# An observation made on Lemma 4.1.1 of “The heat equation shrinking convex plane curves”

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!

$\textbf{EDIT:}$

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