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homotopic closed curves same starting point

dwsmith

Well-known member
Feb 1, 2012
1,673
Let $\gamma_0,\gamma_1$ and $\delta_0,\delta_1$ be four closed curves in U, all of which have the same initial point $z_0$ (for some parameterizations). Assume that $\gamma_0$ and $\delta_0$ are homotopic in U with homotophy given by $\Gamma_0$; and assume that $\gamma_1$ and $\delta_1$ are homotopic in U with homotophy given by $\Gamma_1$. Prove that the product curves $\gamma_0\gamma_1$ and $\delta_0\delta_1$ are also homotopic in U by exhibiting an explicit homotophy in terms of $\Gamma_0$ and $\Gamma_1$ (and showing briefly why it is a homotophy).

I am pretty lost on this problem.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Ok so I can intuitively see that this can work even if U isn't simply connected and wlog gamma_0 and delta_0 go around a doughnut. Due to the concatenation, the curves can be deformed to each other.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Are these homotopies correct?

$$
\Gamma_0=\begin{cases}\gamma_0(t,2s), & 0\leq s\leq\frac{1}{2}\\
\delta_0(t,2s-1), & \frac{1}{2}\leq s\leq 1
\end{cases}
$$

$$
\Gamma_1=\begin{cases}\gamma_1(t,2s), & 0\leq s\leq\frac{1}{2}\\
\delta_1(t,2s-1), & \frac{1}{2}\leq s\leq 1
\end{cases}
$$

So the product curve should be?

$$
\gamma_0\gamma_1=\begin{cases}\gamma_0(2t), & 0\leq t\leq\frac{1}{2}\\
\underbrace{\gamma_1(2t)}_{2t \ \text{as well?}}, & \frac{1}{2}\leq t\leq 1
\end{cases}
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I believe this should work and make the product curves homotopic to a point.
Given that $\Gamma_0$ and $\Gamma_1$ are homotopies let
$$
\gamma_0\gamma_1(t,s) =
\begin{cases}\gamma_0(2ts), & 0\leq t\leq\frac{1}{2}\\
\gamma_1(2s - 2ts), & \frac{1}{2}\leq t\leq 1
\end{cases}
\quad\text{and}\quad
\delta_0\delta_1(t,s) =
\begin{cases}\delta_0(2ts), & 0\leq t\leq\frac{1}{2}\\
\delta_1(2s - 2ts), & \frac{1}{2}\leq t\leq 1
\end{cases}.
$$
Since the direction of neither $\gamma_0$ and $\gamma_1$ nor $\delta_0$ and $\delta_1$ were specified, we can have $\gamma_1$ and $\delta_1$ run in the reverse direction as $\gamma_0$ and $\delta_1$ but they aren't specifically the same path.
Then $\gamma_0\gamma_1(0,s) = \gamma_0\gamma_1(1,s) = z_0$, $\gamma_0\gamma_1(t,0) = \gamma_0$, and $\gamma_0\gamma_1(t,1) = \gamma_1$.
Also, $\delta_0\delta_1(0,s) = \delta_0\delta_1(1,s) = z_0$, $\delta_0\delta_1(t,0) = \delta_0$, and $\delta_0\delta_1(t,1) = \delta_1$.
If $U$ is simply connected, then the product curves would be homotopic to a point.
If $U$ isn't simply connected such that two curves wrapped around a doughnut, then the product curves would be path homotopic.
 
Last edited:

Tinyboss

New member
Mar 3, 2012
7
I think you just want $\Gamma(t,s):=\begin{cases}\Gamma_0(t,2s)&&s\le \frac12\\\Gamma_1(t,2s-1)&&s\ge \frac12\end{cases}$, where $t$ is time (the homotopy parameter) and $s$ is going around the curve. In other words, at time $t$ what you have is the concatenation of something between $\gamma_0$ and $\delta_0$ with something between $\gamma_1$ and $\delta_1$. Graphically, if each homotopy is a function on the square $I\times I$, then what we did was shrink them horizontally so they are functions on $[0,\frac12]\times I$ and $[\frac12,1]\times I$, then we glued them together along $t=\frac12$, where they have the same values (because all the loops have the same base point). So to explain why the result is a homotopy, you just need to invoke the gluing lemma.

PS: My first post, and I'm already in love with this forum because I can use dollar-signs for TeX! Sweet freedom!
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
I think you just want $\Gamma(t,s):=\begin{cases}\Gamma_0(t,2s)&&s\le \frac12\\\Gamma_1(t,2s-1)&&s\ge \frac12\end{cases}$, where $t$ is time (the homotopy parameter) and $s$ is going around the curve. In other words, at time $t$ what you have is the concatenation of something between $\gamma_0$ and $\delta_0$ with something between $\gamma_1$ and $\delta_1$. Graphically, if each homotopy is a function on the square $I\times I$, then what we did was shrink them horizontally so they are functions on $[0,\frac12]\times I$ and $[\frac12,1]\times I$, then we glued them together along $t=\frac12$, where they have the same values (because all the loops have the same base point). So to explain why the result is a homotopy, you just need to invoke the gluing lemma.

PS: My first post, and I'm already in love with this forum because I can use dollar-signs for TeX! Sweet freedom!
Even by using the pasting lemma, how does that show the product curves are homotopic?
 

Tinyboss

New member
Mar 3, 2012
7
By construction, $\Gamma:I\times I\to U$ satisfies $\Gamma(0,\cdot)=\gamma_0\delta_0$ and $\Gamma(1,\cdot)=\gamma_1\delta_1$. This, together with continuity, is the definition of path homotopy.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
By construction, $\Gamma:I\times I\to U$ satisfies $\Gamma(0,\cdot)=\gamma_0\delta_0$ and $\Gamma(1,\cdot)=\gamma_1\delta_1$. This, together with continuity, is the definition of path homotopy.
But we want $\gamma_0\gamma_1$ to be the homotopic. We are given that those are homotopic already.
 

Tinyboss

New member
Mar 3, 2012
7
Oops, I just got my symbols mixed up in that last post. What I meant to say is:

By construction, $\Gamma:I\times I\to U$ satisfies $\Gamma(0,\cdot)=\gamma_0\gamma_1$ and $\Gamma(1,\cdot)=\delta_0\delta_1$. This, together with continuity, is the definition of path homotopy.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Oops, I just got my symbols mixed up in that last post. What I meant to say is:

By construction, $\Gamma:I\times I\to U$ satisfies $\Gamma(0,\cdot)=\gamma_0\gamma_1$ and $\Gamma(1,\cdot)=\delta_0\delta_1$. This, together with continuity, is the definition of path homotopy.
So this part is true via the pasting lemma?
 

Tinyboss

New member
Mar 3, 2012
7
No, that part is true by the way I constructed $\Gamma$: if you just step through the definitions (and if I didn't mess something up), you'll be able to verify that $\Gamma(0,s)=(\gamma_0\gamma_1)(s)$ and $\Gamma(1,s)=(\delta_0\delta_1)(s)$.

So $\Gamma$ is the homotopy we're after, if it's continuous. The pasting lemma gives you a sufficient condition for such a "piecewise-defined" function to be continuous, and $\Gamma$ satisfies the condition.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
No, that part is true by the way I constructed $\Gamma$: if you just step through the definitions (and if I didn't mess something up), you'll be able to verify that $\Gamma(0,s)=(\gamma_0\gamma_1)(s)$ and $\Gamma(1,s)=(\delta_0\delta_1)(s)$.

So $\Gamma$ is the homotopy we're after, if it's continuous. The pasting lemma gives you a sufficient condition for such a "piecewise-defined" function to be continuous, and $\Gamma$ satisfies the condition.
Shouldn't that be the other way around. If t = 0, shouldn't we be on $z_0$ and when s = 0, we should be on the first curve?
 

Tinyboss

New member
Mar 3, 2012
7
The role of $s$ is to move us around the curve, and the role of $t$ is to move us through the homotopy (think of $t$ as time, as you watch a movie where one curve is continuously deforming into another). So whenever $s\in\{0,\frac12,1\}$ we are on $z_0$, because at any time $t$, our homotopy $\Gamma$ is giving us the concatenation of two paths with basepoint $z_0$. Those two paths are exactly the ones given to us by $\Gamma_0$ and $\Gamma_1$ for that same value of $t$.

We just concatenated the homotopies, by concatenating each path in one with the path in the other corresponding to the same $t$ value.