# $\hat{H} \cong G/H^{\perp}$?

Let $G$ be a locally compact Hausdorff abelian group, and $H$ a closed subgroup of $G$. Let $\hat{G}$ denote the Pontraygin dual of $G$, i.e. the group of coninuous homomorphisms $G \rightarrow S^1$ in the compact-open topology. The notes I'm reading claims there is an isomorphism of topological groups $\hat{G} / H^{ \perp} \cong \hat{H}$, where $H^{\perp}$ denotes the subgroup of $\chi \in \hat{G}$ for which $\chi(H) = \{1\}$.

From the inclusion map $\phi: H \rightarrow G$, I know that we get a corresponding continuous map $\hat{\phi}: \hat{G} \rightarrow \hat{H}$ given by $\hat{\phi}(\chi) = \chi \circ \phi = \chi |_H$. The kernel of this map is $H^{\perp}$, so $H^{\perp}$ is a closed subgroup of $\hat{G}$, and this induces a continuous injective map $$\overline{\hat{\phi}}: \hat{G}/H^{\perp} \rightarrow \hat{H}$$ But why is this map surjective? In other words, why is every character on $H$ the restriction of a character on $G$? Also, this map $\overline{\hat{\phi}}$ is continuous, but in order for there to be an isomorphism of topological groups, it would need to be bicontinuous (an open map). Why is this?

• There is an argument here using annihilators. – Slade Oct 29 '15 at 22:10
• Thank you, but do you know if it's possible to do without pontraygin duality? this exercise comes from Terrence Tao's notes terrytao.wordpress.com/2009/04/06/the-fourier-transform and I don't think the exercise assumes such machinery. – D_S Oct 29 '15 at 22:29
• @D_S. The essential argument that $\widehat{G}$ separates the points of $G$ can be proved by Pontryagin duality as done here, but can also proved alternatively. In fact, it follows directly from the Gelfand-Raikov theorem. Unfortunately, I am not aware of a proof without using the fact that $\widehat{G}$ separates points of $G$. – user342207 Dec 18 '16 at 9:54