Stiefel Whitney and Euler Classes by Universal Constructions — Examples

The base is a CW complex by [F4], hence paracompact Hausdorff CGWH and admissible, and its tautological line is numerable by [F4]. The projective bundle of ${\gamma }_1$ is ${\mathbb{R}\mathbb{P}}^{\mathrm{\infty }}$ itself, since a point of $P({\gamma }_1)$ is a line in a line, and its tautological line is ${\gamma }_1$; by [F2] the defining relation is $x_{{\gamma }_1}+w_1({\gamma }_1)=0$ in $H^1({\mathbb{R}\mathbb{P}}^{\mathrm{\infty }};{\mathbb{F}}_2)$.

By [F3], $x_{{\gamma }_1}$ is the pullback of $a$ along any classifying map of ${\gamma }_1$. The identity map of ${\mathbb{R}\mathbb{P}}^{\mathrm{\infty }}$ classifies ${\gamma }_1$, so $x_{{\gamma }_1}={\mathrm{id}}^{\ast }a=a$. Substituting in step 1.1 and using that $-1=1$ in ${\mathbb{F}}_2$ gives $w_1({\gamma }_1)=a$; by the rank convention $w_i({\gamma }_1)=0$ for $i\ge 2$ and $w_0=1$, so $w({\gamma }_1)=1+a$.

Boundary cases. The degree-zero class is the unit $1$ in $H^0$ of the connected space ${\mathbb{R}\mathbb{P}}^{\mathrm{\infty }}$, matching the degree-zero convention. The bundle is a line, so every Stiefel–Whitney class of degree at least two vanishes, while $w_1=a$ is nonzero by [F1]. All these classes have ${\mathbb{F}}_2$ coefficients. The base is nonempty, so no empty-base convention is exercised, and no choice beyond [A1] is made.

The cohomology ring is polynomial. By [F2] applied to the $n$-fold product and [F1], the cross product identifies $H^{\ast }(({\mathbb{R}\mathbb{P}}^{\mathrm{\infty }}{)}^n;{\mathbb{F}}_2)$ with the $n$-fold graded tensor product of ${\mathbb{F}}_2[a]$, which is the polynomial ring ${\mathbb{F}}_2[a_1,\dots ,a_n]$ under the identifications $a_i={\mathrm{pr}}_i^{\ast }a$. The construction is by iterated Künneth over the finite product, and each factor's homology is ${\mathbb{F}}_2$ in each degree, so the finite-freeness hypothesis of [F2] holds factor by factor.

The total class of the sum. Each $L_i={\mathrm{pr}}_i^{\ast }{\gamma }_1$ is a line bundle, so naturality [F4] applied to the universal computation [F3] gives $w(L_i)={\mathrm{pr}}_i^{\ast }w({\gamma }_1)=1+{\mathrm{pr}}_i^{\ast }a=1+a_i\mathrm{.}$ The Whitney product formula [F4] applied to the finite sum gives $w(L_1\oplus \cdots \oplus L_n)={\prod }_{i=1}^n(1+a_i)$ in ${\mathbb{F}}_2[a_1,\dots ,a_n]$.

The individual classes. Expanding the product in the polynomial ring, the coefficient of degree $k$ is the sum of all products of $k$ distinct variables, that is the elementary symmetric polynomial ${\sigma }_k(a_1,\dots ,a_n)$; the coefficient of degree zero is $1=w_0$. Since the $a_i$ are algebraically independent by step 1.1, each ${\sigma }_k$ is nonzero for $k\le n$, which is the standard algebraic-independence input used to see that a rank-$n$ bundle can have all $n$ positive classes nonzero.

Boundary cases. For $n=0$ the product is a point, the empty sum of lines is the zero bundle, the empty product is $1$, and $w(0)=1$ by the rank-zero convention. For $n=1$ the statements reduce to $w({\gamma }_1)=1+a$ and ${\sigma }_1=a$. The polynomial ring is commutative, so the order of the factors does not matter, and the displayed class is independent of the chosen ordering of the coordinates. AC is used through Künneth and the Whitney formula, as recorded.

Base hypotheses. The real Grassmannian and $C$ are CW complexes by [F1]. The double cover $B\to {\mathrm{Gr}}_2({\mathbb{R}}^{\mathrm{\infty }})$ is numerable by a partition subordinate to evenly covered neighborhoods. Its fiber is the compact two-point CW complex, and the CW base is paracompact Hausdorff and CGWH. The separate paracompactness, compact-generation, and CW-type clauses of [F3] therefore make $B$ paracompact Hausdorff CGWH of CW type; Hausdorff also makes it weak Hausdorff. Thus both $B$ and $C$ satisfy the classification and Thom base hypotheses. Partitions subordinate to the tautological linear charts make their bundles numerable; the same holds for their pullbacks. These are the uses of AC and its consequence DC.

Complex structures on planes. On an oriented Euclidean plane let $J$ be the positive quarter turn. In each positive orthonormal frame it has the same matrix, which commutes with every transition rotation in $\mathrm{SO}(2)$; hence $J$ is continuous and makes ${\gamma }_2^{+}$ a complex line $\eta$. If $T$ is any orientation-preserving real isomorphism between complex lines, its complex-linear part is $T^{1,0}=(T-J_{\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{t}}T{J}_{\mathrm{s}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{c}\mathrm{e}})\mathrm{/}2$. In complex coordinates $T(z)=az+b\bar{z}$, and ${\det }_{\mathbb{R}}T=\mathrm{\mid }a{\mathrm{\mid }}^2-\mathrm{\mid }b{\mathrm{\mid }}^2>0$ implies $a\mathrm{\ne }0$. Therefore $T^{1,0}$ is a complex-linear isomorphism. This formula is continuous and independent of frames, so it also applies to bundle isomorphisms.

Classifying maps. By [F2], choose $f:C\to B$ classifying ${\gamma }_{\mathbb{R}}$ with its complex orientation, and $g:B\to C$ classifying $\eta$. The complex charts of $\eta$ constructed in step 1.2 admit a subordinate partition by step 1.1. The composite $fg$ classifies ${\eta }_{\mathbb{R}}={\gamma }_2^{+}$, so $fg\simeq 1_B$. The oriented isomorphism $f^{\ast }{\eta }_{\mathbb{R}}\cong {\gamma }_{\mathbb{R}}$ yields a complex isomorphism $f^{\ast }\eta \cong \gamma$ by step 1.2; therefore $gf\simeq 1_C$. Composition of pullbacks is identified fiberwise by $(x,(f(x),v))\mapsto (x,v)$. Thus $f$ is a homotopy equivalence, without asserting a homeomorphism between the chosen Grassmannian models.

Integral generator. By [F7], $f^{\ast }:H^2(B;\mathbb{Z})\to H^2(C;\mathbb{Z})$ is an isomorphism. Naturality [F4] gives $f^{\ast }e({\gamma }_2^{+})=e({\gamma }_{\mathbb{R}})=u$, which generates the target by [F4]. Hence $e({\gamma }_2^{+})$ generates the source.

Sign on ${\mathbb{C}\mathbb{P}}^1$. Put $M={\mathbb{C}\mathbb{P}}^1$, $L={\gamma }^{\ast }{\mathrm{\mid }}_M$, and $p=[1:0]$. The functional $(z_0,z_1)\mapsto z_1$ restricts on each line to a section $s$ of $L$, vanishing only at $p$. In the affine coordinate $w=z_1\mathrm{/}{z}_0$ and the dual tautological frame $e(w)$, this section is $s(w)=we(w)$. Choose a small closed coordinate disk $D=\{\mathrm{\mid }w\mathrm{\mid }\le \epsilon \}$ and write $a(w)=\mathrm{\parallel }e(w)\mathrm{\parallel }>0$. Define a section $\tilde{s}$ of the unit disk bundle by $\tilde{s}(q)=s(q)\mathrm{/}(\epsilon a(q))$ for $q\in D$ and $\tilde{s}(q)=s(q)\mathrm{/}\mathrm{\parallel }s(q)\mathrm{\parallel }$ for $q\mathrm{\notin }\mathrm{int}D$. The two formulas agree on $\mathrm{\partial }D$, where $\mathrm{\parallel }s(w)\mathrm{\parallel }=\epsilon a(w)$, so $\tilde{s}$ is continuous; it is sphere-valued on $A=M\setminus \mathrm{int}D$. Hence it is a genuine map of pairs $(M,A)\to (D(L),S(L))$ and pulls the normalized Thom class $U$ back to $\alpha \in H^2(M,A;\mathbb{Z})$. The absolute image of $\alpha$ is ${\tilde{s}}^{\ast }{j}^{\ast }U=e(L_{\mathbb{R}})=c_1(L)$ by [F5]: as maps to $D(L)$, $\tilde{s}$ and the zero section are joined by the fiberwise straight-line homotopy $(1-t)\tilde{s}$. Excision (equivalently the quotient identification $M\mathrm{/}A\cong D\mathrm{/}\mathrm{\partial }D$) restricts $\alpha$ to the class induced on $(D,\mathrm{\partial }D)$ by $w\mapsto w\mathrm{/}(\epsilon a(w))$ in the frame $e(w)$. The positive scalar factor preserves the complex orientation and the boundary map has degree $+1$, so this is the positive local orientation class. The complex-oriented fundamental class restricts to that generator, and the evaluation pairing gives $⟨c_1(L),[M]⟩=+1$. Thus the positive generator is $x=c_1({\gamma }^{\ast })$, while [F5] gives $u{\mathrm{\mid }}_M=c_1(\gamma )=-x$. Complex orientation of the bundle fiber therefore does not make its Euler number positive on the complex-oriented base.

Reduction and boundary. The admissibility and numerability checked in step 1.1 allow [F6] to give ${\rho }_2(e({\gamma }_2^{+}))=w_2({\gamma }_2^{+})$. Its pullback and then restriction to ${\mathbb{C}\mathbb{P}}^1$ is the nonzero reduction of $-x$ by [F6] and [F7], proving nonvanishing. This is a positive-rank computation on a nonempty base. For comparison, the trivial oriented two-plane over a point has Euler class zero since $H^2(\mathrm{p}\mathrm{t};\mathbb{Z})=0$ (its normalized singular cochain complex has no positive degrees); the rank-zero Euler unit lies instead in degree zero.

The rank-zero value. Give $0_B$ the standard unit orientation. Its disk bundle is $B$, its sphere bundle is empty, and the zero section and relative-to-absolute map are identities; fiber normalization makes the Thom class the unit $1$ in degree zero. Hence $e(0_B,1)=s^{\ast }{j}^{\ast }(1)=1\in H^0(B;R)$ by [F1]. For another supplied rank-zero orientation $o$, the same calculation instead gives $e(0_B,o)=o$.

The trivial bundle of positive rank. For $n\ge 1$ the constant section $b\mapsto (b,e_1)$ of ${\epsilon }_B^n$ is nowhere zero and continuous. Its global product chart, together with the constant partition of unity $1$, makes the bundle numerable, while [F3] supplies its standard product $R$-orientation. The standing hypothesis places $B$ in the general Thom scope. Thus every hypothesis of [F2] holds, and $e({\epsilon }_B^n)=0\in H^n(B;R)$.

Boundary cases. For the empty base the unique cohomology class in every degree is zero, and the standard rank-zero convention reads $e(0,1)=1=0$ in the zero ring; the two displayed identities remain consistent. For a point base, ${\epsilon }^0$ with unit orientation contributes $e=1\in H^0(\ast ;R)=R$ and ${\epsilon }^n$ with $n\ge 1$ contributes $e=0\in H^n(\ast ;R)=0$. For the zero ring both classes coincide with the unique element, so the identities hold. The section of step 1.2 is a specified function and no choice is made in it; AC is used only through the Thom suppliers that give the Euler class its value.

The witness. Via the clutching bijection [F1] with $k=4$ and $n=3$, let $E\to S^4$ be the oriented rank-three real bundle clutched by the map $\rho :S^3\to \mathrm{SO}(3)$ of [F2]. This is the witness; it is an oriented numerable bundle since $S^4$ is a CW complex.

The Euler class vanishes. The bundle $E$ has rank three, so $e(E)\in H^3(S^4;\mathbb{Z})$ by [F8]; this group is zero by [F5]. Hence $e(E)=0$.

The witness is nontrivial. Clause 4 of [F2] is exactly the assertion that the clutching construction over the equatorial $S^3$ with clutching map $\rho$ produces a nontrivial oriented rank-three bundle $E_{\rho }\to S^4$. The witness of step 1.1 is this bundle, so it is nontrivial. Equivalently, clauses 2 and 3 of [F2] identify $[\rho ]$ with a generator of ${\pi }_3(SO(3))\cong \mathbb{Z}$, hence with a nonconstant based homotopy class. No conversion from an unbased nullhomotopy to a based one is used here.

There is no nowhere-zero section. Suppose, for contradiction, that $\sigma$ is a nowhere-zero section of $E$. It spans a trivial line subbundle ${\epsilon }^1\subseteq E$, and a bundle metric on the paracompact Hausdorff base $S^4$ splits the resulting sequence, so [F6] gives $E\cong {\epsilon }^1\oplus F$ with $F$ an oriented rank-two real bundle over $S^4$. By [F1] the bundle $F$ is clutched by a map $S^3\to \mathrm{SO}(2)\cong S^1$, which is nullhomotopic by [F7], since every map from the simply connected $S^3$ to $S^1$ lifts through the contractible universal cover. Hence $F$ is trivial by [F1] and $E\cong {\epsilon }^1\oplus F\cong {\epsilon }^3$ is trivial, contradicting step 2.1. Therefore no nowhere-zero section exists.

Conclusion. The bundle $E$ of step 1.1 is an oriented rank-three real bundle over $S^4$ with $e(E)=0$ by step 1.2 and no nowhere-zero section by step 3.1. This refutes the displayed implication and completes the counterexample.

The line bundles. By [F1] choose line bundles $L_a,L_b,L_{a+b}$ over $B$ with $w_1(L_a)=a$, $w_1(L_b)=b$ and $w_1(L_{a+b})=a+b$; such bundles exist because the classification bijection is surjective and the three displayed classes lie in $H^1(B;{\mathbb{F}}_2)$.

The witness is orientable. Let $E=L_a\oplus L_b\oplus L_{a+b}$, a real rank-three bundle over $B$. By [F3] and step 1.1, $w_1(E)=w_1(L_a)+w_1(L_b)+w_1(L_{a+b})=a+b+(a+b)=0$ in ${\mathbb{F}}_2[a,b]$. Hence $E$ is orientable by [F4]; fix an orientation $o$.

Its top class does not vanish. Again by [F3], $w_3(E)=w_1(L_a)w_1(L_b)w_1(L_{a+b})=ab(a+b)\in {\mathbb{F}}_2[a,b],$ which is a nonzero polynomial since it is a sum of the two distinct monomials $a^{2}b$ and $a{b}^2$ of degree three. Hence $w_3(E)\mathrm{\ne }0$ in $H^3(B;{\mathbb{F}}_2)$ by [F2].

The Euler class is nonzero of order two. By [F5] the mod-two reduction of the integral Euler class is the top Stiefel–Whitney class, so ${\rho }_2(e(E,o))=w_3(E)\mathrm{\ne }0$ by step 2.2; in particular $e(E,o)\mathrm{\ne }0$ in $H^3(B;\mathbb{Z})$. Also by [F5] the odd rank three gives $2e(E,o)=0$. Therefore $e(E,o)$ is a nonzero element of order two, and the slogan of the statement refuted is false.

Boundary remarks. The construction uses three line bundles of rank one, so the sum has odd rank three and the two-torsion conclusion applies; if the three line classes summed to a nonzero class the bundle would not be orientable and no integral Euler class would be defined. For the base with the second factor replaced by a point the same computation gives $w_3=0$, so the two factors are both needed for the witness. The orientation $o$ is determined only up to sign, and the statement is sign-independent because both $e\mathrm{\ne }0$ and $2e=0$ are invariant under negation. AC is used through the classification of line bundles and the Thom class, as recorded.