Computing de Rham Cohomology of Connected Sums of Objects

• A
• Calabi
In summary, the author tried to compute the de Rham cohomology of a connected sum of reals projectives plans, but he ran into a problem because he didn't know the dimension of the space. He then tried to solve the problem by assuming that there was a closed form on the space, but he didn't get an answer.
Calabi
Hi, I'd like to compute the de Rham cohomology of the 3 following objects :

-A connected sum of ##g \in \mathbb{N}## reals projectives plans ##P_{2}(\mathbb{R})##.

-A connected sum of ##g \in \mathbb{N}## torus without ##n## points ##\mathbb{T}^{2} - \{x_{1}, x_{2}, ..., x_{n}\}##.

- A connected sum of ##g \in \mathbb{N}## reals projectves plans without ##n## points ##P_{2}(\mathbb{R}) - \{x_{1}, x_{2}, ..., x_{n}\}##.

I'll wrote ##[]## to talk about the class of an element in ##M = P_{2}(\mathbb{R})##.
By connexity I get : ##H^{0}(M) = \mathbb{R}## and by dimension I get ##H^{p}(M) = \{0\}## for ## p > 2##.

Here is what I tried to : let's focus on the first sum :

If ##g = 1## I consider the 2 opens ##U = P_{2}(\mathbb{R}) - [((1, 0, 0)]## : it has the same homotopy as ##P_{1}(\mathbb{R}) \simeq \mathbb{S}_{1}## so the same de Rham cohomology. I consider ##V = \{ [(x, y, z)] \in P^{2}(\mathbb{R}) | x \neq 0\}## which as the same homotopy as ##\mathbb{R}^{2}##. We have that ##U \cup V = M## and ##U \cap V## has the same cohomology than a plan miness a point so the cohomology
than a circle.

Thanks to Mayer-Vietoris, I got the following exact sequences :

##0 \mapsto \mathbb{R} \mapsto \mathbb{R} \times \mathbb{R} \mapsto \mathbb{R} \mapsto H^{1}(M) \mapsto \mathbb{R} \mapsto \mathbb{R} \mapsto H^{2}(M) \mapsto
0 ~ (1)##

The 3th ##\mapsto## is the application ##(x, y) \mapsto x - y## which is surjective. By exactitude we get that ##H^{1}(M) \hookrightarrow \mathbb{R}## so it's ##\mathbb{R}## or ##\{0\}##.

I also know that if ##0 \mapsto E_{1} \mapsto ... \mapsto E_{n} \mapsto 0## is an exacte seqeunces between finite dimension spaces I got ##\sum_{i = 0}^{n} (-1)^{i}dim(E_{i})## but I have to unkown dimension in by sequence ##(1)##.

So how solve this? Whatever the open ##U, V## I take, I always got twoo unkown.

Feel free to move the thread in homework if you think I'll get more answer.

The problem comes from here : http://www.math.u-psud.fr/~paulin/notescours/cours_geodiff.pdf : page 246 exercice 141.

Notice that the method the author use above to look for the de Rham cohomology of a connected sum of thorus is such that he has twoo unknown dimension. But he claims he compute ##H^{2}## with another method or by knowing the morphism (but if I look Mayer-Vietoris theorem proof, we only know the morphism between ##H(M), H(U) \times H(V)##).

I wish you a good day.

I have Ideas with the thorus but I still got 2 unknown dimensions.

I'm sorry if i did many mistakes, I'm french. I corrected the most I saw.

If I didn't have answer could I post this on MathStack?

I ask because ask a same question on several forum is a little bit rude.

Calabi said:
I'm sorry if i did many mistakes, I'm french. I corrected the most I saw.

If I didn't have answer could I post this on MathStack?

I ask because ask a same question on several forum is a little bit rude.
Go ahead. Your question is very special and except @lavinia and @[URL='https://www.physicsforums.com/insights/author/urs-schreiber/']Urs Schreiber[/URL] I don't know members who could help you here right from the spot. It's ok to increase the potential audience. In case you will get a satisfactory answer, you could post it here as well (the answer, not a link) such that this thread can stand for future readers.

Let ##U## a small open disk in ##P^2## and let ##V## be ##P^2## minus a smaller sub disk . By "small" is meant that the pull back of ##U## to the 2 sphere under the antipodal map is two non-intersecting disks around two antipodal points. The intersection ##U∩V## retracts smoothly onto a circle so its de Rham cohomology is ##R## in dimensions ##0## and ##1## and ##0## otherwise. The cohomology of ##U## is zero except in dimension ##0##. ##V## retracts smoothly onto a circle as well. The Meyer-Vietoris sequence is

##0→H^1(P)→H^1(U)⊕H^1(V)→H^1(U∩V)→H^2(P)→0## which reduces to

##0→H^1(P)→0⊕R→R→H^2(P)→0##.

But if you know that the de Rham cohomology of any non-orientable closed manifold is zero in the top dimension then the sequence is

##0→H^1(P)→0⊕R→R→0→0## and exactness implies that ##H^1(P)## de Rham is zero.

You might like to try to show that ##H^2(P)## is zero by assuming that there is a closed 2 form on ##P## that is not exact and then pull it back to the 2 sphere and examine its integral. If ##A:S→S## is the antipodal map and ##π:S→P## is the projection then ##π^{*}ω=A^{*}π^{*}ω## for any differential form on ##P##. This is because ##π = π \circ A##.

- Or if you know that the map ##H^1(V)→H^1(U∩V)## is multiplication by ##2## (which is easy to prove) then in de Rham cohomology this is an isomorphism so the sequence becomes

##0→0⊕R→ R→0## so the cohomology of ##P## is zero in dimensions ##1## and ##2##.

-Note that for the 2 sphere ##U## and ##V## are both disks so their cohomology is zero but their intersection still deforms onto a circle. So the Meyer-Vietoris sequence for the sphere is

##0→H^1(S)→0⊕0→R→H^2(S)→0##. So ##H^1(S)=0## and ##H^2(S)=R##.

-Try removing a small disk from the torus.

Last edited:
Spinnor and Urs Schreiber
Hello, thank you fresh_42 for uping the message and thank you Lavina. I find what you were talking about here : http://www.i2m.univ-amu.fr/perso/jean-baptiste.campesato/docs/memoireM2.pdf page 57 ans at the moment I wanted answer you already answer to me. Thanks.

In my curse the orientation of a manifold is defined after my exercise.

With what you told me, I got 2 questions remained :

-How did you know that ##H^{1}(V) \mapsto H^{1}(U \cap V)## is the multiplication by 2? In the Mayer-Vietoris theorem, I only know the morphism between ##H(M)## and ##H(U) \times H(V)## and the morphism between ##H(U) \times H(V)## and ##H(U \cap V)## (I use the notation above).

-A torus is orientable. So a torus miness ##n \in \mathbb{N}## is orientable, so a connected sum of a such object is orientable. It's also connected and not compact so my second manifold ##N## is such that ##H^{2}(N) = \{0\}##. Is it OK? The 2 others manifolds are non orientable I suppose. How to prove it?
If I know how, I will have one unknown dimension in my Mayer-Vietoris (##H^{1}##) sequence and by reccurence on ##g## I would be able to compute it.

Calabi said:
-How did you know that ##H^{1}(V) \mapsto H^{1}(U \cap V)## is the multiplication by 2? In the Mayer-Vietoris theorem, I only know the morphism between ##H(M)## and ##H(U) \times H(V)## and the morphism between ##H(U) \times H(V)## and ##H(U \cap V)## (I use the notation above).

##V## is a Mobius band and its first cohomology is generated along its equatorial circle. ##U∩V## deforms onto the boundary of the Mobius band which is a circle that wraps around the equator twice.

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OK, not very clear with the link, but I get the main idea above.

And the connected sum of projective plan miness point or not is it orientable or not?

Calabi said:
Hello, thank you fresh_42 for uping the message and thank you Lavina. I find what you were talking about here : http://www.i2m.univ-amu.fr/perso/jean-baptiste.campesato/docs/memoireM2.pdf page 57 ans at the moment I wanted answer you already answer to me. Thanks.

In my curse the orientation of a manifold is defined after my exercise.

With what you told me, I got 2 questions remained :

-How did you know that ##H^{1}(V) \mapsto H^{1}(U \cap V)## is the multiplication by 2? In the Mayer-Vietoris theorem, I only know the morphism between ##H(M)## and ##H(U) \times H(V)## and the morphism between ##H(U) \times H(V)## and ##H(U \cap V)## (I use the notation above).

-A torus is orientable. So a torus miness ##n \in \mathbb{N}## is orientable, so a connected sum of a such object is orientable. It's also connected and not compact so my second manifold ##N## is such that ##H^{2}(N) = \{0\}##. Is it OK? The 2 others manifolds are non orientable I suppose. How to prove it?
If I know how, I will have one unknown dimension in my Mayer-Vietoris (##H^{1}##) sequence and by reccurence on ##g## I would be able to compute it.
If the 2 dimensional manifold - orientable or not - is not compact ##H^2(M)## is zero.

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OK. I still reading my curse, maybe I'll saw this result. Because I don't see how to demonstrate it.

Thank you for your precious help.

Calabi said:
OK, not very clear with the link, but I get the main idea above.

And the connected sum of projective plan miness point or not is it orientable or not?

The preimage of ##V## on the sphere is the sphere minus two polar caps - two disks - and so is homeomorphic to a cylinder, The projection of this cylinder onto the projective plane identifies antipodal points. This is the same as rotating the cylinder by 180 degrees then reflecting it along its vertical axis. Note that the equator is wrapped around itself twice.

The preimage of ##U∩V## is two antipodal cylinders , one in the northern hemisphere, the other in the southern hemisphere. Their equatorial circles are mapped 1-1 onto a single circle in the projective plane. They are not wrapped onto themselves twice as is the equator of the sphere. They become the boundary of a Mobius band contained in ##V##. The boundary of a Mobius band wraps around its equatorial circle twice.

Last edited:
Spinnor
Calabi said:
OK. I still reading my curse, maybe I'll saw this result. Because I don't see how to demonstrate it.

Thank you for your precious help.
BTW: If you use de Rham cohomology with compact supports then ##H^2## of a non-compact surface can be non-zero; For instance the de Rham cohomology with compact supports of the Euclidean plane is ##R## in dimension 2.

Last edited:
Calabi said:
OK. I still reading my curse, maybe I'll saw this result. Because I don't see how to demonstrate it.

Thank you for your precious help.

I think your course is driving at a fundamental theorem for surfaces which is that every surface is the quotient of a polygon with edge identifications. I would guess that the idea of the exercise is to show the new polygon for the connected sum.

If you remove a point from a surface it retracts smoothly into its boundary polygon with identifications. These are a finite set of closed loops and thus have no cohomology above dimension 1. This automatically implies that the second cohomology of the surface minus a point is zero.

OK I think it's OK.

Thank you again both of you.

1. What is the definition of "computing de Rham cohomology"?

De Rham cohomology is a mathematical theory used to study the topology of manifolds, which are geometric objects that locally resemble Euclidean space. It involves using differential forms and their associated operations to define cohomology classes, which are used to measure the "holes" in a manifold. Computing de Rham cohomology involves determining the cohomology groups of a given manifold, which can provide valuable information about its geometric properties.

2. What are connected sums of objects and how are they related to de Rham cohomology?

A connected sum is a way of combining two manifolds by cutting out a small open ball from each and gluing the resulting boundaries together. This process creates a new manifold that is the sum of the two original manifolds. The de Rham cohomology of a connected sum is related to the individual de Rham cohomologies of the two manifolds that were combined, and can be computed by using a Mayer-Vietoris sequence.

3. Is computing de Rham cohomology of connected sums of objects a difficult task?

The difficulty of computing de Rham cohomology of connected sums of objects depends on the specific manifolds involved. In general, it can be a challenging task and often involves using advanced mathematical techniques and computer algorithms. However, for simpler manifolds, the process may be more straightforward and can be done by hand.

4. What are some applications of computing de Rham cohomology of connected sums of objects?

De Rham cohomology has applications in various areas of mathematics and physics, such as topology, differential geometry, and string theory. In particular, computing the de Rham cohomology of connected sums can help classify different types of manifolds and understand their topological properties. It can also be used to study the behavior of differential equations on manifolds.

5. Are there any limitations or drawbacks to using de Rham cohomology to study connected sums of objects?

While de Rham cohomology is a powerful tool for studying manifolds, it does have some limitations. For example, it only applies to smooth manifolds, and may not provide complete information about the topology of a manifold. Additionally, computing de Rham cohomology can be a time-consuming process, especially for more complex manifolds. As such, it is important to use it in conjunction with other techniques and approaches to get a comprehensive understanding of a manifold.

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