Orientation of a vector bundle

This is a fiberwise orientation compatible with local trivializations of a vector bundle. An $R$-orientation of a rank-$d$ vector bundle is a coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ for every fiber, equivalently a Thom class with the corresponding fiberwise restriction.

The total space of a real vector bundle $E\to M$ has the exact sequence $0\to {\pi }^{\ast }E\to TE\to {\pi }^{\ast }TM\to 0$. Taking determinant line bundles gives the displayed identity. If $M$ is an orientable smooth manifold, the total space is orientable exactly when $\det E$ is trivial. Sufficiency follows by pullback; necessity follows by restricting the total-space determinant identity to the zero section. For the cotangent bundle, base and fibre coordinate determinants cancel, giving an orientation even when the base is not orientable.

An ordered complex basis of a rank-$m$ complex vector space gives the real basis $(v_1,i{v}_1,\dots ,v_m,i{v}_m)$. Every complex change-of-basis matrix has positive real determinant $|{\det }_{\mathbb{C}}A{|}^2$, so these bases define a canonical integral orientation and hence an $R$-orientation for every commutative coefficient ring $R$.

A Thom class of an $R$-oriented rank-$d$ vector bundle is a class $u_E\in H^d(D(E),S(E);R)$ restricting to the chosen generator on every fiber pair $(D^d,S^{d-1})$.

An oriented normal bundle of a compact embedded submanifold has a Thom class. A tubular neighborhood and excision identify it with a relative class in $H^d(X,X\setminus Y)$. Its image in absolute cohomology is ${\epsilon }_Y$. It restricts to zero off Y and, in an oriented closed ambient manifold, represents the Poincare dual of Y with the normal-first induced orientation.

For an $h$-oriented rank-$r$ real vector bundle, its Thom class restricts on each fiber to the suspension of the coefficient unit under the oriented identification $h^r(D^r,S^{r-1})\cong h^0(\mathrm{pt})$. Cup product with this class gives the Thom isomorphism. The associated Euler class in a generalized cohomology theory is the pullback along the zero section after forgetting relative supports.

// Target: geometry-and-topology.bigb

For an oriented rank-$r$ vector bundle, let $U\in H^r(D(E),S(E);\mathbb{Z})$ be its Thom class. Forgetting relative supports sends $U$ to ${\pi }^{\ast }e(E)$, where $e(E)$ is the Euler class. Compatibility of relative and mixed cup products givesFor odd $r$, graded commutativity of the cup product makes $2U^2=0$. The Thom isomorphism theorem is injective, so $2e(E)=0$. This explains why an odd-rank Euler class can have only two-torsion.

For an $R$-oriented rank-$d$ vector bundle, multiplication by the Thom class gives isomorphisms

For a closed smooth submanifold $M\subseteq N$ of codimension $c>0$, a tubular neighborhood and the homological Thom isomorphism theorem identifySubstitution in the long exact sequence in relative homology gives the displayed embedding Gysin sequence. It does not require orientability over ${\mathbb{F}}_2$.

For a closed smooth embedding $i:N\hookrightarrow M$ of codimension $r$ whose normal bundle is oriented over a coefficient ring $R$, a tubular neighborhood, excision and the Thom isomorphism theorem identifyThe relative-to-absolute map defines $i_{!}:H^j(N;R)\to H^{j+r}(M;R)$. For closed oriented $M,N$, it is characterized bywith compatible orientations. It obeys $i_{!}(b⌣i^{\ast }c)=i_{!}b⌣c$. If $j:M\setminus N\hookrightarrow M$ and $\delta :H^q(M\setminus N;R)\to H^{q+1-r}(N;R)$ is the connecting map followed by the inverse Thom isomorphism, then, over ${\mathbb{F}}_2$,These identities follow from naturality of the relative cup product and show how the exact sequence determines multiplication in complements. Over ${\mathbb{F}}_2$ every real normal bundle has the required orientation.

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 $(S^2\times S^2)\times S^1$.