Characteristic Class Obstructions to Immersions and Embeddings — Examples

For $m=9={1001}_2$ the condition $i\wedge 9=0$ for $0\le i\le 9$ excludes exactly the binary digit positions $0$ and $3$, so $i$ ranges over the numbers with digits only among positions $1$ and $2$, that is $i\in \{0,2,4,6\}$; hence $S_9=\{0,2,4,6\}$ and, by [F1], $\bar{w}({\mathbb{R}\mathbb{P}}^9)=(1+a{)}^{-10}=1+a^2+a^4+a^6\mathrm{.}$

The top nonzero term is ${\bar{w}}_6({\mathbb{R}\mathbb{P}}^9)=a^6$, nonzero by [F2]; hence $d(9)=6$ and [F1] forbids an immersion of ${\mathbb{R}\mathbb{P}}^9$ into ${\mathbb{R}}^{9+k}$ for every $k<6$, in particular $k=5$: there is no immersion into ${\mathbb{R}}^{14}$.

For $m=4={100}_2$ the condition $i\wedge 4=0$ for $0\le i\le 4$ allows exactly $i\in \{0,1,2,3\}$, so $S_4=\{0,1,2,3\}$, $d(4)=3$, and $\bar{w}({\mathbb{R}\mathbb{P}}^4)=(1+a{)}^{-5}=1+a+a^2+a^3\mathrm{.}$ The top nonzero term is ${\bar{w}}_3=a^3\mathrm{\ne }0$, so ${\mathbb{R}\mathbb{P}}^4$ does not immerse in ${\mathbb{R}}^{4+k}$ for $k<3$, in particular not in ${\mathbb{R}}^5$ or ${\mathbb{R}}^6$. The computations concern the class calculations only; no assertion is made about higher-codimension immersions or about embeddings.

We first compute the inverse. Since $m=2^r$, [F2] gives $(1+a{)}^m=1+a^m$, so the telescoping identity in ${\mathbb{F}}_2[a]$ reads $(1+a)(1+a+\cdots +a^{m-1})=1+a^m$. Multiplying by the same factor $(1+a{)}^m=1+a^m$ gives $(1+a{)}^{m+1}{\sum }_{i=0}^{m-1}{a}^i=(1+a^m{)}^2=1+a^{2m}\text{in }{\mathbb{F}}_2[a],$ using $(u+v{)}^2=u^2+v^2$ in characteristic two. Because $2m\ge m+1$, the term $a^{2m}$ vanishes in $R={\mathbb{F}}_2[a]\mathrm{/}(a^{m+1})$, so $(1+a{)}^{m+1}{\sum }_{i=0}^{m-1}{a}^i=1$ in $R$: the polynomial ${\sum }_{i=0}^{m-1}{a}^i$ is the inverse of $(1+a{)}^{m+1}$, and by [F1] $\bar{w}({\mathbb{R}\mathbb{P}}^m)=(1+a{)}^{-(m+1)}={\sum }_{i=0}^{m-1}{a}^i=1+a+\cdots +a^{m-1}\mathrm{.}$

The top coefficient is ${\bar{w}}_{m-1}({\mathbb{R}\mathbb{P}}^m)=a^{m-1}$, which is nonzero in ${\mathbb{F}}_2[a]\mathrm{/}(a^{m+1})$ because $m-1<m+1$ and the powers $1,a,\dots ,a^m$ are linearly independent. Suppose ${\mathbb{R}\mathbb{P}}^m$ embedded smoothly in ${\mathbb{R}}^{2m-1}$; here $m\ge 2$ and $k=m-1\ge 1$, so [F3] with this codimension forces ${\bar{w}}_{m-1}(T{\mathbb{R}\mathbb{P}}^m)=0$, contradicting the computed nonzero class. Hence no such embedding exists.

The cases $r=1$ and $r=2$ give $m=2$ and $m=4$: ${\mathbb{R}\mathbb{P}}^2$ does not embed in ${\mathbb{R}}^3$, and ${\mathbb{R}\mathbb{P}}^4$ does not embed in ${\mathbb{R}}^7$. The argument proves only non-embeddability: no assertion is made about the existence of an immersion of ${\mathbb{R}\mathbb{P}}^m$ in ${\mathbb{R}}^{2m-1}$, nor about embeddability in ${\mathbb{R}}^{2m}$ or in larger codimension. The only choice used is the AC assumed by the class and embedding suppliers.

By [F1] the torus $T^n$ is a compact connected abelian Lie group of dimension $n$. Choose a basis $v_1,\dots ,v_n$ of the tangent space $T_e{T}^n$ at the identity; by [F2] the corresponding left-invariant vector fields $X_1,\dots ,X_n$ are smooth global sections of $T{T}^n$ whose values at $e$ form a basis, and left invariance carries this basis to a basis of every tangent space $T_g{T}^n$ (translation by $g$ is a diffeomorphism and identifies $T_e{T}^n$ with $T_g{T}^n$). Hence $(X_1,\dots ,X_n)$ is a global frame of $T{T}^n$, smooth by [F2].

By [F3] the existence of the global frame of step 1.1 makes $T{T}^n$ trivial: $T{T}^n\cong {\epsilon }^n$ over $T^n$, with the trivialization determined by the frame.

By [F4] applied to the closed manifold $T^n$ with trivial tangent bundle, the trivial bundle is a rank-$k$ stable normal inverse of $T^n$ for every $k\ge 1$; consequently $T^n$ immerses in ${\mathbb{R}}^{n+k}$ for every $k\ge 1$, in particular in ${\mathbb{R}}^{n+1}$, and its normal classes are trivial: $\bar{w}(T^n)=w(T{T}^n{)}^{-1}=1^{-1}=1,\bar{p}(T^n)=p(T{T}^n{)}^{-1}=1^{-1}=1.$

Therefore no Stiefel-Whitney or Pontryagin class test of this page obstructs a Euclidean immersion of $T^n$: every class ${\bar{w}}_i(T^n)$, ${\bar{p}}_i(T^n)$ with $i\ge 1$ vanishes, and the Euler class of the trivial normal bundle vanishes as well. The example exhibits a Euclidean formal immersion with trivial normal class and says nothing about embeddability of tori; in particular it does not assert that $T^n$ embeds in ${\mathbb{R}}^{n+1}$ or in any other specific Euclidean space, nor does it identify a minimal immersion dimension below $n+1$. The only choice used is the AC assumed by the parallelizable-manifold proposition and its countable-choice input.

By [F1] the standard orientation of ${\mathbb{R}}^{m+1}$ together with the orientation of $M$ orients the normal line bundle $\nu$: the orientation of the ambient bundle and of the tangent bundle determine that of the rank-one transverse normal bundle. This is the coorientation of the hypersurface, and it is a datum determined by the two given orientations.

Use the Euclidean metric on the orthogonal normal line of $M$. By [F2] its positive unit vectors form a smooth nowhere-zero section: on overlaps two positive frames differ by a positive smooth function, which cancels on normalization. This is a global frame, so $\nu \cong {\epsilon }^1$ smoothly.

Consequently the normal bundle of the embedding is trivial, and by [F3] the normal line bundle realizes the normal classes of $M$: ${\bar{w}}_1(M)=w_1(\nu )=0$ because the trivial bundle has trivial total class, and ${\bar{w}}_i(M)=w_i(\nu )=0$ for every $i\ge 2$ by the rank convention for a rank-one bundle; likewise $\bar{p}(M)=p(\nu )=1$. Thus the degree-one and higher normal classes give no obstruction to a codimension-one immersion or embedding of $M$.

The Euler class of the normal line also vanishes: the section exhibited in step 2.1 is nowhere zero, so [F4] gives $e(\nu )=0$. Hence a nonzero normal Euler class is likewise no obstruction here, and every rank-one characteristic-class test of the page returns zero for an oriented hypersurface.

For the unit sphere $S^m$ the outward normal field $x\mapsto x$ of [F5] is a nowhere-zero global section of the normal line, hence a global frame, so the normal bundle of the inclusion is trivial by [F2] and [F5]; combined with the embedding normal identity [F3] this gives $T{S}^m\oplus {\epsilon }^1\cong {\epsilon }^{m+1}$ for the sphere's stable normal inverse, whence $\bar{w}(S^m)=1$ and $\bar{p}(S^m)=1$ as in step 3.1. The argument applies to every oriented hypersurface, uses AC through the Euler and characteristic-class suppliers and its countable-choice consequence through the normal-bundle identifications, and asserts nothing about embeddability in higher codimension or about uniqueness of embeddings.

For the standard embedding $i(p)=p$ the radial field $n_i(p)=p$ spans its orthogonal normal line and is smooth and nowhere zero. For the reflected embedding $r$, the field $n_r(p)=r(p)$ is normal because reflection preserves inner products: $⟨r(p),d{r}_p(v)⟩=⟨p,v⟩=0$. It is again a smooth nowhere-zero frame. Thus both normal lines are trivial. The embedding normal-bundle identity identifies their characteristic classes with the stable normal classes of $S^2$, and the trivial line has total Stiefel--Whitney and Pontryagin class $1$ and Euler class $0$.

The two embeddings have identical stable characteristic classes. Both are embeddings of the same closed oriented surface $S^2$ into ${\mathbb{R}}^3$ with trivial normal line bundle by [F1], so their normal bundles are isomorphic (both trivial). The normal classes $\bar{w}(S^2)=1$ and $\bar{p}(S^2)=1$ of [F1] are classes of the manifold $S^2$ and therefore the same for both embeddings, and the codimension-one Euler class of the trivial normal line is zero. Consequently every characteristic-class test considered on this page — the mod-two normal classes, the rational normal Pontryagin classes and the oriented Euler class — evaluates trivially for both $i$ and $r$, so no such test distinguishes them.

Suppose, for contradiction, that $i$ and $r$ were isotopic: there is a smooth isotopy of embeddings $F:S^2\times I\to {\mathbb{R}}^3$ with $F_0=i$ and $F_1=r$. Replacing $F$ by a reparametrisation in $t$ that is constant near the ends, the isotopy is constant near the ends without changing its endpoints; by [F2] and [F5] it extends to an ambient isotopy $H:{\mathbb{R}}^3\times I\to {\mathbb{R}}^3$ with $H_0={\mathrm{id}}_{{\mathbb{R}}^3}$, every $H_t$ a diffeomorphism, and $H_t\circ i=F_t$ for every $t$. In particular $H_1\circ i=r$.

The time-one map $H_1$ is an orientation-preserving diffeomorphism of ${\mathbb{R}}^3$ by [F3]. Since $H_1\circ i=r$ and $i,r$ have image $S^2$, it satisfies $H_1(S^2)=S^2$. As a diffeomorphism it maps the two connected components of ${\mathbb{R}}^3\setminus S^2$ onto the two components, and $H_1(\mathrm{int}{B}^3)$ is a component whose closure $H_1(B^3)$ is compact because $B^3$ is compact and $H_1$ is continuous: hence $H_1(\mathrm{int}{B}^3)$ is the bounded component $\mathrm{int}{B}^3$ and $H_1(B^3)=B^3$.

Since $H_1$ is an orientation-preserving diffeomorphism of ${\mathbb{R}}^3$ mapping $B^3$ onto itself, it maps the boundary $S^2$ to itself and its differential carries outward-pointing boundary vectors to outward-pointing boundary vectors; therefore the restriction $H_1{\mathrm{\mid }}_{S^2}:S^2\to S^2$ preserves the outward-normal-first boundary orientation. By [F3] and [F4], an orientation-preserving diffeomorphism of the connected closed oriented surface $S^2$ has degree $+1$.

On the other hand $H_1\circ i=r$ means $H_1{\mathrm{\mid }}_{S^2}=r{\mathrm{\mid }}_{S^2}$ as maps $S^2\to S^2$, and by [F4] the reflection has degree $-1$. This contradicts the degree $+1$ computed in step 3.1, so no isotopy from $i$ to $r$ exists. Hence two embeddings with identical (indeed trivial) normal bundles and identical stable characteristic classes need not be isotopic, and these characteristic classes do not determine isotopy. AC is inherited from the characteristic-class suppliers and supplies the countable choice used by isotopy extension.