Second cohomology group up to isoclinism: Difference between revisions

Latest revision as of 20:13, 28 May 2013

Definition

Suppose $G$ is a group and $A$ is an abelian group. The second cohomology group up to isoclinism is the quotient of the usual second cohomology group by the following equivalence relations: two extensions $E_{1}, E_{2}$ are isoclinic as extensions if there is an isoclinism between them that is compatible with the identity maps for $G$ and $A$ .

Explicitly, given two extensions:

$0 \to A \to E_{1} \to G \to 1$

$0 \to A \to E_{2} \to G \to 1$

we want an isomorphism $φ : [E_{1}, E_{1}] \to [E_{2}, E_{2}]$ such that both these conditions hold:

If $γ_{1} : G \times G \to [E_{1}, E_{1}], γ_{2} : G \times G \to [E_{2}, E_{2}]$ are the set maps arising from the commutator map, then $φ \circ γ_{1} = γ_{2}$ . Equivalently, if $Γ_{1} : G \land G \to [E_{1}, E_{1}], Γ_{2} : G \land G \to [E_{2}, E_{2}]$ are the commutator map homomorphisms from the exterior square, then $φ \circ Γ_{1} = Γ_{2}$ as group homomorphisms.
Suppose $B$ is the inverse image in $A$ of $[E_{1}, E_{1}]$ . Then, $B$ is also the inverse image in $A$ of $[E_{2}, E_{2}]$ . Moreover, composing $φ$ with the inclusion of $B$ in $[E_{1}, E_{1}]$ must give the inclusion of $B$ in $[E_{2}, E_{2}]$ .

Relation with formula for second cohomology group

Consider the case that the action is trivial, i.e., we are looking at the second cohomology group for trivial group action.

Then, the second cohomology group up to isoclinism can be identified as the group of homomorphisms:

$H o m (H_{2} (G; Z), A)$

where $H_{2} (G; Z) = M (G)$ is the Schur multiplier.

Further, the second cohomology group up to isoclinism can be viewed as the quotient part of the short exact sequence arising from the formula for second cohomology group for trivial group action in terms of second homology group and abelianization:

$0 \to E_{x}^{t} (G, A) \to H^{2} (G; A) \to H o m (H_{2} (G; Z), A) \to 0$

In other words, it is the group $H o m (H_{2} (G; Z), A)$ . Note that $H_{2} (G; Z)$ is the Schur multiplier, so this is in fact the group of homomorphisms from the Schur multiplier to $A$ .

Explicit justification

Further information: Commutator map in central extension defines homomorphism from Schur multiplier of quotient group to central subgroup

Consider the two short exact sequences below:

$\begin{matrix} 0 & \to & M (G) & \to & G \land G & \to & [G, G] & \to & 1 \\ ↓ & ↓ & ↓ & ↓ & ↓ \\ 0 & \to & A & \to & E & \to & G & \to & 1 \end{matrix}$

Suppose $B$ is the subgroup of $A$ that arises as the image of the homomorphism from $M (G)$ . We then have the following two short exact sequences:

$\begin{matrix} 0 & \to & M (G) & \to & G \land G & \to & [G, G] & \to & 1 \\ ↓ & ↓ & ↓ & ↓ & ↓ \\ 0 & \to & B & \to & [E, E] & \to & [G, G] & \to & 1 \end{matrix}$

The right map is the identity map. Now, both the left and right maps are surjective. It is easy to see from this that the middle map is surjective and is determined by the left and right maps. In other words, we have determined from the element of $H o m (M (G), A)$ the map $G \land G \to [E, E]$ , even though the congruence type of extension group $E$ is unknown. (Note that this assertion is a special case of the nine lemma). From this, it is clear that the extension has been determined up to isoclinism.

Related notions

Second cohomology group up to isologism is a generalization from the variety of abelian groups.
Formula for second cohomology group for trivial group action in terms of Baer invariant and verbal factor group

@@ Line 2: / Line 2: @@
 Suppose <math>G</math> is a [[group]] and <math>A</math> is an [[abelian group]]. The '''second cohomology group up to isoclinism''' is the quotient of the usual [[second cohomology group]] by the following equivalence relations: two extensions <math>E_1, E_2</math> are '''isoclinic''' as extensions if there is an [[isoclinism]] between them that is compatible with the identity maps for <math>G</math> and <math>A</math>.
+Explicitly, given two extensions:
+<math>0 \to A \to E_1 \to G \to 1</math>
+<math>0 \to A \to E_2 \to G \to 1</math>
+we want an isomorphism <math>\varphi:[E_1,E_1] \to [E_2,E_2]</math> such that ''both'' these conditions hold:
+* If <math>\gamma_1:G \times G \to [E_1,E_1], \gamma_2:G \times G \to [E_2,E_2]</math> are the set maps arising from the commutator map, then <math>\varphi \circ \gamma_1 = \gamma_2</math>. Equivalently, if <math>\Gamma_1:G \wedge G \to [E_1,E_1],\Gamma_2:G \wedge G \to [E_2,E_2]</math> are the [[commutator map is homomorphism from exterior square to derived subgroup of central extension|commutator map homomorphisms from the exterior square]], then <math>\varphi \circ \Gamma_1 = \Gamma_2</math> as group homomorphisms.
+* Suppose <math>B</math> is the inverse image in <math>A</math> of <math>[E_1,E_1]</math>. Then, <math>B</math> is also the inverse image in <math>A</math> of <math>[E_2,E_2]</math>. Moreover, composing <math>\varphi</math> with the inclusion of <math>B</math> in <math>[E_1,E_1]</math> must give the inclusion of <math>B</math> in <math>[E_2,E_2]</math>.
 ==Relation with formula for second cohomology group==
-===Case of abelian group===
+Consider the case that the action is trivial, i.e., we are looking at the [[second cohomology group for trivial group action]].
-Consider the case that <math>G</math> is abelian and the action is trivial, i.e., we are looking at the [[second cohomology group for trivial group action]].
 Then, the second cohomology group up to isoclinism can be identified as the group of homomorphisms:
@@ Line 20: / Line 29: @@
 In other words, it is the group <math>\operatorname{Hom}(H_2(G;\mathbb{Z}),A)</math>. Note that <math>H_2(G;\mathbb{Z})</math> is the [[Schur multiplier]], so this is in fact the group of homomorphisms from the Schur multiplier to <math>A</math>.
+===Explicit justification===
+{{further|[[Commutator map in central extension defines homomorphism from Schur multiplier of quotient group to central subgroup]]}}
+Consider the two short exact sequences below:
+<math>\begin{array}{ccccccccc}
+& \to & M(G) & \to & G \wedge G & \to & [G,G] & \to & 1\\
+\downarrow &&   \downarrow  &&  \downarrow     && \downarrow&& \downarrow\\
+& \to & A &\to & E & \to & G & \to & 1\\
+\end{array}</math>
+Suppose <math>B</math> is the subgroup of <math>A</math> that arises as the image of the homomorphism from <math>M(G)</math>. We then have the following two short exact sequences:
+<math>\begin{array}{ccccccccc}
+& \to & M(G) & \to & G \wedge G & \to & [G,G] & \to & 1\\
+\downarrow &&   \downarrow  &&  \downarrow     && \downarrow&& \downarrow\\
+& \to & B &\to & [E,E] & \to & [G,G] & \to & 1\\
+\end{array}</math>
+The right map is the identity map. Now, both the left and right maps are surjective. It is easy to see from this that the middle map is surjective and is determined by the left and right maps. In other words, we have determined from the element of <math>\operatorname{Hom}(M(G),A)</math> the map <math>G \wedge G \to [E,E]</math>, even though the congruence type of extension group <math>E</math> is unknown. (Note that this assertion is a special case of the [[wikipedia:nine lemma|nine lemma]]). From this, it is clear that the extension has been determined up to isoclinism.
 ==Related notions==