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A Question on Cohomology

  1. Feb 20, 2012 #1
    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!
     
  2. jcsd
  3. Feb 20, 2012 #2

    morphism

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    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: Feb 20, 2012
  4. Feb 21, 2012 #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!
     
  5. Feb 21, 2012 #4

    morphism

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    The definition of a basis says you only take finite linear combinations, even if the basis itself is infinite.
     
  6. Feb 21, 2012 #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.
     
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