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Silviu

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In summary: So assuming we have a 2D manifold an element of a cohomology ring would be...Two elements of a cohomology ring are the product of their cohomology classes.

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Silviu

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lavinia

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The De Rham cohomology group ##H^{i}(M)## is the quotient group of closed ##i##-forms modulo exact ##i##-forms.

Two closed ##{i}##-forms ##α_1## and ##α_2## represent the same cohomology class if ##α_1 = α_2 + dω## where ##ω## is an ##i-1## form and ##dω## is its exterior derivative.

Since these cohomology classes are elements of a quotient group one cannot take their wedge product. The best one can do is is to take the wedge product of differential forms that represent the cohomology classes. If ##a## and ##b## are elements of ##H^{i}(M)## and ##H^{j}(M)## and if ##α## and ##β## are differential forms representing ##a## and ##b## then one can take their wedge product ##α∧β## to get an ##i+j## form.

Since ##α## and ##β## are closed, ##α∧β## is also closed so it represents some cohomology class in ##H^{i+j}(M)##. One defines the "cup product" on the cohomology groups as ##a∪b= [α∧β]## where ##[α∧β]## is the cohomology class of ##α∧β##.

One must show that this product is well defined. If ##α+dω## is another differential form representing the cohomology class ##a## one must show that ##[α∧β]= [(α+dω)∧β]##.

- If ##i+j## is greater than the dimension of the manifold ##M## then ##α∧β## is zero. This is because the tangent space to ##M## is an n-dimensional vector space at each point of ##M##.

Two closed ##{i}##-forms ##α_1## and ##α_2## represent the same cohomology class if ##α_1 = α_2 + dω## where ##ω## is an ##i-1## form and ##dω## is its exterior derivative.

Since these cohomology classes are elements of a quotient group one cannot take their wedge product. The best one can do is is to take the wedge product of differential forms that represent the cohomology classes. If ##a## and ##b## are elements of ##H^{i}(M)## and ##H^{j}(M)## and if ##α## and ##β## are differential forms representing ##a## and ##b## then one can take their wedge product ##α∧β## to get an ##i+j## form.

Since ##α## and ##β## are closed, ##α∧β## is also closed so it represents some cohomology class in ##H^{i+j}(M)##. One defines the "cup product" on the cohomology groups as ##a∪b= [α∧β]## where ##[α∧β]## is the cohomology class of ##α∧β##.

One must show that this product is well defined. If ##α+dω## is another differential form representing the cohomology class ##a## one must show that ##[α∧β]= [(α+dω)∧β]##.

- If ##i+j## is greater than the dimension of the manifold ##M## then ##α∧β## is zero. This is because the tangent space to ##M## is an n-dimensional vector space at each point of ##M##.

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Silviu

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Thank you for your reply. However I understood this part. What I am confuse about is the actual multiplication of 2 elements of the ring. what is the results of ##a \wedge c##, for a and c ##\in H^*(M)##lavinia said:

Two closed ##{i}##-forms ##α_1## and ##α_2## represent the same cohomology class if ##α_1 = α_2 + dω## where ##ω## is an ##i-1## form and ##dω## is its exterior derivative.

Since these cohomology classes are elements of a quotient group one cannot take their wedge product. The best one can do is is to take the wedge product of differential forms that represent the cohomology classes. If ##a## and ##b## are elements of ##H^{i}(M)## and ##H^{j}(M)## and if ##α## and ##β## are differential forms representing ##a## and ##b## then one can take their wedge product ##α∧β## to get an ##i+j## form.

Since ##α## and ##β## are closed, ##α∧β## is also closed so it represents some cohomology class in ##H^{i+j}(M)##. One defines the "cup product" on the cohomology groups as ##a∪b= [α∧β]## where ##[α∧β]## is the cohomology class of ##α∧β##.

One must show that this product is well defined. If ##α+dω## is another differential form representing the cohomology class ##a## one must show that ##[α∧β]= [(α+dω)∧β]##.

- If ##i+j## is greater than the dimension of the manifold ##M## then ##α∧β## is zero. This is because the tangent space to ##M## is an n-dimensional vector space at each point of ##M##.

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lavinia

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I explained the multiplication.Silviu said:Thank you for your reply. However I understood this part. What I am confuse about is the actual multiplication of 2 elements of the ring. what is the results of ##a \wedge c##, for a and c ##\in H^*(M)##

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Silviu

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You explained the multiplication of 2 elements of 2 cohomology groups. I would like to know the multiplication of 2 elements of a cohomology ring i.e. the elements of a cohomology ring are not just cohomology classes but direct sum of these and I am not sure how does that looks explicitly.lavinia said:I explained the multiplication.

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lavinia

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Silviu said:You explained the multiplication of 2 elements of 2 cohomology groups. I would like to know the multiplication of 2 elements of a cohomology ring i.e. the elements of a cohomology ring are not just cohomology classes but direct sum of these and I am not sure how does that looks explicitly.

Whenever one has a ring the product ##(Σ_{i}a_{i})⋅(Σ_{j}b_{j})## equals ## Σ_{ij}a_{i}⋅b_{j}##. This rule tells you how to multiply arbitrary elements of the De Rham cohomology ring.

- #7

Silviu

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So assuming we have a 2D manifold an element of a cohomology ring would be ##a_1+b_1dx+c_1dx\wedge dy## and multiplying 2 elements would give ##(a_1+b_1dx+c_1dx\wedge dy)\wedge (a_2+b_2dx+c_2dx\wedge dy) = a_1a_2+a_1b_2dx+a_1c_2dx\wedge dy + b_1a_2dx+c_1a_2dx\wedge dy##, while all the other terms get to 0? Is this correct?lavinia said:Whenever one has a ring the product ##(Σ_{i}a_{i})⋅(Σ_{j}b_{j}) = Σ_{ij}a_{i}⋅b_{j}##. This rule tells you how to multiply arbitrary elements of the De Rham cohomology group.

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lavinia

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Silviu said:So assuming we have a 2D manifold an element of a cohomology ring would be ##a_1+b_1dx+c_1dx\wedge dy## and multiplying 2 elements would give ##(a_1+b_1dx+c_1dx\wedge dy)\wedge (a_2+b_2dx+c_2dx\wedge dy) = a_1a_2+a_1b_2dx+a_1c_2dx\wedge dy + b_1a_2dx+c_1a_2dx\wedge dy##, while all the other terms get to 0? Is this correct?

Unless you made an algebraic mistake this looks right.

The wedge product is a mathematical operation used in differential geometry to define a new operation on differential forms. On a 3-dimensional manifold, the wedge product combines two differential forms to create a new differential form, which is a mathematical object that describes the properties of a given space.

The wedge product is unique in that it is an antisymmetric operation, meaning that the order of the terms in the product does not change the result. This is different from other operations on manifolds, such as the tensor product, which are symmetric and depend on the order of the terms. Additionally, the wedge product is used specifically to define differential forms, which have important applications in physics and engineering.

The wedge product can be thought of as a way to measure the "oriented area" or "oriented volume" of a given space. In other words, it describes how much of one differential form is contained in another differential form, taking into account their orientation. This geometric interpretation is useful in understanding the properties of manifolds and their applications in various fields.

The wedge product has many practical applications in fields such as physics, engineering, and computer science. It is used to define important mathematical objects like differential forms, which are essential in the study of electromagnetism, fluid dynamics, and general relativity. The wedge product is also used in computer graphics and computer vision to describe and manipulate geometric shapes and surfaces.

While the wedge product is a useful operation on 3-dimensional manifolds, it is important to note that it is not defined for all types of manifolds. For example, the wedge product is not defined on manifolds with an odd number of dimensions, and it can only be used on manifolds that are smooth and orientable. Additionally, the wedge product is limited in its ability to fully describe the geometry of a given space and may require other mathematical tools to fully understand a manifold's properties.

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