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Write $L\alpha =\omega \wedge \alpha$, the Lefschetz operator of a Kähler manifold. The metric and volume form define the $L^2$ inner product on smooth complex forms, and $\Lambda =L^{\ast }$ is its formal adjoint. Similarly, ${\bar{\partial }}^{\ast }$ is the formal adjoint of the Dolbeault operator, characterized by $⟨\bar{\partial }\beta ,\alpha ⟩=⟨\beta ,{\bar{\partial }}^{\ast }\alpha ⟩$. The Dolbeault Laplacian isOn the compact manifold without boundary, integration by parts givesIf the Laplacian vanishes, both terms are zero. Conversely, if both operators annihilate $\alpha$, the defining formula annihilates it. Thus harmonicity is equivalent to being both $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed.

Because $\omega$ is closed and has type $(1,1)$, $\bar{\partial }\omega =0$ and $[L,\bar{\partial }]=0$. The supplied identity from the Kähler identities gives $[L,{\bar{\partial }}^{\ast }]=-i\partial$. With ordinary commutators for the even-degree operator $L$,The last identity follows from $d^2=0$. This also proves that the Lefschetz operator preserves harmonic forms.

For the cohomology map one can work directly with forms: $\bar{\partial }({\omega }^k\wedge \alpha )={\omega }^k\wedge \bar{\partial }\alpha$, since $\omega$ has even degree. It takes closed forms to closed forms and exact forms to exact forms. Therefore the $k$th power of $L$ inducesThe bidegree is $(p+k,q+k)$, including the zero groups outside the dimension range. No isomorphism claim is needed here.

Finally put $\theta =-i\partial f$, so $\bar{\partial }\theta =i\partial \bar{\partial }f={\omega }^{\prime }-\omega$. LetBoth Kähler forms are $\bar{\partial }$-closed, so $\bar{\partial }{T}_k=0$. For a closed representative $\alpha$,This primitive has type $(p+k,q+k-1)$; the degree-one sign produces no additional term because $T_k\wedge \alpha$ is closed. HenceThis is the dependence of Lefschetz maps on the Dolbeault class.

The Chern connection decomposes on vector-bundle-valued differential forms as $D=D^{\prime }+D^{\prime \prime }$, where $D^{\prime }={\nabla }_h^{\prime }$ raises holomorphic degree and $D^{\prime \prime }={\nabla }_h^{\prime \prime }={\bar{\partial }}_E$ raises antiholomorphic degree. In a holomorphic local frame, $D^{\prime }=\partial +A\wedge$ and $D^{\prime \prime }=\bar{\partial }$. The type of the curvature form of a connection givesThe Kähler metric and the Hermitian metric define the $L^2$ inner product and the formal adjoints $D^{\prime \ast },D^{\prime \prime \ast }$. The Dolbeault Laplacians areLet $L=\omega \wedge$ be the Lefschetz operator of a Kähler manifold and $\Lambda =L^{\ast }$ its adjoint Lefschetz operator. With the ordinary commutator convention $[A,B]=AB-BA$, the printed Kähler identities giveSubstitution into the two Dolbeault Laplacians, followed by expansion, yieldsThe middle line follows by cancelling the terms with $D^{\prime }\Lambda D^{\prime \prime }$ and $D^{\prime \prime }\Lambda D^{\prime }$; the remaining terms collect the anticommutator of the two differentials. Hence $\boxed{{\Delta }^{\prime \prime }={\Delta }^{\prime }+[i{F}_h,\Lambda ]}$, where $i{F}_h$ denotes its wedge action. This is the Bochner-Kodaira-Nakano identity with the paper's sign convention.

The Kähler form and the Hermitian metric give the $L^2$ inner product on vector-bundle-valued differential forms, using volume ${\omega }^n/n!$. Define the formal adjoint ${\bar{\partial }}_E^{\ast }$ and the elliptic, self-adjoint, nonnegative Dolbeault LaplacianIts harmonic space isThe equality follows from $⟨{\Delta }_E^{\prime \prime }\alpha ,\alpha ⟩=\Vert {\bar{\partial }}_E\alpha {\Vert }^2+\Vert {\bar{\partial }}_E^{\ast }\alpha {\Vert }^2$. On compact $X$, the bundle-valued Dolbeault Hodge decomposition states that this space is finite dimensional and thatThe sum is orthogonal for the $L^2$ inner product and all summands here consist of smooth forms. Every Dolbeault cohomology class has a unique harmonic representative, giving $H_{\bar{\partial }}^{p,q}(X,E)\cong {\mathcal{H}}^{p,q}(X,E)$ and, by the Dolbeault theorem, the corresponding sheaf cohomology isomorphism. This is a decomposition for ${\bar{\partial }}_E$, whose square is zero; it does not require the full Chern connection to be flat.

The formal adjoint of $\mathcal{L}=\Delta -k^2$ is itself. With the bilinear convention used in Green second identity, take any complex-valued $v$ satisfying $\mathcal{L}v=0$. The product rule for divergence gives the familyEquivalently, the differential one-formis a closed differential form. Arbitrary linear combinations of adjoint solutions give further members of this family. One can also write it as $\nabla \cdot [v^2\nabla (u/v)]=0$ wherever $v\ne 0$, but the first formula has no division and remains valid at zeros of $v$. For a sesquilinear convention replace $v$ by $\overline{v}$; no such conjugation is built into the bilinear identity used below.

Repeated integration by parts against a compactly supported test function gives the bilinear formal adjointThere is no boundary remainder because $v$ has compact support. The product rule makes this an order-at-most-$m$ linear differential operator with real analytic coefficients. The displayed integral pairing has no complex conjugation, so this is the formal transpose; for a sesquilinear pairing the coefficients would be conjugated.

For uniqueness, suppose $Q$ is the difference of two candidates. Then $\int uQv=0$ for every smooth compactly supported $u$ and $v$, so $Qv=0$ for every such $v$. At an interior point, a polynomial times a cutoff can prescribe every derivative of $v$ through order $m$ independently. Hence each coefficient of $Q$ is zero. This proves uniqueness as a differential operator, not only uniqueness of its integrated action.

The order-$m$ principal symbol of the formal adjoint is $p_m^{\ast }=(-1{)}^m{p}_m$. Derivatives falling on coefficients contribute only lower-order terms. Thus the original plane remains a non-characteristic hypersurface.

For the curved face use the defining function ${\rho }_{\epsilon }=x+y^2-\epsilon$, whose conormal is $(1,2y)$. On the face with $x>0$, $|y|<\sqrt{\epsilon }$ and $0<x\le \epsilon$. Therefore $(x,y,1,2y)$ tends uniformly to $(0,0,1,0)$ as $\epsilon \downarrow 0$. Since $p_m^{\ast }(0,0;1,0)\ne 0$, continuity givesThe same argument supplies a common neighborhood where the transformed coefficient of the highest normal derivative is nonzero, permitting the real analytic normal form used below.

In the flattened coordinates, ${\partial }_x={\partial }_{\tilde{x}}$ and ${\partial }_y=2\tilde{y}{\partial }_{\tilde{x}}+{\partial }_{\tilde{y}}$. The transformed formal adjoint consequently has coefficients obtained from finitely many real analytic coefficient pullbacks and polynomial factors. They have uniform analyticity bounds on a fixed small ball. By the preceding non-characteristic argument, its coefficient of ${\partial }_{\tilde{x}}^m$ has modulus bounded away from zero there, uniformly for small $\epsilon$. Division by it preserves uniform real analytic bounds.

The transformed forcing ${\tilde{g}}_{\epsilon }$ has analogous bounds, with constants that may depend on $g$. Apply the Cauchy-Kovalevskaya theorem in normal form with zero normal data through order $m-1$ at $\tilde{x}=0$. The supplied majorant series radius result yields a common solution radius $r_g>0$ for all sufficiently small parameters. Tangential differentiation of zero normal data shows that the entire boundary jet through order $m-1$ is zero; changing coordinates preserves this vanishing.

For $(x,y)$ in the closed cap, $|\tilde{x}|\le \epsilon$ and $|\tilde{y}|\le \sqrt{\epsilon }$. Choose $\epsilon$ so small that $\sqrt{{\epsilon }^2+\epsilon }<r_g$, with a strict margin. The whole closed cap then lies in the common real analytic solution neighborhood. Pulling the solution back provesIn fact $v$ is real analytic on a neighborhood of its closure, which is useful for the boundary integration below.

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.

On the zero-boundary Sobolev space $H_0^1(U)$, the associated bilinear form isThe homogeneous Dirichlet boundary condition removes the boundary term, so $(Lu,v{)}_{L^2}=B[u,v]$ for sufficiently regular $u$ and $v\in H_0^1$.

Formal self-adjointness means that $L$ equals its formal adjoint, or equivalently $(L\varphi ,\chi )=(\varphi ,L\chi )$ for compactly supported smooth functions. HereWith the symmetric real principal coefficients in the question, $L=L^{\ast }$ is equivalent to $b=0$. It makes $B$ a symmetric bilinear form.

For the assertions in part (b), positivity must mean strict Dirichlet positivity:This is the positive-definite operator convention. The weaker positive semidefinite operator convention $B[u,u]\ge 0$ does not suffice: $L=-\Delta -{\lambda }_1$ has zero energy on a first Dirichlet eigenfunction. Under strict positivity, the compactness argument in part (b)(i) gives a uniform lower bound as well.

Let $D=d/dx$. Repeated integration by parts gives the formal adjoint rulesThereforeEquality of the coefficients of $D$ in $L$ and $L^{\ast }$ requiresso $q=p^{\prime }$. This also makes the zeroth-order coefficients equal because $p^{\prime \prime }-q^{\prime }=0$. Hence, under boundary conditions that remove the boundary terms,This is the self-adjoint fourth-order scalar differential operator $D^4+p{D}^2+p^{\prime }D+r$.

For the formal adjoints $d^{\ast }$, ${\partial }^{\ast }$ and ${\bar{\partial }}^{\ast }$ defined by the Kähler metric,andThese are respectively the Hodge Laplacian and the two Dolbeault Laplacians.

The Kähler manifold structure gives a Riemannian metric, its volume form, the complex orientation, and a Hermitian inner product on complex differential forms. The complex Hodge star operator is the complex-linear map characterized byOn $r$-forms in real dimension $m=2n$, the codifferential isequivalently the formal adjoint of $d$ for the $L^2$ inner product. Similarly ${\bar{\partial }}^{\ast }$ is the formal adjoint of $\bar{\partial }$. Define the Hodge Laplacian and Dolbeault Laplacian by

Expanding $d=\partial +\bar{\partial }$, the Kähler identities make the mixed anticommutators vanish and imply ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Hence the Kähler Laplacian identity is

Let $L\gamma =\omega \wedge \gamma$ be the Lefschetz operator of a Kähler manifold. The Kähler identities also implyThus, if ${\Delta }_{\bar{\partial }}\alpha =0$, thenThis is the fact that the Lefschetz operator preserves harmonic forms.

The Dolbeault Hodge decomposition on a compact Hermitian manifold states thatan orthogonal direct sum, where ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$.

Suppose $\eta =\bar{\partial }\gamma$ has type $(p,q)$. Apply this decomposition to $\gamma \in {\Omega }^{p,q-1}$. The harmonic and $\bar{\partial }$-exact pieces disappear after applying $\bar{\partial }$, so for some $\beta \in {\Omega }^{p,q}$,Put $\theta ={\bar{\partial }}^{\ast }\beta$. If also $\partial \eta =0$, thenThe Kähler anticommutation identity ${\bar{\partial }}^{\ast }\partial +\partial {\bar{\partial }}^{\ast }=0$ and $({\bar{\partial }}^{\ast }{)}^2=0$ giveTherefore $\partial \theta$ is ${\Delta }_{\bar{\partial }}$-harmonic. By ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$ it is also $\partial$-harmonic, but it is $\partial$-exact; orthogonality of harmonic and exact forms forcesThis proves both requested claims: $\partial {\bar{\partial }}^{\ast }\beta =0$ is harmonic, and ${\bar{\partial }}^{\ast }\beta =\theta$ is $\partial$-closed.

Finally, $\theta \in \mathrm{im}{\bar{\partial }}^{\ast }$ is orthogonal to $\ker \bar{\partial }$, and hence to every $\bar{\partial }$-harmonic form. Since the $\partial$- and $\bar{\partial }$-harmonic spaces agree on a compact Kähler manifold, the $\partial$-closed form $\theta$ has zero harmonic component in its $\partial$-Hodge decomposition. It follows that $\theta =\partial \varphi$ for some $\varphi \in {\Omega }^{p-1,q-1}(X)$. Hencewhich is the ddbar lemma in this case.

The Poisson equation for differential forms on a closed oriented Riemannian manifold is solvable exactly when its source is $L^2$-orthogonal to every harmonic differential form of the same degree. Necessity follows from the formal adjoint identity $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. For sufficiency, the Green operator of the Hodge Laplacian gives $\alpha =G\beta$, since $\Delta G=I-P_{\mathcal{H}}$.