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Let $G$ be an abelian group, $A$ a trivial $G$-module. We know that $\text{Ext}(G,A)$ classifies abelian extensions of $G$ by $A$, whereas $H^2(G,A)$ classifies central extensions of $G$ by $A$. So we have a canonical inclusion $\text{Ext}(G,A)\hookrightarrow H^2(G,A)$. Is there some naturally arising exact sequence/spectral sequence which realizes this injection?

Usually this kind of thing can be explained by constructing a clever short exact sequence, but here I have no idea how one might compare $R^1\text{Hom}_\mathbb{Z}(G,\underline{\quad})$ with $R^2\text{Hom}_G(\mathbb{Z},\underline{\quad})$.

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  • $\begingroup$ One general procedure to get the comparison map you needed is to look at the change-of-rings spectral sequence for the map $\mathbb Z\to G$. $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '10 at 16:58
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You get a description from the universal coefficient theorem which gives a (split) exact sequence $$ 0\to \mathrm{Ext}(H_1(G),A) \to H^2(G,A) \to \mathrm{Hom}(H_2(G),A) \to 0 $$ and the fact that $H_1(G)=G$. We have that $H_2(G)=\Lambda^2G$ and the map $H^2(G,A) \to \mathrm{Hom}(H_2(G),A)$ associates to an extension its commutator map.

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Let $G$ and $A$ be groups and assume that $R \rightarrowtail F \twoheadrightarrow G$ is a presentation of $G$.

Then, by a general theorem of MacLane, there is an exact sequence

$\textrm{Hom}(F_{ab}, A) \to \textrm{Hom}(R/[R,F], A) \to H^2(G, A) \to 0$.

Moreover, if $G$ and $A$ are Abelian there is an exact sequence of Abelian groups

$\textrm{Hom}(F_{ab}, A) \to \textrm{Hom}(R/[F,F], A) \to \textrm{Ext}(G, A) \to 0$.

Therefore the inclusion $\textrm{Ext}(G, A) \to H^2(G,A)$ is induced by the natural homomorphism

$\textrm{Hom}(R/[F,F], A) \to \textrm{Hom}(R/[R,F], A)$,

i.e. by the natural homomorphism

$R/[R, F] \to R/[F, F]$.

See [Robinson, A Course in the Theory of Groups, Chapter 11] for more details.

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