2
$\begingroup$

Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $\mathbb{R}^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$

Am i right in imagining the given homotopy as something like this?

enter image description here

such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $s\in [0,1]$) as drawn in the picture?

Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.

Thanks for any kind of feedback!

$\endgroup$
  • 1
    $\begingroup$ It's correct. Furthermore, any point $s$ moves along its straight line at constant speed. $\endgroup$ – Lukas Kofler Apr 21 at 22:53
  • $\begingroup$ isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$ $\endgroup$ – Zest Apr 21 at 22:57
  • 1
    $\begingroup$ That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies. $\endgroup$ – Lukas Kofler Apr 21 at 22:59
4
$\begingroup$

Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $\bar{s}$ and call $a := f_0(\bar{s})$ and $b := f_1(\bar{s})$. Examining the homotopy, $$ H_{\bar{s}}(t) := F(\bar{s}, t) = a(1-t) + bt $$

we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $\mathbb{R}^n$. If you draw this line for each $\bar{s}$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?

Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant, $$ \dfrac{dH_{\bar{s}}}{dt} = b - a $$

$\endgroup$
  • $\begingroup$ this is very helpful @jnez71. Thanks a lot! $\endgroup$ – Zest Apr 21 at 22:58
  • 1
    $\begingroup$ No problem! I added a bit more to cover some of the comments about speed $\endgroup$ – jnez71 Apr 21 at 23:07
1
$\begingroup$

Yes, it is absolutely correct.

$\endgroup$
  • $\begingroup$ thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"? $\endgroup$ – Zest Apr 21 at 22:52
  • $\begingroup$ See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation. $\endgroup$ – Paul Frost Apr 21 at 23:00
  • $\begingroup$ thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me? $\endgroup$ – Zest Apr 21 at 23:02
  • 1
    $\begingroup$ No problem - it is okay! $\endgroup$ – Paul Frost Apr 21 at 23:04
  • $\begingroup$ By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it. $\endgroup$ – Paul Frost Apr 21 at 23:17

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.