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Let $(C_i)_{i \in \mathbb{Z}}$ be a chain complex of free abelian groups. Does the rank of the homology and cohomology groups of $(C_i)_{i \in \mathbb{Z}}$ always coincide, i.e. is $$\operatorname{rank}(H_i(C_*))=\operatorname{rank}(H^i(C_*))$$ for every integer i?

If every homology group $H_i(C_*)$ is finitely generated, we can use a combination of the universal coefficients theorem and the fundamental theorem for finitely generated abelian groups to show this fact. But is it also true in the case where the homology groups are not finitely generated?

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  • $\begingroup$ What is the definition of $H_i$ versus $H^i$ here? In my experience, they both mean the same thing, but you use $H_i$ when the arrows of the complex are pointing from high to low, and $H^i$ when they are pointing from low to high, and it doesn't really make sense to write both. $\endgroup$ – user14972 Aug 9 '14 at 12:03
  • $\begingroup$ $H^i$ are the cohomology groups obtained from the cochain complex $hom(C_*, \mathbb{Z})$, while $H_i$ are the homology groups obtained from the chain complex $C_*$. $\endgroup$ – Tom Bombadil Aug 9 '14 at 12:08
  • $\begingroup$ Ah, thank you very much. $\endgroup$ – user14972 Aug 9 '14 at 12:11
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    $\begingroup$ @ThomasAndrews: I don't know, I'm just not used to this convention (and I'm not the only one it seems). I've had the same experience as Hurkyl. In fact I'm used to writing just $H(C)$, and everything is $\mathbb{Z}$-graded (nonpositively or nonnegatively, depending). $\endgroup$ – Najib Idrissi Aug 9 '14 at 12:47
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    $\begingroup$ Yeah, there seems to be a regional difference in homology and cohomology notation. @NajibIdrissi $\endgroup$ – Thomas Andrews Aug 9 '14 at 12:48
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Here is a counterexample: let $$C_i = \begin{cases} \bigoplus_{n \in \mathbb{N}} \mathbb{Z} & i = 0 \\ 0 & i \neq 0 \end{cases}$$ with the zero differential. Then

  • $H_0(C)=\bigoplus_{n \in \mathbb{N}} \mathbb{Z}$ has countably infinite rank;
  • $H^0(C) = \prod_{n \in \mathbb{N}} \mathbb{Z}$ has a rank $2^{\aleph_0}$ by a theorem of Nöbeling.

It's possible even when the $H_i$ have finite rank that the cohomology has bigger rank. Let: $$C_i = \begin{cases} \bigoplus_{n \in \mathbb{N}} \mathbb{Z} & i = 0,1 \\ 0 & i \neq 0,1 \end{cases}$$ and the differential $d : C_1 \to C_0$ is given by multiplication by $2^n$ on the $n$th factor (so that $d(C_1) = \bigoplus 2^n \mathbb{Z}$).

  • $H_0(C) = \bigoplus_{n \in \mathbb{N}} \mathbb{Z}/2^n$, and $H_i(C) = 0$ elsewhere, so all homology groups have rank zero (because every element in $H_0$ is torsion);
  • We can apply the universal coefficient theorem, because every $C_i$ and $d(C_i)$ is projective (and even free abelian). Therefore $$H^1(C) = \operatorname{Ext}^1_\mathbb{Z}(H_0(C), \mathbb{Z}) = \prod_{n \in \mathbb{N}} \mathbb{Z}/2^n$$ (because $\hom(\mathbb{Z}/2^n, \mathbb{Z}) = 0$ and $\operatorname{Ext}$ sends arbitrary direct sums in the first factor to products). But this group has rank at least one (and probably even infinite), because $x = (1,1,1,\dots)$ has infinite order.
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  • $\begingroup$ Thank you for your answer! I'm still unclear about one thing though: If we're not interested in whether the ranks are countably or uncountably infinite and just want to know if one rank is infinite than so is the other one, how does it follow that $rank(H^i)=rank(H_i)$ if $H_i$ is not finitely generated? The argument in one of your comments above shows that if $hom(H_i, \mathbb{Z})$ is not finitely generated, then so isn't $H^i$. But what does this mean for the ranks of $H_i$ and $H^i$? Does an infinitely generated group always have infinite rank? $\endgroup$ – Tom Bombadil Aug 10 '14 at 7:10
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    $\begingroup$ In this case it's simpler, because by the universal coefficient theorem, the canonical map $H^i \to \hom(H_i, \mathbb{Z})$ is surjective. So if $H_i$ has infinite rank, so does $\hom(H_i, \mathbb{Z})$, so it contains an infinite linearly independent family. This family lifts to $H^i$, and it's still linearly independent (because the image of a linearly dependent family is linearly dependent). This also proves that the rank of $H^i$ is always at least equal to the rank of $H_i$. $\endgroup$ – Najib Idrissi Aug 10 '14 at 7:15
  • $\begingroup$ Okay, I understand that. I'm sorry for being so ignorant, but one part is still unclear to me. If $H_i$ is not finitely generated, does it have infinite rank? $\endgroup$ – Tom Bombadil Aug 10 '14 at 7:24
  • $\begingroup$ No, not necessarily. $\bigoplus_{n \in \mathbb{N}} \mathbb{Z}/2$ is not finitely generated, but it has rank zero (because everything is torsion). $\endgroup$ – Najib Idrissi Aug 10 '14 at 8:04

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