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For the Levi-Civita connection of a Riemannian metric, the divergence of $X$ is ${\mathrm{div}}_{g}X=\mathrm{tr}(Y\mapsto {\nabla }_{Y}X)$. In coordinates it equals $(\det g{)}^{-1/2}{\partial }_i((\det g{)}^{1/2}{X}^i)$. On an oriented manifold it satisfies ${\mathcal{L}}_X{\omega }_g=({\mathrm{div}}_{g}X){\omega }_g$. The codifferential on a one-form satisfies $\delta \theta =-{\mathrm{div}}_g({\theta }^{♯})$, where $♯$ is the musical isomorphism; this follows by integration by parts or the Hodge star formula.

A Riemannian metric determines a unique torsion-free connection that is a metric connection. Metric compatibility and the torsion identity imply the Koszul formula, which proves uniqueness. Conversely the right-hand side of the Koszul formula is a smooth one-form in its last vector-field argument. The musical isomorphism defines $D_{X}Y$ from it; expanding the Lie bracket of vector fields verifies the connection rules, metric compatibility and zero torsion. Thus it constructs the Levi-Civita connection intrinsically.

For relativistic dust, $p=0$, the relativistic Euler equation gives $\rho a_b=0$. The nowhere-vanishing density therefore makes the four-acceleration zero: $u$ is a timelike geodesic vector field. Let $\alpha =u^{♭}$ be the differential one-form obtained by the musical isomorphism. Its exterior derivative satisfiesHere ${\iota }_u$ is the interior product. By hypersurface orthogonality and the Frobenius theorem, $\alpha \wedge d\alpha =0$. Taking the interior product and using $\alpha (u)=-1$ givesThus $\boxed{d{u}^{♭}=0}$. By the Poincaré lemma, locally $\alpha =-dt$, and then $u(t)=-\alpha (u)=1$. Choose spatial coordinate functions $x^i$ on one level set of $t$ and extend them constantly along the integral curves of a vector field $u$, using its smooth flow. This constructs $u={\partial }_t$. Since $dt({\partial }_i)=0$,The metric on the orthogonal complement of a timelike vector is positive-definite, so the synchronous coordinates for irrotational dust haveThis construction is local; crossing geodesics can prevent these synchronous coordinates from covering a larger region.

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.

Yes, the two real smooth vector bundles are isomorphic, although the choice is not canonical. A standard smooth manifold is Hausdorff and second countable, hence admits a smooth locally finite partition of unity. Choose Euclidean Riemannian metrics on coordinate neighborhoods and take their partition-of-unity sum. At each point at least one positive coefficient occurs, so the resulting smooth symmetric tensor $g$ is positive definite.

The resulting musical isomorphism isIt covers the identity of $M$ and is linear on every fibre. In coordinates it is $(x,v)\mapsto (x,G(x)v)$, where $G=(g_{ij})$ is a smooth positive-definite matrix. Its inverse uses the smooth inverse matrix $G^{-1}$, giving the map ${♯}_g$. Thus it is a smooth vector bundle isomorphism.

The Riemannian metric supplies the extra choice: no positive-dimensional real vector space has a linear identification with its dual invariant under all changes of basis. Indeed a scalar change $tI$ acts by $t$ on vectors and $t^{-1}$ on covectors, so an equivariant map $L$ would satisfy $tL(v)=t^{-1}L(v)$ for every $t>0$, forcing $L=0$. An isomorphism therefore exists without being a metric-free canonical one.

The musical isomorphisms are the pointwise maps induced by the Riemannian metric:In local coordinates they are $(X^{♭}{)}_i=g_{ij}{X}^j$ and $({\alpha }^{♯}{)}^i=g^{ij}{\alpha }_j$. The inverse metric makes these inverse bundle maps, hence inverse maps between smooth vector fields and smooth differential one-forms.

For a covariant two-tensor $\beta$, its metric trace is ${\mathrm{tr}}_g\beta =g^{ij}{\beta }_{ij}={\sum }_i\beta (e_i,e_i)$ in a Riemannian orthonormal frame. The contraction is independent of the chosen frame. There is a terminology distinction here: if $\beta$ is literally an alternating differential two-form, this trace is zero. The object needed below is the covariant two-tensor $D(X^{♭})$, generally not alternating. Interpreting that tensor as an alternating form would make the stated divergence construction impossible.

Let $D$ be the Levi-Civita connection. The Riemannian gradient, divergence and positive Laplace-Beltrami operator areThus $g({\mathrm{grad}}_{g}f,Y)=Yf$, and in coordinatesThis choice makes the Laplacian eigenvalues nonnegative on a closed Riemannian manifold. The alternative convention $+\mathrm{div}\mathrm{grad}$ reverses all eigenvalue signs, but none of the comparisons below.

For the Riemannian submersion, let $F=f\circ \pi$ and choose orthonormal horizontal lifts $E_i$ of a base frame and an orthonormal vertical frame $V_{\alpha }$. The horizontal part of $D_{E_i}{E}_i$ projects to the base Levi-Civita connection; this follows directly from its Koszul formula applied to basic horizontal fields, because their inner products and projected brackets are pulled back from the base. Therefore the horizontal trace of the Riemannian Hessian of $F$ is the pulled-back base Hessian trace. Since $V_{\alpha }F=0$, its vertical terms areWriting $H={\sum }_{\alpha }(D_{V_{\alpha }}{V}_{\alpha }{)}^{\mathrm{hor}}$ for the unnormalised fibre mean curvature vector givesThe totally geodesic submanifold condition on the fibres makes $H=0$. Consequently eigenfunction pullback by a Riemannian submersion givesHere the Riemannian submersion is onto, so the map is injective. Its image is exactly the members of the upstairs eigenspace constant on each whole fibre: such a function descends smoothly by local sections of the submersion, and the displayed identity forces the descended function to have the same eigenvalue. In particular, whenever these eigenspaces are finite dimensional, the downstairs multiplicity is no larger than the upstairs multiplicity. The upstairs eigenspace may also contain functions varying along fibres.

A flat torus is ${\mathbb{R}}^d/\Lambda$ with the metric induced by the Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its dual lattice is ${\Lambda }^{\ast }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. The characters $e_w(x)=e^{2\pi i⟨w,x⟩}$ descend to the quotient and form an orthogonal Fourier basis of $L^2$ on it. Direct differentiation gives the spectrum of a flat torus:with multiplicity equal to the number of dual lattice vectors of each length. In particular zero has multiplicity one.

Let the given four-dimensional isospectral manifolds be ${\mathbb{R}}^4/{\Lambda }_1$ and ${\mathbb{R}}^4/{\Lambda }_2$. For $d=4+k$, take their products with the same $k$ circles of length $\epsilon$, equivalently the lattices ${\Lambda }_i\oplus \epsilon {\mathbb{Z}}^k$. The dual lattices are ${\Lambda }_i^{\ast }\oplus {\epsilon }^{-1}{\mathbb{Z}}^k$, so their Laplacian eigenvalues areThe given equality of base spectra with multiplicities makes these product spectra identical.

Choose $\epsilon$ smaller than the shortest nonzero vector length in either original Euclidean lattice. By short-vector cancellation for flat tori, the shortest nonzero vectors in the product lattice are exactly $\pm \epsilon e_j$ in its new factor. To spell out the nonisometry argument, any isometry of the product flat tori lifts to an isometry of their universal Euclidean covers. This is affine, and its orthogonal linear part maps one product lattice onto the other. It must map the span of their shortest vectors, namely the new $k$-dimensional factor, onto itself; orthogonality then preserves the original four-dimensional factor and maps ${\Lambda }_1$ onto ${\Lambda }_2$. That would give an isometry of the original flat tori, contrary to the assumption. Isospectral nonisometric flat tori therefore exist in every dimension $d\ge 4$.

The divergence of a Riemannian vector field $X$ is the trace of the covariant derivative map $Y\mapsto {\nabla }_{Y}X$:It is characterized by ${\mathcal{L}}_X{\omega }_g=(\mathrm{div}X){\omega }_g$ in the oriented case. The metric-dual vector field $X_{\theta }={\theta }^{♯}$ is defined by $g(X_{\theta },Y)=\theta (Y)$ for every $Y$, using the musical isomorphism.

On a compact oriented manifold without boundary, the permitted exactness assertion and Stokes theorem imply ${\int }_M\mathrm{div}Y{\omega }_g=0$ for every smooth vector field $Y$. For a smooth function $a$, the product rule givesSince $⟨da,\theta ⟩=X_{\theta }(a)$, the formal-adjoint identity yieldsThis holds for all $a$; taking $a=\delta \theta +\mathrm{div}{X}_{\theta }$ proves the pointwise formulaIt agrees with $\delta \theta =-\ast d\ast \theta$, since $\ast \theta ={\iota }_{X_{\theta }}{\omega }_g$ and $d{\iota }_{X_{\theta }}{\omega }_g={\mathcal{L}}_{X_{\theta }}{\omega }_g$.

The Bochner formula for one-forms, with the positive Hodge Laplacian, isHere $\Delta =d\delta +\delta d$ acts on one-forms. The covariant derivative $\nabla \theta$ is a section of $T^{\ast }M\otimes T^{\ast }M$, and its formal-adjoint composition, the rough Laplacian, isThis expression is independent of the local orthonormal frame. The Ricci term is the zero-order bundle endomorphism specified byThus the identity separates the differential energy of the form from the curvature contribution; the sign of the Ricci term matches positive sectional curvature on the sphere.

Let $\theta$ be a harmonic one-form on a compact connected manifold with nonnegative Ricci curvature. Pair the formula with $\theta$ and integrate:Both integrands are nonnegative and continuous, so each vanishes everywhere. In particular $\nabla \theta =0$, proving that harmonic one-forms are parallel under nonnegative Ricci curvature. This integration uses the Riemannian volume density and formal adjoints, so orientation is not necessary for the final conclusion.

Fix $p\in M$. The evaluation mapis injective: a parallel form with zero value at $p$ stays zero under parallel transport along every path, and connectedness makes every point reachable from $p$. ThereforeThis is the dimension bound for harmonic one-forms under nonnegative Ricci curvature. The bound is sharp: on a flat $n$-torus the $n$ constant coordinate one-forms are parallel and harmonic. If Ricci curvature is positive definite at even one point, the same integral and parallelism argument force every harmonic one-form to vanish.

A Riemannian metric is a smooth field of positive-definite symmetric bilinear forms $g_x$ on the tangent spaces, equivalently a smooth section of ${\mathrm{Sym}}^2{T}^{\ast }M$ with that positivity. Its musical isomorphism isNondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix $g_{ij}$, and its inverse ${♯}_g$ has matrix $g^{ij}$. Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing determinant. Both maps cover the identity on $M$, giving a smooth vector bundle isomorphism.

For local flattening of a Riemannian metric, choose a manifold chart neighborhood $V$ around $p$ with closure contained in $U$, and let $h={\varphi }^{\ast }\delta$ on $V$. Choose a smooth bump function $0\le \chi \le 1$, with compact support in $V$ and equal to one on a smaller neighborhood $W$ of $p$. SetThe formulas glue smoothly because the modification is supported strictly inside $V$. A convex combination of positive-definite forms is positive definite. On $W$ the metric is exactly $h$, so $\varphi$ gives an isometry to a Euclidean open subset. Outside $U$ the original metric remains unchanged.

Geodesic completeness means that every geodesic with arbitrary initial point and tangent vector extends for every real affine parameter. On each connected component, the Hopf-Rinow theorem equates this with completeness of the Riemannian distance.

The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, $M=\{x\in {\mathbb{R}}^n:|x|>1\}$ with $n\ge 2$, $X=\varnothing$ and $g=\delta$ satisfy those hypotheses. The geodesic $\gamma (t)=(2-t)e_1$ reaches the missing unit sphere at $t=1$ and cannot continue within $M$. Thus the printed assumptions alone do not prove completeness.

Here is the intended compact-core completeness for a Euclidean end. Write $\Phi :M\setminus X\to \{|x|>1\}$ for the end coordinates and assume additionally thatis compact for every sufficiently large $R$. Choose such an $R$ beyond the metric transition and large enough to include the initial point of a geodesic. Its speed $s$ is constant. On every segment outside $K_R$, the Euclidean radius changes at a rate at most $s$, so over a finite parameter interval it cannot exceed $R+sT$. The full segment therefore stays in the compact set $K_{R+sT}$. Its velocities also stay in a compact subset of $TM$, since their metric norm is fixed and the base set is compact. The smooth geodesic ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.

For upward stability of Riemannian completeness, lengths of all curves satisfy $L_{\tilde{g}}\ge L_g$, hence $d_{\tilde{g}}\ge d_g$. A $d_{\tilde{g}}$-Cauchy sequence is therefore $d_g$-Cauchy and has a limit $p$ because $g$ is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around $p$, $\tilde{g}$ has a bounded largest matrix eigenvalue, so the coordinate straight segment gives $d_{\tilde{g}}(p,q)\le C|\varphi (q)-\varphi (p)|$. Thus convergence to $p$ is also in $d_{\tilde{g}}$. That distance is complete, and Hopf-Rinow yieldsFor disconnected $M$, apply this argument on each component.

Combining metric compatibility with the torsion-free condition forces the Koszul formula:Nondegeneracy of $g$ proves uniqueness of the Levi-Civita connection. For existence, use the right side to define $2g({\nabla }_{X}Y,Z)$. Expanding brackets shows that this expression is $C^{\infty }$-linear in $Z$, so it defines a smooth one-form; the musical isomorphism gives the required vector field. The same expansion shows ${\nabla }_{aX}Y=a{\nabla }_{X}Y$, additivity, and ${\nabla }_X(aY)=X(a)Y+a{\nabla }_{X}Y$, establishing the connection rules. Subtracting the formulas with $X,Y$ exchanged gives ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. Adding the formulas pairing ${\nabla }_{X}Y$ with $Z$ and ${\nabla }_{X}Z$ with $Y$ gives metric compatibility. Thus this construction has both required properties.

In coordinate vector fields the brackets vanish. Consequently the Christoffel symbols and coordinate derivative areRepeated indices are summed. The displayed connection type in the source is best understood in its standard form on two vector fields; if the first input is an individual tangent vector at a point, the output lies in the tangent fiber at that point rather than in the space of global sections.

For the parallel metrics with a common Levi-Civita connection, let $\gamma$ be a piecewise smooth path from the point $x$ of equality to any $y$. Connected smooth manifolds admit such paths because coordinate balls are path connected. Parallel transport $P_{\gamma }$ for the common connection is invertible and preserves both metrics. ThereforeEvery pair of tangent vectors at $y$ arises this way, so $\boxed{\tilde{g}=g}$ everywhere.

Dropping the agreement at one point removes the conclusion. For any constant $c>0$, $\tilde{g}=cg$ has the same Christoffel symbols. Nor must the two metrics be proportional: on ${\mathbb{R}}^n$, $n\ge 2$, the constant metrics ${\sum }_i(d{x}^i{)}^2$ and $2(d{x}^1{)}^2+{\sum }_{i\ge 2}(d{x}^i{)}^2$ both have zero connection coefficients. In general write $\tilde{g}(u,v)=g(Au,v)$. Since both metrics are parallel, $0=({\nabla }_X\tilde{g})(Y,Z)=g(({\nabla }_{X}A)Y,Z)$, so $\nabla A=0$. Conversely, a positive $g$-self-adjoint parallel $A$ makes the same torsion-free connection compatible with $\tilde{g}$, proving equality of their Levi-Civita connections. Thus one-point agreement specifies $A=I$ and forces it everywhere; without it, nontrivial parallel choices can remain.

A fiber metric on a real vector bundle $E\to M$ is a smoothly varying positive-definite inner product on each fiber. Choose a trivializing open cover $\{U_i\}$ and a smooth partition of unity $\{{\rho }_i\}$ subordinate to it. The partition theorem gives nonnegative functions summing to one, a locally finite family of subsets of supports, and $\mathrm{supp}{\rho }_i\subset U_i$; the usual Hausdorff second-countable smooth manifold hypotheses ensure this theorem applies. Transfer the Euclidean inner product to each local vector bundle trivialization, obtaining $h_i$, and setEach weighted term extends smoothly by zero outside $U_i$, and local finiteness makes the sum smooth in every vector bundle trivialization. At each $x$ some weight is positive, so $h_x(v,v)>0$ for every nonzero $v$. This proves existence of a fiber metric, with no orientability or triviality assumption.

A vector bundle morphism covering the identity is a smooth map $F:E^{\prime }\to E^{\prime \prime }$ that preserves base points and is a linear map on each fiber. Its induced map on the module of smooth sections is $s\mapsto F\circ s$, and is $C^{\infty }(M)$-linear. We prove the converse by constructing bundle morphisms from maps of smooth sections.

First the given map $\alpha$ is local. If a global section $s$ vanishes on a neighborhood of $x$, take a smooth bump function $\chi$ supported there with $\chi =1$ near $x$. Then $\chi s=0$, so $\chi \alpha (s)=\alpha (\chi s)=0$, and hence $\alpha (s)(x)=0$. Thus sections agreeing near $x$ have images agreeing at $x$.

Choose a local frame $e_1,\dots ,e_r$ on $U$, and a bump function equal to one on a smaller neighborhood $V$ of $x$ and supported in $U$. Multiplying the frame by that bump and extending by zero gives global smooth sections $t_j$ whose restrictions to $V$ are the frame. If $s(x)=0$, write $s=\sum a_j{e}_j$ on $V$. A second bump extends each $a_j$ to a global smooth function ${\tilde{a}}_j$ agreeing near $x$. By locality and $C^{\infty }(M)$-linearity,Every fiber vector $v\in E_x^{\prime }$ is the value of a global smooth section, by the same bumped-frame construction. Define $F_x(v)=\alpha (s)(x)$ for any such section. The just-proved vanishing statement makes this well-defined. The maps $F_x$ are linear maps, and locally their matrix columns are the smooth sections $\alpha (t_j)$ in a frame of $E^{\prime \prime }$. Thus $F$ is smooth, is a vector bundle morphism, and satisfies $\alpha (s)=F\circ s$. Fiberwise evaluation also proves uniqueness.

Finally apply the fiber metric construction to the tangent bundle. A Riemannian metric $g$ defines the musical isomorphismPositive definiteness makes it a fiberwise bijection; its inverse is smooth because inverse metric matrices vary smoothly. Hence $TM$ and $T^{\ast }M$ are isomorphic as real smooth vector bundles on every such manifold. The isomorphism depends on the chosen metric and is not canonical.

A Levi-Civita connection on a Riemannian manifold is a covariant derivative $D$ that is a torsion-free connection, $D_{X}Y-D_{Y}X=[X,Y]$, and a metric connection, $Xg(Y,Z)=g(D_{X}Y,Z)+g(Y,D_{X}Z)$. A covariant derivative is $C^{\infty }(M)$-linear in $X$, real linear in $Y$, and obeys $D_X(fY)=X(f)Y+f{D}_{X}Y$.

Combining metric compatibility for the cyclic triples and eliminating reversed derivatives with the torsion identity gives the Koszul formulaSince $g$ is nondegenerate, this determines $D_{X}Y$ uniquely. For existence, use the right-hand side to define $D_{X}Y$: expansion of the Lie bracket of vector fields shows it is $C^{\infty }(M)$-linear in $Z$, hence defines a smooth one-form, which the musical isomorphism converts to a smooth vector field. The same expansion shows $D_{fX}Y=f{D}_{X}Y$ and $D_X(fY)=X(f)Y+f{D}_{X}Y$. Subtracting the formula with $X,Y$ exchanged gives torsion zero; adding its versions for $Y,Z$ exchanged gives metric compatibility. Thus it is a Levi-Civita connection. Equivalently its coefficients in a manifold chart are the Christoffel symbolsThe intrinsically defined Koszul formula ensures these local expressions fit together. This proves the fundamental theorem of Riemannian geometry.

The product Riemannian metric on $M\times N$ isThe two factor tangent spaces are orthogonal, and the sum is positive definite. Lift $X$ from $M$ and $Y$ from $N$. We check $D_{X}Y$ against lifted local frame fields from both factors, which together span each product tangent space.

If $Z$ is lifted from $M$, then $g(Y,Z)=g(X,Y)=0$, while $g(Z,X)$ depends only on $m$, so $Yg(Z,X)=0$. Also $[Y,Z]=[X,Y]=0$ and $[Z,X]$ is horizontal, hence orthogonal to $Y$. Every term in the Koszul formula is zero. If $W$ is lifted from $N$, $g(W,X)=g(X,Y)=0$ and $g(Y,W)$ depends only on $n$, so $Xg(Y,W)=0$. Now $[W,X]=[X,Y]=0$ and $[Y,W]$ is vertical, hence orthogonal to $X$. Again every term is zero. Thus $D_{X}Y$ is orthogonal to both spanning frame families, and positive definiteness yieldsThis calculation uses the lifts' independence of the other factor coordinates; a vector field with varying coefficients in those coordinates can have a nonzero mixed derivative.

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

Use the nonnegative Hodge-Laplacian convention from Question 4. The Bochner-Weitzenbock formula for one-forms isHere $\nabla$ is the Levi-Civita connection induced on the cotangent bundle, and the rough Laplacian in a local orthonormal frame isIt is the composition of covariant derivative with its formal adjoint. The Ricci endomorphism is defined by $g({\mathrm{Ric}}^{♯}X,Y)=\mathrm{Ric}(X,Y)$ and acts on one-forms by $(\mathrm{Ric}\cdot \alpha )(X)=\alpha ({\mathrm{Ric}}^{♯}X)$. The musical isomorphism defines ${\alpha }^{♯}$ by $g({\alpha }^{♯},X)=\alpha (X)$. The associated scalar formula isThese signs make the integrated rough-Laplacian term $\Vert \nabla \alpha {\Vert }^2$ on a closed manifold.

For a connected manifold, the full Riemannian holonomy group at $p$ is the subgroup of $O(T_{p}M,g_p)$ consisting of parallel transports around all piecewise smooth loops based at $p$. Its natural action on $T_{p}M$ is the holonomy representation; it induces actions on cotangent spaces and all tensor spaces. The holonomy representation is an irreducible representation when it has no nonzero proper invariant subspace.

The fundamental principle of Riemannian holonomy identifies parallel tensor fields with tensors at $p$ fixed by full holonomy. A parallel field returns to its value under every loop. Conversely, transport a fixed tensor along a path from $p$ to each point. Any two paths differ by a loop, so the result is independent of the path; local smooth parallel transport yields a smooth parallel field. Evaluation and construction are inverse. Full holonomy, not merely the contractible-loop subgroup, is required for this global correspondence.

On a compact manifold without boundary and with nonnegative Ricci curvature, a harmonic one-form satisfies the integrated Bochner identityBoth terms are nonnegative; therefore $\nabla \alpha =0$. This proves that harmonic one-forms are parallel under nonnegative Ricci curvature. Integration uses the Riemannian volume density and does not require an orientation.

If $n=\dim M\ge 2$, a nonzero such form would give a nonzero fixed tangent vector via metric duality and hence a proper invariant line, contradicting irreducibility. This is precisely the irreducible holonomy in dimension at least two has no parallel one-form criterion. Since the question permits harmonic representatives of all classes, it gives $b_1(M)=0$, where $b_1$ is the first Betti number.

The covering $\pi :X\times T^k\to M$ is finite: its fibre over a point is a closed discrete subset of the compact total space and hence finite. Use the explicitly permitted equality of the Betti numbers of $M$ and its finite cover, in particular $b_1(X\times T^k)=b_1(M)=0$. This equality is a permission of this question, not a general theorem about finite covers.

The $k$ angular one-forms on $T^k$, pulled back to $X\times T^k$, represent linearly independent de Rham classes: their periods on the $k$ coordinate circles are the standard basis vectors. exact differential forms have zero periods. Thus $b_1(X\times T^k)\ge k$, forcing $k=0$. Now $\pi :X\to M$ is a finite universal covering map because $X$ is simply connected. The fundamental group acts freely and transitively on a fibre, or equivalently loop-lifting identifies its elements with the finite set of possible endpoints upstairs. Consequently the intended conclusion isThe printed statement needs the dimension qualification under the usual definition of irreducibility. Take the standard circle $M=S^1$, with $X$ a point, $k=1$, and the identity covering $X\times T^1\to S^1$. Its Ricci curvature is zero. parallel transport fixes its global unit tangent, so its holonomy representation is the trivial representation on a one-dimensional real vector space, which is irreducible. The allowed Betti-number equality holds for this identity cover, yetThus there is no proof of the unqualified literal assertion in dimension one. If “irreducible holonomy” is instead intended to exclude the one-dimensional trivial representation, the preceding intended proof applies. A connected zero-dimensional manifold is a point and has trivial fundamental group.