Cohomology of Z^+ and Infinite Dimensional de Rham Cohomology

  • Thread starter kakarotyjn
  • Start date
In summary, the de Rham Cohomology H^0(M) and H^0(F) are both infinite dimensional. However, the tensor product of these two spaces, H^0(M)\otimes H^0(F), is made up of finite sums of matrices of rank 1. This is due to the fact that the bases for H^0(M) and H^0(F) are both infinite dimensional, but the definition of a basis only allows for finite linear combinations. This is similar to the isomorphism V \otimes V^\ast = \text{End}V for finite-dimensional vector spaces.
  • #1
kakarotyjn
98
0
Let M and F each be the set [tex]Z^+[/tex] of all positive integers.So the de Rham Cohomolgoy [tex]H^0(M)[/tex] and [tex]H^0(F)[/tex] is infinite dimensional.

But why does [tex]H^0(M)\otimes H^0(F)[/tex] consist of finite sums of matrices [tex](a_{ij})[/tex] of rank 1?

Thank you!
 
Physics news on Phys.org
  • #2
You have to be a bit careful when you write [itex](a_{ij})[/itex]. Presumably here the indices i and j are allowed to run over an (uncountably) infinite set, because otherwise what you're saying doesn't make much sense (to me).

To see what's going on, let [itex]\{e_i\}_{i \in I}[/itex] and [itex]\{f_j\}_{j\in J}[/itex] be bases for H^0(M) and H^0(F). (Note that [itex]|I|=|J|=|\mathbb R|[/itex]. This is because [itex]H^0(M) = H^0(F) = \prod_{i \in \mathbb Z^+} \mathbb R = \mathbb R^{\mathbb Z^+}[/itex], which has dimension [itex]|\mathbb R|^{|\mathbb Z^+|} = |\mathbb R|[/itex].) Then [itex]\{e_i \otimes f_j\}[/itex] is a basis for [itex]H^0(M) \otimes H^0(F)[/itex], and therefore an element of this latter space is a finite linear combination [itex]\sum a_{ij} \, e_i \otimes f_j[/itex]. We can think of [itex]e_i \otimes f_j[/itex] as being a rank 1 |I|x|J| matrix with an entry of 1 in the (i,j)th position and zeros elsewhere.

Does this help? Note that this kind of construction ought to be reminiscent of the isomorphism [itex]V \otimes V^\ast = \text{End}V[/itex] (for finite-dimensional V), where if you fix a basis {e_i} for V with corresponding dual basis {e_i*} then the element [itex]e_i \otimes e_j^*[/itex] is literally the rank 1 matrix with 1 in the (i,j)th position and zeros elsewhere.
 
Last edited:
  • #3
Hi morphism! I'm still not clear why it is a finite linear combination [itex]\sum a_{ij} \, e_i \otimes f_j[/itex],since the bases [itex]\{ e_i \}[/itex] and [itex]\{ f_i \}[/itex] are infinite dimensional,so [itex]\{ e_i \otimes f_i \}[/itex] should be infinite dimensional,isn't it?

Thank you!
 
  • #4
The definition of a basis says you only take finite linear combinations, even if the basis itself is infinite.
 
  • #5
Oh,I see.Every element is a finite linear combination of bases,so [itex]H^0(M)\otimes H^0(F)
[/itex] consist of finite sum of matrices.
 

1. What is cohomology?

Cohomology is a mathematical concept used in topology and algebraic geometry to study geometric objects and their properties. It involves assigning algebraic structures to these objects and then studying how these structures behave under certain transformations.

2. How is cohomology different from homology?

While both cohomology and homology are used to study geometric objects, they differ in terms of the algebraic structures they assign. Homology assigns abelian groups to these objects, while cohomology assigns cohomology groups, which are dual to homology groups.

3. What is the importance of cohomology in mathematics?

Cohomology has many applications in mathematics, including in topology, algebraic geometry, and differential geometry. It allows for the study of geometric objects through their algebraic structures, making it a powerful tool in understanding these objects and their properties.

4. How is cohomology used in physics?

In physics, cohomology is used to study the properties of physical systems, such as in quantum field theory and string theory. It allows for the analysis of symmetries and conservation laws in these systems, providing a deeper understanding of their behavior.

5. Are there different types of cohomology?

Yes, there are different types of cohomology, including singular cohomology, Čech cohomology, de Rham cohomology, and sheaf cohomology. Each type is used to study different types of geometric objects and has its own unique properties and applications.

Similar threads

  • Differential Geometry
Replies
13
Views
2K
  • Differential Geometry
Replies
5
Views
1K
  • Differential Geometry
Replies
9
Views
4K
Replies
4
Views
3K
  • Differential Geometry
Replies
7
Views
3K
  • Linear and Abstract Algebra
Replies
8
Views
800
  • Math Proof Training and Practice
2
Replies
69
Views
4K
  • Differential Geometry
Replies
19
Views
5K
  • Quantum Physics
Replies
2
Views
978
  • Differential Geometry
Replies
1
Views
2K
Back
Top