Whitney Embedding Tubular Neighbourhoods and Approximation — Examples

Every vector orthogonal to $T_p{S}^1$ is a scalar multiple of $p$, so the displayed identification of the normal bundle is correct. Under that identification the normal addition map is $E(p,t)=p+tp=(1+t)p$.

If $\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}2$, then $1\mathrm{/}2<1+t<3\mathrm{/}2$, so $E$ lands in the stated annulus. Conversely, every nonzero $x$ in that annulus can be written uniquely as $x=\mathrm{\parallel }x\mathrm{\parallel }\frac{x}{\mathrm{\parallel }x\mathrm{\parallel }}=E\text{\hspace{0.17em}⁣}(\frac{x}{\mathrm{\parallel }x\mathrm{\parallel }},\mathrm{\parallel }x\mathrm{\parallel }-1)\mathrm{.}$ Hence this is exactly the tubular neighbourhood promised by The Euclidean tubular neighbourhood theorem.

The inverse formula in step 2.1 sends $x$ to $(x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel },\mathrm{\parallel }x\mathrm{\parallel }-1)$, so the associated retraction is $r(x)=x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }$, in agreement with A closed Euclidean submanifold has a smooth neighborhood retraction.

Since $T_p{S}^n=\{v\in {\mathbb{R}}^{n+1}:⟨v,p⟩=0\}$, the orthogonal complement is the one-dimensional span of $p$. Thus $N^{\perp }{S}^n\cong S^n\times \mathbb{R}$.

Under that identification, normal addition is $E(p,t)=p+tp=(1+t)p$. Since $1+t>0$ for $\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}2$, its image is exactly the shell $1\mathrm{/}2<\mathrm{\parallel }x\mathrm{\parallel }<3\mathrm{/}2$: the inverse is $x\mapsto (x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel },\mathrm{\parallel }x\mathrm{\parallel }-1)$, which is smooth there. Thus this is an explicit two-sided tubular neighbourhood.

The first coordinate of $\gamma$ is $t$, so $\gamma$ is injective with a continuous inverse on its image; also ${\gamma }^{\mathrm{\prime }}(t)=(1,f^{\mathrm{\prime }}(t))\mathrm{\ne }0$. Thus it is a smooth embedding, and its image is noncompact because its first-coordinate projection is all of $\mathbb{R}$.

For $k\ge 0$ put $t_k=\sqrt{\pi \mathrm{/}2+2\pi k}$. Direct differentiation gives $f^{\mathrm{\prime }}(t_k)=0$ and $f^{\mathrm{\prime }\mathrm{\prime }}(t_k)=-4t_k^2$, so $\mathrm{\mid }{f}^{\mathrm{\prime }\mathrm{\prime }}(t_k)\mathrm{\mid }\to \mathrm{\infty }$.

In the normal-bundle coordinates $(t,v)$, normal addition is $E(t,v)=\gamma (t)+v\nu (t)$. At $t_k$, $\nu (t_k)=(0,1)$ and ${\nu }^{\mathrm{\prime }}(t_k)=(-f^{\mathrm{\prime }\mathrm{\prime }}(t_k),0)$, so ${\mathrm{\partial }}_{t}E(t_k,v)=(1-v{f}^{\mathrm{\prime }\mathrm{\prime }}(t_k),0)$ and ${\mathrm{\partial }}_{v}E(t_k,v)=(0,1)$. At $v_k=1\mathrm{/}{f}^{\mathrm{\prime }\mathrm{\prime }}(t_k)$ the differential of $E$ is singular.

Given $r>0$, choose $k$ with $\mathrm{\mid }{v}_k\mathrm{\mid }=1\mathrm{/}(4t_k^2)<r$. Then $(t_k,v_k)$ lies inside the constant-radius normal tube, but its normal addition map is not a local diffeomorphism there. Hence that tube cannot be a tubular diffeomorphism for any fixed $r>0$.

On any point of the smaller arc where ${\varphi }_{+}=1$, the first two coordinates of $F$ recover the chart coordinate ${\theta }_{+}$. The same is true for ${\varphi }_{-}=1$ on the other smaller arc.

Because those smaller arcs cover $S^1$, equality $F(p)=F(q)$ forces equality in one active chart coordinate, hence $p=q$. The active chart coordinates also show that $dF$ is injective at each point. This is exactly the compact coordinate-bump mechanism of A finite coordinate-bump map embeds a compact manifold in some Euclidean space.

The curve $\gamma$ is smooth and embedded: the first two coordinates already parametrize the standard unit circle, so the third coordinate only lifts that circle into space.

The projection $\pi$ forgets the $y$-coordinate. Since $\cos (-t)=\cos t$ and $\cos (-2t)=\cos 2t$, one has $\pi (\gamma ([-t]))=\pi (\gamma ([t]))$ for every $t$, while $\gamma ([-t])\mathrm{\ne }\gamma ([t])$ away from $[0]$ and $[\pi ]$. Thus the projected curve has double points.

This is a concrete bad direction of the type excluded by A generic linear projection preserves injectivity and immersion, so it witnesses the false claim FALSE: an arbitrary linear projection of an embedding is an embedding.

The function $f$ is continuous on $\mathbb{R}$ and smooth on the open neighbourhood $(-\mathrm{\infty },-1\mathrm{/}2)$ of $A$.

Let $\epsilon :\mathbb{R}\to (0,\mathrm{\infty })$ be any continuous error function. By continuity at $0$, choose $0<\delta <\min \{1\mathrm{/}2,\epsilon (0)\mathrm{/}2\}$ such that $\epsilon (x)>\epsilon (0)\mathrm{/}2$ whenever $\mathrm{\mid }x\mathrm{\mid }\le \delta$.

Let $\sigma$ be the standard smooth step from The standard smooth step function and set $\tilde{f}(x)=x\sigma ((x+\delta )\mathrm{/}(2\delta ))$. This is smooth, equals $0=f(x)$ for $x\le -\delta$, and equals $x=f(x)$ for $x\ge \delta$. In particular it agrees with $f$ on the open neighbourhood $(-\mathrm{\infty },-\delta )$ of $A$.

For $\mathrm{\mid }x\mathrm{\mid }\le \delta$, the bound $0\le \sigma \le 1$ gives $\mathrm{\mid }\tilde{f}(x)-f(x)\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }\le \delta <\epsilon (0)\mathrm{/}2<\epsilon (x)$; outside this interval the error is $0<\epsilon (x)$. Thus this explicit smoothing has the stated relative agreement and arbitrary positive continuous error control.

The map $F$ is continuous and well defined because the endpoint values and derivatives at $t=\pm \pi$ agree. Near $t=0$, $\mathrm{\mid }\sin (t\mathrm{/}2)\mathrm{\mid }$ has an absolute-value cusp, so $F$ fails to be smooth there.

Fix $0<\delta <1\mathrm{/}2$ and put $g_{\delta }(t)=\sqrt{{\sin }^2(t\mathrm{/}2)+{\delta }^2}-\delta$ and $H_{\delta }(e^{it})=(1-\delta )e^{i{g}_{\delta }(t)}$. Because ${\sin }^2(t\mathrm{/}2)=(1-\cos t)\mathrm{/}2$, the function $g_{\delta }$ is smooth and $2\pi$-periodic, so $H_{\delta }:S^1\to {\mathbb{R}}^2$ is smooth. Moreover, for $u=\mathrm{\mid }\sin (t\mathrm{/}2)\mathrm{\mid }$ one has $0\le u-g_{\delta }(t)\le \delta$, giving $\mathrm{\mid }{H}_{\delta }(e^{it})-F(e^{it})\mathrm{\mid }\le 2\delta$. Thus $H_{\delta }$ uniformly approximates the planar representative of $F$ as $\delta \downarrow 0$. Its image lies in the annulus $1\mathrm{/}2<\mathrm{\mid }z\mathrm{\mid }<3\mathrm{/}2$.

The radial retraction $R(z)=z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ is smooth on that annulus, and $\tilde{F}=R\circ H_{\delta }=e^{i{g}_{\delta }(t)}$ is a smooth circle-valued map. The formula $\mathcal{H}(e^{it},u)=\exp \text{\hspace{0.17em}⁣}\left(i\right((1-u)\mathrm{\mid }\sin (t\mathrm{/}2)\mathrm{\mid }+u{g}_{\delta }(t)\left)\right)$ for $0\le u\le 1$ is continuous and $2\pi$-periodic in $t$, so it defines a homotopy on $S^1$ from $F$ to $\tilde{F}$. This constructs the claimed smoothing and annular retraction directly.

The derivative of the scalar function $s(x,y)=x$ is $d{s}_{(x,y)}(u,v)=u$. At every zero $(0,y)$ this is surjective onto the fibre $\mathbb{R}$, so $s$ is transverse to the zero section.

Therefore the conclusion of A smooth section transverse to the zero section has a submanifold zero set holds in this concrete case, and the zero locus is the hypersurface $\{x=0\}$.

Every point of $S^1$ is distance $1$ from the origin. Hence the origin has infinitely many nearest points on $S^1$.

Therefore nearest-point projection is not uniquely defined at the origin, which lies outside every sufficiently small annular tubular neighbourhood of the circle.

For every $ϵ>0$, the smooth function $g_{ϵ}(x):=ϵe^{-x^2}$ satisfies ${\sup }_x\mathrm{\mid }{g}_{ϵ}(x)-f(x)\mathrm{\mid }\le ϵ$. So $g_{ϵ}$ is an arbitrarily close smooth approximation to $f$.

But $g_{ϵ}(0)=ϵ\mathrm{\ne }0=f(0)$, so this approximation does not preserve the prescribed value on $A$. Thus closeness alone is weaker than the relative conclusion in [L1].