Hopf's formula and inflation map

Question

Suppose we have two finite groups given by presentations $G=F/N_1 , H=F/N_2$

where $N_1 \subset N_2$ and $F$ is a free group of finite rank.

The canonical map $\pi: G \rightarrow H $ induces the inflation map between Schur Multipliers, $Inf: M(H) \rightarrow M(G)$. (Take a cocyle in M(H) and compose with $\pi \times \pi$ to get a cocyle in M(G))

By Hops formula we have $M(G)\cong (N_1\cap F')/[N_1,F]$ and $M(H)\cong (N_2\cap F')/[N_2,F] $

What is the description of the inflation map in terms of Hopf's formula?

i.e does $$Inf: (N_2\cap F')/[N_2,F] \rightarrow (N_1\cap F')/[N_1,F] $$ have a nice description?

What can we say about its kernel?

Thanks in advance.

Answer 1

I am very far from an expert, but I suspect you have written the map you want going in the wrong direction.

Have a look, for example, at the early part of the book review by Van der Kallen in The Bulletin of the AMS, Vol 10, Number 2 (1984), pages 330-333

He explains that "There is now some confusion as to what the term "Schur multipliers" refers to. The usual meaning is described by the formula $M(Q)=H_2(Q,Z)$, which is also used when $Q$ is infinite. The competing candidate is its dual $Hom(H_2(Q,Z),C^*)= H^2(Q,C^*)$..."

For finite groups these dual homology and cohomology groups are isomorphic, but not canonically isomorphic.

Your earler reference to cocycles suggests you begin with the cohomology version, but then the Hopf formula you write down is definitely a formula for $H_2(G,Z)$

One version will be covariant, one contravariant.

To get from one to the other, you have to dualise, which changes the direction of the morphisms.

In cohomology, the inflation map goes from the cohomology of the quotient group to the cohomology of the original group.

In homology, the inflation map goes in the other direction - the opposite direction to the one you have written your arrow. in particular, if I understand it right, with Van der Kallen's "usual meaning" the inflation map should go from $M(G)$ to $M(H)$, not the way around you have it.

And there is certainly an obvious map in that direction.

.......................................

(Added after your comment). O.K. Let me re-express my answer in terms of the definition you are using.

With $M(G)=H^2(G,C^*)$, I certainly agree that the inflation map goes from $M(H)$ to $M(G)$

Since $C^*$ is divisible, the Universal Coefficient theorem shows that

$H^2(G,C^*) \approx Hom(H_2(G,Z),C^*)$

And this isomorphism is natural.

We know that, for a finite group, $H_2(G,Z)$ is a finite abelian group.

So $H_2(G,Z)$ is isomorphic to $Hom(H_2(G,Z),C^*)$.

But this isomorphism is not natural- it depends on expressing $H_2(G,Z)$as a direct sum of cyclic groups. A different sum will give rise to a different isomorphism. (Compare with the fact that there is no natural isomorphism between a finite-dimensional vector space and its dual).

Now the Hopf isomorphism $H_2(G,Z) \approx (N_1 \cap F^')/[N_1,F]$ depends only on $F$ and $N_1$

So the moral of that is that you are using the wrong isomorphic copies to ask your question. You don't have good enough control of the isomorphisms to the copies you are using.

If you want to relate inflation to Hopf's formula, you should be asking

What is the map from $Hom((N_2 \cap F^')/[N _2,F],C^*)$ to $Hom((N_1 \cap F^')/[N _1,F],C^*)$ ?

And there is then an obvious candidate for the answer:

the dual of the obvious map from $(N_1 \cap F^')/[N _1,F]$ to $(N_2 \cap F^')/[N _2,F]$.