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Vanishing differential forms in cohomology

No, this is not always possible. One can use differential forms to define higher order cohomology operations called Massey products and if they don't vanish, then you have an obstruction for the ...
levap's user avatar
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26 votes
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Can $\pi$ be defined in a p-adic context?

In a very flattering note @saulspatz has asked me to weigh in on this issue. But many of the commenters to your question have already evinced rather greater familiarity with deep $p$-adic matters than ...
Lubin's user avatar
  • 63k
24 votes

So what is Cohomology?

Some very basic answers, with the aim of giving you an idea of the big picture: On the most basic level, you can think of cohomology as a fancy way of counting/classifying holes in an underlying space ...
Ben Grossmann's user avatar
13 votes

Top deRham cohomology group of a compact orientable manifold is 1-dimensional

$\def\RR{\mathbb{R}}$The following is what I think of as the standard argument; I don't know whether it counts as simple. I'll be showing that, for $M$ a connected, oriented $n$-manifold, if $\omega$ ...
David E Speyer's user avatar
13 votes
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Cap product and de Rham cohomology

After a few days of pondering I think I understood the answer: Let $M$ be a $d$-dimensional manifold as above, and let $i: N\hookrightarrow M$ be the embedding of a submanifold of dimension $\ell$. ...
atabler's user avatar
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11 votes
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Cauchy's Theorem, Stokes' Theorem, de Rham Cohomology

There are easy and hard things to see here. Let me see if I can help. Complex valued differential forms First of all, throughout this answer, I'll want to work with differential forms that take ...
David E Speyer's user avatar
11 votes
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How is de Rham cohomology useful?

The one-sentence explanation is that cohomology on a space $X$ answers the question of when you can promote local solutions of a problem to global solutions; that is, if you can solve a problem on ...
anomaly's user avatar
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10 votes
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De Rham cohomology of compact manifold minus one point

The trick is to look at the actual arrows involved. Also for convenience sake, I will assume $N$ is actually $M$ delete an open ball, since they're homotopy equivalent. The $0$, $1$ exact sequence ...
jgon's user avatar
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9 votes
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Is there a name for this constant associated to smooth maps between spheres? (not degree)

As Eric says, this integral equals the Hopf Invariant. This equality was first proven by Whitehead in Whitehead, J. H. C., An expression of Hopf’s invariant as an integral, Proc. Natl. Acad. Sci. USA ...
Moishe Kohan's user avatar
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9 votes
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Square differential forms in cohomology

No, this is not always possible. I'll analyze the case $n = 1$ because it is related to symplectic geometry. Note that the sign of $\omega' \wedge \omega'$ is not well-defined without choosing a ...
levap's user avatar
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8 votes
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Is there a topological point of view for Liouville's theorem about elliptic functions?

If you know that an elliptic function has the same number of zeroes and poles, then there is indeed a simple topological proof. If $f$ is an elliptic function with just one simple pole, then it also ...
Eric Wofsey's user avatar
7 votes
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De Rham cohomology groups of projective real space

The following approach has the benefit that it is elementary in the sense that it only uses basic (homological and linear) algebra and the tools you allude to. More on this later*. I will just expand ...
Aloizio Macedo's user avatar
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7 votes
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Cup and wedge product in singular and de Rham cohomology

The deRham map gives an arrow $\theta:H_{dR}(X) \to H(X)^*$, and the algebra structure on the right is induced from the coalgebra structue on $H(X)$ given by the Alexander-Whitney coproduct: $\Delta(\...
Pedro's user avatar
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7 votes

How does one introduce characteristic classes

I was always confused by characteristic classes until I understood the definition of characteristic classes via the classifying map. Corresponding to a vector bundle with structure group $G$ there ...
Thomas Rot's user avatar
  • 10.1k
7 votes

De Rham cohomology of punctured manifold

For $p=1$, the Mayer-Vietoris sequence gives us the following exact sequence: $$ \cdots \to H_{dR}^0(U) \oplus H_{dR}^0(M \setminus \{x\}) \to H_{dR}^0(\mathbb S^{n-1}) \to H_{dR}^1(M) \to H_{dR}^1(M ...
D Ford's user avatar
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7 votes
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First de Rham cohomology vs fundamental group of an open subset of the Euclidean space

The claim as stated is false. What is true is that if $U \subset \mathbb{R}^3$ is open, then $H_1(U;\mathbb{Z})=0$ if and only if $H^1_{dR}(U)=0$. Basically, the error lies in statement $2.$ So let's ...
Aloizio Macedo's user avatar
  • 34.4k
6 votes

De Rham cohomology of $S^1$

Hint: Define $p: \mathbb R \to S^1$ by $p(t) = e^{it}$. This map induces a pullback homomorphism $p^*: \Omega^1(S^1) \to \Omega^1(\mathbb R)$ so that if $\alpha \in \Omega^1(S^1)$, then $p^*(\alpha) = ...
William Stagner's user avatar
6 votes
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Question Regarding the Coordinate Independent Form of the Exterior Derivative

If you're trying to avoid coordinates, the way to verify that an $\mathbb R$-multilinear expression involving vector fields is "pointwise" (i.e. depends only on the value of the vector fields at the ...
Anthony Carapetis's user avatar
6 votes
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The Converse of Poincare Lemma

No. For instance, $\mathbb{RP}^2$ has trivial de Rham cohomology, but it is not contractible (its fundamental group is nontrivial, for instance).
Eric Wofsey's user avatar
6 votes
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Two forms related by an automorphism are in the same cohomology class?

No. One example: take the torus $X = \mathbb{R}^2/\mathbb{Z}^2$. The flip-flop on the factors interchanges the closed forms $dx$ and $dy$ which are linearly independent in $H^1(X)$.
hunter's user avatar
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6 votes
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$S^{n + m}$ is not homeomorphic to $M \times N$

I assume that $M,N$ are manifolds of dimensions $m,n>0$ (this part is obviously necessary, otherwise we would get trivial counterexamples as $S^n\simeq \{0\}\times S^n)$. If $M\times N\cong S^{n+m}$...
leoli1's user avatar
  • 6,989
6 votes

Poincaré's take on Poincaré duality before the advent of cohomology?

I have no idea about question 1. But, regarding question 2, the new proof, which is still taught in topology textbooks, goes like this. Let $M$ be an closed, connected, oriented $n$-manifold, which we ...
Lee Mosher's user avatar
  • 122k
6 votes
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Is true that a closed $k$-form defined on an open connected set is exact iff its integral on a compact $k$-manifold without boundary is zero?

This is true but as far as I know it is a hard theorem without any elementary proof. By the de Rham theorem, $\omega$ is exact iff its integral over every smooth $k$-cycle is $0$. So, your question ...
Eric Wofsey's user avatar
6 votes

Why the De Rham cohomology just a topological invariant?

It doesn't really vanish. You need the differential structure to define it. Once you do have a differential structure there is one crucial point to understand. The point is the celebrated poincare ...
Elad's user avatar
  • 3,192
6 votes
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Two definitions of a degree of map $f: S^n \to S^n$ equivalent?

There is a Hurewicz homomorphism $\Phi:\pi_n(S^n) \to H_n(S^n)$ given by taking a class of a map $\pi_n(S^n)\ni[\alpha]:S^n \to S^n$ and sending it to $\alpha_*([S^n])$ where $[S^n]$ is the ...
Andres Mejia's user avatar
6 votes
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Can we make homology from interior derivative (interior product)?

This is more of a comment that is far too long for a comment. But I have also wondered about this, so I thought I would get started with the simplest possible examples. If anyone has a good ...
Quaere Verum's user avatar
  • 3,168
5 votes
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Proof of De Rham Theorem

According with "Introduction to Smooth Manifolds" by J.M. Lee every smoth manifold has a basis of subsets diffeomorphic to $\mathbb{R}^n$ and this basis is obvsliuly a De Rham basis, then what is ...
Jack Lee's user avatar
  • 47.3k
5 votes

The Converse of Poincare Lemma

No. The Poincare homology 3-sphere, for instance, is a smooth 3-manifold whose (integer) homology groups are those of $S^3$. If you remove a point from it, you get a homologically trivial manifold, ...
John Hughes's user avatar

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