Questions tagged [de-rham-cohomology]

This a cohomology theory for smooth manifolds, where the (co)chain complex is defined by differential forms on a smooth manifold with differential given by exterior derivative. Then $n^{th}$ de Rham cohomology group is the quotient "closed $n$-forms/exact $n$-forms". Use in conjunction with other algebraic topology and differential geometry related tags if necessary.

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Mayer-Vietoris for smooth manifold with boundary

Disclaimer: I am aware that this question is entirely moot, since any smooth manifold with boundary is homotopic to its interior! Nonetheless, I am still interested in learning what goes wrong if we ...
ArchimedeezNuts's user avatar
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1 answer
61 views

Compute the induced homomorphism on deRham cohomology

Consider the smooth map on the torus to itself $f: S^1 \times S^1 \rightarrow S^1 \times S^1$ defined by $f(z_1, z_2) = (z_1^2z_2, z_1^{-1}z_2)$ (here we identify $S^1$ as the unit circle on the ...
mathlover314's user avatar
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How to prove the existence of differential forms on a manifold using de Rham cohomology?

Let $S^3$ be the 3-sphere, and $\Sigma$ be a 2-dimension manifold. Let $\omega$ be a 2-form on $\Sigma$. $f:S^3\rightarrow \Sigma$ is a $C^{\infty}$ map.Then there is a 1-form $\alpha$ on $S^3$ such ...
Amemiya's user avatar
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Second De Rham Cohomology of $\mathbb{P}^2_\mathbb{R}$

I'm trying to show that $H^2(\mathbb{P}^2_\mathbb{R})=0$ by pulling back closed 2-forms $\omega$ on $\mathbb{P}^2_\mathbb{R}$ to $S^2$ using the fact that $\pi^*\omega$ is exact (where $\pi:S^2\...
LiminalSpace's user avatar
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De Rham Cohomology of $\mathbb{P}_\mathbb{R}^2$

I'm trying to show that every closed 1-form on $\mathbb{P}_\mathbb{R}^2$ is exact. Towards this end, let $\alpha:S^2\longrightarrow S^2$ be the antipodal map, and $\pi:S^2\longrightarrow \mathbb{P}_\...
LiminalSpace's user avatar
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What is "torsion" in the context of cohomology, and why is it important?

I searched for some answers, but most answers discussed the meaning of torsion, instead of its definition. Not knowing how the torsion is defined (in cohomology) I couldn't understand those answers at ...
Youngsub Yoon's user avatar
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Question about the Euler class

Let $M$ be a compact orientable $n$ dimensional smooth manifold without boundary, and $e\in H^n_{dR}(M)$ denote the Euler class of the tangent bundle $TM$. We have that \begin{align*} \int_Me=\chi(M) \...
Chris's user avatar
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Gauss-Manin connection of a family over a formal disk

Let $X$ be a smooth proper scheme over $\mathbb{C}[[t]]$ and $X_0$ its special fiber. The cohomology $H^*_{dR} (X/\mathbb{C}[[t]])$ admit the Gauss-Manin connection which is determined by the action ...
iou's user avatar
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Contravariant functor $\Omega^*$ from category of smooth manifolds to commutative differential graded algebras

I thought I understood the Mayer-Vietoris Sequence in the context of De Rham cohomology, but I have realized that I have taken something for granted in the set up of the proof, and I can't quite ...
Chris's user avatar
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Big confusion about cohomology of complex manifold

First, I never learned complex manifold properly (I am more familiar with complex variety) hence the questions will be very elementary. I apologise for this. Let $X$ be a complex smooth manifold of ...
Molang's user avatar
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Do we still use vector potential in Maxwell's equations?

In Maxwell's equations, since field $\mathbf{B}$ is divergence free, we can find vector potential $\mathbf{A}$ such that $\mathbf{B}=\nabla\times\mathbf{A}$. However, this identity holds only when ...
Eric's user avatar
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Understanding the calculation of Rham’s first cohomology group of the puntured plane

I have a question: How is the first Rham cohomology group computed? I revised the answer in 2, but I don't understand the proof. Let M be the puntured plane and $\alpha \in \Omega^1(M)$ be a closed ...
Samuel QG's user avatar
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First cohomology group of Kähler-Einstein manifolds [closed]

I have read that the first cohomology group of a Kahler-Einstein manifold of dimension four, with positive scalar curvature, is zero. But I cannot find a proof. How can I prove it? Where can I find a ...
Oxido's user avatar
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de Rham cohomology of dual abelian variety

Suppose $X$ is an abelian variety over a field $k$. Consider the de Rham cohomology of $X^{\vee}$ (denoted by $H_{dR}^i(X^{\vee})$). I want to ask is $H_{dR}^i(X^{\vee})$ naturally dual to $H_{dR}^i(X)...
Richard's user avatar
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Cohomology of mixed degree differential forms

$\newcommand{\d}{\mathrm{d}}\newcommand{\H}{\mathrm{H}}$Consider the space, $\Omega^n(M)$, of differential $n$-forms on a smooth, torsion-free manifold, $M$, without boundary, of dimension $D$, and ...
ɪdɪət strəʊlə's user avatar
1 vote
1 answer
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A nonzero integral of a closed differential form over the torus implies it is not exact

Let $\omega$ be a closed differential form over the torus $T$. Suppose $$\int_T \omega \neq 0.$$ Why does it follow that $\omega$ represents a nonzero cohomology class on $T$? I assume this somehow ...
CBBAM's user avatar
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Homotopy equivalence and top deRham cohomology group of closed disk

On one hand, a closed disk $D^2$ is contractible so that it has the same cohomology as a point, so $H^2(D^2)=0$. On the other hand, $D^2$ is a compact orientable manifold (isn't it?). According to Top ...
jizhou's user avatar
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Computing a generator for $H^1(S^1)$

In Section 26.2 of Tu's Introduction to Manifolds he computes the cohomology of the circle. I understand the proof of why $H^1(S^1) \cong \mathbb{R}$, but I am having some difficulty understanding his ...
CBBAM's user avatar
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Proof of Zig-Zag Lemma

In Tu's book on manifolds he gives the Zig-Zag Lemma as: A short exact sequence of cochain complexes $$0 \to \mathcal{A} \xrightarrow{i} \mathcal{B} \xrightarrow{j} \mathcal{C} \to 0$$ gives rise to ...
CBBAM's user avatar
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Questions about Gauss Manin connection.

Suppose $X,Y$ are two algebraic varieties over $\mathbb C$, and $f:X\to Y$ is a homomorphism. Then we can consider the relative de Rham complex $\mathcal H^i(X/Y):=\mathbf R^if_*(\Omega^{\bullet}_{X/\...
Richard's user avatar
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Counterexamples to "irrotational implies potential"

I'm trying to figure out some counterexamples of the following: If $F$ is an irrotational vector field then there exists some $U$ such that $F = \text{grad}(U)$ (or more generally, using the language ...
groupoid's user avatar
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On a certain "anticurl" operator

I recently found myself curious what explicit formula I would get if I traced through the de Rham cohomology proof that if $\mathbf{F}$ is a vector field defined on all of $\mathbb{R}^3$ which has ...
Daniel Schepler's user avatar
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Is there a smooth global section from projective space to the sphere?

I'm having a hard time trying to find (if there is any) global section from $\mathbb{RP}^n$ to the sphere $\mathbb{S}^n$ or $\mathbb{R}^{n+1}$ (global sections of the the natural projection map $\pi$ ...
Jp_'s user avatar
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1 answer
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Computing the Euler Characteristic of $\mathbb{CP}^2 \# \mathbb{CP}^2$

I'm very new at algebraic topology. I'm trying to compute the Euler characteristic of $\mathbb{CP}^2 \# \mathbb{CP}^2$ by using the following known facts. ($\mathbb{CP}^2 \# \mathbb{CP}^2$ represents ...
HeroZhang001's user avatar
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Is it possible to study de Rham Cohomology of surfaces through Riemann surfaces?

I am studying de Rham Cohomology of surfaces. I am following the book "An Introduction to Manifolds" by Loring Tu. I found that it uses the "Mayer Veitoris Sequence" and notions ...
user1138886's user avatar
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What is the product on Kunneth formula for deRham cohomology?

The Künneth theorem states that for every k>0, $$ H_{dR}^k(M \times N) = \bigoplus_{i+j=k} H_{dR}^i (M) \otimes H_{dR}^j (N).$$ I would just like to know: what is this product inside the sum? Is it ...
Jp_'s user avatar
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Hochschild cohomology and de Rham cohomology

I know that given a real finite dimensional associative algebra $A$, its automorphism group $\text{Aut}(A)$ is a Lie group. Is there some relation between the de Rham cohomology of this Lie group and ...
Kandinskij's user avatar
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2 votes
1 answer
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About naturality of Godbillon-vey class [closed]

This is a problem from Lawrence Conlon's differential manifolds a first course. I do not know how to prove in the following problem If $f: N \rightarrow M$ is transverse to $\mathcal{F}$, prove that $$...
user473085's user avatar
2 votes
1 answer
109 views

Generalised Mayer-Vietoris long exact sequence

In chapter 8 of Bott/Tu, the authors generalise the standard Mayer-Vietoris sequence to the setting of a countable open cover of $X$. Let's fix a countable cover $\{ U_i\}$ of $X$. According to Prop 8....
ArchimedeezNuts's user avatar
4 votes
2 answers
193 views

Can we make homology from interior derivative (interior product)?

I have a question that I have been curious about for years. In differential geometry, since the exterior derivative satisfies property $d^2=0$, we can make a de Rham cohomology from it. Then if we ...
Taewan Kim's user avatar
1 vote
0 answers
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Hodge inner product of the compact support de Rham cohomologies

Let us consider de Rham chain complex $\Omega^{*}_c(\mathbb{R}^n)$ with compact support. There is a well-defined inner product on such a complex - the Hodge inner product, i.e. \begin{equation} \int_{\...
DGeometry's user avatar
3 votes
0 answers
114 views

Cohomology of submanifolds

Suppose I have a manifold $M$ and a submanifold or a boundary $N\subset M$. By the natural inclusion $\iota:N\hookrightarrow M$ we can easily see that $$\omega\in\mathrm{H}^k(M) \quad\implies\quad \...
brick's user avatar
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2 votes
1 answer
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Correspondence between $H^1_{dR}$ and harmonic 1-forms

Let $S$ be a compact oriented Riemannian surface. And let $\mathcal{H}^1(S)$ be the space of the harmonic $1$-forms of $S$. I’m trying to prove that there is a linear isomorphism: $$H^1_{dR}(S)\to \...
Kandinskij's user avatar
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5 votes
1 answer
81 views

Is the cohomology of a pushout of open sets a pullback of cohomologies?

I recently found an interesting question, and I'd like to ask the co-question. I will differ from what is linked by using de Rham cohomology, since its the easiest way for me to articulate my ...
ArchimedeezNuts's user avatar
3 votes
2 answers
351 views

Computing the De Rham Cohomology Group of Connected Sum $H_{\rm dR}^p(M_1 \# M_2)$

I'm reading Lee's Introduction to Smooth Manifolds. I have a question about problem 17-7. Problem 17-7 Let $M_1$, $M_2$ be connected smooth manifolds of dimension $n\geq3$, and let $M_1\# M_2$ denote ...
HeroZhang001's user avatar
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De Rham cohomology of $\mathbb{CP}^n$ by Mayer-Vietoris sequences?

I wonder if one can compute the De Rham cohomology of $\mathbb{CP}^n$ only via Mayer-Vietoris sequences, here I mean we make use only of the usual open covering of $\mathbb{CP}^n,$ i.e., $U_i = \{[z_0:...
Pedro's user avatar
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1 answer
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De Rham cohomology of connected components

I am studying a bit about De Rham cohomology and, in the reference I am using, it says that is clear that: If M is a compact, orientable and differenciable manifold of dimension $n \geq 1$ with ...
Anyway142's user avatar
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2 votes
1 answer
188 views

Calculate the de Rham cohomology of $S^1 \times S^2$

For our homework sheet it says: calculate the de Rham cohomology of $M = S^1 \times S^2$ or check a book if you are unsure. And then it continues. Since I didn't find the calculation in a book I ...
Paul Joh's user avatar
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Top de rham degree of the total space of a vector bundle

By homotopy invariance of the de Rham cohomology and the fact that the total space of every smooth vector bundle $E\to M$ is homotopy equivalent to its base space $M$ ($M$ is a smooth manifold), we ...
Flavius Aetius's user avatar
2 votes
1 answer
128 views

interior product and cohomology

We know that we can define the de-Rham cohomology because $d^2=0$. However, we also have the property $\iota_X \iota_X=0$ for interior product, so I am wondering can we also define a cohomology for $\...
Light man's user avatar
1 vote
0 answers
39 views

Does an equivariant map (/deformation retract) which induces an isomorphism in cohomology also induce an isomorphism in equivariant cohomology?

Let $G$ be a compact Lie group acting smoothly on two manifolds $M$ and $N$ and suppose we have an equivariant map $f: M \rightarrow N$ which induces an isomorphism in cohomology $f^*: H^*(N) \...
rosecabbage's user avatar
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2 votes
1 answer
83 views

Generator of $H^1(S^1)$ via integration of a bump $1$-form on $S^1$

I have a statement in Bott & Tu's Differential form in Algebraic topology (p.36) that I can't understand. We say previously that a generator of $H^1(S^1)$ is a bump $1$-form on $S^1$ which gives ...
one potato two potato's user avatar
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Is there a manifold $M\subset\mathbb{R}^3$ s.t. $H^k(M)=\mathbb{R}$ for $k=1,2,3$? (deRham cohomology groups)

I was thinking the following: First, note that $\mathbb{S}^1\subset\mathbb{R}^2$ and we have the deRham cohomology groups $H^{k}(\mathbb{S}^1)=\mathbb{R}$ for $0\leq k<2$ and $0$ otherwise, but $\...
Jacaré's user avatar
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For a geom. integral curve $X/k$ of char $p$, do the constants $\mathrm{ker}\left[K(X) \to \Omega^1_{X/k}\otimes K(X)\right]$ equal $K(X)^{(p)}$?

The motivations for this question are related to Constants of universal derivation on an $R$-algebra $A$ under localization ($d: S^{-1}A \to S^{-1}\Omega_{A/R}$) Let $X$ be a geometrically integral ...
Somatic Custard's user avatar
6 votes
1 answer
136 views

Is there a topological point of view for Liouville's theorem about elliptic functions?

Good evening! One of Liouville's theorems about elliptic functions states that there is no such function that has only one pole of order 1. This result is very well known and easily proven using the ...
Nuke_Gunray's user avatar
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Exterior Algebra with Jets instead of Forms?

For a smooth, $n$-dimensional manifold $M$ over $\mathbb{R}$, I would like to know if sections of the dual space of the order-$k$ jet bundle sit within a commutative differential graded algebra ...
richokicked800goals's user avatar
4 votes
0 answers
154 views

Different comparisons between de Rham and singular cohomology

Let $X$ be a smooth manifold. I am aware of two comparison isomorphisms $H^*_{dR}(X,\mathbb{R}) \rightarrow H^*_{sing}(X,\mathbb{R})$ between de Rham cohomology and singular cohomology (with real ...
654897419's user avatar
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1 vote
1 answer
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Poincare duality for algebraic de Rham cohomology with integrable connection coefficients

I am reading "The Gauss-Manin Connection and Tannaka Duality" (here is the link to the paper). I am specifically interested in the proof of Proposition 2.2. In this proof, the authors use ...
Khainq's user avatar
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Confusion about the following diagram

I was reading the book here, and in 2.1.11, there is a diagram which involves an arrow looking like this: $$ H^0_{\mathrm{dR}}(M) \hookrightarrow \Omega^0(M)/\Omega^0_{\text{cl}}(M)_{\mathbb{Z}} $$ ...
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"Pushforward" of de Rham cohomology along "normal" surjective local diffeomorphisms?

Notation: for $X$ a smooth manifold, let $T^n(X)$ denote the space of (global, smooth) covariant rank-$n$ tensors on $X$, i.e. sections of $(T^*X)^{\otimes n}$. Let $X\xrightarrow{Q}Y$ be a smooth, ...
I.A.S. Tambe's user avatar
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