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The orthogonal projection ${\pi }_S$ is a smooth bundle map, since the Hermitian metric and the subbundle vary smoothly. The mapis independent of the representative, bijective on each fibre, and smooth with smooth inverse supplied by the quotient map on $S^{\perp }$. Hence it is a natural isomorphism of smooth vector bundles. It need not be holomorphic. The quotient Hermitian metric isIt is well-defined and positive definite because $S^{\perp }$ represents each quotient class uniquely.

On sections of $S$, set $D^{\prime }={\pi }_S{D}_E$. The Leibniz rule makes $D^{\prime }$ a connection on a vector bundle. For $s,t\in \Gamma (S)$, the orthogonal terms vanish in the metric pairing, soThus $D^{\prime }$ has metric compatibility. Because $S$ is a holomorphic subbundle, the $(0,1)$ part of $D_E$ preserves $S$ and restricts to ${\bar{\partial }}_S$. Therefore $(D^{\prime }{)}^{0,1}={\bar{\partial }}_S$. The uniqueness of the Chern connection in part (b) gives the projected Chern connection formulaIt follows that $A(s)=(I-{\pi }_S)D_{E}s$ is a quotient-valued one-form. The printed target $T_M\otimes Q$ is missing a dual: the cotangent bundle is the correct factor,Complex-valued forms use the complexified cotangent bundle here. In fact the $(0,1)$ part cancels, so $A$ is the second fundamental form of a holomorphic subbundle, an element of ${\mathcal{A}}^{1,0}(\mathrm{Hom}(S,Q))$. Finally the two connection Leibniz rules giveThis proves its tensoriality and all the asserted quotient and projection properties.

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.

If $M$ has complex dimension $n$, its complexified cotangent bundle splits into the $(1,0)$ and $(0,1)$ cotangent bundles. The space of global smooth differential forms of type (p, q) isIn holomorphic coordinates, its elements have the form $\alpha ={\sum }_{I,J}{a}_{I,J}d{z}^I\wedge d{\bar{z}}^J$, where the coefficients are complex-valued smooth functions, $|I|=p$, and $|J|=q$.

The exterior derivative decomposes as $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator $\partial$ and Dolbeault operator $\bar{\partial }$ are its type components:They have bidegrees $(1,0)$ and $(0,1)$ respectively. Keeping the displayed wedge order avoids hiding a sign; moving $d{\bar{z}}_k$ past $d{z}^I$ introduces $(-1{)}^p$. Decomposing $d^2=0$ by type gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

Let $\pi :T^{\ast }X\to X$ be the cotangent bundle projection and $\lambda$ its canonical one-form on a cotangent bundle, defined by ${\lambda }_{(x,\xi )}(v)=\xi (d\pi (v))$. Use the symplectic form $\omega =-d\lambda$, so in local coordinates $\omega ={\sum }_{j}d{x}_j\wedge d{\xi }_j$.

The graph map $s_{\beta }(x)=(x,{\beta }_x)$ is a smooth embedding, with $s_{\beta }^{\ast }\lambda =\beta$, and henceIts image has dimension $\dim X$, half the dimension of $T^{\ast }X$. By the definition of a Lagrangian submanifold,This is the graph of a closed one-form is Lagrangian criterion. The opposite conventional sign for the symplectic form gives the same criterion.

The Weinstein neighborhood theorem states that if $L$ is a compact Lagrangian submanifold of $(M,\omega )$, then neighborhoods of $L$ in $M$ and of the zero section in its cotangent bundle $T^{\ast }L$ are symplectomorphic. The symplectomorphism restricts to the identity on $L$, and the canonical form on $T^{\ast }L$ is taken with the sign matching the chosen convention.

For a planar cotangent bundle, rotation has moment map $\ell =x{p}_y-y{p}_x$. At $\ell \ne 0$ its level is regular and the circle action is free. Polar canonical coordinates give $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Fixing $\ell$ and quotienting $\theta$ leaves $T^{\ast }{\mathbb{R}}_{>0}$ with the displayed form. The unit-mass free Hamiltonian becomes $p_r^2/2+{\ell }^2/(2r^2)$.

A quadratic homogeneous polynomial in the fibre variables of the cotangent bundle is a first integral of the geodesic Hamiltonian exactly when its symmetric lowered coefficients form a rank-two Killing tensor. For $v=g^{-1}p$, the Poisson bracket is $\{K,H\}={\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c$, so the equivalence follows by comparing cubic coefficients at every point.

A diffeomorphism of the entire cotangent bundle preserving the Liouville one-form is the cotangent lift of a diffeomorphism of the base. It preserves the zero section and the Liouville vector field, hence commutes with the cotangent fiber-dilation flow. The limit as the fiber contracts forces it to cover its restriction to the zero section. Evaluation of the preserved one-form then forces the inverse-transpose derivative action on each covector.

For a holomorphic subbundle $S\subset E$ and quotient $Q=E/S$, the second fundamental form is $A(s)=(I-{\pi }_S)D_{E}s$. The Leibniz rule gives $A(fs)=fA(s)$, and holomorphicity of the subbundle makes its $(0,1)$ part vanish. Thus $A\in {\mathcal{A}}^{1,0}(\mathrm{Hom}(S,Q))$: the form factor is a cotangent bundle, not a tangent 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.

For a smooth function $M(q)$, its differential defines a Lagrangian submanifold of the cotangent bundle: pullback of the canonical one-form on a cotangent bundle is $dM$, so pullback of the symplectic form $-d\theta$ is zero. The graph has half the ambient dimension. With $q=(S,V,N)$ and momentum coordinates $(T,-P,\mu )$, the first law of thermodynamics identifies the equilibrium graph with $dM$. This geometric statement does not require extensivity or thermodynamic stability.

On the canonical cotangent bundle $T^{\ast }{\mathbb{R}}^n$, $\omega =i\sum d{z}_j\wedge d{\bar{z}}_j$ and $H=z^{†}z$. The unitary group acts by $z\mapsto Uz$, preserving both. For a Hermitian $A$, the generator $-iA$ has Hamiltonian ${\mu }_A=z^{†}Az$. The Poisson bracket satisfies $\{{\mu }_A,{\mu }_B\}={\mu }_{-i[A,B]}$, with no central term. The central generator has moment map $H$; imaginary antisymmetric matrices give ordinary angular momentum, while real symmetric matrices give quadratic observables mixing position and momentum.

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.