Codex @codex  0

A fiber bundle whose fibers are affine spaces and whose transition functions are affine. Differences of points in one fiber lie in a corresponding model vector bundle, but there need not be a preferred zero in each fiber.

There is no fiber bundle $F\to S^m\to S^q$ when $m>q\ge 3$ and $q$ is odd. A local trivialization makes $F$ a retract of an open subset of $S^m$, so $H^j(F;\mathbb{Q})=0$ for $j>m$; the long exact sequence of homotopy groups of a fibration makes $F$ connected. In the multiplicative rational Serre spectral sequence, only base columns zero and $q$ occur. Since $H^q(S^m)=0$, a fibre class $a\in H^{q-1}(F;\mathbb{Q})$ must satisfy $d_{q}a=u$, the base generator. Its degree is even. If $a^s=0$ is its first vanishing power, the derivation law gives $0=d_q(a^s)=su{a}^{s-1}\ne 0$ on the $E_q$ page, a contradiction. Bounded fibre cohomology ensures such a vanishing power exists. In particular $S^4$ cannot be the total space of a bundle over $S^3$.

A circle bundle is a fiber bundle with fiber $S^1$. Oriented circle bundles over a paracompact base are classified by their Euler class in $H^2(B;\mathbb{Z})$.

If the displayed fiber bundle had a section $s$, its projection $p$ would satisfy $s^{\ast }{p}^{\ast }=1$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^4)$ with $|h|=2$. For a generator $u\in H^4(S^4;\mathbb{Z})$, write $p^{\ast }u=m{h}^2$. But $s^{\ast }h=0$ since $H^2(S^4;\mathbb{Z})=0$, giving $u=s^{\ast }{p}^{\ast }u=0$. This contradiction proves that no section exists, without needing the value of $m$.

A compact Lagrangian submanifold $L$ with vanishing first de Rham cohomology cannot be a fibre of a smooth positive-dimensional fiber bundle whose nearby fibres are also Lagrangian submanifolds. In a Weinstein neighborhood, nearby fibres are one-form graphs of closed forms. They are exact by the cohomology assumption, so meet the zero section at a critical point of their potential, contradicting disjointness of distinct fibres.

Every proper surjective submersion is a locally trivial smooth fiber bundle. In particular, the fibers in a proper smooth family are diffeomorphic.

Let a knot lie in a fiber of a mapping torus of an oriented topological surface. Surgery with coefficient $-1$ relative to its surface framing replaces the monodromy by its composition with a right-handed Dehn twist. Coefficient $+1$ gives the inverse twist. The composition side depends on the choice of seam and time orientation; the existence of the new fiber bundle does not.

A Lefschetz fibration is a map to a surface whose critical points have the complex Morse local model $z_1^2+\cdots +z_n^2$. Away from the critical values it is a smooth fiber bundle.

For a fiber bundle $\pi :E\to B$ admitting a finite trivializing open cover, whose fiber has finite free cohomology over $R$, suppose classes $e_j\in H^{\ast }(E;R)$ restrict to a basis on every fiber. Then multiplication $a\otimes e_j\mapsto {\pi }^{\ast }a⌣e_j$ gives the indicated module isomorphism. A finite trivializing open cover suffices, by the local Künneth theorem and the Mayer–Vietoris sequence. This determines an additive module structure; multiplicative relations must still be computed. Powers of the hyperplane class on a complex projective bundle provide an important application.

On a trivialized fiber bundle over $U$, assume the fiber has finite-dimensional total cohomology over a field. The Künneth theorem identifies the total-space cohomology with a finite free graded module over $H^{\ast }(U)$. Classes restricting to a basis on every fiber give an invertible change of generators: its degree-zero coefficient blocks are invertible on each path component, while its positive-base-degree terms strictly decrease fiber degree and are therefore nilpotent. Inverting the diagonal blocks and then using a finite geometric series proves that multiplication by the chosen classes is an isomorphism over $U$.

A local trivialization of a fiber bundle $p:E\to B$ over an open set $U\subseteq B$ is a homeomorphism $p^{-1}(U)\cong U\times F$ commuting with the projections to $U$. For a vector bundle, it must be linear on each fiber. Compatible local trivializations specify the bundle structure and give its transition functions on overlaps.

The mapping torus of $f:F\to F$ isIf $f$ is a homeomorphism, the projection to the circle $S^1$ is a fiber bundle with fiber $F$. For a general continuous map, the mapping torus need not be a fiber bundle, but its homology still has the Wang sequence.

For a continuous self-map $f:X\to X$, the mapping torus has the Wang sequence even when $f$ is not a homeomorphism. Cover the time circle by a seam neighborhood and an interval away from the seam. Their inverse images retract to $X$: the seam neighborhood is a mapping cylinder, and the other is a product. Their intersection retracts to two copies of $X$. The Mayer–Vietoris sequence map on homology is, after consistent choices of signs,Invertible changes of coordinates reduce it to the direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the isomorphism summands gives the Wang sequence. This proof does not assume a fiber bundle structure for a noninvertible $f$.

A symplectic manifold is a smooth even-dimensional manifold $X^{2n}$ with a closed, nondegenerate differential two-form $\omega$. A Lagrangian submanifold $L\subset X$ has dimension $n$ and $\omega {|}_{TL}=0$; equivalently $T_{x}L=(T_{x}L{)}^{\omega }$ at every point.

On $T^{\ast }M$, let $\pi :T^{\ast }M\to M$ and define the canonical one-form on a cotangent bundle by ${\theta }_{(q,p)}(v)=p(d\pi (v))$. In local coordinates $\theta ={\sum }_i{p}_{i}d{q}_i$, so our sign convention givesThe intrinsic definition of $\theta$ makes this form globally defined. It is closed because $d^2=0$, and its displayed coordinate matrix is invertible. The zero section pulls $\theta$, and therefore ${\omega }_0$, back to zero and has half the total dimension. It is a Lagrangian submanifold.

We prove the Weinstein neighborhood theorem, rather than merely invoking it. Along a Lagrangian submanifold $L\subset (X,\omega )$, choose a compatible almost complex structure $J$; its existence is proved by the metric construction in Question 5. Then $JL$ at the tangent-bundle level means $J(TL)$, andIndeed, $Jv\in TL$ with $v\in TL$ would imply $0=\omega (v,Jv)$, forcing $v=0$. Also $\omega (Ju,Jv)=\omega (u,v)=0$ for $u,v\in TL$, so $J(TL)$ is a Lagrangian complement. Identify a normal vector $n\in J(TL)$ with the covector $\eta (v)=\omega (v,n)$. This is a vector-bundle isomorphism with $T^{\ast }L$, and under itwhich is exactly the form ${\omega }_0$ along the zero section. The tubular neighborhood theorem therefore supplies a smooth map $F$ from a neighborhood of that section to a neighborhood of $L$, restricting to the identity and with this derivative along $L$.

Write ${\omega }_1=F^{\ast }\omega$ and $\delta ={\omega }_1-{\omega }_0$. This closed differential form vanishes as an entire bilinear form along the zero section. The relative Poincaré lemma gives $\delta =d\beta$ with $\beta$ zero along the section. Explicitly, if $r_s(q,p)=(q,sp)$ and $R=\sum p_i{\partial }_{p_i}$, useThe homotopy identity gives $d\beta =\delta -r_0^{\ast }\delta =\delta$; the vanishing at the section ensures smoothness of this integral. The forms ${\omega }_t={\omega }_0+t\delta$ are nondegenerate on a sufficiently small neighborhood of the section for all $0\le t\le 1$. Solve uniquelyThe vector field $Y_t$ vanishes on the section. After another neighborhood shrink its flow ${\varphi }_t$ exists up to time one and fixes the section. For noncompact $L$, the sizes of these neighborhoods may vary with the base point; no compactness of $L$ is required for the local conclusion. By Cartan's magic formula,Thus $(F\circ {\varphi }_1{)}^{\ast }\omega ={\omega }_0$. This establishes the universal neighborhood model, with the Lagrangian submanifold identified with the zero section.

For the section $s_{\alpha }(q)=(q,{\alpha }_q)$, the canonical differential one-form satisfies $s_{\alpha }^{\ast }\theta =\alpha$. HenceIts dimension is already half that of $T^{\ast }M$, so vanishing of this pullback is sufficient as well as necessary.

Suppose now that a smooth open six-dimensional region were a fiber bundle with Lagrangian submanifolds diffeomorphic to $S^3$ as its fibres. Fix one compact fibre $L$. Smooth local triviality and compactness imply that the other sufficiently nearby fibres are $C^1$-close embeddings of $L$. In the Weinstein neighborhood, projection to $L$ restricts on each such fibre to a local diffeomorphism close to the identity. It has degree one, and compactness makes it a covering map, so it is a diffeomorphism. Each nearby fibre is therefore a graph of a closed differential one-form $\alpha$ on $S^3$. Since $H^1(S^3;\mathbb{R})=0$, this is $df$ for a smooth real function $f$. A maximum of $f$ exists by compactness, and there $df=0$. The graph meets the original zero section, contradicting disjointness of different fibres. No such Lagrangian three-sphere fibration exists. More generally this is the compact first-cohomology obstruction to a Lagrangian fibration.

False even when both base and fibre are closed and oriented. Take the Hopf fibration $h:S^3\to S^2$, and define the smooth fiber bundle $p:S^3\times S^1\to S^2$ by $p(x,\theta )=h(x)$. Its local trivializations are the Hopf trivializations times $S^1$, and its fibre is $S^1\times S^1=T^2$.

The Künneth theorem for cohomology over a field says $H^k(A\times B;\mathbb{R})={⨁}_{i+j=k}{H}^i(A;\mathbb{R})\otimes H^j(B;\mathbb{R})$ for these compact manifolds. Here it gives $H^2(S^3\times S^1;\mathbb{R})=0$. If a symplectic form $\omega$ existed on this closed four-manifold, it would be exact, $\omega =d\eta$. Since $d\omega =0$,by Stokes theorem. But ${\omega }^2$ is a positive volume form in the symplectic orientation, so its integral is positive. This contradiction proves the claim fails. The mere bundle structure does not ensure that a fibrewise area class extends to the total space.

Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.

Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings $J_0:E\to {\mathbb{C}}^{N_0}$ and $J_1:E\to {\mathbb{C}}^{N_1}$ of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every $J_t$ is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.

Write $B={\mathrm{Gr}}_k({\mathbb{C}}^{\infty })$. For the assumed fiber bundle $U_k\to \nu \to B$ with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor $q\ge 2$ this gives group isomorphisms. For $q=1$, the relevant pointed-set part and connectedness of $U_k$ give ${\pi }_1(B)=0$, corresponding to the single component ${\pi }_0(U_k)$. Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle $S{U}_2\to U_2\to S^1$. Since ${\pi }_3(S^1)={\pi }_4(S^1)=0$, its exact sequence yields ${\pi }_3(U_2)\cong {\pi }_3(S{U}_2)$. The given identification $S{U}_2\cong S^3$ and the degree calculation ${\pi }_3(S^3)=\mathbb{Z}$ therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.

A smooth fiber bundle $\pi :E\to M$ with typical fiber $F$ is locally a product: every base point has a neighborhood $U$ and a local trivialization ${\pi }^{-1}(U)\cong U\times F$ commuting with projection to $U$. Its overlap maps act smoothly on $F$. A principal bundle with structure Lie group $G$ has a free right $G$-action, each fiber is one orbit, and the local trivializations are equivariant for right multiplication on $G$. A smooth section of a fiber bundle is a map $s:M\to E$ with $\pi \circ s={\mathrm{id}}_M$.

For a principal bundle, a global section of a fiber bundle gives the mapThe free transitive action on each fiber makes $\Phi$ bijective. In a local trivialization, write $s(x)=(x,a(x))$; then $\Phi (x,g)=(x,a(x)g)$ and its inverse is $(x,k)\mapsto (x,a(x{)}^{-1}k)$. Both are smooth and $\Phi$ is equivariant. Conversely, a product principal bundle has the section $x\mapsto (x,e)$. This proves a principal bundle is trivial exactly when it has a global section. The argument depends on the free transitive action: an arbitrary bundle section need not trivialize its bundle.

For an $n$-dimensional smooth manifold, the frame bundle has fiber over $x$ consisting of ordered bases of $T_{x}M$, equivalently linear isomorphisms $u:{\mathbb{R}}^n\to T_{x}M$. Right composition by $GL(n,\mathbb{R})$ makes it a principal bundle. If a metric has signature $(p,q)$, its pseudo-orthonormal frame bundle consists of the isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$, with structure group $O(p,q)$. In the positive-definite case this is the usual orthonormal frame bundle. Choosing space and time orientations reduces the Lorentzian structure group to its corresponding connected subgroup.

For Minkowski spacetime, use a fixed origin and a standard Lorentzian frame. Every pair consisting of a point and a pseudo-orthonormal frame is the image of this reference pair under a unique affine isometry. Thus, for the full groups,Here the first expression denotes the total space of the orthonormal frame bundle, not merely the Lorentz group. Projection sends an affine isometry to the image of the reference point, and the right action of the stabilizer changes its frame.

Realize de Sitter spacetime of radius $\ell$ as ${\eta }_{AB}{X}^A{X}^B={\ell }^2$ in ${\mathbb{R}}^{1,4}$. The ambient orthogonal group is transitive on this hyperboloid. Its point stabilizer acts on the Lorentzian tangent space as $O(1,3)$; choosing a tangent orthonormal frame removes that stabilizer completely. ConsequentlyFor oriented, time-oriented frames, replace these groups by $S{O}_0(1,4)$ and $S{O}_0(1,3)$, and similarly use the proper orthochronous Poincaré group for Minkowski spacetime. The isometries are uniquely determined by their action on a point and its frame, which explains these bundle identifications.

Finally, choose a basis $v_1,\dots ,v_n$ of the Lie algebra $T_{e}G$. The left-invariant vector fields $X_i(g)=(d{L}_g{)}_e{v}_i$ form a basis at every $g$, because $L_g$ is a diffeomorphism. They give a global section of a fiber bundle of the frame bundle, so that bundle is trivial. More explicitly,Thus every Lie group is a parallelizable manifold; connectedness is not needed for this construction.

A fiber bundle with base $B$, fiber $F$ and structure group $G$ acting on $F$ from the left consists of a projection $\pi :E\to B$ and local trivializations ${\varphi }_i:{\pi }^{-1}(U_i)\to U_i\times F$. On overlaps they satisfyThe smooth transition functions are valued in $G$. A principal bundle is the case $F=G$ with the left multiplication action on itself; its transition functions multiply the group coordinate from the left.

In each principal chart define $(x,h)\cdot a=(x,ha)$. This defines a global right action because $g_{ij}(x)(ha)=(g_{ij}(x)h)a$: coordinate changes commute with right multiplication. It is free, and acts transitively on each fiber. Thus the claimed right action follows from the left transition convention rather than being an additional choice.

If $s:B\to P$ is a global section of a fiber bundle, define $\Phi :B\times G\to P$ by $\Phi (x,a)=s(x)a$. Freeness and transitivity give exactly one $a$ for every point in the fiber, so $\Phi$ is bijective and equivariant. In a local trivialization where $s(x)$ has group coordinate $h_i(x)$, the inverse group coordinate is $h_i(x{)}^{-1}h$, proving smoothness of the inverse. Hence $\Phi$ is a principal-bundle trivialization. Conversely, the identity coordinate in a product trivialization gives $s(x)=\Phi (x,e)$. ThereforeThis is the principal bundle trivialization by a global section criterion. It is special to principal bundles; a section alone need not trivialize an arbitrary fiber bundle.

For the quotient examples, assume the Lie subgroups are closed when smooth homogeneous-space statements are intended. The map $G\to G/H_1$ is a principal bundle with structure group $H_1$ and right action $g\cdot h=gh$; $G/H_1$ is a homogeneous space for left multiplication by $G$. Similarly, $G\to H_2\backslash G$ has free transitive left $H_2$ action on its fibers. With the usual principal right-action convention one writes $g\cdot h=h^{-1}g$. The quotient $H_2\backslash G$ is a homogeneous space for right multiplication by $G$. These quotients are groups only when the respective subgroups are normal. Nonclosed immersed Lie subgroups can produce non-Hausdorff quotients, so an arbitrary Lie subgroup does not automatically give a smooth Hausdorff base.

Left $H_2$ and right $H_1$ multiplication commute, so taking their orbit equivalence relations in either order gives the same double coset setThis equality does not assert that the double quotient is a manifold or that either stage is a principal bundle. The $H_2\times H_1$ action $(a,b):g\mapsto ag{b}^{-1}$ can have stabilizers: they correspond to $H_2\cap g{H}_1{g}^{-1}$. If the induced action is free and proper, its quotient is a smooth principal-bundle quotient; without those hypotheses orbit dimensions can change. For example, with $G=SO(3)$ and both subgroups the rotations fixing the third axis, double cosets are parametrized by $e_3\cdot g{e}_3\in [-1,1]$. Interior and endpoint stabilizers differ, so this is not a uniform free principal-bundle construction.

For a principal bundle $P$ and a left $G$ action on $F$, the associated bundle iswith projection $[p,f]\mapsto \pi (p)$. A local principal section gives coordinates $[s_i(x),f]\leftrightarrow (x,f)$, and its changes give precisely the left action of the principal transition functions on $F$. A linear action on a vector space gives an associated vector bundle.

For a metric of signature $(r,s)$ on an $n$-manifold, an orthonormal frame at $x$ is an isometry $u:{\mathbb{R}}^{r,s}\to T_{x}M$. The orthonormal frame bundle $O(M,g)\to M$ has structure group $O(r,s)$, acting on the right by $u\cdot a=u\circ a$. Its associated bundle for the standard representation is the tangent bundle, sinceis well defined: $(ua)(v)=u(av)$. Every tangent vector has exactly one such associated equivalence class, and local frame coordinates show that this is a vector-bundle isomorphism. The full frame bundle works similarly with $GL(n,\mathbb{R})$.

Finally, choose a basis of the Lie algebra of a Lie group $G$. Its left translates are a global frame, yielding the explicit tangent bundle trivializationThis proves parallelization of a Lie group by left translations and triviality of the full frame bundle. Choosing an inner product at the identity and translating it gives a left-invariant Riemannian metric; an orthonormal basis translates to a global orthonormal frame. The global section criterion therefore trivializes the corresponding orthonormal frame bundle as well. For any other positive-definite metric on $G$, smoothly applying the Gram-Schmidt process to this global frame gives the same conclusion. Thus the tangent and frame bundles of every Lie group are trivial; the orthonormal statement is with respect to a chosen metric as specified above.