Cohomology of Z^+ and Infinite Dimensional de Rham Cohomology

[kakarotyjn]

Let M and F each be the set Z^+ of all positive integers.So the de Rham Cohomolgoy H^0(M) and H^0(F) is infinite dimensional.

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

Thank you!

 

[morphism]

You have to be a bit careful when you write (a_{ij}). 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 \{e_i\}_{i \in I} and \{f_j\}_{j\in J} be bases for H^0(M) and H^0(F). (Note that |I|=|J|=|\mathbb R|. This is because H^0(M) = H^0(F) = \prod_{i \in \mathbb Z^+} \mathbb R = \mathbb R^{\mathbb Z^+}, which has dimension |\mathbb R|^{|\mathbb Z^+|} = |\mathbb R|.) Then \{e_i \otimes f_j\} is a basis for H^0(M) \otimes H^0(F), and therefore an element of this latter space is a finite linear combination \sum a_{ij} \, e_i \otimes f_j. We can think of e_i \otimes f_j 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 V \otimes V^\ast = \text{End}V (for finite-dimensional V), where if you fix a basis {e_i} for V with corresponding dual basis {e_i*} then the element e_i \otimes e_j^* is literally the rank 1 matrix with 1 in the (i,j)th position and zeros elsewhere.

 

[kakarotyjn]

Hi morphism! I'm still not clear why it is a finite linear combination \sum a_{ij} \, e_i \otimes f_j,since the bases \{ e_i \} and \{ f_i \} are infinite dimensional,so \{ e_i \otimes f_i \} should be infinite dimensional,isn't it?

Thank you!

 

[morphism]

The definition of a basis says you only take finite linear combinations, even if the basis itself is infinite.

 

[kakarotyjn]

Oh,I see.Every element is a finite linear combination of bases,so H^0(M)\otimes H^0(F)<br /> consist of finite sum of matrices.