2

I have to answer the following question:

Given $f:\Bbb{S}^2\to \Bbb{S}^2$ with $(x,y,z)\mapsto (\cos(30^{\circ} )x+\sin(30^{\circ})y,\sin(30^{\circ})x-\cos(30^{\circ})y,z)$ show that $df_p(v)\neq 0$ for all $v\in T_p \Bbb{S}^2$

I am a bit confused: What is the $df_p(v)$ here? I know this definition of the $df_p$:

enter image description here

But sometimes, some books call the Jacobian of the parametrization of the surface of "differential". Doing it as this definition with curve $\alpha(t)=(x(t),y(t))$, we get for each coordinate (I'm writing each coordinate in a line because the expressions got very long).

  • $\frac{1}{2} \left(\sin (x(0)) \left(y'(0) \cos (y(0))-\sqrt{3} x'(0) \sin (y(0))\right)+\cos (x(0)) \left(x'(0) \sin (y(0))+\sqrt{3} y'(0) \cos (y(0))\right)\right)$

  • $\frac{1}{2} \left(\cos (x(0)) \left(y'(0) \cos (y(0))-\sqrt{3} x'(0) \sin (y(0))\right)-\sin (x(0)) \left(x'(0) \sin (y(0))+\sqrt{3} y'(0) \cos (y(0))\right)\right)$

  • $y'(0) (-\sin (y(0)))$

Now I must choose a convenient curve and show it's different of $0$?

Red Banana
  • 23,956
  • 20
  • 91
  • 192

1 Answers1

3

First, the statement needs to be slightly amended. Recall that the differential $df_{p}\colon \operatorname{T}_{p}S^{2} \to \operatorname{T}_{f(p)}S^{2}$ is a linear map. In particular, $df_{p}(\mathbf{0}) = \mathbf{0}$. We thus have a trivial counterexample to your assertion. What is true, however, is that this is the only counterexample, i.e., that $df_{p}(v) \neq \mathbf{0}$ provided that $v \neq \mathbf{0}$. I prove this as follows.

Let us consider the circumstances under which $(f \circ \alpha)'(0) = \mathbf{0}$. I first substitute the formula for $f$ to obtain a formula for $f \circ \alpha$. $$ t \mapsto \begin{pmatrix} \cos(30^{\circ}) \alpha_{1}(t) + \sin(30^{\circ}) \alpha_{2}(t) \\ \sin(30^{\circ}) \alpha_{1}(t) - \cos(30^{\circ}) \alpha_{2}(t) \\ \alpha_{3}(t) \end{pmatrix}. $$ The key is to the rest of this computation is to note that $f$ is actually linear; the trigonometric coefficients are constant. The derivative is therefore a quick application of the chain rule. $$ (f \circ \alpha)'(0) = \begin{pmatrix} \cos(30^{\circ}) \alpha_{1}'(0) + \sin(30^{\circ}) \alpha_{2}'(0) \\ \sin(30^{\circ}) \alpha_{1}'(0) - \cos(30^{\circ}) \alpha_{2}'(0) \\ \alpha_{3}'(0) \end{pmatrix} = \begin{pmatrix} \cos(30^{\circ}) & \sin(30^{\circ}) & 0 \\ \sin(30^{\circ}) & -\cos(30^{\circ}) & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \alpha_{1}'(0) \\ \alpha_{2}'(0) \\ \alpha_{3}'(0) \end{pmatrix}. $$ Since the matrix has determinant $-1$, the derivative $(f \circ \alpha)'(0)$ vanishes if and only if $\alpha'(0) = \mathbf{0}$, i.e., $v = \mathbf{0}$. This proves the amended assertion.

In closing, let me add two remarks.

  1. We expect the result because $f$ is a rotation of the sphere about the $z$-axis by 30 degrees composed with a reflection across the $x$-$z$ plane, and so regular curves remain regular.

$$ \begin{pmatrix} \cos(30^{\circ}) & \sin(30^{\circ}) & 0 \\ \sin(30^{\circ}) & -\cos(30^{\circ}) & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos(-30^{\circ}) & -\sin(-30^{\circ}) & 0 \\ \sin (-30^{\circ}) & -\cos(-30^{\circ}) & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$

  1. Note that the matrix with determinant $-1$ above is the Jacobian matrix of $f$ viewed as a map from $\mathbf{R}^{3} \to \mathbf{R}^{3}$.
destine
  • 288
  • 2
    the determinant is $-1$ – peek-a-boo Apr 11 '23 at 03:51
  • @peek-a-boo Thanks! – destine Apr 11 '23 at 03:52
  • @destine Yeah, it seems someone who was doing the exercise with me found the problem when $v=0.$ – Red Banana Apr 11 '23 at 04:45
  • Im confused with this: This is a map from the sphere to the sphere, why don't we need to pick a parametrization for the sphere to compute this? The formulas I obtained were made by picking a parametrization for the sphere $X: R^2 \to S^2$, taking a curve $\alpha(t)$ and then computing $X(f(\alpha(t)))$, differentiating this I obtained the formulas I came up with. – Red Banana Apr 11 '23 at 06:05
  • 1
    @RedBanana You should be computing $f \circ X \circ \alpha$. Notice $\alpha\colon (-\epsilon,\epsilon) \to \mathbf{R}^{2}$ is a curve in the plane, $X \circ \alpha\colon (-\epsilon,\epsilon) \to S^{2}$ is the corresponding curve in the sphere, and $f \circ X \circ \alpha$ is the image of that curve under the self-map of $S^{2}$. – destine Apr 11 '23 at 14:19
  • 1
    @RedBanana Re: your first question. I do not know your definition for parametrization, but I assume parametrizations are immersions. If this is the case, then the argument above still follows through, for the curve $X \circ \alpha$ is but a curve in $\mathbf{R}^{3}$ and regular curves in the plane correspond via $X$ to regular curves in $\mathbf{R}^{3}$. – destine Apr 11 '23 at 17:00
  • @destine A parametrization is a map $X: R^2 \to S$ where $S$ is a surface. I was talking to my professor today, he explained me that we do not need to assume we have such a parametrization. It's a still a bit confusing to me but it seems this simplifies the computations A LOT. – Red Banana Apr 11 '23 at 18:14