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A differential one-form $\theta$ is a smooth section of a vector bundle $T^{\ast }X\to X$. Its graph is its image as a section, and the projection restricts to a diffeomorphism onto $X$. Pullback of the Liouville one-form by that section equals $\theta$, making the twisted Lagrangian graph criterion immediate.

For an embedded submanifold $i:M\hookrightarrow N$, a smooth section of a vector bundle $X\in \Gamma (i^{\ast }TN)$ locally extends to an ambient vector field. In an adapted chart $M=\{y=0\}$, extend each component $X^a(x)$ independently of the transverse coordinates $y$. If $X$ takes values in $TM$, the extension criterion is $(\tilde{X}F){|}_M=X(F{|}_M)$ for every local smooth function $F$ on $N$. Testing coordinate functions suffices. For tangent fields $X,Y$, this criterion also proves $[\tilde{X},\tilde{Y}]{|}_M=[X,Y]$. General fields with normal components have no extension-independent ambient Lie bracket.

// Target: riemannian-geometry.bigb

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 smooth vector field on a smooth manifold $M$ is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle projection $\pi$, so $\pi \circ X={\mathrm{id}}_M$. In a coordinate chart it has the form $X={\sum }_i{X}^i{\partial }_i$ with smooth component functions $X^i$.

Suppose a smooth vector bundle isomorphism $\Phi :M\times {\mathbb{R}}^n\to TM$ covering the identity is given. For the standard basis $e_1,\dots ,e_n$ of ${\mathbb{R}}^n$, put $X_i(p)=\Phi (p,e_i)$. These are smooth vector fields, and the linear isomorphism on each fibre makes $X_1(p),\dots ,X_n(p)$ a basis of $T_{p}M$.

Conversely, given such a global frame of a vector bundle, defineIt is smooth, covers the identity and is a linear isomorphism on each fibre. Its inverse is smooth as well: in any coordinate chart, the component columns of the $X_i$ form an invertible smooth matrix $C(p)$, and the fibre coordinates of the inverse are $C(p{)}^{-1}v$. matrix inversion is smooth on the invertible matrices. Thus the tangent bundle is trivial exactly when a global frame exists, the defining property of a parallelizable manifold.

For a Lie group $G$, write $L_h(g)=hg$. A left-invariant vector field is a field satisfyingSuch a field is determined by $v=X(e)\in T_{e}G$, since $X(g)=(d{L}_g{)}_{e}v$. Conversely, this formula defines a left-invariant field, by the chain rule and $L_h{L}_g=L_{hg}$.

It also proves smoothness even if smoothness was not initially assumed. Choose a smooth curve $c(t)$ with $c(0)=e$, $c^{\prime }(0)=v$. ThenMultiplication is a smooth map on a Lie group, so differentiating its coordinate functions in the second variable yields a smooth function of $g$. This gives smooth components for $X$ in every local chart.

Choose a basis $v_1,\dots ,v_m$ of $T_{e}G$, where $m=\dim G$, and left translate it. Each $L_g$ is a diffeomorphism, with inverse $L_{g^{-1}}$, so $(d{L}_g{)}_e$ is a linear isomorphism and the translated fields are a basis at every $g$. The explicit parallelization of a Lie group by left translations isIts inverse sends $w\in T_{g}G$ to $(g,(d{L}_{g^{-1}}{)}_{g}w)$. Identifying $T_{e}G$ with ${\mathbb{R}}^m$ gives the required product bundle. The differential facts used are linearity, the chain rule, $d(\mathrm{id})=\mathrm{id}$ and smooth dependence of the differential of a smooth map on its base point.

Take the Möbius line bundle over $S^1$,A continuous section of a vector bundle corresponds to a continuous real-valued function $s:[0,1]\to \mathbb{R}$ with $s(1)=-s(0)$. If $s(0)=0$, it already vanishes. Otherwise its endpoint values have opposite signs, so the intermediate value theorem gives $s(t_0)=0$ for some $t_0\in (0,1)$. Thus this vector bundle has no nowhere-zero section. Every continuous section must vanish.

Realize the pullback vector bundle asIts fibre over $p$ is canonically $E_{\varphi (p)}$, with the same vector space operations. If a vector bundle trivialization over $U\subset M$ is ${\tau }_U(v)=(\pi (v),\hat{v})$, give the pullback vector bundle the local trivializationDeclare these maps to be local homeomorphisms and use coordinate charts on $M^{\prime }$ to obtain a smooth atlas on the total space. On overlaps the smooth transition maps are $(p,w)\mapsto (p,h(\varphi (p))w)$, hence smooth and fibrewise linear. They satisfy the same cocycle identities as the original transition maps. This gives a well-defined smooth vector bundle, with smooth projection and rank $r$. It also gives the subspace topology inherited from $M^{\prime }\times E$: inside each product ${\varphi }^{-1}(U)\times {\pi }^{-1}(U)$ the defining condition is the graph of $\varphi$ in the base coordinates. Thus the total space is Hausdorff and second countable, as required for a smooth manifold.

For a smooth section of a vector bundle $s$ over $U$, its pulled-back section is more precisely $p\mapsto (p,s(\varphi (p)))$. If $s$ has component column $u:U\to {\mathbb{R}}^r$ in the original vector bundle trivialization, its component column in $T_U$ is $u\circ \varphi$, which is smooth. The pullback is a smooth rank-$r$ vector bundle, and every local section pulls back smoothly.

A smooth rank-$r$ vector bundle over a smooth manifold $M$ consists of a smooth total space $E$, a smooth projection $\pi :E\to M$, and an $r$-dimensional real vector space structure on each fibre $E_p={\pi }^{-1}(p)$. Each point has a neighborhood $U$ with a vector bundle trivializationthat is a diffeomorphism, commutes with projection to $U$, and is a linear isomorphism on every fibre. The overlap maps have the form $(p,v)\mapsto (p,h(p)v)$ with $h:U\cap V\to G{L}_r(\mathbb{R})$ smooth. For a complex vector bundle, replace $\mathbb{R}$ by $\mathbb{C}$ and require complex-linear fibre maps.

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhere $\Gamma (E)$ is the module of smooth sections, ${\Omega }^1(M;E)$ consists of vector-bundle-valued differential forms of degree one, and $f$ is a smooth function. Equivalently, evaluating on a vector field $X$ gives operators ${\nabla }_X$ with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$. A connection on a vector bundle is local: its value at a point depends on the germ of the section of a vector bundle, so the same definition applies to local sections.

A vector field along a map $\gamma$ assigns $V(t)\in T_{\gamma (t)}M$ smoothly to each parameter value. Equivalently, it is a section of a vector bundle ${\gamma }^{\ast }TM$. The pullback connection defines its covariant derivative along a curve, $D_{t}V$. To make the definition explicit, use connection coefficients with the conventionFor $V=V^k(t){\partial }_k{|}_{\gamma (t)}$ the derivative isThe change of frame of a vector-bundle connection ensures this expression is independent of the coordinate chart. It remains meaningful when $\dot{\gamma }=0$ and does not require extending $V$ to one ambient vector field.

The field is parallel exactly when $\boxed{D_{t}V=0}$. A geodesic for $\nabla$, with its specified affine parameter, is a smooth curve whose own tangent is parallel:This definition applies to any affine connection, whether or not it is a metric connection. A non-affine change of parameter generally gives a pregeodesic instead of this parametrized geodesic.

The differential one-form $\theta$ defines a smooth section of a vector bundle, and its one-form graph is an embedded copy of $X$ because $\pi \circ \theta ={\mathrm{id}}_X$. The defining property of the Liouville one-form gives ${\theta }^{\ast }\alpha =\theta$. ConsequentlyThe one-form graph has dimension $n$, half the dimension of the cotangent bundle. It is therefore a Lagrangian submanifold precisely when this pullback of a differential form vanishes:This is the twisted Lagrangian graph criterion.

Now suppose the projection restricts to a diffeomorphism on a Lagrangian submanifold $L$. Its inverse followed by the inclusion defines a smooth section $s:X\to T^{\ast }X$, hence a global differential one-form $\theta$ with image $L$. The criterion forces $\sigma =d\theta$, so its class in de Rham cohomology is zero. Thus $[\sigma ]\ne 0$ rules out every such Lagrangian section. This cohomological obstruction to a Lagrangian section concerns graphs over the entire base; it does not claim that all Lagrangian submanifolds are absent.

The Hermitian metric on cotangent vectors is the dual metric, and its exterior-power metric is defined on decomposable vectors by the determinant of their pairwise inner products. Tensor this with the metric on $E$. In a unitary coframe and unitary frame of $E$, the forms obtained by wedging distinct holomorphic and antiholomorphic coframe elements and tensoring with a frame vector form an orthonormal basis. Let $dV={\omega }^n/n!$, and use inner products linear in the first variable.

The bundle-valued conjugate-linear Hodge star is characterized by contraction of the $E,E^{\ast }$ coefficients inIt is conjugate-linear, maps type $(p,q)$ to type $(n-p,n-q)$, and uses the metric identification of $E$ with its dual. For total degree $k=p+q$, the ordinary star-square rule giveswhere $E^{\ast \ast }$ is identified with $E$. The conjugate-linear convention is essential for the bidegrees specified here.

If $\psi$ has degree $k-2$ and $\eta$ degree $k$, the even degree of the real form $\omega$ givesSince ${\ast }_E\Lambda \eta =L{\ast }_E\eta$ for $\Lambda \eta =(-1{)}^k{\ast }_{E^{\ast }}L{\ast }_E\eta$, it follows that $(L\psi ,\eta )=(\psi ,\Lambda \eta )$ pointwise. Thus $\Lambda$ is the adjoint Lefschetz operator.

For holomorphic $E$, ${\bar{\partial }}_E$ differentiates coefficients in holomorphic local frames and uses the graded Leibniz rule on forms. Its bundle-valued Dolbeault adjoint is the formal differential operatorIt lowers the antiholomorphic degree by one. To verify this on compact $M$, define $⟨\psi ,\eta {⟩}_{L^2}={\int }_M(\psi ,\eta )dV$. For $\psi \in A^{p,q-1}(E)$ and $\eta \in A^{p,q}(E)$, Stokes applied to the scalar $(n,n-1)$-form $\psi \wedge {\ast }_E\eta$ yieldsThe star-square identity converts the right side into $⟨\psi ,{\bar{\partial }}_E^{\ast }\eta {⟩}_{L^2}$. Compactness and absence of boundary remove boundary terms. This establishes the asserted global adjoint relation on smooth sections.

The Dolbeault Laplacian ${\Delta }_{\bar{\partial }}=B{B}^{\ast }+B^{\ast }B$, with $B={\bar{\partial }}_E$, is formally self-adjoint. Consequently the adjoint of a commutator satisfiesFor the $(1,0)$ part $D^{\prime }$ of the Chern connection, the explicit corresponding adjoint iswhere $D_{E^{\ast }}^{\prime }$ is the $(1,0)$ part of the dual Chern connection. This is the same integration-by-parts formula in holomorphic degree.

Now assume the metric is Kähler. Since $\bar{\partial }\omega =0$, $B$ commutes with $L$. Taking adjoints provesThe scalar Kähler identities needed areThey extend to the bundle-valued Kähler identitiesHere is a local justification of the extension. At any chosen point, take holomorphic normal coordinates for the Kähler metric and a holomorphic local frame of $E$ with $H=I$ and $\partial H=0$ at that point. This is a Chern connection in a normal holomorphic frame. Such a frame is obtained first by normalizing the metric matrix at the point and then prescribing the holomorphic local frame change's first derivatives to cancel $A=H^{-1}\partial H$. The first-order operators and their formal adjoints at the point are then the scalar operators acting componentwise. The scalar identities therefore give the two displayed bundle identities there. Since the point was arbitrary, they hold globally.

Let $A=D^{\prime }$. Its square and $B^2$ vanish, whilebecause the Chern curvature has pure type $(1,1)$. Use the two commutators already proved to calculateTaking adjoints, including conjugation of the scalar $-i$, gives the Lefschetz-Dolbeault curvature commutatorIf $\Theta =0$, commutation follows. Conversely, vanishing of the commutator on all $E$-valued forms implies $\Theta \wedge s=0$ for every smooth section of a vector bundle $s$. At any point, a bump-supported local section can have any prescribed fiber value. Thus $\Theta$ annihilates every vector in every fiber, forcing $\Theta =0$ identically. This proves the flatness criterion from Lefschetz commutation; the test on degree-zero forms is why the assertion is made on the entire form space.

A smooth vector field is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.

A smooth tensor field of type $(r,s)$ is a smooth section of $(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$, where the factors are the tangent bundle and cotangent bundle. The flow definition of the Lie derivative of a tensor field isHere pullback applies $d{\varphi }_t^{-1}$ to each vector factor and the dual of $d{\varphi }_t$ to each covector factor, evaluating $T$ at ${\varphi }_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.

For a smooth function, ${\varphi }_t^{\ast }f=f\circ {\varphi }_t$, and the chain rule givesFor a vector field $Y$, in local coordinates the expansions ${\varphi }_t(x)=x+tX(x)+O(t^2)$ and $d{\varphi }_t(x)=I+td{X}_x+O(t^2)$ giveConsequentlyPullback preserves tensor products and tensor contractions, so differentiation makes this definition a tensor derivation. It therefore agrees on every tensor field with the Lie derivative of a tensor field determined by these two formulas.

Now put ${\psi }_t=\alpha \circ {\varphi }_t\circ {\alpha }^{-1}$. The chain rule givesThus ${\psi }_t$ is the local flow of ${\alpha }_{\ast }X$. In the case $M=N$, if ${\alpha }_{\ast }X=X$, uniqueness of integral curves of a vector field makes ${\psi }_t={\varphi }_t$ on their common domains, or $\alpha \circ {\varphi }_t={\varphi }_t\circ \alpha$. Conversely, differentiating this commuting identity at zero gives $d{\alpha }_p{X}_p=X_{\alpha (p)}$, hence ${\alpha }_{\ast }X=X$. This proves that diffeomorphism invariance of a vector field is equivalent to commuting with its local flow, with every identity understood on the domain where its compositions exist.

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.

Under the standard smooth manifold hypotheses, the first answer is yes, although the isomorphism requires a choice. A partition of unity combines Euclidean inner products in manifold charts to produce a Riemannian metric $g$. Nonnegative local weights summing to one preserve positivity, and local finiteness preserves smoothness. The musical isomorphism isIt is smooth and fiberwise invertible, with inverse having local matrix $(g_{ij}{)}^{-1}$. It is therefore a vector bundle isomorphism. There is no preferred such map before a metric or equivalent additional structure is chosen.

The second answer is no. On $S^1=\mathbb{R}/2\pi \mathbb{Z}$, compare the trivial vector bundle $S^1\times \mathbb{R}$ with the Möbius line bundleBoth have rank one. A continuous section of a vector bundle of $L$ is represented on $[0,2\pi ]$ by a continuous function $s$ satisfying $s(2\pi )=-s(0)$. If it were nowhere zero, the intermediate value theorem would prevent its endpoint signs from being opposite. Thus $L$ has no nowhere-zero section. The trivial bundle has the section $1$, and a bundle isomorphism would carry that section to a nowhere-zero section of $L$, a contradiction. Hence

The zero locus of a holomorphic section of a vector bundle of a rank-$r$ bundle is regular when its derivative has rank $r$ along the locus. Its kernel is then the tangent bundle of the smooth zero set. Smoothness of the underlying set alone is insufficient: the square of a reduced defining function has the same set of zeros but zero derivative there.

A section of a fiber bundle $\pi :E\to B$ is a continuous map $s:B\to E$ with $\pi \circ s={\mathrm{id}}_B$. It chooses a point of each fibre continuously. A section of a vector bundle is the special case for vector fibres; a projective bundle section chooses a line in each fibre of the underlying vector bundle.

A tensor field is a smooth section of a vector bundle formed from tensor products of the tangent bundle and cotangent bundle. With $k$ covector factors and $l$ vector factors it belongs toIts coefficients in any local frame vary smoothly. Different conventions list the two counts in different orders, so the factor order should be specified. Tensor contractions use the canonical evaluation of a covector on a vector and are independent of the frame.

A vector field along a curve $\gamma :I\to M$ is a smooth section of a vector bundle with $V(t)\in T_{\gamma (t)}M$. Its values may differ at distinct times mapping to the same base point. The covariant derivative along a curve differentiates its time-dependent coefficients using the pulled-back connection on a vector bundle; it does not require extending it to one ambient vector field.

The zero section sends each base point to the zero vector in its fiber. It is a smooth section of a vector bundle and an embedded copy of the base. The Liouville one-form vanishes as an ambient covector precisely along the zero section of a cotangent bundle.