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A Riemannian metric is a smooth positive-definite symmetric inner product $g_x$ on every tangent space. A Levi-Civita connection is a connection on the tangent bundle which is torsion-free, ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$, and metric-compatible,Apply this identity with the cyclic triples $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and replace differences of covariant derivatives by Lie brackets. The result is the Koszul formulaIts right side depends only on $g$ and the vector fields; nondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely. This proves uniqueness.

For existence, in a coordinate chart defineThis is linear over smooth functions in $X$ and obeys the Leibniz rule in $Y$. Symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ proves zero torsion, and direct substitution giveswhich proves metric compatibility. On chart overlaps the two resulting connections satisfy the same intrinsic Koszul identity and therefore agree by uniqueness. They glue globally. Thus the existence and uniqueness of the Levi-Civita connection holds on every Riemannian manifold.

Fix the curvature conventionFor coordinate vectors $e_i$, define the Riemann curvature tensor components by $R_{ij,kl}=g(R(e_i,e_j)e_l,e_k)$. In these indices the Ricci and scalar contractions areExplicitly, $R_{ij,kl}=g_{ka}{R}^a{}_{lij}$, whereThese expressions define Ricci curvature and scalar curvature without ambiguity about which indices are contracted. With this convention a round unit $m$-sphere has $\mathrm{Ric}=(m-1)g$ and $s=m(m-1)$.

If $\mathrm{Ric}=\lambda g$ with a constant real $\lambda$, tracing gives $\boxed{s=m\lambda }$, so every Einstein metric has constant scalar curvature. The converse is not valid in arbitrary dimension. On $S^2\times S^1$ with the product Riemannian metric of the unit round sphere and a circle, the connection splits between factors. The mixed curvatures and circle curvature vanish, while the sphere has curvature one. HenceA scalar multiple of the whole metric would require $\lambda =1$ on the sphere directions and $\lambda =0$ on the circle direction, which is impossible. This proves that constant scalar curvature does not imply an Einstein metric, and supplies a counterexample to the unrestricted requested equivalence.

In dimension two, the curvature tensor is determined by the Gaussian curvature $K$:There the valid equivalence isIn dimension one both curvature and scalar curvature vanish identically. Without a two-dimensional restriction or another added geometric hypothesis, the general converse cannot be proved.

A principal bundle with structural Lie group $G$ is locally a product $U\times G$, with changes of trivialization of the form $(x,g)\mapsto (x,t(x)g)$. Define a right action locally by $(x,g)\cdot h=(x,gh)$. It is independent of trivialization because $t(x)(gh)=(t(x)g)h$. It is smooth and obeys the group-action law. Right multiplication in $G$ is free and transitive, so the global action is free and each fiber is a single orbit. Equivalently these properties and equivariant local triviality can be included in the definition of a principal bundle.

For a representation $\rho :G\to GL(V)$, form the associated vector bundleProjection sends $[p,v]$ to the base point of $p$. A local section $s(x)$ gives coordinates $[s(x),v]$ and hence a local product $U\times V$; on overlaps the principal transition functions act through $\rho$. Because these maps are linear in $v$, the fibers have well-defined vector-space structures.

For an $n$-dimensional manifold, its frame bundle has points $u:{\mathbb{R}}^n\to T_{x}M$ that are linear isomorphisms. Right action is composition $u\cdot A=u\circ A$ for $A\in GL(n,\mathbb{R})$. Coordinate frames provide local trivializations. Its associated bundle for the defining representation is the tangent bundle, by the well-defined map $[u,v]\mapsto u(v)$. A metric restricts the frames to linear isometries and reduces the group to $O(r,s)$; choosing orientation gives the corresponding special-orthogonal reduction.

In Minkowski spacetime choose a reference oriented pseudo-orthonormal frame at the origin. Every such frame is specified by its base point $a\in {\mathbb{R}}^{3,1}$ and a Lorentz transformation $L\in SO(3,1)$. The Poincaré group has multiplicationand its right Lorentz action is $(a,L)\cdot h=(a,Lh)$. Henceis the oriented orthonormal frame bundle of Minkowski spacetime. The same argument with $O(3,1)$ gives the full unoriented bundle. If time orientation is fixed, use the orthochronous components throughout; the use of $SO$ in the printed quotient is an orientation reduction.

For four-dimensional de Sitter spacetime, use the spacelike hyperboloid $x\cdot x={\ell }^2$ in ${\mathbb{R}}^{4,1}$. An element of $SO(4,1)$ maps a fixed unit spacelike vector to $x/\ell$, and its other columns give an oriented pseudo-orthonormal tangent frame. The stabilizer of that unit vector is $SO(3,1)$. This identifies the oriented frame bundle of de Sitter spacetime and its projection:The right stabilizer action changes the tangent frame without changing $x$.

For Anti-de Sitter spacetime, take $x\cdot x=-{\ell }^2$ in ${\mathbb{R}}^{3,2}$. A unit timelike normal has orthogonal complement of signature $(3,1)$, so its stabilizer is again $SO(3,1)$. Appending this normal to a tangent frame realizes the oriented orthonormal frame bundle of anti-de Sitter spacetime asConnected components or orientation/time-orientation restrictions must be chosen consistently. This quotient is the anti-de Sitter quadric, which has a periodic timelike coordinate. If “anti-de Sitter spacetime” denotes its universal cover, pull the frame bundle back to that cover, rather than identify the covered spacetime with the unmodified quotient.

Let $D$ be the Levi-Civita connection of the Riemannian metric. We use the curvature sign specified in the question, the negative of the usual vector-bundle curvature convention in Question 4. Extend $D$ to the covariant exterior derivative on tangent-bundle-valued forms and define the curvature two-form by $R=-D^2$. In a local frame with connection matrix $A$, it is the endomorphism-valued two-form $-(dA+A\wedge A)$. The transformation law for $A$ makes this transform by conjugation, so it is an intrinsic tensor. It measures the infinitesimal failure of parallel derivatives to commute, with this chosen sign.

For a tangent-bundle-valued one-form $\eta$, the covariant exterior derivative isApply this to the one-form $DZ$, for which $(DZ)(Y)=D_{Y}Z$. This calculation proves, rather than just renames, the operator expressionIt is linear over smooth functions in $Z$: the extra first- and second-derivative terms in $R(X,Y)(fZ)$ cancel by $[X,Y]f=X(Yf)-Y(Xf)$. The connection's linearity in its vector-field argument similarly shows linearity over smooth functions in $X$ and $Y$. Thus $R(X,Y)$ is an endomorphism of the tangent bundle, not a differential operator on its sections, and $R$ is a well-defined endomorphism-valued two-form.

The torsion-free identity is $D_{X}Y-D_{Y}X=[X,Y]$. Expand the cyclic curvature sum and use this identity to obtainThe last equality is the Jacobi identity for the Lie bracket of vector fields. This proves the first Bianchi identity

Lower the output index with the metric and define the Riemann curvature tensor byThis is a covariant four-tensor, because the curvature endomorphism is tensorial in all three inputs. Skew-symmetry of the curvature two-form immediately gives $R_{ji,kl}=-R_{ij,kl}$.

Metric compatibility gives the second skew-symmetry. Apply the scalar operator $[X,Y]-XY+YX$, which is zero, to $g(Z,W)$. Expanding each derivative using $Dg=0$ cancels the mixed products of first derivatives and leavesThus $R_{ij,lk}=-R_{ij,kl}$.

For a direct proof of pair interchange, choose the geodesic coordinates of Question 5 at an arbitrary point $p$. There $\Gamma (p)=0$ and $\partial g(p)=0$. Write $D_{{\partial }_i}{\partial }_k={\Gamma }^a{}_{ik}{\partial }_a$. The operator formula gives, at $p$,The Levi-Civita formula is $g_{al}{\Gamma }^a{}_{jk}=\frac{1}{2}({\partial }_j{g}_{kl}+{\partial }_k{g}_{jl}-{\partial }_l{g}_{jk})$. Differentiating at $p$, the products involving first derivatives of the metric and Christoffel coefficients vanish. The terms ${\partial }_i{\partial }_j{g}_{kl}$ cancel, leavingHere a comma denotes ordinary coordinate derivatives at the normal-coordinate center. Swapping $(i,j)$ with $(k,l)$ leaves this expression unchanged, by symmetry of $g$ and commutation of its second partial derivatives. This is the pair symmetry from the normal-coordinate curvature formula. Since $p$ was arbitrary and the equality is tensorial, it holds in every frame. We have proved all the requested symmetries:Changing the overall curvature sign would change the displayed coordinate formula's sign, but would not change the first Bianchi identity or any of these algebraic symmetries.

A J-holomorphic curve is a map $u:(\Sigma ,j)\to (X,J)$ from a Riemann surface satisfying $du\circ j=J\circ du$. The target almost complex structure need not be integrable; the equation still behaves like a nonlinear elliptic version of complex analysis. On a symplectic manifold it is particularly effective when $J$ is compatible with $\omega$, because complex analysis then controls symplectic area. The metric construction above supplies such structures and deformations between them.

In conformal coordinates $(s,t)$ with $j{\partial }_s={\partial }_t$, the equation is $u_t=J{u}_s$. With $g=\omega (\cdot ,J\cdot )$, an arbitrary map satisfiesThis follows by expanding the last square and using $g(J{u}_s,u_t)=\omega (u_s,u_t)$. Integrating proves the Energy identity for a J-holomorphic curve. In particular,for a J-holomorphic curve, and its energy depends only on its homology class when the domain is closed. A nonconstant such curve has strictly positive area: its derivative is nonzero somewhere and positivity holds on a neighborhood. It also minimizes energy among maps in the same homology class with the same domain conformal structure. Thus analytic bounds can be obtained from topological information.

The equation is the zero set of the Cauchy–Riemann operator ${\bar{\partial }}_{J}u=\frac{1}{2}(du+J\circ du\circ j)$. Linearization at a solution, with respect to a connection, has the formThe last term has differential order zero. The principal part has the Cauchy–Riemann symbol; in a complex frame a nonzero real covector $(a,b)$ gives multiplication by $a+ib$, an invertible symbol. This establishes ellipticity. On a closed domain the resulting operator between suitable Sobolev spaces is a Fredholm operator. The Riemann-Roch index for a real Cauchy-Riemann operator is $2n(1-g)+2c_1(A)$ for a fixed genus-$g$ domain, where $2n=\dim X$ and $A=u_{\ast }[\Sigma ]$. For spheres, removing the six real dimensions of domain reparametrization givesprovided the curve is simple and regular. Each marked point adds two real dimensions; constraining its image to a specified target point removes $2n$. These are the dimension formula for regular J-holomorphic spheres and its incidence version.

Here regular means that $D_u$ is surjective. Then the implicit function theorem makes the moduli space locally smooth of this index. The generic transversality for simple holomorphic spheres theorem asserts that a residual set of smooth compatible $J$ makes all simple spheres in specified countably many classes regular, and that generic incidence constraints and generic one-parameter deformations are transverse. It does not assert regularity for all multiple covers. This distinction matters whenever one tries to count curves. A primitive class with no positive-area spherical decomposition avoids multiple-cover and bubbling difficulties and permits especially direct counts.

The other central ingredient is Gromov compactness for spheres. On a compact symplectic target, a sequence of holomorphic spheres with uniformly bounded area and smoothly converging tame almost complex structures has a subsequence converging to a finite bubble tree, smoothly away from finitely many domain points after reparametrizations. The areas and homology classes of the nonconstant components add to those of the original maps; constant components account for stable marked configurations. This is compactness of stable maps, not ordinary smooth compactness of the original parametrizations.

The proof mechanism explains why bubbling is indispensable. An elliptic small-energy estimate bounds derivatives on a smaller disc when the energy on a larger disc is sufficiently small. Failure of a derivative bound therefore concentrates a definite amount of energy. Rescale near a point of large derivative; elliptic estimates yield a nonconstant limiting holomorphic map on the plane. Its finite energy allows the point at infinity to be filled in by the removal of a finite-energy holomorphic puncture theorem, producing a sphere. Repeating extracts further spheres. Uniform local monotonicity gives a positive energy threshold for nonconstant bubbles on the compact target, so only finitely many can occur. Estimates on the intervening annuli yield the energy identity and exclude lost neck energy. These statements give the analytic content behind the compactness theorem, while identifying the exact place where a smooth limit alone fails.

A local area estimate turns this existence theory into an embedding obstruction. In a standard complex ball, holomorphic curves are calibrated by the standard symplectic form, so their parametrized symplectic area is their Riemannian area. They are stationary minimal surfaces. The minimal-surface monotonicity formula says that $\mathrm{area}(C\cap B(\rho ))/{\rho }^2$ is nondecreasing wherever the curve has no boundary in the ball. At a point of multiplicity $m$, its limiting density is $m\pi$. Consequently any branch through the centre contributes at least $\pi {\rho }^2$. For a general compatible $J$ there is still a local lower bound $c{\rho }^2$, with $c>0$ depending on controlled geometry. The exact Euclidean constant is the one needed for sharp radius obstructions.

Here is a global existence argument that feeds this estimate. Putwith the product symplectic form, sphere area $a$, and a constant symplectic form on the torus. Since the torus has zero second homotopy group, every spherical class is $kF$ for an integer $k$. A nonconstant holomorphic sphere in this class has area $ka>0$, so $k\ge 1$. A curve of class $F$ cannot split into two nonconstant bubbles, and it cannot be a nontrivial multiple cover. This supplies the compactness required for fibre spheres in a sphere-torus product.

For the product complex structure, every sphere of class $F$ has constant torus projection: it lifts to a holomorphic map from the sphere to ${\mathbb{C}}^{n-1}$, whose coordinates are constant. Its sphere projection has degree one. Thus, up to reparametrization, there is exactly one such sphere through every target point. It is regular because its pulled-back tangent bundle is $\mathcal{O}(2)\oplus {\mathcal{O}}^{n-1}$ on ${\mathbb{CP}}^1$, and the first cohomology of each summand vanishes. Its linearized Dolbeault operator therefore has zero cokernel.

Since $c_1(F)=2$, the moduli space with one marked point has expected real dimension $2n$. Imposing passage through a fixed point gives dimension zero. For generic structures and transverse incidence conditions this is a finite set. Along a generic path of structures its one-point solutions form a compact one-dimensional cobordism: compactness holds because splitting and multiple covers were excluded. Its boundary contains the endpoint solutions. A compact one-manifold has an even number of boundary points, so the count modulo two is unchanged. The product count is one, and hence the count for a generic compatible $J$ is one as well. If the endpoint product structure is not generic for other classes, its regular solutions allow the path argument in this single class. Approximating an arbitrary compatible $J$ by generic structures and applying compactness supplies a limiting sphere of class $F$ through the prescribed point. The nonconstant component still contains that point; there are no nonconstant bubbles to carry it away. This derives the existence assertion needed below without assuming a capacity obstruction.

Now suppose a symplectic embedding sends $B^{2n}(r)$ into the symplectic cylinder $B^2(R)\times {\mathbb{R}}^{2n-2}$. Choose $0<\rho <{\rho }^{\prime }<r$. The image of the closed radius-${\rho }^{\prime }$ ball is compact. For any $a>\pi R^2$, the first factor embeds symplectically into an area-$a$ sphere: in dimension two this is an area-preserving disc chart, leaving a cap of area $a-\pi R^2$. Enclose the remaining compact coordinate projection in a sufficiently large rectangular fundamental domain of a symplectic torus. The compact ball image thereby embeds into $S^2(a)\times T^{2n-2}$.

Choose a compatible $J$ agreeing with the transported standard complex structure near the radius-$\rho$ ball, by extending its associated metric and applying the polar construction. The sphere of class $F$ through the centre has area $a$. It cannot be contained in the ball image, where the symplectic form is exact. Euclidean monotonicity therefore gives $a\ge \pi {\rho }^2$. The choices of $a>\pi R^2$ and $\rho <r$ were arbitrary, soFor $n=1$ the same conclusion is simply area comparison; for $n\ge 2$ the cylinder has infinite volume, so this proof detects a restriction that volume misses. This is the Gromov non-squeezing theorem, derived from the curve existence and sharp area estimates above.

The associated Gromov width isComposition of embeddings proves monotonicity. Rescaling coordinates in a ball proves $c_G(U,c\omega )=c{c}_G(U,\omega )$ for $c>0$. Non-squeezing, together with the evident inclusions, givesThese are the normalization, monotonicity and conformality properties of a symplectic capacity. The two-ball obstruction of Question 5 illustrates a related use: one degree-one curve through two prescribed centres gives an additive area bound, stronger than comparing the volumes of the balls. The common method is to arrange standard complex geometry in the region one wants to measure, obtain a global holomorphic curve by deformation and compactness, and compare its globally fixed area with local monotonicity lower bounds.

Identify the ambient Euclidean space ${\mathbb{R}}^{2m+2}$ with ${\mathbb{C}}^{m+1}$ and let $J$ be multiplication by $i$. Equivalently, in real coordinates,This linear map preserves the norm and satisfies $⟨x,Jx⟩=0$. On the unit sphere $S^{2m+1}$ defineIt is continuous, tangent because it lies in $x^{\perp }=T_x{S}^{2m+1}$, and has norm one everywhere. It is therefore a nowhere-zero section of the tangent bundle, including when $m=0$ and the sphere is $S^1$. This supplies a nonvanishing tangent field on an odd-dimensional sphere explicitly.

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.

A Levi-Civita connection is a connection on the tangent bundle which is torsion-free and compatible with the Riemannian metric. Its defining identities areTo obtain uniqueness, write the metric identity for $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and use the torsion identity to interchange the derivatives. This gives the Koszul formulaThe metric is nondegenerate, so these inner products determine ${\nabla }_{X}Y$ uniquely.

For existence, let $K(X,Y,Z)$ denote the right-hand side. The bracket rules $[X,fY]=f[X,Y]+X(f)Y$ and $[fX,Y]=f[X,Y]-Y(f)X$ give, by cancellation of all extra derivatives,Thus the nondegenerate metric defines a unique smooth vector field ${\nabla }_{X}Y$ from $2g({\nabla }_{X}Y,Z)=K(X,Y,Z)$, and these identities give the connection rules. In coordinates the construction readsThe coefficients are smooth and symmetric in $i,j$, proving zero torsion. Substituting them giveswhich is exactly metric compatibility. Since the construction was given intrinsically by $K$, the local formulas agree on overlaps. This proves existence and uniqueness of the Levi-Civita connection on every Riemannian manifold.

Define the Riemann curvature tensor byThe connection rules make this expression tensorial in all three arguments. To fix the component convention, throughout this solution useWith this convention, the unit sphere has $R_{ijkl}=g_{ik}{g}_{jl}-g_{il}{g}_{jk}$ and positive Ricci curvature. In a coordinate frame, the components are obtained fromThe algebraic curvature symmetries aretogether with the first Bianchi identityThe first antisymmetry follows from the definition. Metric compatibility gives $g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0$, giving the second. Torsion-freeness reduces the cyclic sum to the Jacobi identity for Lie brackets of vector fields; pair interchange follows algebraically from these antisymmetries and the cyclic identity. These explain the identities rather than choosing unrelated sign conventions.

Define the Ricci curvature by the tensor contractionThis is a bilinear form, since $R$ is tensorial and trace is linear. At a point choose an orthonormal basis. Pair interchange givesso Ricci curvature is symmetric.

For the final determination in dimension three, an algebraic curvature tensor is determined by its entries in the three unordered index pairs $12,13,23$: antisymmetry removes repeated indices, and pair interchange makes the resulting $3\times 3$ array symmetric. It therefore has at most six independent entries. WriteThe diagonal Ricci components giveso, for example, $K_{12}=({\mathrm{Ric}}_{11}+{\mathrm{Ric}}_{22}-{\mathrm{Ric}}_{33})/2$, and the other two follow cyclically. The three off-diagonal entries are also recovered explicitly:Thus every independent curvature entry is determined by Ricci; this proves injectivity, not merely a dimension count.

An invariant formula making the reconstruction explicit is, with $S={\mathrm{tr}}_g\mathrm{Ric}$,To verify it, the right-hand side has the two antisymmetries, pair interchange and the cyclic identity by symmetry of $g$ and Ricci. Contracting its first and third indices in dimension three givesSubtract it from the original tensor. The difference has zero Ricci contraction, and the six recovered-entry formulas above force it to vanish. This completes the proof of three-dimensional curvature from the Ricci tensor in arbitrary coordinates.

An immersion $M^4\to {\mathbb{R}}^5$ would inject $TM$ into the trivial rank-five bundle. Taking its orthogonal complement gives a normal real line bundle $\nu$ with $TM\oplus \nu \cong {\epsilon }^5$. The Whitney product formula for Stiefel–Whitney classes requiresThe real line bundle need not be orientable, but it has no Stiefel–Whitney classes above degree one.

Let $a,b$ be the pulled-back degree-one generators from the two real projective planes. Their mod-two cohomology ring, by the Künneth theorem, is ${\mathbb{F}}_2[a,b]/(a^3,b^3)$. To compute the tangent classes, at a line $L$ the projective tangent space is $\mathrm{Hom}(L,L^{\perp })$. Adding $\mathrm{Hom}(L,L)\cong \mathbb{R}$ gives $\mathrm{Hom}(L,{\mathbb{R}}^3)\cong 3L$, with the real line identified with its dual using a metric. Thus the stabilized tangent bundle of real projective space gives $T{\mathbb{RP}}^2\oplus {\epsilon }^1\cong 3\gamma$ andThe tangent bundle of the product is the direct sum of the pulled-back tangent bundles. Since $(1+a+a^2)(1+a)=1$ in this truncated ring, and similarly for $b$,The degree-two class $ab$ is nonzero by the product-ring basis. It cannot be a class of the rank-one bundle $\nu$. Equivalently $w_1(TM{)}^2+w_2(TM)=ab\ne 0$, violating the codimension-one Stiefel–Whitney immersion obstruction. Hence ${\mathbb{RP}}^2\times {\mathbb{RP}}^2$ admits no immersion in ${\mathbb{R}}^5$.

A smooth vector field on a smooth manifold $M$ is a smooth section of its tangent bundle: it assigns $v(p)\in T_{p}M$ smoothly to each point. On a Lie group, it is a left-invariant vector field whenFor $X\in T_{e}G$, define $v_X(g)=(d{L}_g{)}_{e}X$. Smoothness of multiplication makes this a smooth vector field, and $L_g{L}_p=L_{gp}$ plus the chain rule proves left invariance. Every left-invariant vector field arises this way from its value at the identity.

For a matrix Lie group, $v_X(g)=gX$, because the derivative at zero of $g\gamma (t)$ is $gX$. The printed group name in this subpart must be read as ${\mathrm{SL}}_2$, rather than the additive vector space ${\mathfrak{sl}}_2$: it is the matrix-group calculation consistent with the preceding subparts. At the specified matrix,A tangent vector at a nonidentity group element need not itself be trace-free; here $g^{-1}{v}_X(g)=X$ is trace-free, as required. If one instead interpreted the printed ${\mathfrak{sl}}_2$ as its additive Lie group, the field would be the constant field $X$, a different problem.

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

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.

Let $F$ be the generator of the smooth flow. The Anosov flow property requires $F$ to be nowhere zero and a continuous splitting of the tangent bundlewhich is invariant under $D{\varphi }_t$. There must be constants $C\ge 1$ and $\kappa >0$ such that, for every $x$, every $t\ge 0$ and every vector in the indicated fibre,The first inequality defines the Anosov stable bundle; the second defines the Anosov unstable bundle and is equivalent to uniform forward expansion there. The remaining one-dimensional direction is the flow direction. The norm can come from any Riemannian metric, since norms are uniformly equivalent on a closed manifold. The essential requirements are a continuous invariant splitting and estimates uniform in both the base point and time, rather than separate exponential rates along individual orbits.

A smooth vector field on a smooth manifold $M$ is a smooth section $X:M\to TM$ of its tangent bundle. It acts on smooth functions by $Xh(p)=d{h}_p(X_p)$. In a manifold chart, $X={\sum }_i{X}^i{\partial }_i$. The Lie bracket of vector fields is the commutator of their actions on functions:Expanding in coordinates makes all second derivatives cancel and givesThus the commutator is again a smooth vector field, not a second-order operator. The coordinate expression represents the intrinsic derivation on functions, so agrees on overlaps.

For a diffeomorphism $f:M\to N$, its pushforward of a vector field isFor every smooth function $h$ on $N$, the chain rule gives $((f_{\ast }X)h)\circ f=X(h\circ f)$. Applying this twice yieldsSmooth functions determine a tangent vector, and $f$ is surjective, so

On a Lie group, a left-invariant vector field satisfies $(L_g{)}_{\ast }X=X$ for every left translation $L_g(h)=gh$. Given $\xi \in T_{e}G$, setThis field is smooth by smoothness of multiplication. The equality $L_a{L}_g=L_{ag}$ proves left invariance. Conversely, left invariance forces $X(g)=(d{L}_g{)}_{e}X(e)$, so $X_{\xi }$ is the unique left-invariant field with value $\xi$ at $e$. This is a linear isomorphism between $T_{e}G$ and the space of left-invariant vector fields.