Local diffeomorphism between unit sphere and paraboloid

Question

Let $S^2$ is the unit sphere in $\mathbb{R}^3$ and $P$ is the paraboloid defined as $$ P = \left\{(x,y,z) \in \mathbb{R}^3 : z = \frac{x^2 + y^2}{2}\right\}. $$ Consider $f : S^2 \to P$ given by $f(x,y,z) = (x-z, x+z, 1-y^2)$. I want to find the points where $f$ is not a local diffeomorphism. I know that, in point of the form $p = (x,y,z)$ with $y \neq 0$, the function $f$ is a local diffeomorphism, because $d f_{p}$ has non trivial Jacobian. However, I don't know how to proceed to find the points where $f$ is not a local diffeomorphism.

Answer 1

Let $i: \mathbb{S}^2 \to \mathbb{R}^3$ denote the inclusion map and $\widetilde{f}: \mathbb{R}^3 \to \mathbb{R}^3$ be the obvious global map which induces $f$. Since $f = \widetilde{f} \circ i$, where $i: \mathbb{S}^2 \to \mathbb{R}^3$ is the inclusion map, we have $\operatorname{ker}(\mathrm{d}f_p) = \left(\mathrm{d} i_p\right)^{-1} \left( \operatorname{ker} d \widetilde{F}_p \right)$, which kills the problem right away. Indeed,

• at any point $p= (a, b, c) \in \mathbb{S}^2$ where $b \neq 0$, $\ker(\mathrm{d}f_p) = \left(\mathrm{d} i_p\right)^{-1} \left( \{0 \} \right) = \{ 0 \}$ (which implies $f$ is a local diffeomorphism at those points)

• at any point $p = (a, b, c)$ where $b = 0$, we have $T_p \mathbb{S}^2 = \{ p\}^{\perp} =\{(x, y, z) \in \mathbb{R}^3 \ \vert \ ax + cz = 0 \}$, so that $\ker(\mathrm{d} f_p) = \left(\mathrm{d} i_p\right)^{-1}\left( \{ (x, y, z) \in \mathbb{R}^3 \ \vert \ x = z = 0 \} \right) = \mathbb{R}(0, 1, 0)$, a one dimensional subspace of $T_p \mathbb{S}^2$, which implies $f$ is not a local diffeomorphism around any point where $b = 0$.

Therefore, the set of points where $f$ is not a local diffeomorphism is precisely the great circle $\{(x, y, z) \in \mathbb{R}^3 \ \vert \ y = 0 \} \cap \mathbb{S}^2 $.

Initially, I posted an answer using charts, which, as pointed out by Ted Shifrin, is very unnecessary. Since the details below are still useful in more general contexts, I won't delete them.

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Before getting into the nitty-gritty of it, let us review a more general context. Let $F: M \to N$ be a smooth map between surfaces. Take charts $\varphi: U \subset \mathbb{R}^2 \to M$, $\psi: V \subset \mathbb{R}^2 \to N$ centered around $p \in M$ and $F(p) \in N$ (i.e. $\varphi(0) = p$, $\psi(0) = F(p)$). We'll want to compute the matrix of $\mathrm{d}F_p : T_p S_1 \to T_{F(p)} S_2$ with respect to the basis $\{\mathrm{d}\varphi_0(e_1), \mathrm{d}\varphi_0(e_2) \} = \{\varphi_u(0), \varphi_v(0) \}$ and $\{ \mathrm{d}\psi_0(e_1), \mathrm{d}\psi_0(e_2) \} = \{\psi_u(0), \psi_v(0) \}$. The coordinate representation of $F$ is given by $\widehat{F} = \left(\widehat{F}_1, \widehat{F}_2\right)\doteq \psi^{-1} \circ F \circ \varphi : U \to V$. Since $F \circ \varphi = \psi \circ \widehat{F}$ (where both are defined), we see by the chain rule that $$\begin{aligned} \mathrm{d}(F\circ \varphi)_0(e_i) &= \mathrm{d}F_p(\varphi_i(0)) \\ &= \mathrm{d}\left(\psi \circ \widehat{F}\right)_0(e_i) \\ &= \mathrm{d} \psi_0 \left( \mathrm{d}\widehat{F}_0(e_i)\right) \\ &= \mathrm{d} \psi_0 \left( \frac{\partial \widehat{F}_1}{\partial x_i}(0) \ e_1 + \frac{\partial \widehat{F}_2}{\partial x_i}(0) \ e_2\right) \\ &= \frac{\partial \widehat{F}_1}{\partial x_i}(0) \ \psi_u(0) + \frac{\partial \widehat{F}_2}{\partial x_i}(0) \ \psi_v(0)\end{aligned}$$

This shows that the matrix representation of $\mathrm{d}F_p$ in the basis $\{\varphi_u(0), \varphi_v(0) \}$ and $ \{\psi_u(0), \psi_v(0) \}$ is given by

$$\begin{pmatrix} \frac{\partial \widehat{F}_1}{\partial u}(0) && \frac{\partial \widehat{F}_1}{\partial v}(0) \\ \frac{\partial \widehat{F}_2}{\partial u}(0) && \frac{\partial \widehat{F}_2}{\partial v}(0)\end{pmatrix}$$

Now, in our specific example, let's take the charts $$\mathbb{R}^2 \ni (x, y) \mapsto \varphi(x, y) = \left(\frac{2x}{1+x^2+y^2}, \frac{2y}{1+x^2 + y^2}, \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} \right)$$ and $$\mathbb{R}^2 \ni (a, b) \mapsto \psi(a, b)= \left(a, b, \frac{a^2 + b^2}{2} \right)$$

Going through all the necessary computations, we get

$$\displaystyle{ \det\left(\mathrm{d} F_{(x, y)}\right) = \frac{16 y}{(x^2 + y^2 + 1)^3} }$$

Now, by the chain rule, the differential of a local diffeomorphism is an isomorphism. Reciprocally, if the differential of a map is an isomorphism at some point, it's a local diffeomorphism around that point (by the inverse function theorem). So to find where $F$ is not a local diffeomorphism, we just need to look at the points where its differential has zero determinant. In our case, we found that $F$ is a not a local diffeomorphism around every point of $$\varphi\left(\mathbb{R} \times \{0\} \right) = \left( \{(x, y, z) \in \mathbb{R}^3 \ \vert \ y = 0 \} \cap \mathbb{S}^2 \right) \setminus \{(0, 0, 1)\}$$ I was careful to leave out $ (0, 0, 1)$ because $\varphi$ doesn't cover it, but you can do the same computations with $\varphi$ being the stereographic projection from the south pole instead of the north one and conclude that, in fact, the set of points where $f$ is not a local diffeomorphism is precisely the great circle $\{(x, y, z) \in \mathbb{R}^3 \ \vert \ y = 0 \} \cap \mathbb{S}^2 $.

Answer 2

It is easy to see that in a point of the form $p = (x,0,z)$, the function $f$ is not a local diffeomorphism. In general, the Jacobian of $f$ in a point $(x,y,z)$ is $4y$. If f was a local diffeomorphism around $p$, then $df_{p}$ would be invertible. But this is not possible.