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 $E(p,t)=p+tp=(1+t)p$. The same computation as for the circle shows that the image for $\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}2$ is exactly the shell $1\mathrm{/}2<\mathrm{\parallel }x\mathrm{\parallel }<3\mathrm{/}2$. This is the Euclidean tubular neighbourhood in the sense of The Euclidean tubular neighbourhood theorem.

Each hairpin occupies a region disjoint from all the previous ones except for one joining segment, and the joins can be smoothed so that the velocity never vanishes. Therefore the concatenated curve is a smooth embedding of $\mathbb{R}$.

Fix $r>0$ and choose $n$ with $2^{-n}<2r$. In the $n$th hairpin the two nearly parallel strands are closer than $2r$, so normal discs of radius $r$ based on opposite strands intersect. Hence no tubular neighbourhood of constant radius $r$ can be injective there.

This realizes the failure asserted in FALSE: every noncompact submanifold has a uniform-radius tubular neighbourhood while remaining compatible with the variable-radius theorem The Euclidean tubular neighbourhood theorem.

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 $A$.

Applying Relative Whitney approximation for Euclidean-valued maps with any positive continuous error function produces a smooth map $\tilde{f}$ that equals $f$ on a smaller neighbourhood of $A$ and smooths the corner near $0$.

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.

The manifold-valued Whitney approximation theorem Whitney approximation for manifold-valued maps applies with target $S^1$, using the annular tubular neighbourhood from the standard circle example. It therefore produces a smooth map $\tilde{F}:S^1\to S^1$ homotopic to $F$.

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. This is exactly the boundary described in [L1].

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].