $z$ is $[w,z] = [-(y-1),x] = [1-y,x]$. Note that $g(x,y) = [1-y,x]$.

Hope that helps,

Answer 2

I guess you define $\mathbb RP^1 = (\mathbb C \setminus \{0\})/ \sim$, where $z \sim z'$ iff $z' = tz$ for some $t \in \mathbb R \setminus \{0\}$. Note that I identified $\mathbb R^2$ with $\mathbb C$. Then your map $f$ is essentially given by $f([z]) = \dfrac{z^2}{\lvert z \rvert^2}$ (note that I exchanged the first and second coordinate of your original $f([z])$ which does not play a role conceptually because the "flip-map" $\tau(x+iy) = i \overline z = y + ix$ is a homeomorphism on $S^1$).

This map is a well-defined continuous bijection, hence a homeomorphism. You ask for its inverse $f^{-1} : S^1 \to \mathbb RP^1$. It is well-known that for each complex number $w \ne 0$ we can choose a square root $\sqrt w$ (there are two possible choices which differ by a factor $-1$, but we shall not be specific how to choose $\sqrt w$). Note that $\sqrt w \in S^1$ for $w \in S^1$. We claim that $$f^{-1}(w) = [\sqrt w] .$$ Note that the RHS does not depend on the choice of $\sqrt w$ because $\sqrt w \sim -\sqrt w$. But now we have $f([\sqrt w]) = \dfrac{\sqrt w ^2}{\lvert \sqrt w \rvert^2} = w$ which proves our claim.

Concerning square roots it may be interesting to have a look at my answer to Analytic Functions - Entire Function .