Intersection Pairings Self Intersection and Euler Classes — Examples

By The zero section is a smooth embedding the zero section is embedded, and in bundle charts the splitting along it is $TE{\mathrm{\mid }}_Z=T{S}^2\oplus E$, so its quotient normal bundle is $E$ with the specified fibre orientation, and $Z$ is compact, so [F1] gives $Z\cdot Z=⟨e({\nu }_Z),[Z]⟩=⟨e(E),[S^2]⟩$.

Case (a): the constant unit section $x\mapsto (x,(1,0))$ is smooth and nowhere zero, so by [F2] both the Euler number and the self-intersection vanish: $⟨e(E),[S^2]⟩=0=Z\cdot Z$ for the trivial bundle.

Case (b): $S^2$ is the regular level set $\mathrm{\mid }p{\mathrm{\mid }}^2=1$ of a smooth function [F4] with $T_p{S}^2=p^{\perp }$, so $X(p)=e_3-zp$ satisfies $X(p)\cdot p=z-z\mathrm{\mid }p{\mathrm{\mid }}^2=0$ and is a smooth section of $T{S}^2$ by [F5]. It vanishes iff $e_3=zp$, i.e. iff $p=\pm e_3$. In the projection charts $(x,y)\mapsto (x,y,\pm \sqrt{1-x^2-y^2})$ near the poles the linearizations are $-(x{\mathrm{\partial }}_x+y{\mathrm{\partial }}_y)+O(\mathrm{\mid }(x,y){\mathrm{\mid }}^2)$ at $e_3$ and $+(x{\mathrm{\partial }}_x+y{\mathrm{\partial }}_y)+O(\mathrm{\mid }(x,y){\mathrm{\mid }}^2)$ at $-e_3$, whose Jacobians $-I$ and $+I$ both have determinant $+1$ in dimension two, and the chart-orientation sign cancels between source and target in the local index [F3]. Hence both zeros are nondegenerate of index $+1$ and the signed zero count is $2$; [F1] and [F3] identify it with $Z\cdot Z$ and with $⟨e(T{S}^2),[S^2]⟩$.

$\mathrm{\Delta }$ is closed embedded of dimension $2$ with $2\cdot 2=4=\dim (S^2\times S^2)$ [F1], so [F3] gives $\mathrm{\Delta }\cdot \mathrm{\Delta }=⟨e({\nu }_{\mathrm{\Delta }}),[\mathrm{\Delta }]⟩$. By [F2] the normal identification is the canonical one and is orientation-preserving with the orientation of $\mathrm{\Delta }$ transported from $S^2$, so $\mathrm{\Delta }\cdot \mathrm{\Delta }=⟨e(T{S}^2),[S^2]⟩$.

In the projection charts of the two poles $(x,y)\mapsto (x,y,\pm \sqrt{1-x^2-y^2})$ the given field $X(p)=e_3-zp$, tangent because $(e_3-zp)\cdot p=z-z\mathrm{\mid }p{\mathrm{\mid }}^2=0$ by [F4], has exactly the two zeros $\pm e_3$, with local components $(-zx,-zy)$ and derivatives $-I_2$ at the north pole and $I_2$ at the south pole. Both determinants are $+1$, so each zero has index $+1$ and the signed zero count of $T{S}^2$ is $2$; [F5] and [F3] identify that count with $⟨e(T{S}^2),[S^2]⟩$. Hence $\mathrm{\Delta }\cdot \mathrm{\Delta }=2$, which previews the Euler characteristic $\chi (S^2)=2$ of the later Euler/index pair (not used here).

Give $Q$ quotient charts by intervals of length less than one: the quotient projection is injective on each interval and open, since the saturation of an open interval is the union of its integer translates. Its restriction is therefore a homeomorphism onto an open set of $Q$. Chart changes on overlap components are integer translations, hence smooth and increasing. The quotient is Hausdorff: distinct classes have lifts whose difference is not an integer, and sufficiently small intervals about them have disjoint integer saturations. Images of rational-endpoint intervals give a countable base because the quotient projection is open. These charts cover $Q$, define the smooth structure and orientation, and identify its tangent frame with ${\mathrm{\partial }}_x$. The image of $[0,1]$ is all of $Q$, so it is compact and boundaryless. In the product charts, $A$ and $B$ are embedded circles cut out by the coordinate projections [F1], [F2], and they meet transversely in the single point $([0],[0])$ with $T_{([0],[0])}A=\mathbb{R}{\mathrm{\partial }}_x$, $T_{([0],[0])}B=\mathbb{R}{\mathrm{\partial }}_y$ and $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ the positive product frame. Hence $⟨A,B⟩=1$ and, by [F3] with $ab=1$, $⟨B,A⟩=-1$.

For the self-intersections: the constant field ${\mathrm{\partial }}_y$ restricted to $A$ is a nowhere-zero section of the normal bundle of $A$ (the normal bundle is identified with the $y$-factor along $A$), and likewise ${\mathrm{\partial }}_x$ for $B$; by [F4], together with A nowhere-zero section forces the Euler data to vanish, the corresponding push-offs are disjoint, so $⟨A,A⟩=0$ and $⟨B,B⟩=0$.

By [F5] the same numbers are $⟨\mathrm{P}\mathrm{D}[x]⌣\mathrm{P}\mathrm{D}[y],[T^2]⟩$ evaluated on the two classes, giving the displayed matrix: the factor order contributes the minus sign in degree one, and the determinant of the matrix on the classes displayed is $1$. For $x=m[A]+n[B]$ and $y=p[A]+q[B]$, bilinearity gives $⟨x,y⟩=mq-np$, which vanishes when $x=y$. This is the usual alternating (symplectic) block; its displayed signs specify the convention completely. No nondegeneracy claim for the whole pairing on $H_1(T^2;\mathbb{Z})$ is made here.

The quotient has local charts obtained by lifting base intervals of length less than one to $\mathbb{R}$; on overlaps the lifted base coordinates differ by an integer $k$ and the fibre changes by $(-1{)}^k$. These charts are smooth and fibre-linear, so [F1] gives a line bundle over the smooth quotient circle. The chart maps are homeomorphisms because their integer translates have open saturation and are disjoint over each lifted interval. Distinct base points are separated by the Hausdorff quotient circle, and distinct points over one base point are separated in a bundle chart; thus the total space is Hausdorff. Rational-endpoint lifted base intervals and fibre intervals give a countable base. The image of $[0,1]$ covers $Q$, making it compact. Its interval charts have integer-translation transitions, so its zero section is a compact boundaryless embedded circle. Along it $TL{\mathrm{\mid }}_Z=TQ\oplus L$, so the normal quotient is $L$. A two-arc presentation has transition $+1$ on one overlap component and $-1$ on the other; putting $-1$ on both components would instead be a trivial bundle.

Nonorientability. An orientation pulled back to the connected covering $\mathbb{R}$ would be a continuous sign $o(t)\in \{+1,-1\}$, hence constant, but the deck change requires $o(t+1)=-o(t)$, a contradiction. The total-space deck map $(t,v)\mapsto (t+1,-v)$ has determinant $-1$; the same argument on its connected covering plane proves that the total space is nonorientable. Thus [F3] gives $w_1(L)\mathrm{\ne }0$.

For $0<\epsilon <1$ the function $f(t)=\epsilon \sin (\pi t)$ obeys $f(t+1)=-f(t)$ and therefore defines a smooth section of $L$. Its zeros are all integers, which give exactly one point of $Q$; its vertical derivative there is $\epsilon \pi \mathrm{\ne }0$ in a lifted chart. Its graph is a small push-off of the zero section and meets it transversely once, so $Z{\cdot }_{2}Z=1$. By [F4] this equals $⟨w_1(L),[Q]{⟩}_2$, without assuming an unproved cohomology computation or using a rank-one projective fibre-generator claim.

The orientation obstruction in step 2.1 violates [F5], so no untwisted integral oriented self-intersection is defined by that construction. Changing a local fibre trivialization can reverse a local zero sign, and there is no continuous global choice making all such signs consistent. The ambiguity is not merely a single overall sign for an arbitrary finite zero set. Modulo two every local sign is one and step 2.2 gives the invariant count. AC is inherited from [F3]–[F4]; the explicit quotient and section use no extra choice.