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In Kronheimer and Mrowka's Monopoles and Three-Manifolds, Section 28.2, right above Lemma 28.2.1, there is talk of the divisibility of a cohomology class $\epsilon\in H^2(Y,\mathbb{Z})$, which is defined to be $0$ if $\epsilon$ is torsion, and as "the divisibility of the image of $\epsilon$ in the free abelian group $H^2(Y,\mathbb{Z})/\text{Torsion}$ otherwise".

I know that $\epsilon$ is divisible by some $n\in\mathbb{N}$, if there exists some $y\in H^2(Y,\mathbb{Z})$ so that $\epsilon = ny$, but what exactly is meant by "the" divisibility of $\epsilon$?

Probably connected to this: Modding out the torsion means that for a fixed $n$, the class $y$ with $x=ny$ becomes unique (since then $x=ny =ny'$ implies $n(y-y')=0$, i.e. $y-y'\equiv 0$ modulo Torsion), but why is this necessary for this definition?

Nuke_Gunray
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    If $\epsilon$ is divisible by $n$ and $m$, then it is also divisible by the least common multiple of $n$ and $m$, so I would assume "the" divisibility is just the largest integer wrt to divisibility that $\epsilon$ is divisible by. – Thorgott Jul 13 '23 at 09:58
  • @Thorgott Thank you, that makes sense. Do you have an idea about the modding out of the torsion? – Nuke_Gunray Jul 13 '23 at 10:24
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    Well, if $\varepsilon$ is $n$-torsion, then $m\varepsilon=(m+n)\varepsilon=(m+2n)\varepsilon=....$, so there is no largest integer by which $m\varepsilon$ is divisible. I assume they're to avoid something like that. – Thorgott Jul 13 '23 at 17:18

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If there is reason to know that $H^2(Y,\mathbb Z)$ is finitely generated (e.g. if $Y$ is a compact manifold or CW complex) then, applying the classification of finitely generated abelian groups, $H^2(Y,\mathbb Z)$ has a unique maximal finite subgroup, its torsion subgroup; furthermore the quotient by the torsion subgroup is a finite rank free abelian group.

For every nonzero element $x \in A$ of a finite rank free abelian group, there exists a unique maximal natural number $n$ such that the equation $x=ny$ has a solution $y \in A$; that number $n$ is called the divisibility of $x$.

Without the knowledge that $H^2(Y,\mathbb Z)$ is finitely generated, this issue does not make too much sense. For example, perhaps $H^2(Y,\mathbb Z)$ is isomorphic to $\mathbb Q$, in which case a numeric measure of "divisibility" is not defined. Allowing for this example, and for a general torsion free abelian group, one could define the "divisibility" of an element $x$ as the unique supremum in $\mathbb N \cup \{\infty\}$ of those natural numbers $n$ such that $x=ny$ has a solution $y$ in the group.

Lee Mosher
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