# Showing that a quadrifolium function from a circle is an immersion

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: $$$$(2\cos2\theta\cos\theta − \sin2\theta\sin\theta)\frac{\partial}{\partial x_1} + (2\cos2\theta\sin\theta + \sin2\theta\cos\theta)\frac{\partial}{\partial x_2}$$$$

... 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

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 $$d\tilde{f}_{\theta}$$.

• 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.
– fr_
Commented Dec 26, 2023 at 14:46