# A Question on Cohomology

1. Feb 20, 2012

### 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!

2. Feb 20, 2012

### 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.

Last edited: Feb 20, 2012
3. Feb 21, 2012

### 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!

4. Feb 21, 2012

### morphism

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

5. Feb 21, 2012

### kakarotyjn

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