18 votes
Accepted

Sign of codifferential

Okay. I've done it. I'll post it here. I think it may be of help to others. The whole purpose of the codifferential is to be the adjoint of the exterior derivative with respect to the Hodge inner ...
Jackozee Hakkiuz's user avatar
10 votes

Calabi-Yau $3$-fold given as elliptically fibered manifold over $\mathbb{C}P^1 \times \mathbb{C}P^1$

Suppose we have a complex manifold $M$ described by the equation$$F(z_1, z_2, \ldots, z_n) = 0.$$Now, without imposing this equation, the holomorphic form locally is just$$dz_1 \wedge dz_2 \wedge \...
Brian Ng's user avatar
  • 1,422
10 votes
Accepted

$h^{p,q}$ of a complex torus.

Since you use the tag Hodge theory: One can give $\mathbb C^n/\Gamma$ the flat metric from that of $\mathbb C^n$. Thus it suffices to find all harmonic forms. Since the complexified cotangent bundle ...
Arctic Char's user avatar
10 votes

Explicit Hodge decomposition on $T^2$

Let $\sigma_1 = S^1\times \{0\}$ and $\sigma_2 = \{0\}\times S^1$ be the canonical basis for $H_1(T^2)$. As you did, I will use $dx_i$ for the basis $1$-forms on $T^2$ (since these forms on $\Bbb R^2$ ...
Ted Shifrin's user avatar
9 votes
Accepted

Are Complex Elliptic Curves Kähler Manifolds?

Every (complex) smooth projective variety is Kähler. Indeed, $\mathbb{CP}^n$ is Kähler using the Fubini-Study metric, and this metric then restricts to a Kähler metric on any complex submanifold of $\...
Eric Wofsey's user avatar
9 votes
Accepted

Hodge theory: $\Delta \alpha = 0$ iff $d\alpha = d^* \alpha = 0$ on a noncompact manifold?

Consider $M = \mathbb{R}^n$ equipped with the standard metric. For a $0$-form $f$, i.e. a real-valued function, we have $d^*f = 0$ for degree reasons, and $$df = \sum_{i=1}^n\frac{\partial f}{\...
Michael Albanese's user avatar
9 votes

Explicit Hodge decomposition on $T^2$

$\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}$This basically comes down to inverting the Laplacian, which is done by the Green's function. Inverting the Laplacian came up in Ted Shifrin's solution, but I ...
David E Speyer's user avatar
8 votes
Accepted

Does the Lie derivative of a harmonic form with respect to a killing vector field vanish

Here some facts those lead you for a solution: For any p-form $\omega$ follows: $d(L_X\omega)=L_X(d\omega)$. $L_X(*\omega)=(\text{div} X)*\omega+*[(L_Xg)\cdot\omega]+*(L_X\omega)$. (See here) $\text{...
DiegoMath's user avatar
  • 4,069
7 votes
Accepted

The Kähler condition on a Riemann surface

You just follow the usual proof, but it's easier. $h$ is a positive (real analytic?) function and you write $h\,dz\otimes d\bar z = \phi\otimes\bar\phi$. Without loss of generality, we assume $h(0)=1$....
Ted Shifrin's user avatar
6 votes
Accepted

Mixed Hodge structure with examples

See the book here by James Carlson, Stefan Müller-Stach, and Chris Peters, especially the first chapter. Carlson, James; Müller-Stach, Stefan; Peters, Chris. Period mappings and period domains. ...
Simone Weil's user avatar
  • 2,140
6 votes
Accepted

Real Hodge structures and representations of the Deligne torus

Yes, you can leave out the minus signs, but at some point Deligne decided it was better with them in, and most people follow him. Deligne's reasons are a bit obscure, but there's an explanation in ...
anon's user avatar
  • 251
6 votes
Accepted

Standard counterexample in Hodge decomposition

Recall that, in general, $H^{p,0}(X)=H^0(X,\Omega^p_X)$. Now $\Omega_\mathbb C^1$, the co-tangent bundle, is the trivial bundle of rank $1$ on $\mathbb C$, the holomorphic sections of which are just ...
Ben's user avatar
  • 6,117
5 votes
Accepted

Where is the error in this proof of the Hodge theorem?

As Daniel Fischer pointed out in the comments, the gap in your proof is the claim that $\operatorname{ker}(d) = \operatorname{im}(d) \oplus \operatorname{im}(d)^\perp$. But it should be noted that ...
Jack Lee's user avatar
  • 46.9k
5 votes
Accepted

Path to learning Hodge Theory via Voisin's text.

This semester we followed Voisin's book for a reading group on Hodge theory. Here is what you need in order to understand the book : The really important topic is real smooth manifolds. For this, I ...
Nicolas Hemelsoet's user avatar
5 votes

What is a Punctured disk?

It just means a disk minus a point (or the disk minus the point). Most often this means $\{(x,y)\in \Bbb R^2:0<x^2+y^2\le 1\}$, but occasionally people mean the open disk $\{(x,y)\in \Bbb R^2:0<...
pancini's user avatar
  • 19.2k
5 votes
Accepted

Question from $p$-adic HodgeTheory, linear algebra data

I always find this notation also a bit confusing and incomplete. Let me spell out, regarding 1. and 2., the construction of linearization explicitly: $\varphi: M \rightarrow M$ is an $\varphi_O$-...
Pavel Čoupek's user avatar
5 votes
Accepted

Double Hodge star property

I will assume you are in the Riemannian setting, where the metric is positive definite and so on. The notation $\varepsilon^{i_1\dots i_p}{}_{j_{p+1}\dots j_n}$ is kind of weird, because the ...
Jackozee Hakkiuz's user avatar
5 votes
Accepted

Hodge star duality and the metric

The answer is no. To see why, note that every self-dual $n$-form actually satisfies a stronger pointwise condition: If $\omega$ is self-dual with respect to some Riemannian metric $g$ and choice of ...
Jack Lee's user avatar
  • 46.9k
5 votes
Accepted

Analytification of a smooth projective variety is a compact Kähler manifold.

Yes, this boils down to two facts, which you should be able to find in e.g. Huybrecht's Complex Geometry or Voisin's Hodge Theory and Complex Algebraic Geometry: I. $\mathbb{P}^n(\mathbb{C})$ is a ...
Notone's user avatar
  • 2,360
5 votes

Nowhere vanishing harmonic 1-forms on 3-manifolds

This is not a full answer, but I just wanted to add holonomy to what you already found and it became to long for a comment. The Bochner Theorems say, that for a Riemann manifold $(M,g)$ with $Ric \geq ...
Jonas's user avatar
  • 942
5 votes

The definition of special cubic fourfolds

Two surfaces in $X$ are "homologous" if their classes in $H_4(X,\mathbb{Z})$ are equal. Hassett shows (in the paper you are citing) that special cubic fourfolds are parameterized by a ...
Sasha's user avatar
  • 17.2k
5 votes

Compactly supported harmonic forms

It's true, as @ArcticChar proved, that if $\alpha$ is harmonic and compactly supported then $d\alpha = \delta\alpha = 0$. But this is not very useful on a noncompact manifold, because if $\alpha$ is ...
Jack Lee's user avatar
  • 46.9k
4 votes
Accepted

Second Stiefel-Whitney class of self-dual two forms of four manifolds

As usual, the splitting principle comes to the rescue. Below I am going to write direct sums as sums and tensor products as products for readability, which will unfortunately introduce some ambiguity ...
Qiaochu Yuan's user avatar
4 votes
Accepted

Hodge duality in relating wedge and cross products?

You're mostly on the right track, but it looks like you may be confused about the various products that are involved. As Jesse Madnick points out, there are already a lot of assumptions built in to ...
Mike's user avatar
  • 386
4 votes

Some questions about the hodge star operator.

The first excerpt you give talks about the Hodge star for an abstract vector space $V$ which has a metric, i.e. a bilinear function $V\times V\to \mathbb R$. For the second excerpt, you set $V = T_p^*...
Keshav's user avatar
  • 1,620
4 votes
Accepted

Why is the Hodge conjecture equivalent to the assertion that $ \mathcal{R}_{ \mathrm{Hodge} } $ is fully faithfull?

So here is a recap of how the category of motives is constructed : First consider the category $\mathcal{V}_k$ of smooth projective varieties over $k$. Add morphisms to this category by adding all ...
Roland's user avatar
  • 12.5k
4 votes
Accepted

Hodge star operator of $\frac{\omega^k}{k!}$ is $\frac{\omega^{n-k}}{(n-k)!}$

This is just algebraic manipulations. On $\mathbb{C}^n$, the standard Kahler form (with the usual $z_j=x_j+iy_j$ identification of $\mathbb{C}^n\cong\mathbb{R}^{2n}$) is $$\omega=\sum_{j=1}^n \frac{i}...
user10354138's user avatar
  • 33.3k
4 votes
Accepted

Albanese map for complex compact manifold

As I mentioned in the comment, it is proved in Voisin's Hodge Theory and Complex Algebraic Geometry, Volumn I, Lemma 12.11. I will just paraphrase the proof here. Let $k\ge 1$ and denote $X^k=X\times\...
AG learner's user avatar
  • 4,533
4 votes
Accepted

Question Regarding Proof of Hodge Index Theorem

A friend of mine pointed out that my example calculation is incorrect. Indeed, because we have an $n-$dimensional complex manifold, and we are manipulating real forms, we must have $0\le a+b\le 2n$ ...
Alekos Robotis's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible