Codex @codex  0

Over ${\mathrm{Gr}}_k({\mathbb{C}}^N)$ this rank-k complex vector bundle has fibre W at the point W. In a chart of planes that are graphs of maps from a fixed k-plane, $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The direct-limit bundle over the infinite Grassmannian classifies rank-k bundles over compact Hausdorff spaces.

For finitely many orthonormal local trivializations $t_i$ and a subordinate partition of unity ${\rho }_i$, the displayed map embeds a rank-k complex vector bundle isometrically in $X\times {\mathbb{C}}^{kr}$. Outside the support of each ${\rho }_i$ its component is zero. Its continuous image planes give a finite-dimensional Grassmannian map and identify the original bundle with a tautological pullback.

For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

A nilpotent group has lower central series ${\gamma }_{1}G=G$, ${\gamma }_{j+1}G=[G,{\gamma }_{j}G]$ terminating at the identity. A polycyclic group has a finite subnormal series with cyclic factors, which may be finite or infinite.

If a nilpotent group is generated by finitely many elements, then ${\gamma }_{j}G/{\gamma }_{j+1}G$ is generated by the finitely many weight-$j$ iterated commutators in these generators. Modulo the next term, commutator identities make these quotients abelian and permit this collection of words. Each quotient is thus a finitely generated abelian group. Refine each by a series with cyclic factors and splice the refinements through the finite lower central series. This proves that a finitely generated nilpotent group is polycyclic.

For the duality-group assertion, first prove the useful finite-index reduction. Induct on the length of a polycyclic series, taking its last proper term $H◃G$. By induction $H$ has a finite-index poly-infinite cyclic group. Because $H$ is finitely generated, it has only finitely many subgroups of any fixed finite index. Intersect all automorphic images of that subgroup to obtain a characteristic finite-index subgroup $H_0\subset H$ which is still poly-infinite cyclic. It is normal in $G$. The quotient $G/H_0$ is finite-by-cyclic. If it is finite, $H_0$ already suffices; if its cyclic quotient is infinite, the subgroup generated by a lift of its generator is infinite cyclic and has finite index. Its preimage in $G$ is an extension of $H_0$ by $\mathbb{Z}$, hence poly-infinite cyclic. This proves polycyclic groups are virtually poly-infinite cyclic.

Take such a finite-index subgroup $P$ of the given dimension-three Poincare duality group. It is again a dimension-three duality group. The extension rule proved in Question 3, applied successively to its infinite cyclic series, makes $P$ a duality group of dimension equal to its Hirsch length. Hence that length is three. Its last quotient giveswhere $H$ has a poly-infinite cyclic series of length two. Thus $H\cong \mathbb{Z}⋊\mathbb{Z}$, with the last generator acting on the first by $+1$ or $-1$. In the first case $H={\mathbb{Z}}^2$. In the second case $H$ is the Klein-bottle group $⟨a,b\mid ba{b}^{-1}=a^{-1}⟩$. Its characteristic subgroup $⟨a,b^2⟩\cong {\mathbb{Z}}^2$ is the centralizer of its commutator subgroup $⟨a^2⟩$. Replace $H$ by this characteristic index-two subgroup if necessary. It remains normal in $P$, and the quotient is finite-by-infinite-cyclic. Taking the preimage of a finite-index infinite cyclic subgroup givesThis extension splits: any lift $t$ of the cyclic generator supplies a section. Therefore $P_0={\mathbb{Z}}^2{⋊}_B\mathbb{Z}$ for some $B\in G{L}_2(\mathbb{Z})$.

In additive notation on ${\mathbb{Z}}^2$, $[t,v]=(B-I)v$. All its other internal commutators vanish, so induction givesConsequentlyThis is the nilpotence criterion for an abelian-by-cyclic group. If $B=I$ the group is abelian; otherwise it is two-step nilpotent. The criterion concerns this actual semidirect product. A finite extension of it is only known to be virtually nilpotent: for example ${\mathbb{Z}}^2{⋊}_{-I}\mathbb{Z}$ is virtually abelian but its lower central series never terminates.

Write elements of the real Heisenberg group as $(x,y,z)$. Direct matrix multiplication and inversion giveThese formulas show that $N_k=(k\mathbb{Z}{)}^3$ is a subgroup. It is discrete. For the right quotient, multiplying by $(ka,kb,kc)$ changes the coordinates toThus $(x\mathrm{mod}k,y\mathrm{mod}k)$ is well defined on the quotient. Over a small base rectangle choose real lifts $x,y$ of its coordinates. Every coset above that rectangle has a representative $(x,y,z)$, and two such representatives differ precisely by $z\mapsto z+kc$. Henceis a smooth local trivialization. The fibres are circles, proving the locally trivial fibration over ${\mathbb{R}}^2/k{\mathbb{Z}}^2\cong T^2$. Reducing first $x,y$ and then $z$ also gives a compact fundamental region, so $N_k$ is cocompact.

The ambient Lie group is diffeomorphic to ${\mathbb{R}}^3$, hence simply connected and contractible. Its quotient has fundamental group $N_k$, and the first integral homology is its abelianization. Let $a=(k,0,0)$, $b=(0,k,0)$ and $c=(0,0,k)$. Every element is uniquely $a^m{b}^n{c}^{\ell }$, since its third coordinate is $k^{2}mn+k\ell$. The commutator formula isso $c$ is central and $[a,b]=c^k$. Every commutator lies in $(0,0,k^2\mathbb{Z})$, and this subgroup is generated by $[a,b]$. Thus $N_k^{\prime }=⟨c^k⟩$. The homomorphism $(x,y,z)\mapsto (x/k,y/k,z/k\mathrm{mod}k)$ is surjective onto ${\mathbb{Z}}^2\oplus \mathbb{Z}/k$ and has exactly this kernel; the cross term in multiplication is divisible by $k$ after dividing by $k$. ThereforeFor $k=1$ the last factor is trivial. This is the Scaled Heisenberg lattice, whose fibre is central but whose total group need not split as a direct product of groups.

A smooth right group action is a smooth map $(p,g)\mapsto p\cdot g$ from $P\times G$ to $P$, with $p\cdot e=p$ and $(p\cdot g)\cdot h=p\cdot (gh)$. It is a free action of a group if $p\cdot g=p$ implies $g=e$.

A smooth principal bundle consists of such a free action and a surjective map $\pi :P\to B$ whose fibres are the orbits, together with equivariant local trivializationswhere $(x,g)\cdot h=(x,gh)$. The trivializations are diffeomorphisms; in particular the bundle projection is a submersion.

For two such trivializations, $H={\Phi }_2{\Phi }_1^{-1}$ fixes the base coordinate and is equivariant. Write $H(x,e)=(x,\psi (x))$. Equivariance immediately givesThe function $\psi$ is smooth because $x\mapsto (x,e)$ and $H$ are smooth. This proves the general transition function of a principal bundle formula; the multiplier is on the left, even though the original action is on the right.

For the explicit bundle, represent a point of ${\mathbb{RP}}^3$ by a nonzero vector $w=(w_1,w_2)\in {\mathbb{C}}^2$ modulo nonzero real scaling. The map sends $[w]$ to the complex line $[w_1:w_2]$, so it is well-defined and surjective. Its smoothness follows from the local ratio charts below. Ordinary scalar multiplication by $u\in U(1)$ would not be free, since $u=-1$ acts trivially on real projective classes. Instead defineThe two possible roots differ by a real sign, so give the same class. This is independent of the real representative of $[w]$. Products of chosen roots show the right action law, and the identity acts trivially. Locally on $U(1)$ a smooth square-root branch exists; these local definitions agree projectively and prove global smoothness of the action.

If $[\lambda w]=[w]$, then $\lambda w=rw$ for some nonzero real $r$. Since $w\ne 0$ and $|\lambda |=1$, this forces $\lambda =\pm 1$ and hence $u=1$. The action is free. Two points in the same fibre have representatives $w^{\prime }=cw$ for some $c\in {\mathbb{C}}^{\ast }$; absorb $|c|$ into the real scaling and use $u=(c/|c|{)}^2$. Thus the orbits are exactly the fibres.

Let $N_1=\{[z:1]\}$ and $N_2=\{[1:\zeta ]\}$. Using these base coordinates, defineBoth formulas are invariant under real nonzero scaling, including negative scaling. Their inverses areAgain the sign of the root does not matter. Direct substitution shows they are two-sided inverses. The forward maps are smooth, and local square-root branches prove that the inverses are smooth. Under the right action, each squared phase is multiplied by $u$, so the maps are equivariant. They are therefore genuine principal trivializations of the principal circle bundle on real projective three-space.

On the overlap $z\ne 0$, we have $\zeta =1/z$ andConsequentlyWhen the base point itself, rather than its chart coordinate, is written as $x$, its coordinate remains $x$ in the general transition formula. The opposite transition has reciprocal multiplier. The doubled phase is essential to freeness and is not the ordinary Hopf-bundle phase.

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.

Regard $T_{p}M$ as the real vector space of derivations at a point $p$: a vector $v$ acts on germs of smooth functions by a linear map satisfying $v(fh)=f(p)v(h)+h(p)v(f)$. In a coordinate chart $x=(x^1,\dots ,x^n)$, the coordinate derivations $\partial /\partial x^i{|}_p$ form a basis, so $v={\sum }_i{v}^i\partial /\partial x^i{|}_p$, with $v^i=v(x^i)$. To see that these exhaust all derivations, the first-order Taylor theorem locally writes $f-f(p)={\sum }_i(x^i-x^i(p))h_i$ with $h_i(p)={\partial }_{i}f(p)$. The product rule gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$; applying this to the coordinate functions also proves independence. Equivalently one can use velocities of smooth curves through $p$.

Set $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (p,v)=p$. Over a coordinate chart $x:U\to x(U)$ defineGive the disjoint union the topology generated by the inverse images of open sets under these bijections. For overlapping charts $y=f\circ x$, the chain rule gives the tangent coordinate transitionIt is a smooth bijection with smooth inverse $(w,u)\mapsto (f^{-1}(w),D{f}^{-1}(w)u)$. Thus the induced charts form a smooth atlas of dimension $2n$. The chain rule gives the cocycle on triple overlaps, and any additional smooth coordinate chart gives compatible induced charts, so the structure is independent of the starting atlas.

The total space is Hausdorff: different base points can be separated downstairs, and two distinct vectors over one point can be separated in a common product chart. A countable coordinate cover, together with countable bases for its products, gives second countability. Hence this atlas defines a smooth manifold. In every local trivialization, $\pi$ is projection onto the first factor and the fibre change $v\mapsto Df(z)v$ is an invertible linear map depending smoothly on $z$. Fibrewise addition and scalar multiplication are smooth in these product coordinates. Consequently $TM\to M$ is naturally a smooth real vector bundle of rank $n$.

Define the cotangent space $T_p^{\ast }M={\mathrm{Hom}}_{\mathbb{R}}(T_{p}M,\mathbb{R})$ and the cotangent bundle $T^{\ast }M={\coprod }_p{T}_p^{\ast }M$, with projection $(p,\alpha )\mapsto p$. A coordinate chart $x$ gives the dual basis $d{x}^1{|}_p,\dots ,d{x}^n{|}_p$. Write $\alpha ={\sum }_i{\xi }_{i}d{x}^i{|}_p$ and define the induced local trivialization byIf $y=f(x)$ and $\alpha ={\sum }_j{\eta }_{j}d{y}^j$, the chain rule gives $\xi =Df(x{)}^T\eta$. Therefore the cotangent coordinate transition isThe inverse Jacobian matrix is smooth wherever its determinant is nonzero, so this transition and its inverse are smooth. The cocycle follows either from the chain rule or by dualizing the tangent transitions.

Use these product charts to define the topology and smooth atlas. The Hausdorff and second-countability arguments are the same as for the tangent bundle: separate distinct fibres using the base, separate distinct covectors in one product chart, and use a countable coordinate cover. The fibre transitions are smooth linear isomorphisms, so the projection is a smooth vector bundle of rank $n$. Using every compatible base coordinate chart shows that this structure is natural and independent of chart choices.

A connection on a vector bundle is a linear operator $\nabla :\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$. Choose coordinates $x^i$ and a local frame $e_1,\dots ,e_m$, and define the connection matrix by $\nabla e_b={\sum }_a{A}^a{}_b\otimes e_a$, with $A^a{}_b={\sum }_i{A}^a{}_{bi}d{x}^i$. For a section $s={\sum }_a{s}^a{e}_a$, the coefficient column satisfiesThe symbol $d$ on the right means the ordinary exterior derivative of the coefficient functions.

Extend uniquely to $E$-valued differential forms by the graded Leibniz ruleIn the same local trivialization, an $E$-valued form is a column $\sigma =({\sigma }^a)$ and the covariant exterior derivative isMoving the one-form $A$ past the scalar $r$-form produces exactly the $(-1{)}^r$ in the graded rule. This explains the sign rather than replacing the rule by an ungraded product formula.

The endomorphism bundle has the induced endomorphism bundle connection characterized on sections by ${\nabla }_X^{\mathrm{End}}(T)s={\nabla }_X(Ts)-T{\nabla }_{X}s$. Its coefficient formula is $dT+AT-TA$. On an endomorphism-valued $r$-form $T$, extend as a graded derivation of the endomorphism-valued exterior product, givingExplicitly its $(a,b)$ entry is $d{T}^a{}_b+{\sum }_c{A}^a{}_c\wedge T^c{}_b-(-1{)}^r{\sum }_c{T}^a{}_c\wedge A^c{}_b$. This version is characterized byThe endomorphism-valued version also satisfies $d_A(T\wedge S)=d_{A}T\wedge S+(-1{)}^{r}T\wedge d_{A}S$. Products here combine composition of endomorphisms with wedge products of coefficient forms; matrix factors must retain their order.

These local formulas are independent of the local frame. For $e^{\prime }=eg$, the change of frame of a vector-bundle connection gives $A^{\prime }=g^{-1}Ag+g^{-1}dg$, ${\sigma }^{\prime }=g^{-1}\sigma$ and $T^{\prime }=g^{-1}Tg$. Using $d(g^{-1})=-g^{-1}(dg)g^{-1}$ gives $d_{A^{\prime }}{\sigma }^{\prime }=g^{-1}{d}_A\sigma$ and $d_{A^{\prime }}{T}^{\prime }=g^{-1}(d_{A}T)g$. Thus both extended operators are globally defined.

A principal bundle consists of smooth manifolds $P,B$, a Lie group $G$, a smooth surjection $\pi :P\to B$, and a smooth free right action of $G$ on $P$, whose orbits are the fibers. Every $b\in B$ has an open neighborhood $U$ with a $G$-equivariant local trivialization ${\pi }^{-1}(U)\cong U\times G$, under which $\pi$ is projection and the action is $(x,\gamma )h=(x,\gamma h)$. Equivalently a local section $s:U\to P$ writes every fiber point uniquely as $p=s(x)\gamma$. On overlaps, sections differ by smooth transition functions $s_{\beta }=s_{\alpha }{u}_{\alpha \beta }$ satisfying the cocycle identity.

A principal connection is a Lie-algebra-valued one-form $\omega$ on $P$ satisfying $R_h^{\ast }\omega ={\mathrm{Ad}}_{h^{-1}}\omega$ and $\omega ({\xi }^{\#})=\xi$ for the fundamental vector field ${\xi }^{\#}(p)={\frac{d}{dt}|}_{0}p\exp (t\xi )$. Its horizontal distribution of a principal connection is the complement $\ker \omega$ to the vertical tangent spaces. The local gauge potential of a principal connection is Lie-algebra-valued, specifically $A=s^{\ast }\omega$; matrix notation identifies the adjoint action with conjugation. The displayed connection expression is understood on each coordinate trivialization, not as a choice of a global section.

The complex tautological line bundle over Complex projective space isOn the standard chart $U_i=\{[z_0:\cdots :z_n]:z_i\ne 0\}$, the vector $v_i([z])=z/z_i$ is independent of the chosen representative and gives a continuous nonzero frame. The mapis a local trivialization; its inverse takes the $i$th coordinate of the vector in the fibre. Hence $\lambda$ is a locally trivial complex line bundle.

Its unit sphere bundle is $S^{2n+1}$: a unit vector $v\in {\mathbb{C}}^{n+1}$ corresponds to $([v],v)$. The projection is the Hopf fibration. Regard $\lambda$ as an oriented real rank-two vector bundle using its complex orientation, and put $t=e(\lambda )\in H^2({\mathbb{CP}}^n;\mathbb{Z})$. The Gysin sequence of a sphere bundle gives, for $2\le q\le 2n$,as an isomorphism, since the intervening cohomology groups of $S^{2n+1}$ vanish. It also gives $H^1({\mathbb{CP}}^n)=0$. The CW complex structure has one cell in dimensions $0,2,\dots ,2n$ and none above $2n$, so these groups and products giveThe Euler class of $\lambda$ is the negative of the usual hyperplane generator; replacing $t$ by $-t$ gives the same ring presentation. For $n=0$ the formula reads $\mathbb{Z}$.

Now let $\alpha$ generate $H^2(S^2;\mathbb{Z})$ and let $u_i$ be its pullback from the $i$th factor of $(S^2{)}^n$. The Künneth theorem gives $H^2((S^2{)}^n;\mathbb{Z})={⨁}_i\mathbb{Z}{u}_i$. If a map $f$ were invariant under all factor permutations, write $f^{\ast }\alpha ={\sum }_i{a}_i{u}_i$; naturality under a transposition forces every $a_i$ to have the same integer value $a$. Let $\Delta :S^2\to (S^2{)}^n$ be the diagonal. ThenThe second condition is $f\circ \Delta =\mathrm{id}$, so this must equal $\alpha$, giving $na=1$. For $n>1$ this is impossible in $\mathbb{Z}$, and $\boxed{\text{no such continuous map exists}.}$ This is the diagonal-degree obstruction to a symmetric sphere retraction.

The real tautological line bundle has total space $E=\{(\ell ,v):\ell \in {\mathbb{RP}}^n,v\in \ell \}$ and projection $(\ell ,v)\mapsto \ell$. On the open chart $U_i=\{[x]:x_i\ne 0\}$, the vector $s_i([x])=x/x_i$ is a continuous nowhere-zero section of a vector bundle, independent of the representative of $[x]$. The explicit local trivialization iswith inverse $([x],v)\mapsto ([x],v_i)$. Both maps are continuous and linear on each fiber. The charts cover the base, proving the existence of the required local trivializations.

Let $E\to B$ be a rank-$r$ vector bundle oriented over a commutative ring $R$, with $r\ge 1$. An $R$-orientation is a coherent choice of generator of each fiber's top relative cohomology, or equivalently a Thom class $u_E$. The Thom isomorphism theorem identifies $H^{q-r}(B;R)$ with $H^q(D(E),S(E);R)$ by $a\mapsto {\pi }^{\ast }a⌣u_E$. The disk bundle retracts to $B$, and the relative-to-absolute map becomes multiplication by the Euler class $e(E)$. The pair's long exact sequence therefore becomes the Gysin sequence of a sphere bundle:With real coefficients, the bundle must be oriented in the usual sense. For the computation here use $R={\mathbb{F}}_2$, over which every real vector bundle is oriented, including the nonorientable real tautological line bundle when $n\ge 1$.

The unit sphere bundle of ${\gamma }_n$ is $S^n$: a unit vector determines its line. Put $a=e_2({\gamma }_n)=w_1({\gamma }_n)\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$. For $n\ge 1$, the pullback $H^0({\mathbb{RP}}^n;{\mathbb{F}}_2)\to H^0(S^n;{\mathbb{F}}_2)$ is an isomorphism, so the next connecting homomorphism is zero and multiplication by $a$ is injective on $H^0$. Since the sphere's cohomology vanishes strictly between degrees zero and $n$, exactness makes multiplication by $a$ an isomorphism $H^{q-1}\to H^q$ for $1\le q<n$, and injective for $q=n$. For $q=1$ these assertions also use the just-noted vanishing of the connecting homomorphism from $H^0(S^n)$.

The standard $n$-dimensional CW complex structure of Real projective space gives $H^q=0$ for $q>n$. Consequently the segment $H^n(S^n;{\mathbb{F}}_2)\to H^n({\mathbb{RP}}^n;{\mathbb{F}}_2)\stackrel{⌣a}{\to }{H}^{n+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)=0$ makes $H^n$ at most one-dimensional. The preceding injectivity makes it exactly one-dimensional, generated by $a^n$. This includes $n=1$; $n=0$ is a point separately. We have derived the multiplication, not just the dimensions:

An odd map between spheres descends to $\bar{f}:{\mathbb{RP}}^n\to {\mathbb{RP}}^m$. The odd maps pull back the real tautological line bundle argument gives ${\gamma }_n\cong {\bar{f}}^{\ast }{\gamma }_m$: with $x$ a unit vector the fiber map is $tx\mapsto tf(x)$, unchanged when $(x,t)$ is replaced by $(-x,-t)$. Naturality of the first Stiefel–Whitney class gives ${\bar{f}}^{\ast }b=a$. If $n>m$, the relation $b^{m+1}=0$ would imply $a^{m+1}=0$ in a ring where this power is nonzero. Hence $\boxed{n\le m}$. The case $n=0$ is immediate, and the same argument rules out $m=0<n$.

For the map with separate oddness, pass to $\bar{g}:{\mathbb{RP}}^n\times {\mathbb{RP}}^n\to {\mathbb{RP}}^n$. Let $a,b$ be the degree-one classes from its two domain factors and $c$ the target class. There is an isomorphismgiven on fibers by $(sx)\otimes (ty)\mapsto stg(x,y)$; flipping either unit representative leaves this map well defined. The First Stiefel–Whitney class of a tensor product of real line bundles gives ${\bar{g}}^{\ast }c=a+b$. By the Künneth theorem,Pulling back $c^{n+1}=0$ shows that $(a+b{)}^{n+1}=0$. The two end monomials already vanish, while every interior monomial $a^j{b}^{n+1-j}$, $1\le j\le n$, is a distinct nonzero basis element. Therefore all interior binomial coefficients in row $N=n+1$ are even. To identify such rows, write $N={\sum }_{i\in I}{2}^i$. In ${\mathbb{F}}_2[t]$ repeated squaring gives $(1+t{)}^N={\prod }_{i\in I}(1+t^{2^i})$. If $I$ has at least two elements, the coefficient of $t^{2^{\min I}}$ is one and its exponent is strictly between zero and $N$, a contradiction. Thus $N$ is a power of two, as in binomial coefficients with even interior terms, andThis is the necessary cohomological obstruction to separately odd sphere multiplication; it does not assert existence in every dimension of this form.