Euler class of a vector bundle ($e(E)$)

The Euler class of an oriented rank-$d$ vector bundle is the pullback $e(E)=s^{\ast }{u}_E\in H^d(B;R)$ of its Thom class along the zero section. Over ${\mathbb{F}}_2$ it equals the top Stiefel-Whitney class.

For an $h$-oriented real bundle $\xi$ of rank $r$, pull its Thom class back along the zero section to obtain its Euler class in $h^r(X)$. It depends on the orientation and the actual bundle, not merely its stable isomorphism class. In complex cobordism the orientation may be supplied by a stable complex structure on a real vector bundle.

Normalize the section to the sphere bundle. Scaling it from zero to unit length gives a homotopy within the disk bundle from the zero section to this sphere-valued section. The image of the relative Thom class in absolute cohomology restricts to zero on the sphere bundle. Homotopy invariance therefore makes its zero-section pullback vanish.

// Target: geometry-and-topology.bigb

A complex rank-$r$ vector bundle has a canonical real orientation, and its Euler class equals its top Chern class. For a complex line this is $e(L_{\mathbb{R}})=c_1(L)$. The splitting principle for complex vector bundles gives an injective cohomology pullback on which the bundle splits into lines. The Whitney product formula for Euler classes and Whitney sum formula for Chern classes then identify both sides with the product of those first Chern classes, proving the general equality.

For an oriented vector bundle $E$ of odd rank, multiplication by $-1$ on each fiber identifies $E$ with the oppositely oriented bundle. Naturality and reversal of the Thom class giveso $2e(E)=0$.

For oriented real vector bundles $E$ and $F$, the ordered direct-sum orientation satisfiesThe formula follows because the Thom class of the direct sum is the fiberwise exterior product of the two Thom classes.

When an oriented rank-$d$ vector bundle $E$ has a closed oriented $d$-dimensional base $B$, its Euler number is the integer obtained by evaluating the Euler class of a vector bundle on the fundamental class of $B$. A nonzero Euler number obstructs a nowhere-zero section and therefore obstructs a framing.

For a compact oriented smooth manifold $M$, the Euler class of its tangent bundle satisfiesEquivalently, the sum of the indices of the isolated zeros of a vector field equals the Euler characteristic.

The complex orientation turns a complex line bundle $L$ into an oriented real rank-two bundle, andMoreover $e(L^{\ast })=-e(L)$ and $e(L_1{\otimes }_{\mathbb{C}}{L}_2)=e(L_1)+e(L_2)$.

A Gysin sequence of a sphere bundle is the long exact sequence induced by the Gysin homomorphism and the Euler class of a vector bundle.

For the unit sphere bundle $S(E)\to B$ of an oriented rank-$d$ vector bundle, the Gysin sequence contains

Let $E$ be an oriented real vector bundle of positive even rank $r$, with zero Euler class, and suppose the integral cohomology of $B$ is torsion-free. The Gysin sequence gives $0\to H^j(B)\to H^j(S(E))\to H^{j-r+1}(B)\to 0$. Choose an odd-degree class $a\in H^{r-1}(S(E))$ with ${\pi }_{!}a=1$. The projection formula for sphere bundle integration identifies the cohomology additively with ${\pi }^{\ast }{H}^{\ast }(B)\oplus {\pi }^{\ast }{H}^{\ast }(B)a$. It is torsion-free, so graded commutativity of the cup product gives $a^2=0$ from $2a^2=0$. These facts establish the displayed ring isomorphism, not just a decomposition of groups.

For the sphere bundle of an oriented rank-$r$ real vector bundle, the integration map in the Gysin sequence lowers degree by $r-1$. Its compatibility with the cup product is the displayed projection formula. The construction is the connecting map of the disk/sphere pair followed by the inverse Thom isomorphism theorem; the connecting map has its graded module sign, and the degree sign in the conventional fibre-integration map gives the displayed left-module formula.

For a real rank-$r$ vector bundle $E\to B$, the Thom isomorphism theorem with ${\mathbb{F}}_2$ coefficients identifies the relative cohomology sequence of its disk bundle and sphere bundle withThe multiplier is the top Stiefel–Whitney class, also called the mod-two Euler class. No orientation of $E$ is required. For the real tautological line bundle, the sphere bundle is the antipodal cover of Real projective space; this sequence proves that all powers of the degree-one generator through the dimension of the base are nonzero.

The diagonal quotient $L(p)=S^3/(\mathbb{Z}/p)$ is the unit circle bundle of a complex line bundle over $S^2$ whose Euler class is $p$ times a generator. Its Gysin sequence giveswhile $H^j(L(p);{\mathbb{F}}_p)\cong {\mathbb{F}}_p$ for $0\le j\le 3$.

Let $a,b$ be the standard basis of $H^2(S^2\times S^2;\mathbb{Z})$ and let $E\to S^2\times S^2$ be an oriented circle bundle with Euler class $pa+qb\ne 0$. If $d=\gcd (|p|,|q|)$, its Gysin sequence givesFor zero Euler class the bundle is trivial and its cohomology is that of $(S^2\times S^2)\times S^1$.