Characteristic Numbers and Cobordism Obstructions — Examples

For $n\ge 1$, [F1] makes $H^j({\mathbb{R}\mathbb{P}}^n;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ for $0\le j\le n$ and zero above, generated by the powers of $x$, and $x^{n+1}=0$; for $n=0$ the space is a point and $x=0$ in $H^1$, with the ring ${\mathbb{F}}_2$. By [F4] and field duality, the evaluation pairing $H^n({\mathbb{R}\mathbb{P}}^n;{\mathbb{F}}_2)\times H_n({\mathbb{R}\mathbb{P}}^n;{\mathbb{F}}_2)\to {\mathbb{F}}_2$ is a nondegenerate pairing of one-dimensional spaces, so the nonzero classes $x^n$ and $[{\mathbb{R}\mathbb{P}}^n]$ pair to $1$, as asserted.

The line bundle computation [F2] gives $w(\gamma )=1+x$ by the last calculation in [F3], and [F3] gives $T{\mathbb{R}\mathbb{P}}^n\oplus {\epsilon }^1\cong (n+1)\gamma$. Applying the Whitney formula and the trivial-summand stability [F2] to this isomorphism yields $w(T{\mathbb{R}\mathbb{P}}^n)=w(T{\mathbb{R}\mathbb{P}}^n\oplus {\epsilon }^1)=w(\gamma {)}^{n+1}=(1+x{)}^{n+1}$, and naturality makes the identification independent of the chosen isomorphism because isomorphic bundles have equal classes. Expanding in ${\mathbb{F}}_2[x]\mathrm{/}(x^{n+1})$ gives $w_i(T{\mathbb{R}\mathbb{P}}^n)=\left(\genfrac{}{}{0px}{}{n+1}{i}\right)x^i$ for $0\le i\le n$.

Let $I=(i_1,\dots ,i_r)$ be a partition of $n$ and $w^I=w_{i_1}\cdots w_{i_r}$. By step 1.2,[F4] $w^I[{\mathbb{R}\mathbb{P}}^n]=⟨{\prod }_j\left(\genfrac{}{}{0px}{}{n+1}{i_j}\right)x^{i_j},[{\mathbb{R}\mathbb{P}}^n]⟩=\left({\prod }_j\left(\genfrac{}{}{0px}{}{n+1}{i_j}\right)\right)⟨x^n,[{\mathbb{R}\mathbb{P}}^n]⟩={\prod }_j\left(\genfrac{}{}{0px}{}{n+1}{i_j}\right)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2,$ using step 1.1 for the top evaluation; for the monomial of total degree $n$ with $r=1$ and $i_1=n$ this gives the top number $w_n[{\mathbb{R}\mathbb{P}}^n]=\left(\genfrac{}{}{0px}{}{n+1}{n}\right)=n+1<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$, which is $1$ for even $n$ and $0$ for odd $n$.

For even $n=2k$ the top number is $1$, so [F5] obstructs null-cobordism. If $n=2^s-1$ with $s\ge 1$, characteristic two gives $(1+x{)}^{2^s}=1+x^{2^s}=1$ in the truncated ring; all positive-degree characteristic numbers vanish. For the stated boundary witnesses put $L={\gamma }^{\otimes 2}\to {\mathbb{C}\mathbb{P}}^k$, with its tensor metric. Its unit disk bundle is a compact smooth manifold with boundary its unit circle bundle: local smooth unitary frames give charts $U\times D^2$, with boundary $U\times S^1$, and finitely many compact trivializing neighbourhoods cover the compact base. The smooth map $S^{2k+1}\to S(L)$, $v\mapsto ([v],v\otimes v)$, is surjective and identifies precisely $v$ and $-v$. In a local unitary frame it is the circle map $z\mapsto z^2$, so the induced bijection ${\mathbb{R}\mathbb{P}}^{2k+1}\to S(L)$ and its local inverses are smooth. Taking $k=0,1,3$ gives the three claimed boundaries.

For $n=0$ the empty characteristic monomial is $1$ and evaluates to $1$ on the point, so this case is nonbounding; it is excluded from the $s\ge 1$ vanishing assertion. For $n=1$ the above witness is the disk over a point. The empty manifold is allowed by the library conventions and has zero numbers, though it is not a member of the projective-space family. The calculation and explicit witnesses use no choice beyond the declared characteristic-class and duality suppliers.

By [F1] the truncated ring is $\mathbb{Z}[x]\mathrm{/}(x^3)$, so $x^3=0$ and $x^2\mathrm{\ne }0$ with $⟨x^2,[{\mathbb{C}\mathbb{P}}^2]⟩=1$. The total Pontryagin class is $p(T{\mathbb{C}\mathbb{P}}^2)=(1+x^2{)}^3=1+3x^2+3x^4+x^6$, and all powers of $x$ with exponent at least three vanish in the truncated ring, so $p(T{\mathbb{C}\mathbb{P}}^2)=1+3x^2$. Thus $p_1=3x^2$ and $p_2=0$, the latter by the cohomological dimension, not the rank cutoff: the real tangent rank is $4$ and $2\cdot 2=4$ does not exceed it.

The only partition of $1$ is $(1)$, so by [F2] the unique degree-four Pontryagin number is $p_1[{\mathbb{C}\mathbb{P}}^2]=⟨3x^2,[{\mathbb{C}\mathbb{P}}^2]⟩=3⟨x^2,[{\mathbb{C}\mathbb{P}}^2]⟩=3,$ where the evaluation uses the normalization of step 1.1 and the linearity of the Kronecker pairing on classes [F2]. No other partition contributes, and classes of the wrong degree evaluate to zero by the conventions of [F2].

Since $p_1[{\mathbb{C}\mathbb{P}}^2]=3\mathrm{\ne }0$, the manifold is not an oriented boundary by [F3], so its class in ${\mathrm{\Omega }}_4^{SO}$ is nonzero: a boundary class has all Pontryagin numbers zero, and here $p_1$ does not vanish. The empty manifold and the zero class are excluded by [F3]; in particular $[{\mathbb{C}\mathbb{P}}^2]\mathrm{\ne }0$ without any appeal to a classification of ${\mathrm{\Omega }}_4^{SO}$.

Normalization and boundary cases. The degree-zero oriented bordism group is ${\mathrm{\Omega }}_0^{SO}\cong \mathbb{Z}$ generated by the positively oriented point [F4], whose only number is $p_{\mathrm{\varnothing }}=⟨1,[\mathrm{p}\mathrm{t}]⟩=1$; this identifies the degree-zero normalization but makes no claim about ${\mathrm{\Omega }}_4^{SO}$, and in particular no generator or rank statement for degree four is asserted here. For dimension zero the example reduces to that point computation, and the empty manifold has value $0$. The formulas of step 1.1 include the degenerate case $x^3=0$ through the truncation, and the cited suppliers carry their own choice declarations, so the verification adds no choice.

On $M={\mathbb{C}\mathbb{P}}^4$ the ring is $\mathbb{Z}[x]\mathrm{/}(x^5)$ with $⟨x^4,[M]⟩=1$ by [F1], so $x^5=0$ and all higher powers vanish. The total Pontryagin class is $p(TM)=(1+x^2{)}^5=1+5x^2+10x^4+10x^6+5x^8+x^{10}$, which truncates to $1+5x^2+10x^4$; hence $p_1=5x^2$, $p_2=10x^4$, and all other positive classes vanish. Therefore $p_1^2=25x^4$ evaluates to $p_1^2[M]=25⟨x^4,[M]⟩=25$ and $p_2[M]=10⟨x^4,[M]⟩=10$.

On $N={\mathbb{C}\mathbb{P}}^2\times {\mathbb{C}\mathbb{P}}^2$ write $x,y$ for the pullbacks of the generators of the two factors; by [F1] and [F2] the ring is $\mathbb{Z}[x,y]\mathrm{/}(x^3,y^3)$ with $⟨x^2{y}^2,[N]⟩=⟨x^2,[{\mathbb{C}\mathbb{P}}^2]⟩⟨y^2,[{\mathbb{C}\mathbb{P}}^2]⟩=1$, and $p(TN)=(1+x^2{)}^3\times (1+y^2{)}^3$ gives $p_1(TN)=3x^2\otimes 1+1\otimes 3y^2,p_2(TN)=9x^2\otimes y^2\mathrm{.}$ Squaring the first, $p_1^2=9x^4\otimes 1+18x^2\otimes y^2+1\otimes 9y^4$, and the outer terms vanish since $x^3=y^3=0$, while the middle term survives, so $p_1^2[N]=18⟨x^2{y}^2,[N]⟩=18$; similarly $p_2[N]=9⟨x^2{y}^2,[N]⟩=9$.

The resulting matrix with rows $M,N$ and columns $p_1^2,p_2$ is $\left(\begin{array}{cc}25& 10\\ 18& 9\end{array}\right)$, whose determinant is $25\cdot 9-10\cdot 18=225-180=45\mathrm{\ne }0$. This exhibits the nonsingularity of the ordinary degree-eight matrix predicted by the Newton comparison of [F4] and shows that the two classes are separated by their Pontryagin numbers: no nontrivial rational relation between the rows can hold.

Boundary and convention remarks. The monomials $p_1^2$ and $p_2$ are the two partitions of $2$, and monomials of any other total degree evaluate to zero by the conventions of [F3]; in particular the higher Pontryagin classes of the factors vanish by the truncation. For degree zero the point has matrix $(1)$ and the products considered here have positive dimension. The example uses the draft A-page items and the cited suppliers, which carry their own choice declarations.

For each partition $J$ of $k$, [F2] gives $p_J(T(-M))=p_J(TM)$ as cohomology classes: reversing the orientation changes only the ray datum of [F1] and leaves the underlying smooth manifold and its tangent bundle unchanged. Consequently, using the definition of the Pontryagin number and the sign rule of [F3], $p_J[-M]=⟨p_J(T(-M)),[-M]⟩=⟨p_J(TM),-[M]⟩=-⟨p_J(TM),[M]⟩=-p_J[M],$ where the middle equality is the linearity of the Kronecker pairing in its second variable [F4].

The Stiefel-Whitney numbers are unchanged: the canonical mod-two fundamental class used in Stiefel-Whitney numbers of a closed manifold depends only on the smooth structure, and the tangent Stiefel-Whitney classes are computed from $TM$ alone, so replacing $o$ by $-o$ alters neither the classes nor the fundamental class [F2, F5]; equivalently, over ${\mathbb{F}}_2$ the orientation sign $-1$ equals $1$.

Specialize to $M={\mathbb{C}\mathbb{P}}^2$ with its complex orientation. By the A-page tangent-bundle lemma [F5], $p_1[{\mathbb{C}\mathbb{P}}^2]=3$ and it is the only Pontryagin number in degree four, the partition being $(1)$; step 1.1 gives $p_1[-{\mathbb{C}\mathbb{P}}^2]=-3$. Both values are nonzero, so neither orientation is an oriented boundary by [F5], and the two orientations are distinguished by $p_1$ even though the underlying unoriented manifold and all its Stiefel-Whitney numbers are the same.

Boundary cases. For $k=0$ the manifold is a finite set of signed points, the only Pontryagin number is $p_{\mathrm{\varnothing }}=⟨1,[M]⟩$, the signed count, and step 1.1 gives the sign reversal of that count; the empty manifold has value $0$. Higher Pontryagin classes of ${\mathbb{C}\mathbb{P}}^2$ vanish because $p(T{\mathbb{C}\mathbb{P}}^2)=1+3x^2$ with $x^3=0$ in the truncated ring, so there is no second number to test; for a general $M$ all partitions $J$ of $k$ are covered by step 1.1. No choice beyond the cited suppliers is used.