In the article "The heat equation shrinking convex plane curves" by M. Gage and R. S. Hamilton, I didn't understand the observation did by the authors in parenthesis, i.e., the Gauss map injective imply that the curve is simple.

$\textbf{My attempt:}$

Let $\alpha:[0,2\pi] \longrightarrow \mathbb{R}^2$ defined by $\alpha(\theta) := \left( \int_0^{\theta} \frac{\cos (\tau)}{k(\tau)} d\tau, \int_0^{\theta} \frac{\sin (\tau)}{k(\tau)} d\tau \right)$.

On the one hand,

$$\frac{d\alpha}{d\theta} = \frac{d\alpha}{ds} \frac{ds}{d\theta} = \frac{1}{k(\theta)} \frac{d\alpha}{ds}.$$

On the other hand,

$$\frac{d\alpha}{d\theta} = \left( \frac{\cos (\theta)}{k(\theta)}, \frac{\sin (\theta)}{k(\theta)} \right) = \frac{1}{k(\theta)} (\cos (\theta), \sin (\theta)).$$

Comparing the two equalities, we have $ \frac{d\alpha}{ds} = (\cos (\theta), \sin (\theta)) = T(\theta)$, then $N(\theta) = (- \sin (\theta), \cos (\theta))$ and $T$ and $N$ are injectives because the functions sine and cosine are injectives on $[0,2\pi]$. As Frenet's frame $\{T, N\}$ forms a basis of dimension $2$, we see that $\{T, N\}$ span $\mathbb{2}$, in particular, span $\alpha([0,2 \pi])$, therefore exist functions $a, b: [0, 2\pi] \longrightarrow \mathbb{R}$ such that

$$\alpha(\theta) = a(\theta) T(\theta) + b(\theta) N(\theta),$$

but $\{T, N\}$ is an orthonormal basis, then $a(\theta) = \langle \alpha(\theta),T(\theta) \rangle$ and $b(\theta) = \langle \alpha(\theta),N(\theta) \rangle$ for each $\theta \in [0,2 \pi]$ so we get

$$\alpha(\theta) = \langle \alpha(\theta),T(\theta) \rangle T(\theta) + \langle \alpha(\theta),N(\theta) \rangle N(\theta)$$

I'm get stuck here. I think that I need suppose that $\alpha$ is not simple and find a contradiction, but I don't have idea how to do this now. Anyone would help me with this?

Thanks in advance!


I forgot to put the integral smybol in the definition of $\alpha$ and saw this only today! Sorry for this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.