Let $M$ be a smooth manifold ($M=\mathbb{T}^2$ is enough for my purposes) and $f,g:\mathbb{S}^{1}\to M$ two smooth loops, such that $f(1)=g(1)$ and $$\left.\frac{\text{d}}{\text{d}t}f\left(e^{2\pi it}\right)\right|_{t=0}=\left.\frac{\text{d}}{\text{d}t}g\left(e^{2\pi it}\right)\right|_{t=0}. $$ If $f$ and $g$ are homotopic relative to $\{f(1) \}$, then it is possible to prove that there is a smooth homotopy $$H:\mathbb{S}^1 \times[0,1] \to M $$ such that $H(s,0)= f(s)$, $H(s,1)=g(s)$ and $H(1,t) = f(1)=g(1)$.

My Doubt: Is it possible guarantee that there is a smooth homotopy $\tilde{H}: \mathbb{S}^1 \times [0,1]\to M$ between $f$ and $g$ (relative to $\{f(1)\}$) such that $$\left.\frac{\text{d}}{\text{d}s}\tilde{H}\left(e^{2\pi is},t\right)\right|_{s=0} = \left.\frac{\text{d}}{\text{d}s}f\left(e^{2\pi is}\right)\right|_{s=0}, \quad\forall \ t\in [0,1] ?$$

I was trying to demonstrate the above possible result using a local chart and "adjusting the slope" of $\frac{d}{ds}H(s,t)$, but I wasn't able to do such thing.

Solution when $M=\mathbb{T}^2$

If $M = \mathbb{T}^2$, then

$$p: \mathbb{R}^2 \to \mathbb{R}^2 /\mathbb{Z}^2 $$ $$x\mapsto [x] $$

is a covering map and a local diffeomorphism.

Lifting the loops $f,g$, we find the functions $\tilde{f}, \tilde{g}: [0,1] \to \mathbb{R}^2$, such that $$p\circ \tilde{f} = f,\ \text{and}\quad p\circ \tilde{g} = g. $$ Once $f$ and $g$ are homotopic, $\tilde f(0) = \tilde g(0)$ and $\tilde f(1) = \tilde g(1),$ moreover $\tilde f ' (0) = \tilde g ' (0)=\tilde f ' (1) = \tilde g ' (1)$ (we are taking the derivative in the extended sense, considering $[0,1]$ as a manifold with boundary).

Then, we can define \begin{align*}H:&[0,1]\times [0,1] \to \mathbb{T}^2\\ &(s,t) \mapsto p \circ ((1-t) \tilde f(s) + t \tilde g (s) ) , \end{align*} using that $H(0,t) = H(1,t)$ and $\frac{\text{d}H}{\text{d}s} (0,t) = \frac{\text{d}H}{\text{d}s} (1,t)$, we can induce the smooth homotopy ( where $\mathbb {S}^1 = [0,1]/\sim$)

\begin{align*}\tilde{H}:&\mathbb{S}^1\times [0,1] \to \mathbb{T}^2\\ &([ s],t) \mapsto p \circ ((1-t) \tilde f(s) + t \tilde g (s) ), \end{align*} which satisfies the requested conditions.

  • 1
    $\begingroup$ Wild guess: Look at the induced maps $\tilde{f},\tilde{g}: S^1\rightarrow TM$ (where $TM$ denotes the tangent bundle to $M$). I believe (but haven't though through the details) that $\tilde{f}$ and $\tilde{g}$ are homotopic if $f$ and $g$ are (homotopy lifting property?). Now, apply your "it is possible to prove...." claim to $\tilde{f}$ and $\tilde{g}$, so get a homotopy rel $\{f(1), f'(1)\}$, and then project this down. $\endgroup$ – Jason DeVito Aug 17 '18 at 21:24
  • $\begingroup$ Nice idea! I will try. $\endgroup$ – Matheus Manzatto Aug 17 '18 at 21:26
  • $\begingroup$ I was able to conclude that $\tilde{f} ,\tilde{g}$ are homotopic rel$\{f(1),f'(1)\}$. Then, there exists $H: \mathbb{S}^1 \times I \to TM$, such that $H(t,0) = \tilde{f}$ and $H(t,1) = \tilde{g}$, then we can write $H(t,s) = (F(t,s) , V(t,s))$, with $V(t,s) \in T_{H(t,s)} M$, do you know how I complete the demonstration? Because I think not necessarily $\pi \circ H (t,s) = F(t,s)$ is a homotopy "fixing the slope" $\endgroup$ – Matheus Manzatto Aug 18 '18 at 17:44
  • $\begingroup$ I agree that there is no reason why $\pi\circ H=F$ fixes the slope, not sure what I was thinking yesterday.... $\endgroup$ – Jason DeVito Aug 18 '18 at 19:40
  • $\begingroup$ Possible duplicate of Smooth homotopy $\endgroup$ – Alex M. Nov 16 '19 at 10:34

I found clear statement in the authoritative literature, I will introduce it for your reference. It seems to be a more general case answer than you want.

The second part of the following proposition seems to be the answer to your problem. All you have to do is substituting $\{ f(1) \}$ for A. See the P258 of ref 1 ( The second edition seems to be published now. I don't know which page has this proposition in the second edition.).  

enter image description here

(ref.1)John M. Lee; "Introduction to Smooth Manifolds (Graduate Texts in Mathematics, 218)" Springer (2002/9/23)

I'm not very good at English, so I'm sorry if I have some impolite or unclear expressions. I welcome any corrections and English review. (You can edit my question and description to improve them)

  • $\begingroup$ Thx for your answer, but, how do you know that $$\left.\frac{\text{d}}{\text{d}s}{H}\left(e^{2\pi is},t\right)\right|_{s=0} = \left.\frac{\text{d}}{\text{d}s}f\left(e^{2\pi is}\right)\right|_{s=0}, \quad\forall \ t\in [0,1] $$ holds? $\endgroup$ – Matheus Manzatto Oct 4 '19 at 8:36
  • $\begingroup$ I didn't mind because I thought that it could be solved by changing the variables appropriately. But think a little more. $\endgroup$ – Blue Various Oct 4 '19 at 10:44

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