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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Laplace-de Rham operator on $\mathbb{R}^n$

Let $\mathbb{R}^n$ have the standard orientation with volume element $dV = dx_1 \wedge...\wedge dx_n > 0$. Show that $\Delta = - \sum_j \partial^2 / dx_j^2$ on 0-forms on $\mathbb{R}^n$. Where ...
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40 views

Apparently meaningless computation with the Hodge star operator

In the last lecture we started speaking about hodge star operator. Let $E$ be a $n$ dimensional vector space with a non degenerate bilinear form $g$. $\mathcal{O}$ the orientation line of $E$, i.e. ...
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22 views

1-forms as a direct sum

Let us define the spaces $C_0, C_1$ and $C_2$ of differential $0,1,2$ forms respectively on the sphere $S^2.$ Is it true that $C_1$ is the direct sum of $d(C_0) \oplus \delta^* (C_2)$? I think this ...
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102 views

Hodge star isomorphism

In Petersen's Riemannian geometry text, he defines the Hodge operator $*: \Omega^k(M) \to \Omega^{n-k} (M)$ in the standard way. He then proves (Lemma 26, Chap 7) that $*^2: \Omega^k(M) \to ...
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34 views

Basic question: irregularity and geometric genus in Hodge theory

Please could you give me a reference to the theorems that give rise to the following equations which are taken from page 8 and 9 of Friedman' book on holomorphic vector bundles and algebraic surfaces ...
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32 views

Basic question: $H^1$ and $H^{0,1}$

Please could you explain why for a smooth projective variety over $\mathbb{C}$ (or - if you prefer the analytic world - compact complex manifold) $T$ we have $H^1(\mathcal{O}_T)\simeq H^{0,1}(T)$ as ...
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19 views

A set of differential forms, uniformly bounded with their Laplacians, is precompact in $L^2$.

Let $M$ be a compact Riemannian manifold and let $\Delta$ be a Hodge Laplacian on $k$-forms. How to show that the if the set $\{u_\alpha\} \subset C^2(M,\Lambda^k)$ of $C^2$ $k$-forms is uniformly ...
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34 views

unirational complex variety has $H^i(X,O_X) = 0$ for i > 0

Let $X/\mathbf{C}$ be a smooth projective connected unirational variety. Why do we then have $H^i(X,O_X) = 0$ for $i > 0$?
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81 views

Book on advanced Hodge Theory

I'm looking for a book on advanced real Hodge Theory. I finished working through Frank Warner's Foundations of Differentiable Manifolds and Lie Groups, which ends with the Hodge Decomposition,the ...
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96 views

Reference request: Some theorems in an article of Grothendieck.

In "Standard conjectures on algebraic cycles" Grothendieck says: "The first is an existence assertion for algebraic cycles (considerably weaker than the Tate conjectures), and is inspired by and ...
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32 views

Integration of Laplacian on Manifold (Hopf's Lemma ?)

Well, Let $M$ a compact manifold and $dM = \star 1$ the volume element, $f : M \to \mathbb{R}$. Define the inner product of $p-forms$ like: $(\omega,\theta) = \int_M \langle \omega, \theta \rangle dM ...
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56 views

Definition of Hodge structure: is torsion allowed?

I am trying to understand the definition of an integral Hodge structure. Apparently, for $X$ a compact Kahler manifold, $H^n(X,\mathbb R)$, the lattice $H^n(X,\mathbb Z)$ and the Hodge filtration give ...
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57 views

Examples of varieties with torsion in their integral Hodge structure

I am not so used to thinking about integral Hodge structures, so this question might be completely trivial. What are easy and interesting examples of smooth projective connected varieties $X$ with ...
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1answer
86 views

relative sign in Hodge star of tensor product

Let $V$ be a vector space of arbitrary (finite) dimension and let $(V, \langle \ ,\ \rangle, I) = (W_1, \langle\ ,\ \rangle_1, I_1) \oplus (W_2, \langle\ ,\ \rangle_2, I_2)$ be a direct sum ...
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49 views

Vector Laplace Beltrami operator on surface tangent and surface normal vector field

Consider a closed, compact, embedded surface $f:M \rightarrow \mathbb{R}^3$ and a vectorfield $X$ on the surface that can be decomposed in the surface frame basis $\{e_1,e_2,e_3\}$, where ...
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1answer
62 views

Star in Serre duality

Why is there a dual bundle in Serre duality? Let $\mathcal E$ be a vector bundle over complex manifold $X$, without any metric anywhere, then one has a pairing $$(\Omega^{0,q} \otimes \mathcal E) ...
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45 views

Well-definedness of a coboundary map between a reduced $L^2$ de Rham cohomology group and a relative cohomology group

I'm working right now with this paper of Carron. And I think I'm stuck at a relatively simple question. On page 11 he is defining a coboundary map $b : H^k_{2, \text{reduced}}(M - K) \to H^{k+1}(K, ...
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A question on harmonic two-forms

Let $(M^4,g)$ be a closed Riemannian four-manifold with $b_2^+>0$ and $b_2^->0$, is it possible to find two harmonic two-forms $\alpha\in H^2_+(M)$ and $\beta\in H^2_-(M)$, such that ...
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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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71 views

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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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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1answer
133 views

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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220 views

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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107 views

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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1answer
176 views

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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175 views

$\dim_k H^1(X, \Omega_X)$, $X \subset \mathbb{P}^3$ 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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103 views

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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36 views

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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100 views

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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94 views

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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85 views

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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111 views

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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320 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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1answer
138 views

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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385 views

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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146 views

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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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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1answer
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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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259 views

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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377 views

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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206 views

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 ...