Cohomology of BO and BSO away from two

Proof of the ring calculation. Induct on the rank, starting from BSO(1). Use the inspected universal oriented sphere-bundle lemma: on its actual total space S_n, p:S_n→BSO(n), the oriented complement map c:S_n→BSO(n−1) is a homotopy equivalence and pγ_n⁺=ε¹⊕cγ_{n−1}⁺. The published Gysin, Euler and Pontryagin interfaces consequently give, under this identification, a restriction j* carrying each p_i to the preceding-rank p_i and e to zero.

If n=2m, induction computes the preceding rank as R[p₁,…,p_{m-1}]. All those generators lift, so j* is surjective degreewise. In the published rank-2m Gysin sequence the actual maps, after identifying the sphere total space with BSO(2m−1), are H^{k−2m}(BSO(2m);R) --·e--> H^k(BSO(2m);R) --j*--> H^k(BSO(2m−1);R) --p_!--> H^{k−2m+1}(BSO(2m);R) --·e--> H^{k+1}(BSO(2m);R). Surjectivity of j* in degree k makes every class in its target a pullback. Exactness gives p_!j*=0, so p_! vanishes in degree k. At the following term, exactness therefore makes multiplication by e injective on H^{k−2m+1}(BSO(2m);R). Taking k=a+2m−1 for every integer a proves that e is a non-zero-divisor in each degree a. Exactness at H^k(BSO(2m);R) independently gives ker(j*:H^k(BSO(2m);R)→H^k(BSO(2m−1);R)) =eH^{k−2m}(BSO(2m);R). Negative-degree groups are zero, so the same statement includes the initial degrees. For a class of degree d subtract a polynomial lift of its restriction, then divide the remainder by e; the resulting class has degree d−2m. Induction on d proves polynomial generation. For a polynomial relation Σ_{a=0}^N e^a P_a(p)=0, restriction gives P₀=0 by the preceding rank's polynomial independence. Injectivity of multiplication by e then repeats the argument to show every P_a=0. The published top-class identity gives p_m=e². This proves the even-rank presentation over R.

If n=2m+1, the odd-rank Euler class vanishes over R because its integral class is killed by two. Gysin makes j* injective. Its image contains R[p₁,…,p_m]=R[p₁,…,p_{m-1},e²] in the preceding even-rank ring. To prove equality, use the actual sphere-bundle involution τ(V,o,v)=(V,o,−v). It fixes p and reverses the orientation of the complement plane: the ordered first vector v changes sign while the orientation o stays fixed. Thus cτ=σc, with σ orientation reversal on BSO(2m). Since τp=p*, the image of j* is σ-invariant. The Euler sign formula gives σe=−e and σp_i=p_i. Every element of the already computed even-rank polynomial ring has a unique expansion Σ e^a P_a(p₁,…,p_{m-1}); since 2 is invertible, its invariants are exactly the polynomials with even a. This proves the odd-rank presentation. No rank/dimension/saturation assertion is needed here.

Finally the finite-cover transfer identifies BO(n) cohomology with invariants of the orientation double cover. In odd rank all generators are fixed. In even rank the preceding even-power calculation gives R[p₁,…,p_m]. Naturality identifies these p_i with the unoriented universal classes.