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38 votes
4 answers
8k views

Relative De Rham cohomologies

as far as I know, there are two main ways to have a relative version of De Rham Cohomology for a pair (M,N), where M and N are smooth manifolds and N is a closed (as a topological subspace) ...
Taladris's user avatar
  • 830
17 votes
1 answer
1k views

Direct proof that Chern-Weil theory yields integral classes

Suppose $E$ is a complex vector bundle of rank $n$ on a compact oriented manifold (both assumed smooth). Let $h$ be a Hermitian metric on $E$, and let $A$ be a Hermitian connection on $E$ and $F_A$ ...
Mohan Swaminathan's user avatar
13 votes
1 answer
725 views

Counterexample showing that G-invariant de Rham cohomology different from cohomology of G-invariant sub-complex?

If $G$ is a discrete or a Lie Group acting smoothly on a manifold $M$, we can define the algebra of $G$-invariant de Rham classes, $H(M)^G$, and we can also consider the cohomology of the sub-complex ...
ychemama's user avatar
  • 1,346
13 votes
1 answer
2k views

de rham model for relative cohomology

In GTM82, I read a model for the relative cohomology of (M,N) with N a submanifold of M. And in the page: Relative De Rham cohomologies, I got to know that there is another model for relative ...
Ryan Du's user avatar
  • 303
9 votes
2 answers
1k views

Hodge dual of de Rham cohomology and singular cohomology

We know that the de Rham cohomology is isomorphic to the singular cohomology, does the Hodge dual of differential forms induce a dual operation on de Rham cohomology, hence also on singular cohomology?...
wonderich's user avatar
  • 10.5k
9 votes
0 answers
640 views

Does Stokes theorem have anything to do with adjoint functors?

I notice some similarity between Stokes theorem in differential geometry and the definition of adjoint functors: in both cases, there is a 2-placed function (the $\operatorname{hom}$ functor, or the ...
Benaya's user avatar
  • 91
9 votes
0 answers
347 views

Is there a Hodge isomorphism theorem for part-tangential, part-normal, harmonic differential forms?

Let $M$ be an oriented compact Riemannian $n$-manifold with boundary $\partial M$. A differential $p$-form $\omega$ on $M$ is normal if $i^* \omega = 0$ holds, tangential if $i^* \star \omega = 0$ ...
Enok's user avatar
  • 91
7 votes
2 answers
429 views

Can one make sense of de Rham cohomology for the complement of a (dense) irrational flow on the torus?

Recent work has led me to consider whether one could define consider the complement of a dense irrational flow on the torus $P_\alpha \subset T^2$ as some kind of generalized smooth space, and ...
xir's user avatar
  • 2,044
6 votes
1 answer
2k views

Integration currents vs Poincaré dual

Let $M$ be a complex manifold of dimension $n$ and $S \subset M$ a closed complex submanifold of complex codimension $r$. Let $[S] \in H_{2r}(S)$ be the fundamental class of $S$. We have the ...
Student85's user avatar
  • 151
6 votes
1 answer
374 views

De Rham cohomology of Lie groupoid

Let $G$ be a Lie group acting on a manifold $M$. Consider the transformation groupoid $\mathcal{G}=(G\times M\rightrightarrows M)$. We have the notion of de Rham cohomology of a Lie groupoid by ...
Praphulla Koushik's user avatar
6 votes
0 answers
156 views

Geometric theory for cohomology groups $H^p(M;\mathbb{Z})$

An excerpt from the book Loop Spaces, Characteristic Classes and Geometric Quantization by Jean-Luc Brylinski is mentioned below: Characteristic classes are certain cohomology classes associated ...
Praphulla Koushik's user avatar
5 votes
2 answers
361 views

Exterior differentiation of foliations

Let $M$ be a differentiable manifold. Let $T^*M$ be the cotangent bundle of $M$. Consider the exterior differentiation $d: A^p(M)\longrightarrow A^{p+1}(M)$, where $A^p(M)=\Gamma(\...
Shiquan Ren's user avatar
5 votes
2 answers
2k views

Global Definition of the Dolbeault Complex of a Vector Bundle

For an $2n$-dimensional complex manifold $M$, and a smooth vector bundle $E$ over $M$, it is well-known (see Voisin, Huybrechts) that there exists an operator $\overline{\partial}$, built locally from ...
Jean Delinez's user avatar
  • 3,399
5 votes
1 answer
634 views

Leafwise de Rham cohomology (A true definition of differential forms along leaves)

For a foliated space $(M, \mathcal{F})$, one associate a leafwise de Rham cohomology. This cohomology and trace-class operators on this cohomology and trace interpretations for closed orbits of ...
Ali Taghavi's user avatar
5 votes
0 answers
248 views

Algebraic de Rham cohomology with torus coefficients

Let $X$ be a smooth projective variety over $\mathbb{C}.$ On page 3 in this preprint of Simpson, it is stated that Notice first of all that the algebraic de Rham theory is not going to work well in ...
lzww's user avatar
  • 123
4 votes
1 answer
211 views

Existence non-trivial parallel $p$-form implies non-triviality of $p$-th cohomology group using De Rham cohomology

Cross-post from MSE. Suppose $(M,g)$ be a closed Riemannian manifold. Because every parallel (nontrivial) $p$-form $\omega$ is harmonic so the $p$-th Betti number should be positive i.e. $b_p\geq 1$. ...
C.F.G's user avatar
  • 4,195
4 votes
0 answers
109 views

Generalized de Rham cohomology on product bundle giving specified cohomology

Given a compact, smooth manifold $M$ and a real vector bundle $E \to M$ (in general not flat). There already have been numerous questions about how to equip the space $\bigoplus_k \Gamma(\Lambda^k T^* ...
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
3 votes
1 answer
167 views

Models for computing cohomology of Lie groupoids

Given a Lie groupoid $\mathcal{G}=[\mathcal{G}_1\rightrightarrows \mathcal{G}_0]$, let $\mathcal{G}_\bullet$ be the associated simplicial manifold. Let $\Omega^\bullet(\mathcal{G}_\bullet)$ be the ...
Praphulla Koushik's user avatar
3 votes
0 answers
96 views

L^1 gradient bounds for potentials of weakly closed forms

Context: The Poincaré-lemma is a central statement in differential geometry. It shows that a k-form is closed iff it is exact. A special case is as follows: Let $\omega\in\Omega^k(U)$ with $\omega=\...
MrVolt16's user avatar
2 votes
1 answer
200 views

Vanishing product of a closed and coclosed form on a Riemannian manifold

For a (compact) Riemannian manifold $(M,g)$, can it happen that for a non-zero form $\text{d}^*\omega$, and a smooth function $f$ such that $\text{d}f \neq 0$, we can have $$ \text{d}f \wedge \text{d}^...
Max Schattman's user avatar
2 votes
1 answer
839 views

Sheaf / de Rham cohomology of a stack with values in a complex of abelian sheaves

I am reading Differentiable Stacks and Gerbes to understand about (hyper) cohomology groups of a stack $\mathcal{X}$ with values in a complex $\mathcal{M}$ of abelian sheaves over $\mathcal{X}$. ...
Praphulla Koushik's user avatar
2 votes
0 answers
241 views

Monodromy group action on de Rham cohomology

Let $f : Y \longrightarrow X := \mathbb{P}^1\setminus\{0,1,\infty\}$ be the smooth proper morphism associated to the Legendre family, which is an elliptic fibration of the punctured line, with fibre ...
kindasorta's user avatar
  • 2,907
2 votes
0 answers
130 views

Interpreting the Higher-order Hodge-Laplace Operator

As an operator on functions, one intuitive way to think about the Laplacian seems to be as an operator that returns the average difference between a function's value at a point and the values of its ...
richokicked800goals's user avatar
2 votes
0 answers
152 views

When are automorphisms of the cohomology ring realized by isometries?

Let $(M,g)$ be a closed smooth Riemannian manifold, and denote by $G$ a closed subgroup of its isometry group. By considering the maps $g^*$ induced by elements $g\in G$ in the (de Rham) cohomology $H^...
Renato G. Bettiol's user avatar
2 votes
0 answers
327 views

Suppose that two cohomologous forms agree on every restriction. Do they agree?

Let $\eta$, $\omega$ be two $(1,1)$-forms on $\mathbb{C}^m \times Y$, where $Y$ is a compact Kahler manifold with vanishing first Chern class, i.e., a Calabi-Yau manifold. Suppose that for all $z \in \...
AmorFati's user avatar
  • 1,379
2 votes
0 answers
190 views

What is known about this type of generalisation of de Rham cohomology?

I will describe a certain generalisation of de Rham cohomology; things could be generalised further but I will stick to a concrete example. A $0$-double-form is a function on the complex plane $\...
Josh Kirklin's user avatar
1 vote
1 answer
239 views

Can every De Rham cohomology class be represented by a closed form $\alpha$ with $L_X \alpha=0$

Assume that $M$ is a manifold and $X$ is a vector field on $M$. Is it true to say that every closed form is De Rham-cohomologue to a closed form $\alpha$ with $L_X \alpha =0$?
Ali Taghavi's user avatar
1 vote
1 answer
788 views

deRham cohomology of a manifold with covering space $S^{n}$

(A qual problem) Let $\pi:S^{n}\rightarrow M$ be a covering map, $M$ being an orientable manifold. Show that $H_{deR}^{k}(M)=0$ for $1\leq k < n $. We can show $H_{deR}^{1}(M)=0$ by the following ...
Yunfeng's user avatar
  • 23
1 vote
0 answers
82 views

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
  • 111
0 votes
0 answers
85 views

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