Thom identity for Stiefel–Whitney classes

An absolute identity. For a rank-$n$ bundle $E$, $n\ge 1$, take its flag map $q$ from [F4], and write $q^{\ast }E={⨁}_{j=1}^n{L}_j$, $t_j=w_1(L_j)$. The rank convention and Whitney formula give $q^{\ast }{w}_n(E)={\prod }_j{t}_j$ and $q^{\ast }w(E)={\prod }_j(1+t_j)$. Since $t_j$ has degree one, [F1] gives $Sq(t_j)=t_j+t_j^2$. Absolute Cartan and commutativity over ${\mathbb{F}}_2$ therefore give $Sq(q^{\ast }{w}_n(E))={\prod }_j(t_j+t_j^2)=({\prod }_j(1+t_j))\left({\prod }_j{t}_j\right)=q^{\ast }(w(E)w_n(E))\mathrm{.}$ Naturality of each square and injectivity of $q^{\ast }$ give $Sq(w_n(E))=w(E)w_n(E)$. All sums and products here are finite and all classes are absolute.

Forgetting the relative subspace. For any bundle in [F2], put $D=D(E)$, $S=S(E)$, let $\pi :D\to B$ be projection, let $s$ be its zero section, and write $j^{\ast }:H^{\ast }(D,S)\to H^{\ast }(D)$ for the forgetful map induced by $(D,\mathrm{\varnothing })\to (D,S)$. Radial contraction gives $s\pi \simeq {\mathrm{id}}_D$ and $\pi s={\mathrm{id}}_B$. Thus [F6] and the Euler definition imply $j^{\ast }u={\pi }^{\ast }{e}_2(E)$. Naturality of the relative module product, with triples $(D;\mathrm{\varnothing },\mathrm{\varnothing })\to (D;\mathrm{\varnothing },S)$, gives $j^{\ast }\mathrm{\Phi }(a)={\pi }^{\ast }a⌣j^{\ast }u={\pi }^{\ast }(a{e}_2(E))\mathrm{.}$ The relevant cup comparisons exist because $\mathrm{\varnothing }$ and $S$ are open in $S$; no Cartan assertion for a relative product is involved.

Universal injectivity. Let $E={\gamma }_n$, $n\ge 1$. The hypotheses of [F2] hold by [F5] and [F7]. In the polynomial ring of [F5], multiplication by the variable $w_n$ is injective: it shifts the exponent of that variable in each distinct monomial. By [F3] it is multiplication by $e_2({\gamma }_n)$. The identity in step 1.2, the isomorphism $\mathrm{\Phi }$ and the isomorphism ${\pi }^{\ast }$ show that $j^{\ast }:H^k(D({\gamma }_n),S({\gamma }_n))\to H^k(D({\gamma }_n))$ is injective in every degree. Explicitly, write any relative class as $\mathrm{\Phi }(a)$; if its image is zero then $a{w}_n=0$, hence $a=0$. This also covers zero groups in degrees below $n$.

Universal Thom identity. Write $u=u_{{\gamma }_n}$. Naturality of the squares for the pair map defining $j^{\ast }$ and for the space map $\pi$ gives $j^{\ast }Sq(u)=Sq(j^{\ast }u)=Sq({\pi }^{\ast }{w}_n)={\pi }^{\ast }Sq(w_n)={\pi }^{\ast }(w({\gamma }_n)w_n)=j^{\ast }({\pi }^{\ast }w({\gamma }_n)⌣u)\mathrm{.}$ The third equality uses step 1.2 and [F3], the fourth uses step 1.1, and the last uses step 1.2. Injectivity in step 2.1, degree by degree, proves the identity for ${\gamma }_n$.

Pull back to the given bundle. For $n\ge 1$, choose a classifying map $c:B\to BO(n)$ and an isomorphism $E\cong c^{\ast }{\gamma }_n$ by [F5]. Use the pulled-back metric through this isomorphism to obtain a map $\tilde{c}:(D(E),S(E))\to (D({\gamma }_n),S({\gamma }_n))$ over $c$. Changing a supplied metric does not change the identity: the fiberwise radial map sending a nonzero vector $v$ to $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{o}\mathrm{l}\mathrm{d}}v\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{n}\mathrm{e}\mathrm{w}}$ is a homeomorphism of the old and new disk-sphere pairs over $B$, extends continuously by zero, and preserves the mod-two fiber generator; its inverse interchanges the two norms. By normalized Thom naturality, ${\tilde{c}}^{\ast }{u}_{{\gamma }_n}=u_E$. Pull back step 3.1, using pair naturality of $Sq$, naturality of $w$, and relative cup naturality. This gives $Sq(u_E)={\pi }^{\ast }w(E)⌣u_E$, with the base pullback suppressed in the statement's usual module notation.

Degrees and boundaries. For $n=0$ the disk-sphere pair is $(B,\mathrm{\varnothing })$, $u=1$ and $w(0_B)=1$. Normalization and instability give $S{q}^0(1)=1$ and $S{q}^i(1)=0$ for $i>0$, proving the identity separately, without multiplying by a nonexistent polynomial variable $w_0$. Over the empty base all groups and classes are zero. For $n\ge 1$ step 4.1 and for $n=0$ the preceding calculation give the total identity; comparison of degree $n+i$ components gives the stated formula for every $i\ge 0$. Both sides are zero for $i>n$ by instability and the rank convention. The sums are finite even for infinite-dimensional bases. AC enters exactly through the Thom, splitting, classification and universal-cohomology suppliers and the numerability assertion; the subsequent polynomial and cohomology calculations require no further choices.