Oriented real two-plane splitting with real-cohomology injection

Fix $n\ge 3$ and write $G_n={\mathrm{Gr}}_2^{+}({\mathbb{R}}^n)$, with [F2, F13] tautological oriented plane $L$ and oriented orthogonal complement $Q$. The quotient definition gives a continuous bijection from the compact Stiefel quotient to the set of unit simple bivectors in ${\mathrm{\Lambda }}^2{\mathbb{R}}^n$; the bivector records both the plane and its orientation. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism here: a closed subset of the compact source is compact, and its image is closed in the Hausdorff target by [F13]. The graph charts make $G_n$ a smooth manifold of dimension $2n-4$; the finite coordinate-plane chart cover makes it second-countable. The group $\mathrm{SO}(n)$ acts transitively on oriented two-planes, and is path-connected: successive plane rotations reduce any special orthogonal matrix to the identity. Hence $G_n$ is path-connected.

Let $W_n={\mathbb{C}\mathbb{P}}^{n-1}\setminus {\mathbb{R}\mathbb{P}}^{n-1}$. Write a point as [F2] $[z]=[x-iy]$ with $x,y\in {\mathbb{R}}^n$ linearly independent. The ordered pair $(x,y)$ orients its real span; multiplying $z$ by a nonzero complex scalar changes $(x,y)$ by a positive-determinant similarity, so this defines a map $W_n\to G_n$. Conversely, in a local oriented orthonormal frame of a plane, the point of $W_n$ is an invertible $2\times 2$ real matrix modulo positive similarities. Polar decomposition writes it uniquely as a positive scalar, an element of $\mathrm{SO}(2)$, and a positive symmetric determinant-one matrix. Thus $W_n$ is the associated bundle over $G_n$ with fiber the positive symmetric determinant-one matrices. The homotopy $A\mapsto A^{1-t}$ contracts that fiber to the identity; it is equivariant under orthogonal conjugation, so it descends through the frame changes and gives a deformation retraction $W_n\simeq G_n$. The real-part map on each complex tautological line sends $z=x-iy$ and $iz=y+ix$ to the positively oriented basis $(x,y)$, identifying the pulled-back oriented plane with the underlying real complex tautological line.

Put $X={\mathbb{C}\mathbb{P}}^{n-1}$, $A={\mathbb{R}\mathbb{P}}^{n-1}$, and $d=2n-2$. [F6, F7, F14] The quotient maps from $S^{2n-1}$ to $X$ and from $S^{n-1}$ to $A$ show that both are compact. By [F7], $A$ is a closed smooth embedded submanifold. The tubular-neighborhood theorem [F7] gives a tubular diffeomorphism from an open neighborhood of the zero section of the normal bundle onto a neighborhood of $A$. By [F14], give the normal bundle a smooth metric. Around each point of the compact zero section, a bundle chart contains a product neighborhood lying inside the tubular domain; finitely many such base patches cover $A$, and the minimum of their positive fiber radii gives a uniform disk neighborhood. Every smaller-radius disk neighborhood deformation retracts radially to $A$. Their images $U$ are nested tubular neighborhoods. Their complements $K=X\setminus U$ are compact subsets of $W_n$ and are cofinal among compact subsets of $W_n$: for compact $C\subset W_n$, the open set $X\setminus C$ contains the compact zero section, so compactness gives a sufficiently small uniform disk neighborhood lying in $X\setminus C$. By [F7] and excision, $H_c^q(W_n;\mathbb{R})\cong {\mathrm{colim}}_U{H}^q(X,U;\mathbb{R})$. The inclusion $A\hookrightarrow U$ is a homotopy equivalence; the natural pair long exact sequences and five lemma identify every $H^q(X,U;\mathbb{R})$ with $H^q(X,A;\mathbb{R})$. Consequently $H_c^q(W_n;\mathbb{R})\cong H^q(X,A;\mathbb{R})$, and the map forgetting support is the relative-to-absolute map $ϵ:H^q(X,A;\mathbb{R})\to H^q(X;\mathbb{R})$.

Apply the natural UCT to the integral homology in [F6]. For $G=\mathbb{Z}$, [F6, F7] $\mathrm{Hom}(G,\mathbb{R})=\mathbb{R}$ and ${\mathrm{Ext}}^1(G,\mathbb{R})=0$. For $G=\mathbb{Z}\mathrm{/}2$ both terms vanish: the Hom group is zero because $\mathbb{R}$ has no 2-torsion, and the free resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$ computes ${\mathrm{Ext}}^1(\mathbb{Z}\mathrm{/}2,\mathbb{R})=\mathbb{R}\mathrm{/}2\mathbb{R}=0$. Thus $H^q(X;\mathbb{R})=\mathbb{R}$ in even degrees $0,2,\dots ,d$ and zero otherwise. For $A$, it is $\mathbb{R}$ in degree zero, and also in degree $n-1$ when $n$ is even; all its other positive-degree groups vanish. The pair long exact sequence now gives $H_c^0(W_n;\mathbb{R})=0$; for positive even $q\le d$ it gives $H_c^q(W_n;\mathbb{R})=\mathbb{R}$ except that $H_c^n(W_n;\mathbb{R})={\mathbb{R}}^2$ when $n$ is even; all odd groups and groups outside $[0,d]$ vanish.

The complex orientation of $X$ restricts to an orientation of the open [F6, F8, F9] manifold $W_n$. Poincaré duality and the deformation retraction in step 1.2 therefore compute $H^{\ast }(G_n;\mathbb{R})$: it is $\mathbb{R}$ in degrees $0,2,\dots ,2n-4$, zero in odd degrees, and has dimension two in degree $n-2$ when $n$ is even. Let $a=e({\gamma }_{\mathbb{C},\mathbb{R}})\in H^2(X;\mathbb{Z})$ be the generator from [F6], let $i:W_n\hookrightarrow X$, and let $x=e(L)\in H^2(G_n;\mathbb{Z})$, with integral classes mapped to real coefficients as needed. By [F6] and [F8], the image of each $a^j$ is a nonzero generator of $H^{2j}(X;\mathbb{R})$. For positive even $q\le d$ except $q=n$ when $n$ is even, the pair sequence shows $ϵ$ of step 1.3 is an isomorphism. PD identifies it with $i_{\ast }:H_{d-q}(W_n;\mathbb{R})\to H_{d-q}(X;\mathbb{R})$. By the natural UCT pairing, the dual restriction $i^{\ast }:H^{d-q}(X;\mathbb{R})\to H^{d-q}(W_n;\mathbb{R})$ is therefore an isomorphism. Step 1.2 identifies the pullback of $x$ with $i^{\ast }a$; since $a^j$ generates $H^{2j}(X;\mathbb{Z})$, it follows that $x^j\mathrm{\ne }0$ for every $0\le j\le n-2$ except possibly $j=(n-2)\mathrm{/}2$ when $n$ is even. In that exceptional case $x^j\mathrm{\ne }0$ follows below from the nonzero square $x^{n-2}$.

Suppose $n=2k\ge 4$. The ordered sum $L\oplus Q$ is the trivial oriented [F9, F10] rank-$2k$ bundle. By [F9], $xe(Q)=e(L)e(Q)=e(L\oplus Q)=0$, since the trivial positive-rank bundle has a nowhere-zero section. To calculate the square, take a homotopy equivalence $h:Y\to G_n$ from a path-connected CW complex. Since $G_n$ is a finite-dimensional second-countable smooth manifold, [F4] makes $Q$ numerable; pulling its numeration back makes $Q_Y=h^{\ast }Q$ numerable on $Y$. Hatcher's CW paracompactness proof, recorded in [F4], gives that $Y$ is paracompact and Hausdorff, so the Pontryagin suppliers apply. Set $L_Y=h^{\ast }L$ and $x_Y=h^{\ast }x$. Euler naturality identifies their Euler classes with pullbacks of those on $G_n$. The top Pontryagin theorem gives $p_1(L_Y)=x_Y^2$, and the rank cutoff gives $p(L_Y)=1+x_Y^2$. Pontryagin multiplicativity over $\mathbb{R}$ and stability under a trivial summand yield $p(L_Y)p(Q_Y)=p(L_Y\oplus Q_Y)=1$. Comparing successive homogeneous degrees in this equation gives $p_j(Q_Y)=(-1{)}^j{x}_Y^{2j}$ for $0\le j\le k-1$. The top Pontryagin theorem for $Q_Y$ then gives $e(Q_Y{)}^2=p_{k-1}(Q_Y)=(-1{)}^{k-1}{x}_Y^{n-2}$. Since $h^{\ast }$ is an isomorphism and Euler classes are natural, this descends to $e(Q{)}^2=(-1{)}^{k-1}{x}^{n-2}$ on $G_n$. Step 1.5 established $x^{n-2}\mathrm{\ne }0$; hence $x^{(n-2)\mathrm{/}2}$ and $e(Q)$ are linearly independent: multiplying a relation $\alpha x^{(n-2)\mathrm{/}2}+\beta e(Q)=0$ by $x$ and using $xe(Q)=0$ gives $\alpha x^{n\mathrm{/}2}=0$, while step 1.5 gives $x^{n\mathrm{/}2}\mathrm{\ne }0$; then $\alpha =0$, and $e(Q{)}^2\mathrm{\ne }0$ forces $\beta =0$. These two classes therefore form a basis of the two-dimensional middle cohomology. For odd $n$, the nonzero powers in step 1.5 already form a basis by their distinct degrees. Thus the full fiber basis is $1,x,\dots ,x^{n-2}$ when $n$ is odd, and $1,x,\dots ,x^{n-2},e(Q)$ when $n$ is even.

For each oriented smooth Euclidean bundle $V\to B$ of rank $n\ge 3$, [F1, F2, F3, F5] form its oriented Grassmann bundle $\pi :G_2^{+}(V)\to B$. Local positive orthonormal frames and the graph charts of [F2] give smooth local product charts with fiber $G_n$. Transition maps act smoothly by $\mathrm{SO}(n)$; their formulas remain smooth on half-space charts by [F5] when $B$ has boundary. Thus the total is a finite-dimensional smooth manifold, with boundary exactly over the boundary of $B$. It is Hausdorff: points over different base points separate by inverse images of base neighborhoods, and points in one fiber separate in a bundle chart. It is second-countable: the trivializing cover has a countable subcover by Lindelöfness; each product chart has a countable basis, and their countable union is a basis. The pulled-back bundle splits orthogonally as ${\pi }^{\ast }V=L_V\oplus Q_V$, with both summands oriented as in [F2].

Each $\pi$ is proper. Let $K\subset B$ be compact. [F13] Around each point of $K$, choose a relatively open coordinate ball or half-ball $U$ whose compact closure lies inside a bundle-trivializing chart. Heine–Borel makes each closure compact. Finitely many such $U_i$ cover $K$. Each $K\cap \overline{U_i}$ is a closed subset of compact $K$, hence compact; in the trivialization, ${\pi }^{-1}(K\cap \overline{U_i})$ is homeomorphic to $(K\cap \overline{U_i})\times G_n$, compact by [F13]. Their finite union is ${\pi }^{-1}(K)$, so it is compact.

Every stage projection is numerable. On its smooth base, apply the [F3, F4] boundaryless or boundary partition theorem in [F3] to its bundle-trivializing cover; the supplied partition satisfies the support and local-finiteness conditions in the definition of numerable fiber bundle. Its total is again a second-countable smooth manifold, so [F4] makes every stage paracompact Hausdorff and of CW homotopy type and makes its smooth finite-rank vector bundles numerable.

We prove real-cohomology injectivity for $\pi :G_2^{+}(V)\to B$ componentwise. [F4, F11] The fiber $G_n$ is path-connected. Since a smooth manifold is locally path connected, its connected components are path components; the local bundle charts and path lifting show that the total-space components are exactly the preimages of base components. Fix one such component $C$ and a path-connected CW complex $Y$ with a homotopy equivalence $h:Y\to C$, available by [F4]. Pull back $\pi$ to $\tilde{G}\to Y$. The pulled-back bundle is numerable because its partition is the pullback of the partition in step 1.9. By [F11] both projections are Hurewicz, hence Serre, fibrations.

The global classes $1,e(L_V),e(L_V{)}^2,\dots ,e(L_V{)}^{n-2}$, [F9, F12] together with $e(Q_V)$ when $n$ is even, restrict to the full fiber basis of step 1.6 by naturality of Euler classes and cup products. Pull these classes back to $\tilde{G}$; on each fiber their restrictions are the same basis. Applying Leray–Hirsch [F12] to $\tilde{G}\to Y$ shows that ${\tilde{\pi }}^{\ast }:H^{\ast }(Y;\mathbb{R})\to H^{\ast }(\tilde{G};\mathbb{R})$ is injective: in the module isomorphism, pullback is exactly the coefficient of the basis element $1$.

The pullback map $\tilde{h}:\tilde{G}\to {\pi }^{-1}(C)$ induces a [F11] weak homotopy equivalence. On fibers it is the identity, and on bases it is the homotopy equivalence $h$. The natural long exact sequences [F11] give isomorphisms on all higher homotopy groups. For $m\ge 3$, the terms in the five-term segment around ${\pi }_m$ are abelian, so the module five lemma applies. In degree two, if $z\in {\pi }_2(\tilde{G})$ maps to zero, its base class is zero because $h_{\ast }:{\pi }_2(Y)\to {\pi }_2(C)$ is injective; hence $z$ comes from ${\pi }_2(G_n)$. Its fiber class is a boundary from ${\pi }_3(C)$, which lifts through the surjection $h_{\ast }:{\pi }_3(Y)\to {\pi }_3(C)$, so exactness makes $z=0$. Conversely, for $z\in {\pi }_2({\pi }^{-1}(C))$, its base class lifts to ${\pi }_2(Y)$; naturality and the identity fiber map make the lifted class have zero boundary in ${\pi }_1(G_n)$, so exactness lifts it to ${\pi }_2(\tilde{G})$. The difference from $z$ lies in the image of ${\pi }_2(G_n)$ and can be corrected there. For ${\pi }_1$ use the group sequence ${\pi }_2(C)\to {\pi }_1(G_n)\to {\pi }_1({\pi }^{-1}C)\to {\pi }_1(C)\to {\pi }_0(G_n)$: injectivity lifts a boundary witness through ${\pi }_2(Y)\to {\pi }_2(C)$, and surjectivity first lifts the base loop through $h_{\ast }$ and then corrects by a loop in the common fiber. Since the fiber and both bases are path-connected, the total spaces are path-connected too, so $\tilde{h}$ is also a bijection on components.

By [F11], $\tilde{h}$ induces an isomorphism on integral homology. [F8, F11] Apply the natural UCT sequence to the free singular chain complexes with coefficient group $\mathbb{R}$. The induced maps on the Hom and Ext terms are isomorphisms because the integral homology maps are; the five lemma therefore makes ${\tilde{h}}^{\ast }:H^{\ast }({\pi }^{-1}C;\mathbb{R})\to H^{\ast }(\tilde{G};\mathbb{R})$ an isomorphism. Also $h^{\ast }:H^{\ast }(C;\mathbb{R})\to H^{\ast }(Y;\mathbb{R})$ is an isomorphism by [F8]. The square ${\tilde{h}}^{\ast }{\pi }^{\ast }={\tilde{\pi }}^{\ast }{h}^{\ast }$ commutes by functoriality. If ${\pi }^{\ast }a=0$, then ${\tilde{\pi }}^{\ast }{h}^{\ast }a=0$; step 1.11 gives $h^{\ast }a=0$, hence $a=0$. Thus ${\pi }^{\ast }$ is injective on every component. Singular cochains on a disjoint union are the product of component cochains; full AC makes the product of component coboundary preimages surjective, so cohomology is the product of component cohomologies. Therefore ${\pi }^{\ast }$ is injective globally.

Start with $B_0=M$ and $V_0=E$. Whenever the current oriented complement [F1] $V_j$ has rank $n_j\ge 3$, set $B_{j+1}=G_2^{+}(V_j)$, pull $V_j$ back, and replace it by its oriented orthogonal complement $V_{j+1}$. Step 1.7 keeps each stage smooth, Hausdorff and second-countable; steps 1.8–1.9 make every projection proper and numerable; step 1.13 proves every cohomology pullback injective. The rank drops by two at each stage, so the process stops after finitely many stages with rank zero, one, or two. A rank-one oriented Euclidean bundle has the unique positive unit section and is the oriented trivial line; a rank-two terminal complement is itself the final oriented two-plane. The composite $q:F(E)=B_s\to M$ is proper by finite composition of proper maps; its cohomology pullback is the composition of the stagewise injections. The tautological planes and terminal rank-two plane, or final line when rank one, give the required ordered orthogonal decomposition.

If $M=\mathrm{\varnothing }$, its tower is empty and all cohomology groups are [A1, F1, F3] zero. If $r=0$, take $q={\mathrm{id}}_M$ and the empty sum. If $r=1$, take the identity and the unique positive unit section. If $r=2$, no Grassmann stage is needed: take the identity and the single oriented plane $E$. These identity maps are proper and induce identity maps in cohomology. At every positive-rank stage the complement is oriented by the rule that $L_j\oplus V_j$ has the pulled-back orientation, so no orientation choice is hidden. The boundary case is included by the half-space chart and partition arguments of steps 1.7–1.9; the fiber calculation uses only closed boundaryless manifolds. The item is a one-way existence statement, so neither direction of an iff is applicable. AC is used only in the supplier and component-product uses recorded in [A1]. Full AC lets us choose cocycle representatives for any family of component cohomology classes and choose coboundary preimages for any family of component boundaries; hence the canonical map from cohomology of the disjoint union to the product of component cohomologies is an isomorphism.