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An Anosov diffeomorphism is a continuously differentiable diffeomorphism of a compact smooth manifold with an invariant splitting of its tangent bundle into stable and unstable vector subbundles satisfying the displayed uniform exponential estimates for $n\ge 0$, with $C,\lambda >0$. Hyperbolic toral automorphisms give examples. An Anosov flow has a further one-dimensional invariant bundle along the flow direction; its definition is a distinct continuous-time version of uniform hyperbolicity.

A rank-$k$ distribution on a smooth manifold is a rank-$k$ vector subbundle of its tangent bundle. It is involutive when the Lie bracket of vector fields of any two local sections is again a section.

A holomorphic subbundle of a holomorphic vector bundle is a constant-rank vector subbundle which is locally spanned by a holomorphic linearly independent family of sections and whose inclusion is a holomorphic bundle map. It has an induced holomorphic vector bundle structure. A varying-rank coherent subsheaf is not generally a subbundle.

A rank-$n$ vector subbundle of a rank-$2n$ symplectic vector bundle is Lagrangian when each fibre is a Lagrangian subspace. If it exists, a compatible almost complex structure supplies a complementary Lagrangian subbundle by applying $J$. Existence is a global condition: the area-oriented tangent bundle of the two-sphere has no real line subbundle, by triviality of real line bundles over the sphere and the Hairy ball theorem.

A fiber metric on a smooth vector bundle $E$ gives a smooth orthogonal projection $P:E\to F$ onto any vector subbundle. In a local frame of $E$, let $G$ be the metric matrix and $A$ the columns of a frame of $F$. Then $P=A(A^{t}GA{)}^{-1}{A}^{t}G$, proving smoothness. The complementary subbundle $F^{\perp }=\ker P$ maps isomorphically onto the quotient vector bundle. Explicitly $v\mapsto (Pv,[v])$ is a vector bundle isomorphism $E\to F\oplus (E/F)$, with inverse $(f,[v])\mapsto f+(I-P)v$. This inverse is independent of the representative $v$, but the splitting depends on the chosen metric.

An almost complex structure is a smooth real vector bundle endomorphism $J:TM\to TM$ satisfying $J^2=-I$. In particular, $M$ has even real dimension $2n$. Extend $J$ complex-linearly to $T_{\mathbb{C}}M=TM{\otimes }_{\mathbb{R}}\mathbb{C}$. Its minimal polynomial divides $(u-i)(u+i)$, whose roots are distinct, so the type decomposition of the complexified tangent bundle isThe smooth projections are $(I-iJ)/2$ and $(I+iJ)/2$. Complex conjugation exchanges the two rank-$n$ eigenbundles. A $(1,0)$-covector annihilates $T^{0,1}M$ and a $(0,1)$-covector annihilates $T^{1,0}M$. Thus the differential forms of type (p, q) are the smooth sections ofHere the tensor product is identified with its image under the wedge product of differential forms. For a real tangent vector $v$, the vectors $v-iJv$ and $v+iJv$ exhibit the two types directly.

We now construct the almost complex structure from a decomposable volume form. Locally write $\Omega ={\theta }_1\wedge \cdots \wedge {\theta }_n$. The nonvanishing of $\Omega \wedge \overline{\Omega }$ says that the $2n$ covectors ${\theta }_1,\dots ,{\theta }_n,{\overline{\theta }}_1,\dots ,{\overline{\theta }}_n$ form a complex coframe. In particular, the span of the ${\theta }_j$ has the intrinsic descriptionIndeed, expansion in that coframe shows that every coefficient of a conjugate factor must vanish if $\alpha \wedge {\Omega }_x=0$. Therefore this span is independent of the chosen local factorization. The local coframes show that $W$ is a smooth rank-$n$ vector subbundle, andDefine $J_{\Omega }^{\ast }$ to be multiplication by $i$ on $W$ and by $-i$ on $\overline{W}$. It commutes with complex conjugation, so is the complexification of a real smooth bundle endomorphism, whose dual defines $J_{\Omega }$ on $TM$. It has square $-I$, and its $(1,0)$-covectors are exactly $W$. Conversely, any such almost complex structure must have this same action on the entire coframe. Thus $J_{\Omega }$ exists globally and is unique, and every factor in every allowed decomposition is a $(1,0)$-form.

Finally suppose $d\Omega =0$. Since ${\theta }_j\wedge \Omega =0$, the exterior derivative product rule givesDecompose $d{\theta }_j$ into types $(2,0)$, $(1,1)$ and $(0,2)$ using $J_{\Omega }$. The first two wedge products with the $(n,0)$-form $\Omega$ vanish by dimension. On $(0,2)$-forms, wedging with $\Omega$ is injective: the monomials $\Omega \wedge {\overline{\theta }}_a\wedge {\overline{\theta }}_b$ are linearly independent. For $n=1$ the $(0,2)$ space is zero, so the statement still holds. Hence every $d{\theta }_j$ has no $(0,2)$ component. For an arbitrary $(1,0)$-form $\alpha ={\sum }_j{a}_j{\theta }_j$, the extra terms $d{a}_j\wedge {\theta }_j$ likewise have only types $(2,0)$ and $(1,1)$. The bracket and differential-form criteria for integrability proved below now giveLocal decomposability has been used essentially; the nonvanishing top-degree product alone does not provide the required coframe.

Work with integral cohomology. For a frame $v=(v_1,\dots ,v_k)$ in the complex Stiefel manifold, defineThese fibres form the complement bundle on a complex Stiefel manifold, a complex vector subbundle of the trivial rank-$n$ bundle. Indeedis a smooth orthogonal projector of constant complex rank $n-k$; locally projecting suitable fixed basis vectors and applying Gram-Schmidt process provides bundle frames. The underlying real vector bundle has rank $r=2(n-k)$ and is oriented by its complex structure: a complex change of frame has real determinant equal to the positive number $|{\det }_{\mathbb{C}}A{|}^2$.

A point of its sphere bundle is $(v_1,\dots ,v_k,w)$ with $\Vert w\Vert =1$ and $w$ orthogonal to the frame. Appending $w$ gives a diffeomorphismwhose inverse forgets the last vector while retaining it as the unit vector in the complement. Thus the asserted homotopy equivalence is actually an identification of sphere bundles, for $0\le k<n$.

Start the induction at $V_0({\mathbb{C}}^n)=\{\mathrm{point}\}$, with cohomology the exterior algebra on no generators. SupposeThis is a free abelian group in every degree, and for $k\ge 1$ its smallest positive degree is $2n-2k+1=r+1$. Hence $H^r(V_k;\mathbb{Z})=0$. This vanishing holds at $k=0$ as well, since the base is a point and $r=2n>0$. The Euler class $e(E)$ belongs to this zero group and is therefore zero.

For this oriented rank-$r$ vector bundle, the Gysin sequence of a sphere bundle has segmentBecause $e(E)=0$, this becomesIn particular $H^{\ast }(V_{k+1};\mathbb{Z})$ is torsion-free. At $j=r-1$, choose a class $a\in H^{r-1}(V_{k+1};\mathbb{Z})$ with ${\pi }_{!}a=1$. Its restriction to the sphere fibre is an integral generator. The projection formula for sphere bundle integration, with the conventional fibre-integration sign, isIt can be obtained from the relative cohomology of the disk/sphere pair and the Thom isomorphism theorem. Consequently every cohomology class of $V_{k+1}$ is uniquely expressible as ${\pi }^{\ast }c+{\pi }^{\ast }b⌣a$: subtract the second term using its fibre integral, then use exactness to identify the remaining term; injectivity of ${\pi }^{\ast }$ and the projection formula prove uniqueness. This establishes an additive splitting by the old monomials and the old monomials multiplied by $a$.

To establish the ring, $|a|=r-1$ is odd, so graded commutativity of the cup product gives $a^2=-a^2$, and hence $2a^2=0$. The torsion-freeness just proved now gives $a^2=0$. Graded commutativity also gives the required anticommutation with each old odd generator, and the old relations persist under the injective ring map ${\pi }^{\ast }$. The additive splitting proves that there are no additional relations. This is exactly Gysin ring splitting for an odd-dimensional sphere bundle, and supplies a new exterior generator of degreeInduction therefore proves the full ring calculationOver $\mathbb{Z}$, graded commutativity by itself only kills twice an odd square; torsion-freeness is the essential extra step in this integral cohomology of a complex Stiefel manifold proof.

Finally an ordered orthonormal $n$-frame is a unitary matrix, so $V_n({\mathbb{C}}^n)=U(n)$. Each exterior generator may be included once or omitted from a basis monomial. The rank of degree $j$ is therefore the number of subsets of the generator degrees $1,3,\dots ,2n-1$ summing to $j$. Multiplication of their two-term generating factors givesThe highest exponent is $1+3+\cdots +(2n-1)=n^2$, agreeing with the real dimension of the unitary group, and the total rank is $2^n$.

Let $r=\mathrm{rank}F$ and $s=\mathrm{rank}E$. Around any $p$, choose a local frame $f_1,\dots ,f_r$ of the vector subbundle $F$. Choose additional smooth sections $e_{r+1},\dots ,e_s$ of $E$ which complement these vectors at $p$. They can be obtained by constant-coordinate sections in a vector bundle trivialization. The determinant of the resulting frame is nonzero at $p$, hence on a sufficiently small neighborhood. Thusis an adapted local frame of $E$ near $p$.

On the fiberwise quotient, the classes $[e_{r+1}],\dots ,[e_s]$ form a basis. Give the quotient vector bundle its local coordinates byAn overlap between adapted frames has a smooth transition function of a vector bundle represented by an invertible block upper triangular matrix,The zero lower-left block expresses that both sets of first $r$ frame vectors span $F$. Its lower-right block $D$ is smooth and invertible. Passing to quotient coordinates removes the upper block, leaving precisely the transition $D$. These lower-right blocks satisfy the cocycle identity because the original transition matrices do. Consequently they define a compatible smooth bundle atlas of rank $s-r$. This proves $E/F$ is a smooth vector bundle, and its fiberwise quotient map $q:E\to E/F$ is a smooth vector bundle morphism. The construction is independent of the adapted frames: any two choices have the same type of compatible transition. The usual quotient topology agrees with this atlas, since in adapted coordinates $q$ is an open linear projection on each fiber.

Choose the smooth fiber metric supplied by part (a), and let $P_p:E_p\to F_p$ be orthogonal projection. These projections vary smoothly. Indeed, in a local frame of $E$ write its metric as a positive-definite matrix $G(p)$, and let the columns of $A(p)$ be a frame of $F$. ThenThe matrix $A^{t}GA$ is positive definite, so its inverse is smooth. This formula proves smoothness of the intrinsic projections in every chart. It also gives $P^2=P$ and $\mathrm{im}P=F$.

The orthogonal complement $F^{\perp }=\ker P$ is a smooth vector subbundle: applying $I-P$ to a complement frame gives a frame of its fibers, of constant rank $s-r$, after shrinking the chart. Fiberwise,and the restriction $q{|}_{F^{\perp }}:F^{\perp }\to E/F$ is a smooth fiberwise bijective vector bundle morphism. In local frames its matrix is invertible, with smooth inverse, so it is a vector bundle isomorphism.

More explicitly, the desired orthogonal splitting of a vector subbundle isIts inverse is $(f,[v])\mapsto f+(I-P)v$. Replacing $v$ by $v+w$ with $w\in F_p$ leaves $(I-P)v$ unchanged, so the inverse is well-defined. Both maps are smooth and linear on fibers, and their compositions are identities. The splitting exists globally but depends on the chosen fiber metric.

For a smooth vector subbundle $F\subset E$, the quotient has fibers $(E/F{)}_p=E_p/F_p$. Extend a local frame of a vector bundle of $F$ to one of $E$; the classes of the complementary frame vectors trivialize the quotient. Adapted transition functions of a vector bundle are block upper triangular, and their lower-right blocks give the quotient transitions. These are smooth and satisfy the cocycle identity, defining a smooth vector bundle. The projection $E\to E/F$ is a smooth vector bundle morphism.