Suppose $G$ and $A$ are abelian groups (I'm setting $G$ abelian to keep the discussion simple, though there are analogues for non-abelian $G$) with $G$ acting trivially on $A$. By the universal coefficients theorem, we have short exact sequences for all positive integers $n$:

$$0 \to \operatorname{Ext}^1(H_{n-1}(G;\mathbb{Z});A) \to H^n(G;A) \to \operatorname{Hom}(H_n(G;\mathbb{Z}),A) \to 0$$

In the case $n = 2$, this becomes:

$$0 \to \operatorname{Ext}^1(G;A) \to H^2(G;A) \to \operatorname{Hom}(H_2(G;\mathbb{Z}),A) \to 0$$

The surjection $H^2(G;A) \to \operatorname{Hom}(H_2(G;\mathbb{Z}),A)$ has a nice interpretation:

$H_2(G;\mathbb{Z})$ is the Schur multiplier of $G$, which (since $G$ is abelian) is the exterior square of $G$, so $\operatorname{Hom}(H_2(G;\mathbb{Z}),A)$ is the alternating bilinear maps from $G$ to $A$. The mapping $H^2(G;A) \to \operatorname{Hom}(H_2(G;\mathbb{Z}),A)$ is the "skew" mapping; it sends a 2-cocycle $f:G \times G \to A$ to the function:

$$(x,y) \mapsto f(x,y) - f(y,x)$$

Even without prior knowledge of the universal coefficients theorem, this short exact sequence makes sense: one can check that the skew of a 2-cocycle for an abelian group is alternating and bilinear, and the mapping descends to cohomology classes because any 2-coboundary is symmetric. The kernel of the mapping corresponds to symmetric cohomology classes, which corresponding to the abelian extension groups, given precisely a $\operatorname{Ext}^1(G;A)$. In group extension terms, the image of a given element of $H^2(G;A)$ under the skew map describes the commutator map of the extension group.

My question is: what's an analogous concrete interpretation for higher $n$?

I'm listing below some references with generalizations of the above to non-abelian $G$ and varietal generalizations:

  • John Burns and Graham Ellis, On the nilpotent multipliers of a group, Math. Zeitschr. 226 (1997) pp. 405–428 doi:10.1007/PL00004348.

  • C. R. Leedham-Green and Susan McKay, Baer-invariants, isologism, varietal laws, and homology, Acta Mathematica 137 (1976) pp. 99–150 doi:10.1007/BF02392415 (ResearchGate copy).

Pages 135–139 (37–41 in the pdf) describe a generalization of universal coefficients to varietal laws (where the usual universal coefficients is with respect to abelian groups inside groups).

  • $\begingroup$ Have you tried assuming that $G$ is finite cyclic, so that cohomology is $2$-periodic? The isomorphism $H^q\cong H^{q+2}$ is given by an explicit cup-product so you might deduce something...but that's only a guess. $\endgroup$ – Filippo Alberto Edoardo Jun 28 '12 at 10:56
  • $\begingroup$ Noting here: the original version of the question had the wrong statement of the universal coefficients theorem for general n (correct for n = 2 though) -- I fixed it today morning. I had said "Ext^{n-1}(G^{ab},A)" and the correct term is "Ext^1(H_{n-1}(G;Z),A)" $\endgroup$ – Vipul Naik Jun 29 '12 at 3:12

This is an extended-comment, but I think no:

In homology, we know that $H_1$ is abelianization, $H_2$ is Schur multiplier, but $H_{n\ge 3}$ is ???
In cohomology, $H^1(G;A)$ is split extensions, and this fits well with $Hom(G,A)$ in the UCT. As you mention, $H^2(G,A)$ corresponds to group extensions of $G$ by $A$, and this fits well with $Hom(H_2G,A)$ in the UCT. But as we see, this all ties in to knowing $H^n(G;A)$ and $H_n(G;A)$ very well...

If we look at $H^3(G;A)$ we get crossed module extensions $0\to A\to N\to E\to G\to 0$, and these are cumbersome. We have no nice interpretation for $H^n(G;A)$ for $n>3$, except more crazy-looking exact sequences. This is why I don't expect a "nice" map (i.e. "interpretation" of it) in the UCT to arise.

  • 2
    $\begingroup$ Let me say that what you describe as 'cumbersome' and 'crazy' looks to me (and to many others) as very elegant and natural constructions. $\endgroup$ – Fernando Muro Jun 29 '12 at 16:08

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