The study of relations between, one the one hand, the topology of a smooth manifold as encoded in the cohomology groups, and on the other, the set of solutions to the Laplace operator on differential forms relative to some Riemannian metric on the manifold.

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Construction of Hodge decomposition

We know Hodge decomposition splits any $k$-form into three $L^2$ components. And I see some proofs, none of them provide an explicit constructive method. Is there any general method to construct one? ...
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Dolbeault cohomology on torus

Let $T=\mathbb{C}/\Gamma$ where $\Gamma$ is a lattice of $\mathbb C$. Given that $H_{dR}^1(T)=\mathbb{C}^2$. Prove that $H^{1,0}_\bar{\partial}(T)=\mathbb{C}$. I have no idea what to do. Can someone ...
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54 views

Elliptic Operators and Continuity

I am reading a book on Hodge theory (Ref: http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/hodge-smf.pdf) or for english ...
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Relation between algebraic hyper de Rham cohomology and hodge theory in positive characteristic

I have recently been looking at algebraic de Rham cohomology of curves in positive characteristic. In particular, I am looking at when the sequence $$0 \rightarrow H^0(X,\Omega_X) \rightarrow ...
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Relation of Hodge Theorem to Eigenfunction Basis of Laplacian

The classical Hodge theorem I know of relates the de Rham cohomology groups isomorphically to the space of harmonic forms and shows that $Id=\pi+\Delta G$, where $\pi$ is the harmonic projection of ...
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Application of Hodge decomposition

Hodge decomposition states any $p$ form can be decomposed into three orthogonal $L^2$ components: exact form, co-exact form and hamonic form. But actually we don't know how to decompose a general one. ...
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Various Proofs for Hodge Decomposition Theorem

This is the version I am referring it to: $H^{k}_{DR}(X,\mathbb{C})=\bigoplus _{p+q=k}H^{p,q}_{DR}(X)$,where X is a Kahler manifold and $H^{k}_{DR}(X,\mathbb{C})=\dfrac{closed~forms}{exact~forms}$ in ...
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Is a curl-free vector field always a gradient?

I tried to prove this problem using the Helmholtz decomposition theorem, but it seems the two are entirely contradictory--thus leaving me with empty hands. Does anyone know how to proceed? Thanks ...
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$\dim_k H^1(X, \Omega_X)$, $X \subset \mathbb{P}^3$ a projective surface

Is it possible to calculate $\dim_k H^1(X, \Omega_X)$, for $X \subseteq \mathbb{P}_\mathbb{C}^3$ a smooth projective surface of degree $d$, without using Chern classes? I've been trying to do this by ...
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Show that $a \wedge * b = g(a,b) \operatorname{vol}$

$\newcommand{\vol}{\operatorname{vol}}$ Let $\omega$ be a $p$-form on a Riemannian manifold $M^n$ with metric $g$ and let $\vol_{i_1,\ldots,i_n}=\sqrt{\lvert g\rvert} \epsilon_{i_1,\ldots,i_n}$ be a ...
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37 views

How to diagonalise the Laplace-de Rham operator on the 5-forms of AdS5 x S5?

I am considering the manifold $M=AdS_5 \times S^5$. I would like to diagonalize the Laplace-de Rham operator $\Delta=(d+\delta)^2$ on the 5-forms of $M$. This yields complicated differential ...
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How do I compute the harmonic component of a given differential form?

The Hodge decomposition theorem tells us that any $r$-form $\omega$ on a Riemannian manifold $M$ (without boundary and compact) may be uniquely decomposed as $$ \omega = d \gamma_1 + d^\dagger ...
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Can there be non trivial self-dual 5-forms on a 10-dimensional compact orientable manifold without boundary?

I am puzzled about the following. Let $(M,g)$ be a compact, orientable Riemannian manifold without boundary. We define the usual inner product $(,)$ for two r-forms $\alpha,\beta\in\Omega^r(M)$ by ...
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Self-dual and anti-self-dual decomposition

Please take a look at the following: Let $(M,g)$ be a four-dimensional oriented Riemannian manifold. The Hodge star operator $*$ obeys $**=Id$ acting on 2-forms. This allow us to decompose the ...
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How to integrate this differential form on the boundary of the cube

The setup. Assume $u = u_1+iu_2: \mathbb{R}^3 \to \mathbb{C}$ and we have the differential 1-forms $$ \star\xi=-x_2 dx_3 + x_3 dx_2 $$ and $$ u \times du = \sum_{i=1}^3 (u \times \partial_i u) dx_i = ...
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Poincare duality on non closed manifolds

For proving the Poincare duality $H_{\mathrm{dR},p}\cong H_\mathrm{dR}^{m-p}$ one can use the bilinear form $B:H_\mathrm{dR}^p\times H_\mathrm{dR}^{m-p}\to\mathbb{R}$ given by ...
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The Moore-Penrose pseudo-inverse of linear operators

I'm reading a paper trying to convert many pairwise orderings into a global ordering. There is an essay on ams.org that briefly introduces the method, and the paper is titled: Xiaoye Jiang, Lek-Heng ...
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Inner product of De Rham cohomology classes

Is there a well-defined inner product between cohomology classes? In particular, is it possible to extend the Hodge inner product? If I try, I obtain this: $$\int *(\omega + d\lambda)\wedge (\sigma + ...
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Weil operator of elliptic curve

Let $V$ be a $1$-dimensional $\mathbb{C}$-vector space and $\Lambda \subset V$ be an elliptic curve (=lattice). Let $C : V \rightarrow V$ be the multiplication by $i$. Consider the two following ...
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I need help understanding the proof of the Lifting Theorem for $H^1$-functions.

The Lifiting Theorem by Béthuel [1,Lem.4.2] states Theorem Assume $\mathbb{T}^3 \simeq \Omega=[-\pi n, \pi n]$ is the 3-dimensional torus obtained by identifying opposing edges. Assume further $v ...
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what to replace Hodge theory with if I don't know what complex numbers are

Q1: How do I prove $h^i(X,\Omega^j_X)=h^j(X,\Omega^i_X)$ for say $X$ smooth geometrically integral projective variety over some field of characteristic zero (I have a strong suspicion this should be ...
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215 views

Why is Griffiths Transversality part of the definition of a variation of Hodge structures?

If $X \to S$ is a family of compact Kahler manifolds, then parallel transport with respect to the Gauss-Manin connection on the relative cohomology bundle does not respect the Hodge filtration, e.g. a ...
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Trouble understanding differential forms. A basic question: what does $w \times dw$ mean?

After reading [1] and [2] I (kind of) understand what differential forms are, but I am still having trouble understanding the following argument from [3,Lem.4.2]: Let $\mathbb{T^3_n}$ be the ...
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Prerequisites for studying Hodge theory and the Hodge conjecture

To what branch of mathematics does the Hodge conjecture belong? I'm aware that it's very advanced, but what kind of prerequisites would one need to understand those problems? Can you suggest some good ...
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Curvature estimates in Hodge theory

I was wondering what is the importance of estimating the curvature of the Hodge Bundles in Hodge theory. I mean is there any real advantage of doing that? And what about curvature of periods domain? ...
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A vanishing theorem for differential forms.

I am trying to prove that for an algebraic surface $X$ (under some extra assumptions that are probably not important) there the space $H^0(X,\Omega_X^1)$ is trivial, i.e. that there exist no globally ...
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61 views

Hodge decomposition on a manifold with a nontrivial connection

I am familiar with the notion of Hodge decomposition of an arbitrary differential form into an exact form, a co-exact form, and a harmonic form. Given a curved space with a connection, could you ...
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About twistor space of a K3 surface

I know that for $X=(M,I)$ , where $I$ is the complex structure, a K3 surfaces and $\alpha \in H^2(X,\mathbb{R})$ a Kähler class, there exist a Kähler metric g and J,K complex structures such that 1) ...
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Algebraic classes in Hodge decomposition.

Let $X$ be a Kaehler manifold. The torsion free part of the singular cohomology $H^n(X,\mathbb{C})$ has a Hodge decomposition $$ H^n(X,\mathbb{C})=\bigoplus_{p+q=n}H^{p,q}(X), $$ where $H^{p,q}(X)$ ...
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Prerequisites for understanding the Hodge conjecture

The Hodge conjecture is a major open mathematical problem that states that on a complex manifold $X$ and its respective Hodge classes, defined as $Hdg^k(X)= H^{2k}(X,\mathbb{Q})\cap H^{k,k}(X)$ that ...
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Can a Non-Compact Manifold have Infinite Dimensional Cohomology?

For compact manifolds, Hodge Theory tells us that (de Rham) cohomology is finite dimensional. What about non-compact manifolds? That is: Can non-compact manifolds have infinite dimensional ...
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Sum of operator and adjoint is self-adjoint

In abstract Hodge theory there is the following lemma: Let $H$ be a Hilbert space and $A \in \mathcal{C}(H)$ a densely defined, closed operator (so possibly unbounded) and $A^*$ its adjoint operator. ...
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Explicit computation of the Hodge codifferential

Question I'm given a Laplacian $\Delta_n=-4y^2 \cdot \frac{\partial^2}{\partial\bar{z} \partial z} + 4 iny \cdot \frac{\partial}{\partial\bar{z}}$, and I want it to be the Laplace operator associated ...
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Why is Hodge more difficult than Tate?

There are strong connections between the Hodge and the Tate conjectures, mainly at the level of similarities and analogies. To quote from an answer of Matthew Emerton on MathOverflow: "[...] we ...
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Reference request for “Hodge Theorem”

I have been told about a theorem (it was called Hodge Theorem), which states the following isomorphism: $H^q(X, E) \simeq Ker(\Delta^q).$ Where $X$ is a Kähler Manifold, $E$ an Hermitian vector ...
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Laplace-Beltrami Operator for Euclidean Space

Consider the space $\mathbb{R}^n$ and let $x_1,\ldots, x_n$ be the coordinates. Fix the orientation $dx_1\wedge dx_2\ldots\wedge dx_n$. Let $E^p$ denote the space of smooth $p$ forms and let $d$ ...
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English translation or summary of “Relevements modulo $p^2$ et decomposition du complexe de de Rham. ”

I'm looking for either an English translation or summary of the article "Relevements modulo $p^2$ et decomposition du complexe de de Rham." by Deligne. I'm attempting to read this for background ...
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Hodge theory for toric varieties

Say we are given a complex smooth projective toric variety $X$. How can one read off hodge theoretic information from combinatorial data? For example I would like to extract dimensions of the various ...