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gr.group concept – Computations in group cohomology retort

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gr.group concept – Computations in group cohomology

EDIT: I used to be explaining this to a grad pupil immediately, and I spotted that I did not give any references. The consequence I painting beneath was first said by Sullivan in

Sullivan, Dennis
On the intersection ring of compact three manifolds.
Topology 14 (1975), no. 3, 275-277.

He claims it’s correct for a 3-manifold, however all he says in regards to the proof is that it’s “a inescapable amount of soul searching classical algebraic topology.”. In truth, the end result is correct for any linked CW-complicated (together with an Eilenberg-MacLane house, as within the group cohomology query I used to be answering). This all painting was later subsumed into Sullivan’s concept of 1-minimal fashions and rational homotopy concept in

Sullivan, Dennis
Infinitesimal computations in topology.
Inst. Hautes Études Sci. Publ. Math. No. 47 (1977), 269-331 (1978).

An accessible reference for that is

Griffiths, Phillip A.; Morgan, John W.
Rational homotopy concept and differential types.
Progress in Mathematics, 16. Birkhäuser, Boston, Mass., 1981. xi+242 pp. ISBN: 3-7643-3041-4

The stuff on elementary teams is in Chapter 13. I do not know the place the proof I gave first appeared (I got here up with it myself, however I doubt I used to be the primary). An alternate and really good-looking geometric proof is in

De Michelis, Stefano,
A comment on cup merchandise in $H^1(X)$,
splinter. Accad. Naz. Sci. XL Mem. Mat. (5) 14 (1990), no. 1, 323-325.


This could be very computable. Let $G^{(ok)}$ breathe the scowl central succession of $G$, ie $G^{(1)}=G$ and $G^{(ok+1)} = [G^{(k)},G]$. There are algorithmic methods to compute the quotients $G^{(ok)}/G^{(ok+1)}$ (eg utilizing the Fox free differential calculus — graze Fox’s succession of papers on the free differential calculus for the main points). The direct sum
$$oplus_{ok=1}^{infty} G^{(ok)} / G^{(ok+1)}$$
has the construction of a graded equivocate algebra with the equivocate bracket induced by conjugation (that is defined in lots of locations — I like to recommend the ultimate chapter of Magnus-Karass-Solitar’s bespeak on combinatorial group concept or Serre’s bespeak “equivocate Algebras and equivocate Groups”). This equivocate algebra is generated by the diploma 1 piece, specifically $G^{(1)} / G^{(2)} cong G^{ab}$. The diploma 2 piece is a quotient of $wedge^2 G^{ab}$ by some subgroup $R$. I pretense that judgement $R$ is precisely what you necessity to know to win the kernel of the cup product map. Namely, we maintain a surjection
$$wedge^2 G^{ab} rightarrow wedge^2 G^{ab} / R$$
and thus a twin injection
$$(wedge^2 G^{ab} / R)^{ast} hookrightarrow wedge^2 (G^{ab})^{ast}.$$
The picture of this injection is precisely the kernel of the cup product map.

Let me sketch a proof. To simplify issues, let’s occupy that every thing in confirm is torsion-free (it would simplify our statements). clique $H = H_1(G)$ and $H^{ast} = H^1(G) = Hom(H,mathbb{Z})$. The above will assist you to compute the kernel of the cup product map $wedge^2 H^{ast} rightarrow H^2(G)$ as follows. assume the quick require sequence

$$1 longrightarrow G^{(2)} longrightarrow G longrightarrow H longrightarrow 1.$$

There is an related 5-term require sequence in group cohomology which takes the design

$$0 longrightarrow H^1(H) longrightarrow H^1(G) longrightarrow (H^1(G^{(2)}))^H longrightarrow H^2(H) longrightarrow H^2(G).$$

Now, the map $H^1(H) rightarrow H^1(G)$ is an isomorphism. too, $H^2(H) = wedge^2 H^{ast}$, and the map $H^2(H) rightarrow H^2(G)$ is definitely seen to breathe the cup product map. What we deduce is that we maintain an require sequence

$$0 longrightarrow (H^1(G^{(2)}))^H longrightarrow wedge^2 H^{ast} longrightarrow H^2(G).$$

In different phrases, the kernel of the cup product map is the subgroup $(H^1(G^{(2)}))^H$ of $wedge^2 H^{ast}$.

Let us now warrant this subspace. It is best to dualize. The twin of the above inclusion is the surjection

$$H_2(H) rightarrow (H_1(G^{(2)}))_H.$$

Now, $H_1(G^{(2)})$ is simply $G^{(2)} / [G^{(2)},G^{(2)}]$, and we’re killing off the motion of $H$, which is the equivalent as killing off the conjugation motion of $G$. In different phrases, we maintain an isomorphism

$$(H_1(G^{(2)}))_H cong G^{(2)} / [G,G^{(2)}] = G^{(2)} / G^{(3)}.$$

The desired pretense is a right away consequence.

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