Calculus of cohomology

[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 $H(M), H(U) \times H(V)$).

Could you help me please?

I wish you a good day.

 

[Calabi]

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

 

[Calabi]

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.

 

[fresh_42]

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.

 

[lavinia]

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

$0→H^1(P)→0⊕R→R→H^2(P)→0$.

Without more information the Meyer-Vietoris sequence cannot tell you anything more.

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.

 

[Calabi]

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.

 

[lavinia]

-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.

 

[Calabi]

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?

 

[lavinia]

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.

 

[Calabi]

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.

 

[lavinia]

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.

 

[lavinia]

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.

 

[lavinia]

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.

 

[Calabi]

OK I think it's OK.

Thank you again both of you.