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Questions tagged [derham-cohomology]

Better spelling "DeRham", not derham... I can't figure out how to change this... moderators? The cohomology of the complex of differential forms on a smooth manifold with differential given by exterior derivative.

12 questions from the last 365 days
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de Rham cohomology relative to a closed subset

I am interested whether there exists a versions of de Rham relative cohomology $H^\bullet(M, N)$ in which $N$ does not need to be a manifold. I know there are two main definitions in literature as ...
Janczar Knurek's user avatar
3 votes
1 answer
203 views

Cohomology of the complex of differential forms with Schwartz coefficients

Let $U$ be an open manifold (say an open subset of $\mathbb{R}^n$ for simplicity). Denote by $\mathscr{S}(U)$ the space of Schwartz functions on $U$. Schwartz functions are defined as usual to be ...
Grisha Taroyan's user avatar
4 votes
1 answer
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de Rham's Theorem using atlases and colimit preservation

$\DeclareMathOperator\DR{DR}\newcommand\SmoothManifold{\mathrm{SmoothManifold}}\newcommand\ChainComplexes{\mathrm{ChainComplexes}}\DeclareMathOperator\Sing{Sing}\newcommand\DifferentialGradedRAlgebras{...
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12 votes
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Is there a differential form which corresponds to an eigenvalue of the homomorphism in cohomology?

Let $M$ be a closed manifold and $f:M\to M$ be a diffeomorphism. Suppose the homomorphism $f^*:H^k(M;\mathbb R)\to H^k(M;\mathbb R)$ has an eigenvalue $\lambda\in\mathbb{R}$. Note that $\lambda$ is ...
Andrey Ryabichev's user avatar
8 votes
0 answers
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Triple comparison of cohomology in algebraic geometry

Let $X$ be a smooth proper variety over $\mathbb{Q}$ and $p$ a prime number. For an integer $k$, we have: a finitely-generated abelian group $H^k(X^{\mathrm{an}}(\mathbb{C});\mathbb{Z})$ a finitely-...
David Corwin's user avatar
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1 vote
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Projection to trivial reduced cohomology class in $L^2(\mathbb{R})$

Given that I have had no success on the mathematics stackexchange (see here), I've decided to try my luck here. I am attempting to solve the following exercise (original formulation here), which to my ...
Pastudent's user avatar
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Stokes' lemma / Yoneda's theorem

Definition: Write SmoothOrientedManifold for the category of smooth oriented real manifolds. Definition: Write ChainComplex for the category of chain complexes, and write $\otimes$ for the tensor ...
Ron Z.'s user avatar
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How does the cohomology theory on de Rham (pre)stack compute de Rham cohomology?

Recently when I'm reading PTVV, Shifted Symplectic Structures, in Sec. 2.1 Mapping stacks, the authors use the identification $H^*(X,E)\simeq H^{*}_{dR}(Y/k,\mathcal{E})$ to show $X=Y_{dR}$ admits an ...
Yining Chen's user avatar
2 votes
1 answer
225 views

Prefactor $2\pi i$ for Tate-Hodge structure

A rather basic question. What was the original reason to consider the underlying $\mathbb{Z}$-module of the - as canonical object regarded - Tate-Hodge structure $\mathbb{Z}(1)$ to be given as $2 \pi ...
user267839's user avatar
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Hodge filtration vs Hodge structure on algebraic de Rham cohomology

I have a basic question on the relation between the definitions of the Hodge structure on the algebraic de Rham of a smooth proper scheme defined over a subfield of $\mathbb{C}$ and the Hodge ...
kindasorta's user avatar
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Nice proof that de Rham complex computes Lie algebra cohomology?

If $G$ is a nice enough group acting on a nice enough space $X$, then the relative de Rham complex $$\Omega^\bullet_{X/(X/G)}\ \simeq\ \mathcal{O}_X\otimes\text{Sym}\,\mathfrak{g}^*[-1]$$ is given by (...
Pulcinella's user avatar
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Existence of covering space with trivial pullback map on $H^1$

I have seen somewhere the following claim (but can't remember where): let $M$ be a connected orientable closed smooth manifold with $b_1(M)=1$, then there exists a connected covering space $p:\tilde{M}...
F. Müller's user avatar