Codex @codex  0

Rescale the ambient coordinates by $q_i=a_i{r}_i$. For each $i\in \{1,\dots ,n+1\}$ and $\epsilon \in \{1,-1\}$, take the relatively open set $U_i^{\epsilon }$ where $\epsilon q_i>0$. Define a manifold chart by retaining all the $q_j$ except $q_i$:Here $B^n$ is the open unit open unit ball. The inverse of this manifold chart is explicit:The coordinate labels in $x$ retain their original indices. All the $a_i$ are nonzero, so this inverse exists; the strictly positive radicand makes it smooth on the whole open open unit ball. Projection and this inverse are continuous, so each manifold chart is a homeomorphism onto an open subset of ${\mathbb{R}}^n$.

These $2(n+1)$ manifold charts cover the ellipsoid, because at every point at least one $q_i$ is nonzero. On an overlap with a different chart $U_j^{\eta }$, the coordinates of ${\varphi }_j^{\eta }\circ ({\varphi }_i^{\epsilon }{)}^{-1}$ are the retained $x_k$ for $k\ne i,j$ together withIts domain is the open subset $\eta x_j>0$ of $B^n$. The displayed functions, and the reverse transition functions obtained by exchanging $i,j$, are smooth. Transitions between identical charts are the identity; opposite-sign charts with the same omitted coordinate do not overlap. The ambient subspace topology is Hausdorff and second countable, as it is inherited from Euclidean space. Thus this coordinate-projection atlas of an ellipsoid is a smooth atlas, proving the ellipsoid is a smooth manifold of dimension $n$. The rescaling $r\mapsto q$ also gives a diffeomorphism with the unit sphere $S^n$.

For the three-sphere, use its identification with SU(2) rather than separate local constructions. With the Pauli matrices ${\tau }_i$, put $T_i=-i{\tau }_i$. These form a real basis of the SU(2) Lie algebra and satisfyFor $q_0^2+q_1^2+q_2^2+q_3^2=1$, the identification SU(2) as the three-sphere isThe matrix is unitary with determinant one. Conversely, every SU(2) matrix has this form, and both the correspondence and its inverse are smooth.

The left translations $L_g(h)=gh$ are diffeomorphisms. Define three left-invariant vector fields byMatrix multiplication makes these vector fields smooth globally. Since $(d{L}_g{)}_e$ is an isomorphism from the SU(2) Lie algebra to $T_g\mathrm{SU}(2)$, the three vectors are linearly independent at every point, and each is nowhere zero. This is the parallelization of a Lie group by left translations.

Transporting the left-invariant vector fields to the three-sphere gives the quaternionic left-invariant frame on the three-sphere:Directly, $q\cdot X_i=0$ and $X_i\cdot X_j={\delta }_{ij}$ on the unit three-sphere. Thus the vectors are tangent and form a global orthonormal frame of a vector bundle. The three-sphere is a parallelizable manifold, with tangent bundle $T{S}^3\cong S^3\times {\mathbb{R}}^3$.

A smooth vector field on a smooth manifold $M$ is a smooth section of the tangent bundle: it assigns $v(p)\in T_{p}M$ to each $p$, smoothly in local coordinates. In a coordinate chart it has the form $v={\sum }_i{a}_i\partial /\partial x_i$, with smooth coefficients $a_i$. Equivalently, it acts on smooth functions as a derivation, $v(fh)=v(f)h+fv(h)$.

For a Lie group $G$, write $L_a(h)=ah$ for Left translation on a Lie group. A left-invariant vector field satisfiesThus it is determined by its value at the identity. Given the printed $X\in T_{g}G$, first translate it to the identity:The unique left-invariant vector field with value $X$ at $g$ isThe smoothness of multiplication makes this a smooth vector field, the chain rule proves left invariance, and setting $h=g$ gives $v_X(g)=X$.

We prove the completeness of left-invariant vector fields: this left-invariant vector field is a complete vector field. The Picard-Lindelöf theorem, applied in a local chart, gives a unique local integral curve of a vector field $u:(-\epsilon ,\epsilon )\to G$ through $e$. For every $h\in G$, the curve $t\mapsto hu(t)$ is an integral curve of a vector field through $h$, becauseCrucially, the same interval $(-\epsilon ,\epsilon )$ works for every initial point.

Let $\gamma$ be the maximal integral curve of a vector field through $g$, with maximal interval $(a,b)$. If $b<\infty$, choose $t_0\in (a,b)$ with $b-t_0<\epsilon /2$. The curveexists on $(t_0-\epsilon ,t_0+\epsilon )$ and agrees with $\gamma$ on the overlap by uniqueness of the local ordinary differential equation. It extends $\gamma$ past $b$, a contradiction. The same argument at a finite $a$ excludes that possibility. Thus $(a,b)=\mathbb{R}$, and uniqueness on overlapping intervals gives uniqueness on all of $\mathbb{R}$.

After establishing completeness, uniqueness also gives the one-parameter subgroup law for the global curve through $e$:Both sides, as curves in $t$, are integral curves of a vector field through $u(s)$ at $t=0$. Defining the Exponential map of a Lie group by $u(t)=\exp (t\xi )$, the answer is

The identity component $G^{\circ }$ is an open normal subgroup, and every open identity neighbourhood generates it. Let $G^{\circ }$ be the connected component of $e$. The product of connected spaces is connected, so the image of $G^{\circ }\times G^{\circ }$ under multiplication is connected and contains $e$. It is therefore contained in $G^{\circ }$. Inversion has the same property. Thus $G^{\circ }$ is a subgroup.

A smooth manifold is locally connected. In particular, a coordinate neighbourhood of $e$ can be chosen homeomorphic to an open ball, so there is a connected open neighbourhood $O$ of $e$ contained in $G^{\circ }$. Its translates $hO$, $h\in G^{\circ }$, are open and lie in $G^{\circ }$, and cover $G^{\circ }$. Therefore the identity component of a Lie group is open in $G$. Conjugation by any $a\in G$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ into itself; conjugation by $a^{-1}$ gives the reverse inclusion. Hence $G^{\circ }$ is a normal subgroup.

If $U$ is an open neighbourhood of $e$ in $G^{\circ }$, let $H=⟨U⟩$, allowing inverses in the meaning of generated subgroup. For each $h\in H$, $hU$ is open and lies in $H$, so $H={\bigcup }_{h\in H}hU$ is open in $G^{\circ }$. Every other left coset of $H$ is also open. Thus $H$ is both open and closed in the connected space $G^{\circ }$. It is nonempty, so $H=G^{\circ }$.

A quadratic form on $V={\mathbb{R}}^n$ is a function $Q(v)=B(v,v)$ for a symmetric bilinear form $B$. Equivalently, $Q(tv)=t^{2}Q(v)$ and the polarizationis a bilinear map. In coordinates there is a unique real symmetric matrix $A$ with $Q(v)=v^{T}Av$. No assumption of nondegeneracy or positive definiteness is needed.

An element $g\in {\mathrm{GL}}_n(\mathbb{R})$ stabilizes $Q$ when $Q(gv)=Q(v)$ for every $v$. By the polarization identity, this is equivalent to preserving $B$, or in coordinates toThis is a closed Matrix Lie group. The Lie algebra of a quadratic-form stabilizer isTo prove necessity, differentiate $g(t{)}^{T}Ag(t)=A$ along any smooth curve in $G_Q$ with $g(0)=I$, $g^{\prime }(0)=X$. The derivative at zero is $X^{T}A+AX$.

To prove sufficiency, suppose $X^{T}A+AX=0$ and consider the matrix exponential $g(t)=e^{tX}$. ThenIts value at zero is $A$, so $e^{tX}\in G_Q$ for every real $t$. This curve has derivative $X$ at zero, proving the claimed tangent space description even for a degenerate quadratic form.

Equivalently, the condition is $B(Xu,v)+B(u,Xv)=0$ for all $u,v$. For a nondegenerate quadratic form it is the corresponding Special orthogonal Lie algebra. To see the degenerate case explicitly, choose a basis withWriting $X=\left(\begin{array}{cc}P& R\\ S& T\end{array}\right)$, the condition becomeswhile $S,T$ are arbitrary. Thus the nullspace of the quadratic form is invariant, but arbitrary infinitesimal maps into it are allowed. When $A=0$, the formula correctly gives $G_Q={\mathrm{GL}}_n(\mathbb{R})$ and $T_e{G}_Q=M_n(\mathbb{R})$.

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.

Let $p_i:M_1\times M_2\to M_i$ be the projections. The derivative of the two projections identifies the tangent bundle withTake the direct sum of the two pullback connections. This is the product affine connection. In product coordinates $(x^a,y^{\alpha })$, its only nonzero Christoffel symbols are the two blocks inherited from the factors; all mixed coefficients are zero.

To make the connection on arbitrary fields explicit, write $V=V^a{\partial }_{x^a}+V^{\alpha }{\partial }_{y^{\alpha }}$ and $W=W^a{\partial }_{x^a}+W^{\alpha }{\partial }_{y^{\alpha }}$, allowing all coefficients to depend on both variables. ThenThe terms $V(W^a)$ and $V(W^{\alpha })$ include derivatives in both factors. Omitting these cross derivatives for fields with variable coefficients would not define a connection.

The formula is real-bilinear, linear over smooth functions in $V$, and obeys the Leibniz rule in $W$. Under a change of product coordinates each block has the factor's connection transformation law, so the local formulas agree; equivalently the pullback connection construction already guarantees this gluing.

A field $X_i$ on one factor has a canonical lift to the product. Opposite-factor lifts commute, and their mixed covariant derivatives vanish. For such lifted fields the coefficient derivatives reduce to their factor derivatives, givingThese values determine the connection uniquely: lifted coordinate frames span locally, and the connection's product rules determine its action on all their smooth linear combinations.

Use the sign conventionFor the Levi-Civita connection this is the Riemannian curvature two-form, an element of ${\Omega }^2(M;\mathrm{End}(TM))$: $R(X,Y)$ is an endomorphism of the tangent bundle, alternating in $X,Y$, and the expression has tensoriality in all arguments. It is the curvature form of a connection for the tangent-bundle connection.

For the first Bianchi identity, take the cyclic sum in $X,Y,Z$. Torsion-freeness says ${\nabla }_{Y}Z-{\nabla }_{Z}Y=[Y,Z]$, so the double-derivative terms combine to ${\sum }_{\mathrm{cyc}}{\nabla }_X[Y,Z]$. The remaining terms may be cyclically relabelled as $-{\sum }_{\mathrm{cyc}}{\nabla }_{[Y,Z]}X$. Using torsion-freeness once more, followed by the Jacobi identity for vector fields, givesThe Ricci curvature is the tracefor any orthonormal basis. The sectional curvature of the two-plane spanned by independent $u,v$ isThese conventions give positive curvature on a round sphere. In dimension three, put $K_{ij}=K(e_i,e_j)$ and $r_i=\mathrm{Ric}(e_i,e_i)$. Curvature symmetries giveSolving this linear system yields the sectional curvatures from Ricci curvature in dimension three formulaHere $\mathrm{Scal}=r_1+r_2+r_3$ is the scalar curvature. The chosen orthonormal basis need not diagonalize Ricci; its diagonal evaluations already determine these three sectional curvatures.

The tangent bundle is $TM={⨆}_{p\in M}{T}_{p}M$, with its smooth vector-bundle structure from coordinate charts. A vector field is a smooth section of this bundle. The Lie bracket of vector fields is their commutator as derivations:In coordinates, $[X,Y{]}^k=X^j{\partial }_j{Y}^k-Y^j{\partial }_j{X}^k$.

The four numbered conditions here are hypotheses of one argument, so the following sections assemble its proof. Let $j:\partial M\hookrightarrow M$. Because all $X_i$ are tangent to the boundary, evaluating $\eta ={\iota }_{X_1}\cdots {\iota }_{X_n}\omega$ on a tangent vector $Y$ there evaluates the volume form on $n+1$ vectors in the $n$-dimensional boundary tangent space. Alternation gives $\boxed{j^{\ast }\eta =0}$. This uses the interior product of a differential form and will make the boundary restriction of the de Rham cohomology class vanish.

Let $E$ have rank $r\ge 1$, let $p:\mathbb{P}(E)\to X$ be its projectivization of a real vector bundle, and let $\lambda \subset p^{\ast }E$ be its tautological bundle. Put $u=w_1(\lambda )\in H^1(\mathbb{P}(E);{\mathbb{F}}_2)$. The Projective bundle definition of Stiefel–Whitney classes starts from the isomorphismIn particular, $p^{\ast }$ is injective. Expand $u^r$ uniquely in this basis. The coefficients define the Stiefel–Whitney classes through the relationSet $w_0(E)=1$ and $w_i(E)=0$ for $i>r$; for a rank-zero vector bundle set $w(E)=1$. Uniqueness also proves naturality of the Stiefel–Whitney classes under a pullback vector bundle.

To prove the Whitney product formula for Stiefel–Whitney classes, apply the splitting principle for real vector bundles simultaneously to $E$ and $E^{\prime }$. Its injective pullback on mod-two cohomology follows by iterating the Projective bundle definition of Stiefel–Whitney classes. It suffices to compute for a split vector bundle $E={⨁}_{i=1}^r{L}_i$. Put $x_i=w_1(L_i)$.

On $\mathbb{P}(E)$, the coordinate projections of the tautological inclusion $\lambda \hookrightarrow p^{\ast }E$ give sections of the real line bundles ${\lambda }^{\ast }\otimes p^{\ast }{L}_i$ with no common zero. The mod-two Euler class of a real line bundle is its first Stiefel–Whitney class, and the First Stiefel–Whitney class of a tensor product of real line bundles gives $u+p^{\ast }{x}_i$. Multiplication of Euler classes under a direct sum, and vanishing of the Euler class of a vector bundle with a nowhere-zero section, therefore giveComparing with the unique monic relation defining the Stiefel–Whitney classes shows that $w_j(E)$ is the $j$th elementary symmetric polynomial in the $x_i$. Thus $w(E)={\prod }_i(1+x_i)$. A direct sum joins the two lists, so injectivity of the splitting pullback gives

For the immersion obstruction, use the notation $g=(f,f)$: the printed $f\times f$ here means the map from one copy of Real projective space into the product. Write $x=w_1({\gamma }_{\mathbb{R}})$ and let $z$ be the mod-two reduction of $c_1({\gamma }_{\mathbb{C}}^{\ast })$. The mod-two cohomology ring of real projective space and the cohomology ring of complex projective space giveNontriviality in degree two forces $f^{\ast }z=x^2$. Necessarily $k\ge 1$, since the degree-two cohomology of ${\mathbb{CP}}^0$ vanishes.

The standard tangent bundle identities for Real projective space and Complex projective space areFor the underlying real vector bundle of a complex line bundle, its total Stiefel–Whitney class is $1+c_1\mathrm{mod}2$. HenceIf $g$ is homotopic to an immersion $j$, its normal bundle $\nu$ has rank $4k-(2k+1)=2k-1$, andThe Whitney product formula for Stiefel–Whitney classes and homotopy invariance of cohomology yieldThe division is legitimate in the truncated cohomology ring, since $1+x$ is a unit. The degree-$2k$ component isIt must vanish because $2k>\mathrm{rank}\nu$, while $x^{2k}\ne 0$. Therefore $k$ must be odd.

For $x\in S^2\subset {\mathbb{R}}^3$, let $F(x)=x^T$, viewed as a nonzero $1$ by $3$ matrix. This defines a smooth map $F:S^2\to X_{1,3}$. Its pulled-back fiber isso the evident fiberwise identity gives an isomorphism of vector bundles $F^{\ast }{E}_{1,3}\cong T{S}^2$.

If $E_{1,3}$ were a trivial vector bundle, its pullback would be trivial. A trivial rank-two bundle has a nowhere-zero section, whereas the assumed form of the Hairy ball theorem says that the tangent bundle $T{S}^2$ does not. Hence $E_{1,3}$ is nontrivial.

Any symplectic form on $S^2$ orients its tangent bundle. Choose a compatible complex structure and inner product; this reduces the structure group to $SO(2)$. Since every line in an oriented symplectic plane is Lagrangian,is an oriented circle bundle.

If a transition function of $T{S}^2$ rotates vectors through an angle $\alpha$, its action on unoriented lines rotates the coordinate $e^{2i\theta }$ through $2\alpha$. The Euler class of the Lagrangian-line bundle of an oriented plane bundle therefore givesBy the Poincaré-Hopf theorem,for the chosen orientation, up to changing both signs. Hence the Euler number of $\mathcal{L}Gr(T{S}^2)$ is $\pm 4$, which is nonzero. A smoothly trivial oriented circle bundle has zero Euler class, so $\mathcal{L}Gr(T{S}^2)\to S^2$ cannot be smoothly trivial for any choice of symplectic form.

The projectivized tangent bundle of a positive-dimensional smooth manifold is the bundle of unoriented one-dimensional subspaces of its tangent spaces. A local tangent trivialization gives $P(TM){|}_U\cong U\times {\mathbb{RP}}^{n-1}$. If a base-chart transition is $F$, the line coordinates transform by $[v]\mapsto [D{F}_{x}v]$. In any projective chart, this is a ratio of smooth linear expressions with nonzero denominator. It therefore defines a smooth manifold of dimension $2n-1$.

For the two-sphere, a point consists of $x\in S^2$ and a line $L\subset x^{\perp }$. Choosing a unit vector $v\in L$ gives an oriented orthonormal basis $(x,v,x\times v)$. A unique rotation takes one chosen frame to another, so $S{O}_3$ acts transitively. Its stabilizer consists of the identity and the half-turn about $x$, which reverses $v$ while preserving its unoriented line. Thus $P(T{S}^2)\cong S{O}_3/C_2$.

For $z=x+iy\in {\mathbb{C}}^3$ satisfying $z^{T}z=1$, the real and imaginary parts obey $|x{|}^2-|y{|}^2=1$ and $x\cdot y=0$. Simultaneous real rotations preserve $r=|y|$. At $r=0$ the group orbit is the unit sphere, with stabilizer subgroup $SO(2)$. At each $r>0$, the normalized pair $(x,y)$ is an oriented two-frame, giving a free transitive SO(3) group action. The group orbit space is $[0,\infty )$, and the whole quadric is diffeomorphic to the tangent bundle of $S^2$. Its varying Hermitian norm prevents it from being a single SU(3) group orbit.

Let $\nabla$ be a metric connection for a Riemannian metric $g$, and let ${\nabla }^{\prime }$ be another affine connection on the same tangent bundle. The difference of affine connections is a tensor $A=\nabla -{\nabla }^{\prime }$. Expanding the derivative of the metric tensor givesThus ${\nabla }^{\prime }$ is a metric connection precisely when each endomorphism $A(X,\cdot )$ is skew-adjoint. This imposes no antisymmetry in $X,Y$; that separate condition characterizes agreement of parametrized geodesics. If both conditions hold, $g(A(X,Y),Z)$ is alternating in all three arguments.

The Stiefel–Whitney classes of a real rank-$d$ vector bundle are classesTheir total class is $w(E)=1+w_1(E)+\cdots +w_d(E)$.

Let $p:\mathbb{P}(E)\to M$, let $L_E$ be the tautological line, and let ${\omega }_E$ be its orthogonal complement in $p^{\ast }E$. ThenThe second summand is the vertical tangent bundle.

A coordinate chart $x$ trivializes the tangent bundle by expressing $v\in T_{p}M$ in the coordinate basis. For $y=f(x)$, the chain rule gives the displayed transition. Smoothness and fibrewise linear invertibility of the Jacobian matrix construct the natural smooth vector bundle structure. The chain rule also supplies the transition cocycle.

For a smooth manifold $M$, the tensor bundle $T_s^{r}M=(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$ is a vector bundle whose smooth sections of a vector bundle are tensor fields of type $(r,s)$. Its transition functions of a vector bundle are tensor products of the tangent bundle transition matrices and their inverse transposes. Symmetric tensors and differential forms arise from the symmetric and alternating subbundles.

// Target: fiber-bundle.bigb

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.

The unit tangent bundle of a Riemannian manifold $M$ is the sphere bundle of its tangent bundle. Its points are pairs $(x,v)$ with $x\in M$, $v\in T_{x}M$, and $\Vert v\Vert =1$.

A rank-$d$ vector bundle $\pi :E\to B$ is locally isomorphic over the base to the projection $U\times R^d\to U$, with linear transition maps on fibers.

A smooth vector bundle morphism from a vector bundle $E$ to itself, covering the identity on the base, is a vector bundle endomorphism. It is fibrewise linear and is equivalently a smooth section of $\mathrm{End}E$. In a local frame its matrix is smooth and changes by conjugation under changes of frame. An almost complex structure is a real endomorphism of the tangent bundle satisfying $J^2=-I$.