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.

Given a smooth $n$-dimensional manifold $M$, there is a cochain complex

$$0 \to \Omega^0(M) \xrightarrow{d} \Omega^1(M) \xrightarrow{d} \Omega^2(M) \xrightarrow{d} \dots \xrightarrow{d} \Omega^{n-1}(M) \xrightarrow{d} \Omega^n(M) \to 0$$

of differential forms with exterior derivative as the differential, called the de Rham complex (named after Georges de Rham). The cohomology of this complex is called de Rham cohomology: $$H^k_{\text{dR}}(M) = H^k(\Omega^{\bullet}(M), d)$$

These quotient abelian groups (in fact, real vector spaces) measures the extent to which closed $k$-forms to be exact. As a consequence of Hodge theorem if $M$ is compact, $H^k_{\text{dR}}(M)$ is a finite-dimensional vector space for every $k$. Also, by Poincaré lemma, every closed differential form is locally exact and therefore contractible spaces have trivial de Rham cohomology.

By Stokes' theorem, integration of differential forms along singular chains induces, for any compact smooth manifold $M,$ a bilinear pairing

$$H^k_{\text{Sing}}(M)\times H^k_{\text{dR}}(M)\to\mathbb{R}$$

de Rham's theorem asserts that this pairing induces an isomorphism between singular cohomology with real coefficients and de Rham cohomology by showing each vector space in above pairing is dual to one another. Moreover, it coincides with the Čech cohomology.

318 questions
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Intuitive Approach to de Rham Cohomology

The intuition behind homology may be summarized in a sentence: to find objects without boundary which are not the boundary of an object. This has geometric meaning and explains the algebraic boundary operator $\partial$ - quotient of vector spaces…
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Vanishing differential forms in cohomology

Let $X$ be a smooth differentiable manifold. Consider on $X$ a closed $p$-form $\eta$ and a closed $q$-form $\omega$, which have associated cohomology classes $[\eta] \in H^p(X)$ and $[\omega] \in H^q(X)$. Now assume their wedge product is zero in…
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Can $\pi$ be defined in a p-adic context?

I am not at all an expert in p-adic analysis, but I was wondering if there is any sensible (or even generally accepted) way to define the number $\pi$ in $\mathbb Q_p$ or $\mathbb C_p$. I think that circles, therefore also angles, are problematic in…
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Is there a name for this constant associated to smooth maps between spheres? (not degree)

Consider the following constant associated to smooth maps $F: S^{2n-1} \to S^n$ for $n \geq 2$: Let $\omega \in \Omega^n(S^n)$ be a volume form with $\int_{S^n} \omega = 1$. Then there exists $\eta \in \Omega^{n-1}\left( S^{2n-1} \right)$ with…
15
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Homotopy invariance of de Rham cohomology

Let $M,N$ be smooth manifolds which are homotopy equivalent i.e., there exists smooth maps $F:M\rightarrow N$ and $G:N\rightarrow M$ such that $F\circ G$ is homotopic to identity map on $N$ and $G\circ F$ is homotopic to identity map on $M$. Then,…
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Complex Analysis with differential forms

I'm studying a little of Complex Analysis and I have seen that I can use the integrals of complex functions as integrals of differential forms in $\mathbb{R}^n.$ For example: Cauchy Theorem for complex analytic functions is a consequence of the fact…
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Cup and wedge product in singular and de Rham cohomology

De Rham's theorem asserts that the map $I: H_{dR}^p(M) \to H_{sing}^p(M, \mathbb{R})$ defined as $$I(\omega)= [\sigma^p] \mapsto\int_{\sigma^p}\omega $$ is an isomorphism ($\sigma^p \in [\sigma^p] $ is a smooth representative). On $H_{sing}^\ast(M,…
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show that $\omega$ is exact if and only if the integral of $\omega$ over every $p$-cycle is $0$

Let $M$ be an oriented smooth manifold and $\omega$ a closed $p$-form on $M$. Show that $\omega$ is exact if and only if the integral of $\omega$ over every $p$-cycle is $0$. In particular, how to prove that if the integral of $\omega$ over every…
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Square differential forms in cohomology

Let $X$ be a differentiable manifold (connected, compact, orientiable) of dimension $4n$. Consider on $X$ a closed $2n$-form $\omega$, with associated cohomology class $[\omega] \in H^{2n}(X,\mathbb{R})$. The integral of its square is some real…
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Normalization of the generator of third cohomology of a compact Lie group

It is proven in "Loop groups" by Pressley and Segal (Prop. 4.4.5, p. 49) that the left invariant 3-form $\sigma$ on a simply-connected compact Lie group $G$ whose value at the identity is given by $$ \sigma(\xi, \eta, \zeta) = \langle [\xi, \eta],…
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Poincaré duality for de Rham cohomology on non-compact manifolds

Let $M$ be an $n$-dimensional orientable non-compact manifold. Is there an isomorphism as follows, and if so how can we construct it? (Or can you provide a reference?) $$ H^{n-i}_{\operatorname{dR},c}(M, \mathbb R) \cong H_i(M,\mathbb R). $$ On…
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First De Rham cohomology group of Cantor ternary

I proved that $\mathcal H^1 (\mathbb R^2)=\mathbb 0$ and if $\{ P_1,...,P_n\ |P_i\neq P_j,\ i\neq j \}$ is an $\mathbb R^2$ subset, $\mathcal H^1 (\mathbb R^2\setminus\{P_1,...,P_n\})=\mathbb R^n $. Thus I proved this fact: $\mathcal H^1 (\mathbb…
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Why is every top form on a non-compact manifold exact? (without invoking Poincare duality)

Let $M$ be a connected, orientable $n$-manifold without boundary. A well known fact is that the top cohomology $H^n(M, \mathbb{R})$ vanishes if and only if $M$ is not compact, but I have not been able to find an intuitive proof. One direction is…
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De Rham cohomology of punctured manifold

I'm trying to solve the following problem (Lee's Intro to Smooth Manifolds, 17-6): Let $M$ be a connected smooth manifold of dimension $n \geq 3$. For any $x \in M$ and $0 \leq p \leq n-2$, prove that the map $H^p_{dR}(M) \to H^p_{dR}(M \setminus…
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Hodge star duality and the metric

Let $X$ be a smooth compact Riemannian manifold of even dimension $2n$. Using the Hodge star $*: \Omega^r(X) \to \Omega^{2n-r}(X)$ one can define self-dual and anti-self-dual $n$-forms on $X$, $$ \omega = * \omega \,, \quad \eta = - * \eta \,. $$ A…
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