0
$\begingroup$

Let $\gamma:[0,1]\to \Bbb C$ be a closed rectifiable curve and consider $\gamma^{-1}:[0,1]\to \Bbb C$ given by $\gamma^{-1}=\gamma(1-t)$. Show that trace($\gamma$)=trace($\gamma^{-1}$).

I tried to show that $\gamma$ and $\gamma^{-1}$ is equivalent, since equivalent paths has the same trace. However I found that they are not equivalent, since the $\gamma^{-1}(t)=\gamma\circ\phi$, where $\phi(t)=1-t$ doesn't satisfy the condition strictly increasing and also does not satisfy $\phi(0)=0,\phi(1)=1$.

Intuitively I believe this is true, but I am having trouble proving it. Could anyone kindly help? Thanks so much!

$\endgroup$
  • 1
    $\begingroup$ Can you define the trace of a curve? $\endgroup$ – mathcounterexamples.net Nov 30 '15 at 7:41
  • $\begingroup$ This is not (complex-analysis). $\endgroup$ – Did Nov 30 '15 at 8:38
2
$\begingroup$

The trace of a curve $\gamma:[0,1]\to \Bbb C$ (as I remember it) is just the range: $$ \text{trace}(\gamma) = \{ \gamma(t) \mid t \in [0, 1] \} \, . $$ Then $\text{trace}(\gamma) = \text{trace}(\gamma^{-1})$ follows directly from $$ t \in [0, 1] \Longleftrightarrow 1- t \in [0, 1] \, . $$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.