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What is $H_2(\mathbb{Q},\mathbb{Z})$ where the action is trivial. Thanks in advance

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    Group cohomology of $\mathbb Q$ with coefficients in $\mathbb Z$?2011-10-12

1 Answers 1

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Group homology commutes with directed colimits. Since $\mathbb Q=\varinjlim_{n\in\mathbb N}\mathbb Z\tfrac1n,$ (the colimit being taken with respect to divisibility of the indices, and inclusions of the subgroups of $\mathbb Q$) it follows that $H_2(\mathbb Q,\mathbb Z)=\varinjlim_{n\in\mathbb N}H_2(\mathbb Z\tfrac1n,\mathbb Z)$ with all the groups acting trivially. Since the subgroup $\mathbb Z\tfrac1n\subseteq\mathbb Q$ is free abelian of rank $1$, it homological dimension $1$, so $H_2(\mathbb Z\tfrac1n,\mathbb Z)=0$ and, therefore, $H_2(\mathbb Q,\mathbb Z)=0$.

N.B. If $G=\varinjlim_{i\in I}G_i$ is a directed colimit of groups, one can check that the bar resolution of $G$ is a directed colimit of complexes of the bar resolutions of the $G_i$, and then tensoring everything in sight with $\mathbb Z$ and working a bit proves the above claim about $H_\bullet$ commuting with $\varinjlim$.

Later. We can also use Hopf's formula to compute this (!). First, notice that we can present $\mathbb Q$ as $G=\langle r_n, n\geq1:r_{nm}^nr_m^{-1}, n,m\geq1\rangle.$ Indeed, in this group $r_a$ an $r_b$ commute, for they are equal to $r_{ab}^b$ and to $r_{ab}^a$, respectively, so that $G$ is commutative. There is a group homomorphism $G\to\mathbb Q$ mapping $r_n$ to $\tfrac1n$, with an evident inverse.

So let $F$ be the free group $\langle r_n, n\geq1\rangle$, and let $R$ be the normal subgroup generated by the set of elements $\{\rho_{n,m}:n,m\geq1\}$ with $\rho_{n,m}=r_{nm}^nr_m^{-1}$, so that $\mathbb Q\cong F/R$. Hopf's formula form the second cohomology group tells us that $H_2(\mathbb Q)=\frac{R\cap[F,F]}{[F,R]}.$ Since $\mathbb Q$ is abelian, $[F,F]\subseteq R$, so that this simplifies to $\frac{[F,F]}{[F,R]}.$ To see that this is trivial, it is enough to show that $[r_a,r_b]\in[F,R]$ for all $a$, $b\geq1$. This follows from the formula $[r_a,r_b]=[r_a,\rho_{a,b}^{-1}][\rho_{a,b}^{-1},\rho_{b,a}^{-1}][r_b,\rho_{b,a}^{-1}]^{-1}$