Computing a Generating Set in Cohomology

In summary, the conversation discusses the computation of deRham cohomology and an explicit generating set for ##H^1(X)## in deRham cohomology of a given space ##X##. The conversation also explores the use of Mayer-Vietoris sequence and deRham's theorem to find generators, and the difficulty in finding an explicit generating set for singular homology.
  • #1
WWGD
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Hi, this issue came up in another site:

We want to compute ( not just ) the deRham cohomology of ## X=\mathbb R^2 - ##{p,q} , but also

an explicit generating set for ## H^1 (X) = \mathbb Z (+) \mathbb Z## in deRham cohomology . Only explicit
generating set I can see here is {(0, +/- 1),(+/-1,0)}. How to map this into a pair of 1-forms that generate the first deRham cohomology?

We can get the actual cohomology using , e.g., Mayer Vietoris, but , without an explicit isomorphism, I don't see how to get a generating set. Maybe we can use the explicit maps in the
MV sequence to get some generators (in the deRham chain complex) ?

I was thinking of this: we compute ## H^1(X) ##, then we know all theories are equivalent, so this gives us singular cohomology , and then we use the explicit isomorphism in deRham's theorem to get some generators? We would see where the generators are sent.
 
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  • #2
A standard example is the 1-form

[tex]\omega = \frac{ydx - xdy}{x^2 + y^2}[/tex]

This defines a closed form on ##\mathbb{R}^2\setminus \{(0,0)\}## that is not exact. So it serves as a basis for the cohomology group ##H^1(\mathbb{R}^2\setminus\{(0,0)\)##.

Then consider the diffeomorphisms ##F,G:\mathbb{R}^2\rightarrow \mathbb{R}^2## by ##F(x,y) = (x,y) - p## and ##G(x,y) = (x,y) - q##, where ##p,q\in\mathbb{R}^2##.

Are then ##F^*\omega## and ##G^*\omega## not generating elements of ##H^1(X)##
 
  • #3
Thanks; if I understood you correctly, I think you need to define maps F,G on or into ## \mathbb R^2 -p ## to be able to pull back the information, because w is not defined in the entire plane.

I was trying something similar, but the problem is that ## \mathbb R^2 -{p,q} ## retracts into a wedge of circles, but is not homeomorphic to it, and I don't see how to use a homotopy equivalence to pullback a form.

I still think deRham's isomorphism , if we can use singular, will send generators to generators. But, again, we would need to know an explicit generating set for singular. Maybe the actual maps in Mayer-Vietoris can help get the explicit generating set. Thanks.
 
  • #4
WWGD said:
Thanks; if I understood you correctly, I think you need to define maps F,G on or into ## \mathbb R^2 -p ## to be able to pull back the information, because w is not defined in the entire plane.

Yes, you can just restrict domain and codomain to something suitable.

I still think deRham's isomorphism , if we can use singular, will send generators to generators.

Sure.

But, again, we would need to know an explicit generating set for singular.

Finding an explicit generating set for singular homology doesn't seem all that difficult. The intuition is that a circle around one of the points cannot be deformed to a point. So the idea is just to take two cycles, one surrounding ##p## and one surrounding ##q##. Then (in theory) we can find 1-forms associated with these cycles which form a basis for the deRham groups, but this seems difficult in practice.
 
  • #5
Never mind, thanks, I found the answer. It is simpler than I thought.
 

1. What is a generating set in cohomology?

A generating set in cohomology is a set of elements that, when combined in various ways, can produce all other elements in the cohomology group. In other words, these elements serve as building blocks for the cohomology group.

2. Why is computing a generating set important in cohomology?

Computing a generating set in cohomology allows us to better understand the structure and properties of the cohomology group. It also helps us to identify and classify different types of cohomology groups, which can have significant implications in various areas of mathematics and physics.

3. How is a generating set computed in cohomology?

The process of computing a generating set in cohomology involves finding a minimal set of elements that, when combined, can generate all other elements in the cohomology group. This is typically done using algebraic techniques such as basis reduction algorithms or spectral sequences.

4. Can a generating set in cohomology be unique?

No, a generating set in cohomology is not unique. There can be multiple different generating sets for a given cohomology group, depending on the chosen basis or method of computation. However, all generating sets will have the same number of elements, known as the rank of the cohomology group.

5. How is a generating set related to the cohomology ring?

A generating set in cohomology is closely related to the cohomology ring, which is a ring structure defined on the cohomology group. The elements in the generating set can be used to form a basis for the cohomology ring, allowing for calculations and manipulations of the ring's structure.

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