# Tag Info

## Hot answers tagged hodge-theory

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### 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 ...
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### 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 ...
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### 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 $\varphi_O$-...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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}...
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