Pontryagin duality and quotient groups

abstract-algebraduality-theoremsgroup-theoryquotient-group

I am studying the Pontryagin duality for LCA groups, and I came across two results in which I am finding some difficulty.

Here I will denote by $G^*$ the dual of the group $G$, i.e. the group of all continuous homomorphisms $G \to S^1$.

  1. If $H$ is a quotient group of $G$, where $H$ and $G$ are either both compact Hausdorff Abelian groups or discrete Abelian groups, then the dual $H^*$ is topologically isomorphic to a subgroup of $G^*$.
  2. If $H$ is a subgroup of $G$, with $H$, $G$ discrete Abelian groups, then $H^*$ is a quotient group of $G^*$.

I already know that, given an LCA group $G$, if $G$ is compact then $G^*$ is discrete, while if $G$ is discrete then $G^*$ is compact.
Moreover, I have proved the following result:

Let $\gamma: H \to G$ be a continuous homomorphism of LCA groups, and define $\gamma^*: G^* \to H^*$ by putting $\gamma^*(f)(h)=f \gamma (h) \ \forall f \in G^*, h \in H$. Then:

  • $\gamma^*$ is a continuous homomorphism;
  • if $\gamma$ is surjective, then $\gamma^*$ is injective;
  • if $\gamma$ is injective and is also open, then $\gamma^*$ is surjective.

This seems a useful result, but if I try to apply it to the open quotient map $q: G \to G/H$ or to the injective $i: H \to G$, it doesn't seem to help.

Any help would be highly appreciated. Thanks!

Best Answer

Let $H$ be a quotient of $G$, denote $\pi : G \to H$ the projection. Using your result, since $\pi$ is surjective the dual $\pi^*$ must be injective and continuous. Denoting its image $\pi^*(H^*)$ by the letter $K$, we get :

  • $K$ is a subgroup of $G^*$ since $\pi^*$ is a group homomorphism ;
  • $\pi^* : H^* \to K$ is injective (see above) and surjective (by definition) :
    • Either $H$ and $G$ are both compact Hausdorff, then $G^*$ and $H^*$ as well, and some theorem from general topology tells us that $\pi^*$ must be a homeomorphism (it is bijective from a compact space to a Hausdorff space) ;
    • Either $H$ and $G$ are both discrete, and every bijective function becomes a homeomorphism.

In both cases, $\pi^*$ is a group isomorphism and a homeomorphism from $H^*$ to some subgroup of $G^*$.

The same argument works for the second question, using the fact that the inclusion $H \to G$ has to be injective (obviously) and open (since $H$ and $G$ are assumed to be discrete).

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