I am trying to prove that a certain function is an immersion, but I am confused as to how to do it.

I have a function $f: S^1 \rightarrow \mathbb{R}^2$ s.t. $(\cos\theta, \sin\theta) \mapsto (\sin2\theta\cos\theta, \sin2\theta\sin\theta)$, i.e. a function that draws a quadrifolium on a plane. I'm supposed to show that it is an immersion, i.e. that $T_pf: T_{p}S^1 \rightarrow T_{f(p)}\mathbb{R}^2$ is injective. The answer sheet suggests that I compute $T_pf$ explicitly by considering its action on a vector $(-\sin\theta\frac{\partial}{\partial x_1}+\cos\theta\frac{\partial}{\partial x_2})$. Then, I should obtain: \begin{equation} (2\cos2\theta\cos\theta − \sin2\theta\sin\theta)\frac{\partial}{\partial x_1} + (2\cos2\theta\sin\theta + \sin2\theta\cos\theta)\frac{\partial}{\partial x_2} \end{equation}

... and then figure out why $T_pf$ is injective.

I have troubles obtaining this formula. In case $g$ is a function between $\mathbb{R^n}$ and $\mathbb{R^m}$, I can think of $T_pg$ as a Jacobian matrix, and compute it. But, I don't know how to transform $f$ to think of it as a function from $\mathbb{R}^n$. I know that $S^1 \subset \mathbb{R^2}$, but thinking of $f$ as a function of $(r, \theta)$ and keeping $r = 1$ didn't work. Or should I parametrize the circle by $h: \mathbb{R} \rightarrow S^1$ and then consider the Jacobian $(\frac{df_i \circ h}{dx})$?


1 Answer 1


It helps to consider what the definition of the tangent space of $S^1$ is and how the tangent map is defined.

If $M$ is a smooth manifold $T_pM$ for $p\in M$ is defined to be the space of equivalence classes of all smooth curves $\gamma: (-1,1)\to M$ with $\gamma(0)=p$ with the equivalence relation being with respect to the derivative in any chart containing $p$. The circle considered as a subset of the plane $S^1\subset \mathbb{R}^2$ has charts given by $\phi:\mathbb{R}\to S^1$ by $\theta\mapsto (\cos(\theta), \sin(\theta))$. This makes $S^1$ into a submanifold of $\mathbb{R}^2$ with tangent space at $\phi(\theta)$ given by the image of the Jacobian $d\phi_{\theta}:\mathbb{R}\to \mathbb{R}^2$ (what is the Jacobian of $\phi$?). The image of $1$ under the Jacobian is $v_{\theta}=-\sin(\theta) e^1+\cos(\theta)e^2$ and since $\mathbb{R}$ is one dimensional, this vector spans $T_{\phi(\theta)}S^1$.

Since $\phi(\theta+2\pi)=\phi(\theta)$ we can identify $S^1$ with $\mathbb{R}/2\pi\mathbb{Z}$ and any map $f:S^1\to \mathbb{R}^2$ is the same as a map from $\tilde{f}:\mathbb{R}\to \mathbb{R}^2$ such that $\tilde{f}(\theta+2\pi)=\tilde{f}(\theta)$. This means that we can calculate $T_\theta f$ by calculating $d\tilde{f}_{\theta}$, the map $d\phi_{\theta}$ gives us an isomorphism and tells us that the image of $v_{\theta}$ under $T_{\phi(\theta)}f$ is the same as the image of $1$ under $d\tilde{f}_{\theta}$.

  • $\begingroup$ Thanks! I guess I was getting there with the idea of "parametrizing" $S^1$ with $h$ (although I should have meant "charting") and then considering the Jacobian of $f \circ h$, but your answer cleared it all up nicely, and was very helpful. $\endgroup$
    – fr_
    Commented Dec 26, 2023 at 14:46

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