Path-connectedness of the set of circle immersions into $\mathbb{R}^n$ when $n\geqslant 3$.

Question

Let $n$ be an integer greater than $3$ and let $I(\mathbb{S}^1,\mathbb{R}^n)$ be the set of immersions from $\mathbb{S}^1$ into $\mathbb{R}^n$.

While messing around with the Whitney-Grauestein theorem, I felt like the following claim is true:

Theorem. Let $f$ and $g$ be in $I(\mathbb{S}^1,\mathbb{R}^n)$, then there exists $F\colon\mathbb{S}^1\times[0,1]\rightarrow\mathbb{R}^n$ such that for all $t\in[0,1]$, $F(\cdot,t)\in I(\mathbb{S}^1,\mathbb{R}^n)$, $F(\cdot,0)=f$ and $F(\cdot,1)=g$.

Actually, taking a closer look at the sketch of a proof I gave here for the Whitney-Grauestein theorem, it can be seen that $\pi_0(I(\mathbb{S}^1,\mathbb{R}^n))\cong\pi_1(\mathbb{R}^n\setminus\{0\})\cong\pi_1(\mathbb{S}^{n-1})$, which gives the theorem right away.

However, I am not too satisfied with this proof, as I find it not enough explicit.

I am convinced that one can perturb $g$ into $h\in I(\mathbb{S}^1,\mathbb{R}^n)$ such that $g$ and $h$ are homotopic in $I(\mathbb{S}^1,\mathbb{R}^n)$ and for all $x\in\mathbb{S}^1$, the family formed by $f'(x)$ and $h'(x)$ is free in $\mathbb{R}^n$. Notice that this way the radial homotopy $(x,t)\mapsto(1-t)f(x)+th(x)$ would be a regular homotopy between $f$ and $g$. Hence, this leads to my:

Question. How to perturb $g$ in such a fashion?

My first guess would have been to use Sard's theorem to find a rotation $r$ of $\mathbb{R}^n$ such that $h:=r\circ g$ would suit my requirements. However, I find myself stuck doing it in details. Perhaps, I should rather consider orthogonal projections as in the Whitney embedding theorem. Nevertheless, I am reluctant to do it as it would not be obvious that the resulting immersion${}^*$ would be regularly homotopic to $g$.

Any enlightenment and/or references will be greatly appreciated!

$^*$ The orthogonal projection of an immersion onto $v^{\perp}$, where $v\in\mathbb{R}^n$ is not necessarily an immersion, but using Sard's theorem this pathological situation only happens for a negligible set of $v$.

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Edit. I found out that it is claimed in Introduction aux variétés différentielles by J. Lafontaine at page $131$ that: "Two immersions of $\mathbb{S}^1$ in $\mathbb{R}^n$ are always isotopic if $n\geqslant 3$" and the reader is encouraged to establish it adapting the proof of Whitney's embedding theorem.

Answer 1

As I finally found an answer to my question, I figured that it might be of interest to post it here.

Please note that I will only prove the claim for smooth immersions of $\mathbb{S}^1$ in $\mathbb{R}^n$, with $n\geqslant 4$. If I happened to find a refinement of my proof for $n=3$, I will share it with you.

First, let me establish the following:

Lemma. Let $n\geqslant 3$ and $f\colon\mathbb{S}^1\rightarrow\mathbb{S}^{n-1}$ be a smooth map, then $f(\mathbb{S}^1)$ is negligeable in $\mathbb{S}^{n-1}$.

Proof. For all $x\in\mathbb{S}^1$, the linear map $\mathrm{d}_xf\colon T_x\mathbb{S}^1\rightarrow T_{f(x)}\mathbb{S}^{n-1}$ is not surjective, indeed: $$\dim(\mathbb{S}^1)=1

From there, I will derive the:

Proposition 1. Let $n\geqslant 3$ and $g\in I(\mathbb{S}^1,\mathbb{R}^n)$ be smooth, then for almost all $v\in\mathbb{S}^{n-1}$, $\pi_v\circ g$ is an immersion, where $\pi_v$ is the orthogonal projection onto $v^{\perp}$.

Proof. Let us introduce the following map: $$\varphi\colon\begin{cases}\mathbb{S}^1 & \rightarrow &\mathbb{S}^{n-1}\\x&\mapsto&\displaystyle\frac{g'(x)}{\|g'(x)\|}\end{cases}.$$ Let $v\in\mathbb{S}^{n-1}$, since $\mathbb{S}^1$ has dimension $1$, $\pi_v\circ g$ is not an immersion if and only if there exists $x\in\mathbb{S}^1$: $$(\pi_v\circ g)'(x)=0.$$ Yet, for all $x\in\mathbb{S}^1$, one has: $$(\pi_v\circ g)'(x)=0\Leftrightarrow g'(x)\in\mathbb{R}v.$$ Finally, $\pi_v\circ g$ is not an immersion if and only if $\pm v\in\varphi(\mathbb{S}^1)$. Whence the result, using the lemma. $\Box$

The last step before the sought result is the:

Proposition 2. Let $g\in I(\mathbb{S}^1,\mathbb{R}^n)$ be smooth and $v\in\mathbb{S}^{n-1}$ such that $\pi_v\circ g\in I(\mathbb{S}^1,\mathbb{R}^n)$, where $\pi_v$ is the orthogonal projection onto $v^\perp$, then $g$ and $\pi_v\circ g$ are regularly homotopic.

Proof. Let define $H\colon\mathbb{S}^1\times[0,1]\rightarrow\mathbb{R}^n$ by: $$H(x,t)=(1-t)g(x)+t\pi_v(g(x)).$$ One checks that $H$ is continuous and such that $H(\cdot,0)=g$, $H(\cdot,1)=\pi_v\circ g$. The only thing left to show is that for all $t\in[0,1]$, $H(\cdot,t)\in I(\mathbb{S}^1,\mathbb{R}^n)$. Assume by contradiction that there exists $t\in[0,1]$ and $x\in\mathbb{S}^1$: $$\frac{\mathrm{d}H}{\mathrm{d}x}(x,t)=(1-t)g'(x)+t(\pi_v\circ g)'(x)=0.$$ Since $g'$ and $(\pi_v\circ g)'$ do not vanish such a $t$ is in $]0,1[$ and one has: $$\pi_v(g'(x))=\frac{t-1}{t}g'(x).$$ Therefore, since $g'(x)\neq 0$, $(t-1)/t$ is an eigenvalue of $\pi_v$ and is equal to $0$ or $1$, which is impossible. Whence the result. $\Box$

We are now able to prove the theorem :

Notice that applying proposition $1$ and $2$, it suffices to prove that any smooth immersions of $\mathbb{S}^1$ in $\mathbb{R}^2$ are regularly homotopic in $\mathbb{R}^n$, with $n\geqslant 4$. Let $g_1:=(g_{1,1},g_{1,2})$ and $g_2=(g_{2,1},g_{2,2})$ be two such immersions and let define the following immersions: $$\begin{align}\overline{g_1}&:=(g_{1,1}(x),g_{1,2}(x),0,\cdots,0)\\\overline{g_2}&:=(0,0,g_{2,1}(x),g_{2,2}(x),0,\cdots,0)\end{align}.$$ Then, let define $H\colon\mathbb{S}^1\times[0,1]\rightarrow\mathbb{S}^1$ by: $$H(x,t)=(1-t)\overline{g_1}(x)+t\overline{g_2}(x).$$ Notice that $H$ is continuous and such that $H(\cdot,0)=\overline{g_1}$, $H(\cdot,1)=\overline{g_2}$. Furthemore, assume by contradiction that there exists $t\in[0,1]$ and $x\in\mathbb{S}^1$ such that the derivative of $H(\cdot,t)$ vanish at $x$, then one has: $$(1-t){g_{1,1}}'(x)=(1-t){g_{1,2}}'(x)=t{g_{2,1}}'(x)=t{g_{2,2}}'(x)=0.$$ Since $g_1$ is an immersion, either ${g_{1,1}}'(x)\neq 0$ or ${g_{1,2}}'(x)\neq 0$, whence $t=1$. Similarly with $g_2$, one gets $t=0$, which is a contradiction. Therefore, at any $t\in[0,1]$, $H(\cdot,t)$ is an immersion. Whence the result.