Smooth Cobordism Relations Groups and Rings — Examples

($D^2$ is a compact smooth surface with boundary $S^1$.) By [F1] with $n=2$, $D^2=B^2$ is a smooth manifold with boundary $S^1$, and $\rho (x)=1-\mathrm{\mid }x{\mathrm{\mid }}^2$ is a boundary-defining function; by compactness of Euclidean balls $D^2$ is compact, hence a compact surface with boundary.

(The induced boundary orientation is counterclockwise.) Let $p\in S^1$. Since $\rho (x)=1-\mathrm{\mid }x{\mathrm{\mid }}^2$ has gradient $\mathrm{\nabla }\rho (p)=-2p$, the function $\rho$ decreases in the radial direction, so the outward normal of $D^2$ at $p$ is the radial vector $p$ (unit length). Let $Jp=(-p_2,p_1)$ be the counterclockwise rotation of $p$ by ${90}^{\circ }$. In the standard orientation of ${\mathbb{R}}^2$ the basis $(p,Jp)$ is positive, because $\det (p,Jp)=\mathrm{\mid }p{\mathrm{\mid }}^2=1>0$. The outward-normal-first rule of [F2] therefore says that $(p,Jp)$ is a positive basis of $T_p{D}^2$ exactly when $Jp$ is a positive basis of $T_p{S}^1$; the unit tangent $Jp$ is the counterclockwise direction of $S^1$, so the induced boundary orientation of $S^1=\mathrm{\partial }{D}^2$ is counterclockwise.

(Null-cobordisms of the two circles.) Let $o$ be the counterclockwise orientation of $S^1$. The map $\theta :[0,1)\times S^1\to D^2$, $\theta (s,p)=(1-s\mathrm{/}2)p$, is a smooth embedding onto the open annulus $\{x:1\mathrm{/}2<\mathrm{\mid }x\mathrm{\mid }\le 1\}$ and satisfies $\theta (0,p)=p$, so it is a supplied collar (Smooth collars of a manifold boundary). With the whole boundary incoming, the standard orientation on $D^2$ gives induced orientation $o$ by step 1.2, so it null-bords $(S^1,-o)$. Reversing the disk orientation gives induced boundary orientation $-o$ and null-bords $(S^1,o)$. Thus both oriented classes and their underlying unoriented classes are zero by [F3].

(The two classes are mutually inverse.) Let $W:=(-D^2)\bigsqcup D^2$ with the disjoint-union orientation, a compact oriented surface whose boundary is the disjoint union of the two circles, and whose induced boundary orientation on $\mathrm{\partial }W$ is (clockwise)$\bigsqcup$(counterclockwise). As a bordism from the closed oriented manifold $S^1\bigsqcup (-S^1)$ to $\mathrm{\varnothing }$ it realises $[S^1]+[-S^1]=[\mathrm{\varnothing }]=0$ in ${\mathrm{\Omega }}_1^{SO}$ by [F3]: the incoming face carries the negative of $o\bigsqcup (-o)$, namely $(-o)\bigsqcup o$, which is exactly the induced orientation of $\mathrm{\partial }W$.

(Assembly.) Steps 1.1–1.2 identify $D^2$ as a compact smooth surface with boundary $S^1$ and compute its induced boundary orientation as counterclockwise; step 2.1 gives the null-cobordisms of both oriented circles, so $[S^1]=0$ in ${\mathrm{\Omega }}_1^{SO}$ and in ${\mathrm{\Omega }}_1^O$; step 2.2 shows that the class of the opposite orientation is also $0$ and is the inverse of $[S^1]$. This is the asserted example.

(Two points bound an interval.) By [F1], $W=[-1,1]$ is a compact smooth one-manifold with boundary $\mathrm{\partial }W=\{-1,1\}$. Define $\theta :[0,1)\times \{-1,1\}\to W$ by $\theta (s,-1)=-1+\frac{s}{2}$ and $\theta (s,1)=1-\frac{s}{2}$; the two images are the disjoint intervals $[-1,-\frac{1}{2})$ and $(\frac{1}{2},1]$, whose union is an open neighbourhood of $\mathrm{\partial }W$, and $\theta (0,-1)=-1$, $\theta (0,1)=1$, so $\theta$ is a collar exhibiting all of $\mathrm{\partial }W$ as the image of the source. Transporting this collar along the bijection $\{a,b\}\to \{-1,1\}$ gives a compact one-manifold whose whole boundary is $\{a,b\}$ with a collar. Hence the two-point manifold $\{a,b\}$ is null-cobordant, its class is zero, and $[\mathrm{p}\mathrm{t}]+[\mathrm{p}\mathrm{t}]=[\mathrm{p}\mathrm{t}\bigsqcup \mathrm{p}\mathrm{t}]=0$ in ${\mathrm{\Omega }}_0^O$.

(A single point is not null-cobordant.) By [F3] the parity of the cardinality is a complete invariant of ${\mathrm{\Omega }}_0^O$ and equals $1$ on a one-point manifold; hence the class of a point is nonzero, and a one-point manifold does not bound a compact one-manifold.

(The generator and the even case.) By [F3] the group ${\mathrm{\Omega }}_0^O$ has exactly two elements; step 1.1 shows that the nonzero class of a point is its own inverse, and step 1.2 shows that it is nonzero, so it is the unique nonzero element and generates ${\mathrm{\Omega }}_0^O\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}$. Finally, a closed zero-manifold with an even number of points has parity zero, so by [F3] it bounds a compact one-manifold and is null-cobordant.

(The interval realizes $[{\mathrm{p}\mathrm{t}}_{+}]+[{\mathrm{p}\mathrm{t}}_{-}]=0$.) Let $x\mathrm{\ne }y$ be two points and orient the zero-manifold $\{x,y\}$ so that $x$ is positive and $y$ is negative. On the compact interval $[-1,1]$ with its standard orientation the induced boundary orientation is $\{1\}-\{-1\}$ by [F1], that is, the point $1$ is positive and the point $-1$ is negative. Define the collar $\theta :[0,1)\times \{x,y\}\to [-1,1]$ by $\theta (s,x)=-1+\frac{s}{2}$ and $\theta (s,y)=1-\frac{s}{2}$; its images are disjoint open intervals around the two endpoints; taking the whole boundary as the incoming part, the induced orientation on the incoming face is $\{y\}-\{x\}$, the negative of the source orientation. Hence $[-1,1]$ with this collar and orientation is an oriented null-cobordism of $\{x,y\}$, and $[{\mathrm{p}\mathrm{t}}_{+}]+[{\mathrm{p}\mathrm{t}}_{-}]=0$ in ${\mathrm{\Omega }}_0^{SO}$.

(The invariant, the generator and the consequences.) By [F3] the signed count is unchanged by oriented cobordism and defines an isomorphism ${\mathrm{\Omega }}_0^{SO}\to \mathbb{Z}$ that sends a positively oriented point to $1$. Step 1.1 shows $[{\mathrm{p}\mathrm{t}}_{-}]=-[{\mathrm{p}\mathrm{t}}_{+}]$ in the group, and this is consistent with the isomorphism because the two signed counts are $+1$ and $-1$. Finally, two finite oriented point sets have equal images under the isomorphism if and only if their signed counts agree, and since the isomorphism is injective this is exactly the condition that they are oriented cobordant; equivalently, their difference has signed count zero and is null-cobordant, again by [F3].

($P$ is a compact smooth surface with boundary the three circles.) Introduce the smooth functions ${\rho }_0(x)=4-\mathrm{\parallel }x{\mathrm{\parallel }}^2$, ${\rho }_{-}(x)=\mathrm{\parallel }x-(1,0){\mathrm{\parallel }}^2-\frac{1}{4}$ and ${\rho }_{+}(x)=\mathrm{\parallel }x+(1,0){\mathrm{\parallel }}^2-\frac{1}{4}$ on ${\mathbb{R}}^2$; then $P=\{x:{\rho }_0(x)\ge 0,{\rho }_{-}(x)\ge 0,{\rho }_{+}(x)\ge 0\}$. On $C_0=\{x:{\rho }_0(x)=0\}$ we have $\mathrm{\mid }x\mathrm{\mid }=2$, so $\mathrm{\parallel }x\mp (1,0)\mathrm{\parallel }\ge 2-1=1>\frac{1}{2}$ and the other two functions are strictly positive; on $C_{\pm }=\{x:{\rho }_{\pm }(x)=0\}$ we have $\mathrm{\parallel }x\pm (1,0)\mathrm{\parallel }=\frac{1}{2}$, so $\mathrm{\parallel }x\mathrm{\parallel }\le 1+\frac{1}{2}=\frac{3}{2}<2$ and, since the two inner centres are at distance $2$ and the radii sum to $1$, $\mathrm{\parallel }x\mp (1,0)\mathrm{\parallel }\ge 2-\frac{1}{2}=\frac{3}{2}>\frac{1}{2}$. Hence the three circles are pairwise disjoint and each boundary point of $P$ lies on exactly one of them, where exactly one defining function vanishes with nonzero gradient. Each such point therefore has a boundary chart obtained from that defining function by the inverse function theorem, so $P$ is a compact smooth surface with boundary $C_0\bigsqcup C_1\bigsqcup C_2$; compactness follows because $P$ is a closed subset of the compact disk $\{\mathrm{\mid }x\mathrm{\mid }\le 2\}$.

(The induced orientations of the three circles.) Give $P$ the orientation induced from the standard orientation of ${\mathbb{R}}^2$. At a point $p\in C_0$ the outward normal of $P$ is the radial unit vector $p\mathrm{/}2$, and $(p\mathrm{/}2,Jp\mathrm{/}2)$ is a positive basis of ${\mathbb{R}}^2$, where $J$ is the quarter-turn; by the outward-normal-first rule the positive tangent direction of $C_0$ is $Jp$, the counterclockwise direction. At a point $p\in C_{\pm }$, let $c$ be the centre of that circle; the removed disk lies outside $P$ in the direction $c-p$, so the outward normal of $P$ at $p$ is the unit vector $(c-p)\mathrm{/}\mathrm{\mid }c-p\mathrm{\mid }=2(c-p)$, and $(2(c-p),2J(c-p))$ is again a positive basis; hence the positive tangent direction is $J(c-p)=-J(p-c)$, which is the clockwise direction of the circle centred at $c$. So the induced orientation of the outer circle is counterclockwise and that of each inner circle is clockwise, i.e. the oriented boundary is $C_0-C_1-C_2=-(C_1\bigsqcup C_2)\bigsqcup C_0$.

($P$ is an oriented bordism from $C_1\bigsqcup C_2$ to $C_0$.) Take the incoming boundary part $(\mathrm{\partial }P{)}_0=C_1\bigsqcup C_2$ with the source orientations counterclockwise on both circles, and the outgoing part $(\mathrm{\partial }P{)}_1=C_0$; the induced orientations computed in step 2.1 are clockwise on $C_1,C_2$, which is the negative of the source orientation on the incoming part, and counterclockwise on $C_0$, which is the target orientation. The radial parametrisations ${\theta }_0(s,p)=c+(1+\frac{s}{4})(p-c)$ for $p$ in an inner circle with centre $c$, and ${\theta }_1(s,p)=(1+\frac{s}{4})p$ for $p\in C_0$, $s\in [0,1)$ respectively $s\in (-1,0]$, are smooth embeddings onto collar neighbourhoods of the corresponding boundary circles: their images have radii $\frac{1}{2}(1+\frac{s}{4})\in [\frac{1}{2},\frac{5}{8})$ and $2+\frac{s}{2}\in (\frac{3}{2},2]$ in the relevant radial directions and lie in $P$ by the estimates of step 1.1. Hence $P$ with these collars and this orientation is an oriented bordism from $C_1\bigsqcup C_2$ to $C_0$.

(The additive relation; all classes vanish.) By step 3.1 the cobordism class of $C_1\bigsqcup C_2$ equals that of $C_0$, that is, $[C_1]+[C_2]=[C_0]$ in ${\mathrm{\Omega }}_1^{SO}$ by [F4]. Each of the three circles is the boundary of a Euclidean disk (with the counterclockwise orientation induced by the standard orientation of the plane, after the outer circle is viewed as the boundary of the disk it encloses and each inner circle as the boundary of the removed disk), so by the disk example, which applies to a circle with either orientation, all three classes are zero in ${\mathrm{\Omega }}_1^{SO}$ and in ${\mathrm{\Omega }}_1^O$; the relation therefore reads $0+0=0$ and exhibits the disjoint-union addition of the group structure rather than an independent invariant.

(The mod-two cohomology of ${\mathbb{R}\mathbb{P}}^2$.) Model ${\mathbb{R}\mathbb{P}}^2$ as the projectivisation of the trivial rank-three bundle over a point. By [F1], $H^{\ast }({\mathbb{R}\mathbb{P}}^2;{\mathbb{F}}_2)$ is free over ${\mathbb{F}}_2$ with basis $1,x,x^2$, where $x$ is the tautological degree-one class, and the only relation is $x^3=0$; in particular $x\mathrm{\ne }0$ and $x^2\mathrm{\ne }0$, and $H^1({\mathbb{R}\mathbb{P}}^2;{\mathbb{F}}_2)={\mathbb{F}}_2\cdot x$ has exactly two elements.

($w_1(T{\mathbb{R}\mathbb{P}}^2)=x\mathrm{\ne }0$.) Suppose $w_1(T{\mathbb{R}\mathbb{P}}^2)=0$. The tangent bundle of a closed smooth manifold is numerable over an admissible base by [F3], so by [F2] the vanishing of $w_1$ would make $T{\mathbb{R}\mathbb{P}}^2$ orientable, and for the closed surface ${\mathbb{R}\mathbb{P}}^2$ the orientation of the tangent determinant lines would give an atlas with positive transition Jacobians, making ${\mathbb{R}\mathbb{P}}^2$ orientable. This contradicts [F2], since ${\mathbb{R}\mathbb{P}}^2$ is not orientable. Hence $w_1(T{\mathbb{R}\mathbb{P}}^2)\mathrm{\ne }0$; by step 1.1 it is the unique nonzero element, $w_1(T{\mathbb{R}\mathbb{P}}^2)=x$.

($w_1^2[{\mathbb{R}\mathbb{P}}^2]=1$.) By step 2.1, $w_1(T{\mathbb{R}\mathbb{P}}^2{)}^2=x^2$, and $x^2\mathrm{\ne }0$ by step 1.1. Apply [F4] with $M={\mathbb{R}\mathbb{P}}^2$: the pairing $H^2\times H^0\to {\mathbb{F}}_2$ is perfect and $H^0={\mathbb{F}}_2\cdot 1$, so the adjoint map $a\mapsto ⟨a\cdot 1,[{\mathbb{R}\mathbb{P}}^2]⟩=⟨a,[{\mathbb{R}\mathbb{P}}^2]⟩$ is an isomorphism $H^2({\mathbb{R}\mathbb{P}}^2;{\mathbb{F}}_2)\to {\mathbb{F}}_2$; a nonzero class therefore has evaluation $1$. Hence $w_1^2[{\mathbb{R}\mathbb{P}}^2]=⟨x^2,[{\mathbb{R}\mathbb{P}}^2]⟩=1\mathrm{\ne }0$.

(Conclusion: ${\mathbb{R}\mathbb{P}}^2$ is not null-cobordant.) The Stiefel-Whitney number $w_1^2[{\mathbb{R}\mathbb{P}}^2]=1$ is nonzero, so by [F5] the closed surface ${\mathbb{R}\mathbb{P}}^2$ is not null-cobordant: it is not the boundary of any compact smooth $3$-manifold, and in particular not every closed surface bounds. Therefore the cobordism class of ${\mathbb{R}\mathbb{P}}^2$ is a nonzero element of ${\mathrm{\Omega }}_2^O$, and the claim that ${\mathbb{R}\mathbb{P}}^2$ is null-cobordant is refuted.