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I wonder whether there is a smooth function $f:\mathbb{S}^{1}\rightarrow \mathbb{R}^{1}$ which may be homotoped to itself by a regular homotopy $H(x,t)$, i.e. by a smooth $H:\mathbb{S}^{1}\times\lbrack0,1]\rightarrow \mathbb{R}^{1}$ such that $H(x,0)=H(x,1)=f(x)$ and $\nabla H(x,t)\neq \mathbf{0}$ for any $(x,t)\in\mathbb{S}^{1}\times\lbrack0,1]$. Here $\nabla H=\left( \frac{\partial H}{\partial x},\frac{\partial H}{\partial t}\right) $ is the gradient.

Note, for example, that a constant function $f(x)\equiv C$ cannot be homotoped this way, since the homotopy $H(x,t)$ should acquire either its minimum, or its maximum in an interior point of $\mathbb{S}^{1}\times\lbrack0,1]$, but then of course $\nabla H$ vanishes there.

sSs01
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1 Answers1

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Yes, there exists such a function. Example: $f(x)=\sin2x(1+\frac12\sin x)$.

Proof. First, take the function $f_1(x)=f(x+\pi)=\sin2x(1-\frac12\sin x)$ and consider the linear homotopy $H_1(x,t)$ between $f_1$ and $f$: $$ H_1(x,t)=tf(x)+(1-t)f_1(x)=\sin2x(1-\frac12\sin x)+t\sin2x\sin x. $$ Then $\frac{\partial H_1}{\partial t}=\sin2x\sin x$ is zero for $x=0,\pi/2,\pi,3\pi/2$. On the other hand, $$ \frac{\partial H_1}{\partial x}=2\cos2x[1+(t-\tfrac12)\sin x] =\pm2[1+(t-\tfrac12)\sin x]\ne0 $$ for these $x$ and for $t\in[0,1]$, and we see that $\nabla H_1\ne0$ everywhere. It remains to set $$ H(x,t)=H_1(x-(1-t)\pi,t). $$ This is a homotopy of $f$ into itself that does the job, because the change of variables preserves the property that the gradient does not vanish.

Vladimir
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