# Local coordinates for a tranversal intersection of a curve with a coordinate axis

Suppose I have a smooth curve $\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$ in the $xy$-plane given by $t \mapsto \gamma(t)=(\gamma_1(t),\gamma_2(t))$ which intersects the $x$-axis transversely. Is it then possible to locally express $\gamma$ in terms of $\gamma_2$?

I have not been able to construct a counter example yet but I have not been able to come up with a proof either. Any help is welcome.

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1
The implicit function theorem should help you out here. – Kris Aug 8 '12 at 21:57

## 2 Answers

To flesh out my comment into an answer: the condition that $\gamma$ intersects the $x$-axis transversely is equivalent to stating that $\frac{\partial \gamma}{\partial \gamma_1}$ is non-zero, and so by the implicit function theorem we can write $\gamma_1$ as a function of $\gamma_2$ and hence $\gamma$ as a function of $\gamma_2$.

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Notice that formally $\partial\gamma/\partial\gamma_1=(1,0) \ne 0$ for any curve $\gamma=(\gamma_1,\gamma_2)$, therefore you are saying that every curve $\gamma=(\gamma_1,\gamma_2)$ intersects transversally the $x$-axis, which of course is not true. – Mercy Aug 8 '12 at 22:33
Guess you mean $\frac{\partial \gamma_2}{\partial \gamma_1}$ is non-zero. However, my problem is that the latter can be undefined. For example consider $\gamma=(t,t^2)$ at $t=0$. Does the implicit function theorem in these cases still hold? – user29751 Aug 9 '12 at 18:14

Let $t_0 \in \mathbb{R}$ be such that $\gamma_2(t_0)=0$. Since $\gamma$ intersects transversally the $x$-axis, we have $\dot{\gamma}_2(t_0) \ne 0$. Thanks to the inverse function theorem, there exist $\epsilon,\delta>0$ such that $\gamma_2: (t_0-\epsilon,t_0+\epsilon)\to (-\delta,\delta)$ is a diffeomorphism.

Hence $\gamma(\gamma_2^{-1}(\tau))=(\gamma_1\circ\gamma_2^{-1}(\tau),\tau) \quad \forall\ \tau \in (-\delta,\delta)$.

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Suppose I have the curve $\gamma:t \mapsto (t^3,t^3)$ then this curve intersects the $x$-axis transversely but $\dot{\gamma}_2(0)=0$! – user29751 Aug 9 '12 at 17:59
It's just a matter of parametrization. The curve $\gamma: t \mapsto (t^3,t^3)$ can be reparametrize as $s \mapsto (s,s)$. – Mercy Aug 9 '12 at 19:15
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