Suppose $C_1$ and $C_2$ are regular curves on a regular surface $S$. Suppose $p$ is a point in $S$ where $C_1$ and $C_2$ are tangent, then if $\varphi:S\rightarrow S$ is a diffeomorphism, prove that $\varphi(C_1)$ and $\varphi(C_2)$ are regular curves which are tangent at $\varphi(p)$.

How would I start this? What would I need to show?

  • $\begingroup$ As a start, can you write out in formulas what it means for $C_1$ and $C_2$ to be tangent at $p$? $\endgroup$ – quarague Feb 11 at 10:55
  • $\begingroup$ @quarague Do you mean they have the same tangent plane at $p$? $\endgroup$ – JB071098 Feb 11 at 10:56
  1. First, we have to show that if $\varphi:S_{1}\rightarrow S_{2}$ is a diffeomorphism, and $C$ is a regular curve on $S_{1}$, then $\varphi(C)$ is a regular curve on $S_{2}$. To do so, let $\alpha:I\rightarrow S_{1}$ be a parametrization of $C_{1}$. Then $\varphi\circ\alpha:I\rightarrow S_{2}$ is a parametrization of $C_{2}$, and we have $$\tag{1}\label{1} (\varphi\circ\alpha)'(t)=d_{\alpha(t)}\varphi(\alpha'(t)). $$ You should check that this expression is nonzero. Can you see why this is the case?

  2. Next, we have two curves $C_{1}$ and $C_{2}$ that are tangent at $p$. This means that their velocity vectors at $p$ are parallel, i.e. there are parametrizations $\alpha_{1}$ and $\alpha_{2}$ of $C_{1}$ resp. $C_{2}$ such that $\alpha_{1}(0)=\alpha_{2}(0)=p$ and $$\tag{2}\label{2} \alpha_{1}'(0)=c\alpha_{2}'(0). $$ To show that $\varphi(C_{1})$ and $\varphi(C_{2})$ are tangent at $\varphi(p)$, we have to show that their velocity vectors $(\varphi\circ\alpha_{1})'(0)$ and $(\varphi\circ\alpha_{2})'(0)$ are parallel. Can you see that this follows from \eqref{1} and \eqref{2}?

  • $\begingroup$ Yes that pretty much answers my question thank you. $\endgroup$ – JB071098 Feb 11 at 12:03
  • $\begingroup$ Just to clarify, the fact that $\varphi\circ \alpha$ is a parametrization allows one to assert that the differential is nonzero correct? $\endgroup$ – JB071098 Feb 11 at 12:12
  • $\begingroup$ @JB071098 The expression (1) above is nonzero because $\alpha'(t)$ is nonzero (since $\alpha$ is regular) and $d_{\alpha(t)}\varphi$ is an isomorphism (since $\varphi$ is a diffeomorphism). $\endgroup$ – studiosus Feb 11 at 13:20

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