All Questions
Tagged with cohomology hodge-theory
30
questions
2
votes
0
answers
132
views
Hodge numbers of a complement
Let $Y\subset X$ be an analytic subvariety of codimension $d$ of a smooth compact complex variety $X$. Denote $U = X\setminus Y$. The relative cohomology exact sequence implies that
$$
H^i(X) \to H^i(...
11
votes
1
answer
1k
views
Hodge conjecture as the equality of arithmetic and algebraic weights of motivic L-functions
Recently I became aware of the following statement given on page 13 of this paper. First, let us recall the following definitions:
Definition 4.1. Suppose $L(s)$ is an analytic $L$-function with ...
2
votes
1
answer
276
views
Geometric Interpretation of absolute Hodge cohomology
$\quad$Let $\mathcal{Sch}/\mathbb C$ denote the category of schemes over $\mathbb C$. For an arbitrary $X\in\mathcal{Ob}(\mathcal{Sch}/\mathbb C)$, Deligne in his Article defined a polarizable Hodge ...
1
vote
0
answers
77
views
Optimality condition of the harmonic form representatives of a homology class
In "Hodge theory on metric spaces, Smale et al." the $d$-th harmonic forms of the Hodge Laplacian $\Delta_d=\delta^* \delta+\delta \delta^*$ satisfying $\Delta_d(f)=0$ are claimed to be ...
3
votes
2
answers
381
views
Generalization of the Leray-Hirsch theorem
We know the classical Leray-Hirsch theorem for fibrations. My question is, whether a similar statement also holds for flat, proper morphism? In particular, consider a faithfully flat, proper morphism $...
2
votes
1
answer
278
views
Middle cohomology of very general hyperplane sections
Let $X$ be a smooth, projective variety over $\mathbb{C}$ of dimension $n$ satisfying the property that for every $i \ge 0$, $H^{i,i}(X,\mathbb{C}) \cap H^{2i}(X,\mathbb{Q})=\mathbb{Q}c_1(\mathcal{O}...
3
votes
0
answers
165
views
Hodge-theoretic criterion for smoothness
Let $k$ be an algebraically closed field of any characteristic. Is it possible to give an equivalent condition for a $k$-variety to be smooth using only the cohomology of the variety (and whatever ...
5
votes
2
answers
808
views
Mixed Hodge Polynomial for Algebraic Stacks
Let $X$ be a complex algebraic variety. The numerical invariants associated with the Mixed Hodge Structure of $X$ can be encoded in a polynomial in three variables called the mixed Hodge polynomial $H(...
56
votes
2
answers
10k
views
What is prismatic cohomology?
Prismatic cohomology is a new theory developed by Bhatt and Scholze; see, for instance, these course notes. For the sake of the community, it would be great if the following question is discussed in ...
26
votes
2
answers
3k
views
Hodge theory (after Deligne)
In an interview with Deligne on the Simons Foundation website, I heard Robert MacPherson say that at the time Deligne's papers on Hodge theory were being published, the results seemed absolutely ...
12
votes
0
answers
593
views
Refinement of Hodge conjecture
This question deals with the classic Hodge conjecture on projective non-singular complex varieties, or in other words, projective Kähler manifolds. In Deligne's writeup for the Clay Foundation he says ...
4
votes
0
answers
413
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Complex manifolds with the same cohomology
Is there an example of two non-homeomorphic projective smooth complex varieties $X$ and $Y$ such that there exists an isomorphism $H^{\ast}(X,\mathbb{C})\rightarrow H^{\ast}(Y,\mathbb{C})$ of ...
5
votes
1
answer
442
views
Hodge decomposition for Bott-Chern cohomology
$\DeclareMathOperator{im}{im}$
I want to prove that Bott-Chern cohomology group $H^{p,q}_{BC}=\frac{\ker\partial\cap \ker\bar{\partial}}{\im\partial\bar{\partial}}$ has finite dimension via Hodge ...
3
votes
1
answer
325
views
Hypersurfaces without variable cohomology
Let $X$ be a smooth projective variety over $\mathbb C$ of dimension $n+1$. If $Y$ is a smooth very ample hypersurface, we know that except $H^n(X;\mathbb Q)\rightarrow H^n(Y;\mathbb Q)$, the ...
6
votes
2
answers
410
views
Hodge map and the Cohomology Ring of a Riemannian Manifold
For a compact Riemannian manifold $M$, we know that the Hodge map $\ast$ and Laplacian $\Delta$ commute. From Hodge decomposition and its implied isomorphism between harmonic forms and cohomology ...