1
vote
1answer
42 views

Does the cup product on de Rham cohomology induce a nondegenerate bilinear form?

I have small issue I came across in the following. Suppose $M$ is a compact, oriented manifold of dimension $4n+2$. I want to prove that the de Rham cohomology group $H^{2n+1}(M)$ are even ...
2
votes
0answers
59 views

cohomology ring of grassmannian

Let $G_k(\mathbb{R}^n)$ be the grassmannian consisting of all $k$-subspaces in $\mathbb{R}^n$. How to compute the cohomology ring $$H^*(G_k(\mathbb{R}^n);\mathbb{Z})$$ and what is the result?
1
vote
1answer
32 views

2-forms on $S^2$

I've read that the group $H^2_{dR}(S^2)=\mathbb{R}$. If I'm not wrong, this implies that one can build closed 2-forms that are not exact. Can somebody show me an example, please? Thanks!
0
votes
1answer
45 views

Hard Lefschetz Thereom and the Cohomology of Flag Varieties

Let $G$ be a compact connected connected Lie group $G$, and $B$ a Borel subgroup containing a maximal torus. Moreover, let $F = G/B$ be the associated flag manifold. Now $F$ is a compact Kahler ...
0
votes
1answer
41 views

Equality in de Rham cohomology

Let $U_1,U_2,...,U_r$ be open sets in $\mathbb{R}^n$ such that $U_i\cap U_j =\emptyset$ for all $i \neq j$. Then prove, $H^k_{dR}(\bigcup_{i=1}^{r} U_i)=\bigoplus_{i=1}^{r} H^k_{dR} (U_i)$
3
votes
0answers
64 views

De Rham cohomology of the pointed plane

i try to work out some examples for de Rham cohomology, but i have some problems: I want to figure out what $H^k(\mathbb{R}^2\setminus\{0\})$ is and want to generalize this to arbitrary finite points ...
3
votes
0answers
58 views

De Rahm Cohomology of Complex Grassmannian

Since the complex Grassmannian $G_k(\mathbb{C}^n)\cong SU(n)/S(U(k)\times U(n-k))$ is connected and simply connected, the first two de Rahm cohomology groups are given by $$ ...
0
votes
0answers
31 views

Dimensions of the cohomology groups of certain complicated space

Let contruct the space $X$. We take the complex projective space $\mathbb{C}P^2$, pick two points $p_1, p_2 \in \mathbb{C}P^2$ and remove two small, disjoint, open $4$-balls $B_j$ centered at $p_j$. ...
3
votes
0answers
45 views

Minimum regularity Of Stoke's theorem to hold in smooth manifold.

Stokes’ Theorem on Manifolds is often express as follows: Given a differential m-form $\omega$ whose support is the $C^{\infty}$ $m$-dimensional compact manifold ${\cal{M}}$ with boundary ...
1
vote
1answer
65 views

Decomposition of cohomology group on $S^{n}$

If we have decomposition of cohomology group on $S^{n}$ it looks like $H^{n}(S^{n})=H^{n}(S^{n})_{+}\oplus H^{n}(S^{n})_{-}$, where $H^{n}(S^{n})_{\pm}$ cohomology of invariant or anti-invariant $n$ ...
3
votes
1answer
44 views

Well-definedness of a coboundary map between a reduced $L^2$ de Rham cohomology group and a relative cohomology group

I'm working right now with this paper of Carron. And I think I'm stuck at a relatively simple question. On page 11 he is defining a coboundary map $b : H^k_{2, \text{reduced}}(M - K) \to H^{k+1}(K, ...
2
votes
1answer
74 views

Poincare lemma via Lie derivative

I found such a beautiful proof of Poincaré lemma here (in Russian): Let $B$ be a star-like neighborhood of 0 in $\mathbb{R}^n$ and $r=\sum x^i \frac{\partial}{\partial x^i}$. Then Lie derivative ...
2
votes
1answer
71 views

De Rham cohomology for $\mathbb{R^2}$

De Rham cohomology groups for $\mathbb{R^2}$. $H^{0}_{dR}(\mathbb{R}^{2})=\mathbb{R}$ since $Z^{0}(\mathbb{R}^{2})$ is the one dimensional space of locally constant functions on $\mathbb{R}^{2}$ and ...
2
votes
1answer
116 views

De Rham cohomology, and forms on manifolds

In String Theory and M-Theory by Becker, Becker and Schwarz, they introduce a group, $$C^{p}(M)$$ which they denote the group of all closed $p$-forms on the manifold $M$. Furthermore, they state ...
11
votes
2answers
133 views

Computing cohomology of hypersurface

I'm taking a course on differential geometry now, and we got the following exercise from the lecturer: compute the (de Rham) cohomology groups $H_{dR}^i(M)$ of your favourite space. In all the ...
6
votes
1answer
85 views

De Rham cohomology of $T^*\mathbb{CP}^n$

I am a bit rusty on my de Rham cohomology, and I'm hoping that someone here could help me. I want to find the cohomology of $T^*\mathbb{CP}^n$ (seen as a real manifold). Now, this should be equal to ...
8
votes
1answer
296 views

Why are de Rham cohomology and Cech cohomology of the constant sheaf the same

I am comfortable with de Rham cohomology, sheaves, sheaf cohomology and Cech cohomology. I am looking to prove the following theorem: If $M$ is a smooth manifold of dimension $m$, then we have ...
2
votes
2answers
73 views

Cochains: terminology

Let a real, smooth manifold $M$ be given. Let $C_k(\mathbb Z, M$) denote the set of $k$-chains with integer coefficients, and let $C_k(\mathbb R, M)$ denote the set of $k$-chains with real ...
3
votes
1answer
64 views

Intuition for chains and cochains

I'd like to get some "geometric," "physical," (or other form of) intuition for chains, cochains, and their relationship to integration on manifolds at an elementary level. In particular, it would be ...
2
votes
0answers
70 views

Relation between algebraic hyper de Rham cohomology and hodge theory in positive characteristic

I have recently been looking at algebraic de Rham cohomology of curves in positive characteristic. In particular, I am looking at when the sequence $$0 \rightarrow H^0(X,\Omega_X) \rightarrow ...
1
vote
0answers
47 views

Chains and cochains: integer versus real coefficients

Let a real, smooth manifold $M$ be given. For each non-negative integer $k$, let a singular $k$-cube on $M$ be a continuous mapping $c:[0,1]\to M$. Let $C_k(M,\mathbb Z)$ denote the set of formal ...
2
votes
1answer
64 views

Exact and closed forms with a vanishing Riemann tensor

I need a result to prove that an closed form is equally exact. I work under the assumption that the Riemann tensor vanishes everywhere on the manifold. (It is the context of general relativity.) The ...
4
votes
0answers
142 views

Top de Rham cohomology

I just realized that I never really understood why $H_{dR}^n(M, \mathbb{R}) = \mathbb{R}$ if $M$ is compact and $H_{dR}^n(M, \mathbb{R}) = \{0\}$ if $M$ is not compact (provided that's true?). I'm ...
7
votes
2answers
843 views

how to compute the de Rham cohomology of the punctured plane just by Calculus?

I have a classmate learning algebra.He ask me how to compute the de Rham cohomology of the punctured plane $\mathbb{R}^2\setminus\{0\}$ by an elementary way,without homotopy type,without ...
3
votes
1answer
77 views

de Rham cohomology of $\mathbb{R}P^n$ via action by $SO(m+1)$

In lecture, my teacher proved the theorem that given a smooth $G$-action by a compact, connected Lie group on a manifold $M$, the de Rham cohomology of the $G$-invariant differential forms $H^p_G(M)$ ...
2
votes
1answer
339 views

Closed but not exact one-form on $S^2$

I would like to know whether there is any nice prescription to give an example of a closed but not exact one-form on $S^2$ (not the $3$-ball). I assume to take some points out of this surface, e.g. 3. ...
1
vote
1answer
58 views

Differential forms as functionals on curves

Please give me a reference to a book or lecture notes where the following stuff is studied. Let $M$ be a Riemann surface with boundary $\partial M$ (but not necessarily, any smooth $n$-dimensional ...
2
votes
0answers
51 views

Definition of the relative de-Rham cohomology and its generalization

Let $M$ be a smooth manifold and $N$ be its smooth submanifold. We say that two closed forms $\omega_1$ and $\omega_2 \in \Lambda^k(M)$ are equivalent if their difference is an exact form from ...
1
vote
0answers
47 views

Relative de Rahm cohomology computation for two disjoint circles embedded in R^2

Consider a submanifold $Y$ of $\mathbb{R}^2$ formed by two disjoint embedded copies of $S^1.$ Compute $H^{\bullet}_{dR}(\mathbb{R}^2,Y).$ In this case the long exact sequence splits, and we can ...
2
votes
1answer
345 views

First proof of Poincaré Lemma

I know that a way of proving Poincare lemma is to use the homotopy invariance and contractibility of the Euclidean space. Is there is a way of doing it directly (without using the contractibility of ...
1
vote
1answer
293 views

The Poincare Lemma for Compactly Supported Cohomology

I´m reading the proof of The Poincare Lemma for Compactly Supported Cohomology there is a part in the proof that said in the text book Bott and Tu: $d \pi_{\ast} = \pi_{\ast} d$ in other words, ...
4
votes
1answer
104 views

The degree of every smooth map $\mathbb{R}^n \to \mathbb{R}^n$ is one…

Let $\varphi : M^n \to N^n$ be a proper smooth map between two connected smooth manifolds. Then $\varphi$ induces a linear map $\varphi^* : H_c^n(N) \simeq \mathbb{R} \to H_c^n(M) \simeq \mathbb{R}$ ...
8
votes
0answers
184 views

De Rham Cohomology of $M \times \mathbb{S}^1$

Let $M$ be a closed (compact, without boundary) $m$-dimensional manifold. I want to prove that $H^{k+1}(M \times \mathbb{S}^1) = H^k(M) \oplus H^{k+1}(M)$. ($H^k$ is the $k$-th De Rham cohomology ...
4
votes
1answer
243 views

Request for companion of Mariano Suárez-Alvarez's proof.

Mariano Suárez-Alvarez's answer to Cohomology of projective plane seems very interesting. However, there are three pieces I could not stitch up for one of his proofs. Wonder if someone may help? ...
1
vote
0answers
54 views

A reference about Dolbeault cohomology

I am looking for a reference about Dolbeault cohomology when the line bundle is not supposed to be positive.
3
votes
1answer
134 views

De Rham cohomology notation

According to http://en.wikipedia.org/wiki/De_Rham_cohomology, one defines the $k$-th de Rham cohomology group $H^{k}_{\mathrm{dR}}(M)$ to be the set of equivalence classes, that is, the set ...
4
votes
3answers
376 views

How do I know when a form represents an integral cohomology class?

Suppose $M$ is an $n$-dimensional manifold, and $\omega \in \Omega^p(M)$ is a closed $p$-form. Moreover, assume that $d\omega = 0$, so that it represents a de Rham cohomology class. I would like to ...
0
votes
1answer
80 views

Restriction map on a compact orientable manifold without a boundary.

I have the following problem: Let $M$ be and $n$-dimensional compact oriented manifold without boundary. Let $p\in M$ be and point and let $M_p=M\backslash\{p\}$. Let $j:S^{n-1}\to M_p$ be the ...
1
vote
0answers
51 views

Cohomologies of $\mathbb R^n$ with rational differential forms

We can consider de Rham complex $0 \to \Omega^1 \to \Omega ^ 2 \to...$ on $\mathbb R^n$, where $\Omega ^r$ are $r$-forms on $\mathbb R^n$ with rational coefficients. What are homologies of this ...
5
votes
1answer
163 views

Simple exercise in cohomology

I know this is a simple exercise but I am stuck unfortunately. Question: Use de Rham cohomology to prove that the sphere $S^2$ is not diffeomorphic to the torus $T$. You may assume that ...
0
votes
0answers
93 views

De Rham Cohomology of Product of Manifold with an Open Interval

Let $X$ be a submanifold of $\mathbb{R}.$ Prove that $H^{k}_{DR} (X) = H^{k}_{DR} (X\times (0,1)).$ I know that we should consider maps $\iota_a: X\to X\times (0,1)$ by $\iota_a(x) = (x,a)$ for ...
7
votes
2answers
183 views

Dimension of de Rham Cohomology groups?

Is there a simple way to prove that the de Rham cohomology groups of a compact manifold $M$ have finite dimension as $\mathbb{R}$-vector spaces?
1
vote
1answer
73 views

Induced sequence of global sections

I'm reading Differential Analysis on Complex Manifolds by Raymond O. Wells. It states the following in the beginning of section 3 of chapter 2 on page 51: Consider a short exact sequence of sheaves: ...
1
vote
0answers
115 views

De Rham cohomology of $\mathbb R^3$ without lines and a circumference

I am trying to calculate De Rham cohomology of the following spaces: $X=\mathbb R^3\setminus r$ where $r$ is a line; $Y=\mathbb R^3\setminus (r \cup \gamma)$ where $r$ is a line and $\gamma$ is a ...
43
votes
1answer
2k views

What is the solution to Nash's problem presented in “A Beautiful Mind”?

I was watching the said movie the other night, and I started thinking about the equation posed by Nash in the movie. More specifically, the one he said would take some students a lifetime to solve ...
4
votes
1answer
50 views

When and *why* should one view the space of forms as one big space?

I wonder why in books about differential geometry there seem to be no relevant results about properties of mixed differential forms like $$\omega_{12}\ \mathbb{d}x^1\land\mathbb{d}x^2+\lambda_{123}\ ...
1
vote
0answers
113 views

group action - compact complex torus with $H^2_{DR}(X,C) = 0$ (de Rham cohomology)

If $\mathbb{Z}$ acts on $\mathbb{C}^n \backslash \{0\}$ by $(m,z) \mapsto 2^m\,z$, I need to show that $H^2_{DR}(X=(\mathbb{C}^n \backslash \{0\})/\mathbb{Z},\mathbb{C})=0$. I start with showing it ...
4
votes
0answers
174 views

de Rham Cohomology of Non-Flat Bundle

Let $E$ be a smooth vector bundle on a smooth manifold $M$. If $E$ is flat, there is a connection $\nabla$ which is a differential which we can use to define the de Rham cohomology of $E$. If $E$ ...
4
votes
2answers
244 views

Can a Non-Compact Manifold have Infinite Dimensional Cohomology?

For compact manifolds, Hodge Theory tells us that (de Rham) cohomology is finite dimensional. What about non-compact manifolds? That is: Can non-compact manifolds have infinite dimensional ...
2
votes
1answer
154 views

ellipticity of the Laplacian associated to the de Rham complex

I am struggeling with the following comment that I read regarding the de Rham complex: Define $(d + \delta)_e : C^\infty(\Lambda^e(T^*M)) + C^\infty(\Lambda^o(T^*M))$ where \begin{equation} ...