Rational cohomology of BO and BSO by Pontryagin and Euler classes

Models and base case. The orientation double cover lifts the Schubert cells of $G_n$ to cells of $B_n$: pull back each characteristic disk, use its two trivial sheets and attach their boundary lifts. The covering topology locally agrees with the lifted weak CW topology; this is the CW structure recorded in [F8]. The universal bundles are in the scope of [F1]–[F3]. For $n>0$, $B_n$ is path-connected because it is the continuous image of the contractible nonempty $V_n({\mathbb{R}}^{\mathrm{\infty }})$ under the frame quotient; the same is true of $G_n$. At rank one, $\mathrm{SO}(1)$ is trivial, so $B_1=V_1({\mathbb{R}}^{\mathrm{\infty }})$ is contractible by [F6]. On a point the rational cochain complex is $\mathbb{Q}\stackrel{0}{\to }\mathbb{Q}\stackrel{1}{\to }\mathbb{Q}\stackrel{0}{\to }\cdots$, by the alternating sum of identical faces in [F7]. Thus its cohomology is $\mathbb{Q}$ in degree zero and zero otherwise. Homotopy invariance proves $P(1)$; $G_0$ and $B_0$ are points as stated separately.

Induction hypothesis. Fix $n\ge 2$ and assume $P(n-1)$. The even-rank case below proves $P(n)$ when $n=2k$; the odd-rank case proves it when $n=2k+1$. Only the immediately preceding rank is assumed in either branch.

Coefficient and sphere-bundle conventions. Postcomposition of integral cochains with $\mathbb{Z}\to \mathbb{Q}$ commutes with the face differential and cup products by [F7]. It carries a normalized integral Thom class to a normalized rational one, hence carries the Euler class to the Euler class used in [F1]. Consequently [F4] makes odd-rank Euler classes zero rationally. For $n\ge 2$, let $s:B_{n-1}\to S_n$ be a homotopy inverse to $r$ from [F2]. The map $j_n=ps$ is a map between the CW bases. Pulling back the actual splitting in [F2] gives $j_n^{\ast }{\gamma }_n^{+}\cong {\epsilon }^1\oplus (rs{)}^{\ast }{\gamma }_{n-1}^{+}$. Naturality and stability on these CW bases and $rs\simeq \mathrm{i}\mathrm{d}$ give $j_n^{\ast }{p}_i=p_i$. Also $j_n^{\ast }e=s^{\ast }{p}^{\ast }e=0$. Under the cohomology isomorphism $s^{\ast }$, the Gysin map $p^{\ast }$ is exactly $j_n^{\ast }$. Thus the Gysin sequence can use $H^{\ast }(B_{n-1})$ without asserting that $S_n$ is literally $B_{n-1}$ or applying a CW-only characteristic-class interface to $S_n$.

Even rank, exactness. Suppose $n=2k$ with $k\ge 1$. By $P(2k-1)$, the target of $j_{2k}^{\ast }$ is $\mathbb{Q}[p_1,\dots ,p_{k-1}]$. Step 2.1 shows each generator lifts, so this map is surjective in every degree. Exactness of [F1], also in the preceding degree, makes multiplication by $e$ injective and gives $\ker j_{2k}^{\ast }=e{H}^{\ast }(B_{2k})$. Explicitly, $0\to H^{j-2k}(B_{2k})\stackrel{⌣e}{\to }{H}^j(B_{2k})\stackrel{j_{2k}^{\ast }}{\to }{H}^j(B_{2k-1})\to 0$.

Odd rank, exactness and subring. Suppose $n=2k+1$ with $k\ge 1$. By step 2.1, the rational Euler class is zero, so [F1] gives $0\to H^j(B_{2k+1})\stackrel{j_{2k+1}^{\ast }}{\to }{H}^j(B_{2k})\to H^{j-2k}(B_{2k+1})\to 0$. The induction hypothesis gives $H^{\ast }(B_{2k})=\mathbb{Q}[p_1,\dots ,p_{k-1},e]$. The image of the injection contains $p_1,\dots ,p_{k-1},p_k=e^2$ by step 2.1 and [F3]. These are algebraically independent: distinct monomials in the formal last variable give distinct even powers of the independent variable $e$. Call their polynomial subring $R$. The target is the graded free $R$-module $R\oplus eR$.

Even rank, polynomial generation and independence. Define $\mathrm{\Phi }:\mathbb{Q}[P_1,\dots ,P_{k-1},E]\to H^{\ast }(B_{2k})$ by $P_i\mapsto p_i$, $E\mapsto e$, with degrees $4i,2k$. For a homogeneous class $x$ of degree $j$, choose a homogeneous polynomial $q(P)$ with $j_{2k}^{\ast }x=q(p)$ by step 3.1; then $x-\mathrm{\Phi }(q)=ey$ with $y$ in degree $j-2k$. Ascending induction on nonnegative degree, with negative groups zero, proves surjectivity. For injectivity write an arbitrary finite polynomial as $R={\sum }_{r=0}^N{E}^r{A}_r(P)$. If $\mathrm{\Phi }(R)=0$, apply $j_{2k}^{\ast }$: algebraic independence in $P(2k-1)$ forces $A_0=0$, because $j_{2k}^{\ast }e=0$. Divide the remaining polynomial formally by $E$; injectivity of multiplication by $e$ from step 3.1 makes its image zero. Finite repetition yields all $A_r=0$. This proves $P(2k)$, with $p_k=e^2$ by [F3]. At $k=1$, the target is $H^{\ast }(B_1)=\mathbb{Q}$ and the same argument starts the induction at $B_2$.

Odd rank, finite-dimensional comparison. Let $r_j={\dim }_{\mathbb{Q}}{R}^j$ and $h_j={\dim }_{\mathbb{Q}}{H}^j(B_{2k+1})$, setting both to zero for $j<0$. Each target degree in step 3.2 is finite-dimensional, since it is a polynomial ring on finitely many positive-degree generators. The injection therefore makes $h_j$ finite too. Exactness gives $h_j+h_{j-2k}=r_j+r_{j-2k}$. Ascending induction on $j\ge 0$ yields $h_j=r_j$. Since $R^j$ is contained in the image and has its full dimension, it equals that image in every degree. The injective ring map thus identifies $H^{\ast }(B_{2k+1})$ with $R=\mathbb{Q}[p_1,\dots ,p_k]$, proving $P(2k+1)$.

Induction conclusion. Starting from $P(1)$, for each $n\ge 2$ exactly its parity branch proves $P(n)$ from $P(n-1)$. In particular the even branch first proves $P(2)$, then the odd branch proves $P(3)$, then the even branch proves $P(4)$; no even-rank result is assumed before it is proved. The canonical rank-zero result was handled separately in step 1.1 rather than by an invalid polynomial formula involving an Euler generator of degree zero.

Forget orientation. For $n\ge 1$ the double cover in [F5] has path-connected total space by step 1.1, and orientation reversal acts transitively on its two-point fibers, so it is regular in the transfer supplier's convention. The reversal is nonidentity and fixes the underlying tautological bundle; hence it fixes every $p_i$ by naturality and negates $e$ when $n$ is even and positive by [F3]. Transfer identifies $H^{\ast }(G_n;\mathbb{Q})$ with the invariant subring. For odd $n=2m+1\ge 3$, all the polynomial generators are fixed. For even $n=2m$, write each polynomial uniquely as $\sum e^r{A}_r(p_1,\dots ,p_{m-1})$. Invariance under $e\mapsto -e$ forces $2A_r=0$ for odd $r$, hence those coefficients vanish over $\mathbb{Q}$. The invariants are exactly $\mathbb{Q}[p_1,\dots ,p_{m-1},e^2]=\mathbb{Q}[p_1,\dots ,p_m]$. Naturality along the cover identifies these with the stated universal Pontryagin classes on $G_n$. This proves both displayed unoriented rings.

Boundaries. The smallest displayed even case is $H^{\ast }(B_2;\mathbb{Q})=\mathbb{Q}[e]$ with $\mathrm{\mid }e\mathrm{\mid }=2$ and $p_1=e^2$; the smallest odd case is $H^{\ast }(B_3;\mathbb{Q})=\mathbb{Q}[p_1]$. Degree zero is $\mathbb{Q}$, negative degrees vanish, and no positive-degree polynomial generator occurs in $G_0,B_0$ or the contractible $B_1$. No double-cover assertion was used in rank zero, and no orientation reversal was applied to its canonical unit orientation. AC is inherited from the stated suppliers; the only division by sheet number is by 2 in rational cohomology.