# Existence of a special homotopy between two smooth loops

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.

• 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. – Jason DeVito Aug 17 '18 at 21:24
• Nice idea! I will try. – Matheus Manzatto Aug 17 '18 at 21:26
• 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" – Matheus Manzatto Aug 18 '18 at 17:44
• I agree that there is no reason why $\pi\circ H=F$ fixes the slope, not sure what I was thinking yesterday.... – Jason DeVito Aug 18 '18 at 19:40
• Possible duplicate of Smooth homotopy – Alex M. Nov 16 '19 at 10:34

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.).
• 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? – Matheus Manzatto Oct 4 '19 at 8:36