Prove that the trajectory of the image of a 3D fixed point by a certain parametric matrix is a smooth curve in $S^2$

Question

For $t \in \Bbb R$ we give the matrix

$R(t)=\begin{pmatrix} \cos(t) & − \sin(t) &0 \\ \sin(t)& \cos(t) & 0 \\ 0 & 0 &1\end{pmatrix}$

If $p \in \Bbb R^3$ (written as a column vector), then we define $\gamma_p : \Bbb R \to \Bbb R^3 $ by $\gamma_p(t)=R(t)p$

Prove that if $p \in S^2$, then $\gamma_p$ is a smooth curve in $S^2$, with $\gamma(0) = p$.

My attempt:

$\gamma_p(0)=R(0)p=Ip=p$. But is this the same as $\gamma(0)=p$?

Let $p=(x,y,z) \in S^2$, so $x^2+y^2+z^2=1$

Since $\gamma_p(t)=R(t)p=\begin{pmatrix} \cos(t) & − \sin(t) &0 \\ \sin(t)& \cos(t) & 0 \\ 0 & 0 &1\end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} x\cos t -y\sin t \\ x\sin t +y\cos t \\ z\end{pmatrix}$

is a function of $t$, I guess $p$ is a constant here, so $\gamma_p$ is obviously smooth since its components are smooth.

This looks suspiciously easy, given that this is an exercise of a manifolds course. (I don't even have to use that $x^2+y^2+z^2=1$ ?) Is this correct or is there something I am missing?

Answer 1

Part of the exercise is for you to show also that for each $t$, we have $\gamma_p(t)\in S^2$. I’ll let you fiddle around with this trivial bit of algebra (this is where you need that for $p=(x,y,z)\in S^2$, we have $x^2+y^2+z^2=1$).

Given your comments here and in the other post, smoothness here should be interpreted in the sense of parametrized curves.

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But now, there is one further small a-priori subtlety whose resolution may/may not be the goal of this exercise (the triviality/non-triviality of what I’m about to say depends on how much you’ve already proved in your course). When talking about smoothness of a function, one needs to know the domain, and the target.

• a-priori, $\gamma_p$ is a function $\Bbb{R}\to\Bbb{R}^3$. Proving smoothness of this function is obvious given that you wrote everything out in components and showed each of them is smooth (yes $p$ is constant).

Let me now assume you’ve also proved that for each $t$, $\gamma_p(t)\in S^2\subset\Bbb{R}^3$.

• At this stage, one might pose another natural question. We know that $S^2$ is a smooth embedded submanifold of $\Bbb{R}^3$. We also know that to talk about smoothness of a map, we simply need both its domain and target to be smooth manifolds. So, one could now wonder if the “target-restricted function” $\sigma_p:\Bbb{R}\to S^2$, $\sigma_p(t)=\gamma_p(t)$ is smooth as a mapping from the manifold $\Bbb{R}$ into the manifold $S^2$. Turns out the answer is yes.

A-priori, these are technically different notions of smoothness, but some basic unwinding of the definitions tells you that smoothness of $\gamma_p$ is equivalent to smoothness of $\sigma_p$ (and the only reason I introduced the two different notations is to highlight that a-priori these are not equivalent ‘by definition’… there’s something (minor) needed to be checked).

The general observation is the following:

Theorem.

Let $M,N$ be smooth manifolds, and $S\subset N$ a smooth embedded submanifold. Let $f:M\to N$ be a map such that its image lies in $S$, and let $\phi:M\to S$ be the map so-obtained by restricting the target space of $f$. Then, $f$ is smooth if and only if $\phi$ is smooth.