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As background, I am trying to do exercise 3.10 in Deitmar's "Principles of Harmonic Analysis." I can do most of the problem but I'm stuck on the third part proving surjectivity.

Given a locally compact abelian group $G,$ a closed subgroup $H,$ and a character $\chi: H \rightarrow S^1,$ I need to construct an extension of $\chi$ to all of $G.$ Any extension will do, but it's not clear to me that one should be able to construct an extension. For example, we can't just define an extension to be 1 outside of $H,$ since elements outside of $H$ can multiply to something within $H.$

Can anyone help me out? Thanks!

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If $h\in H$ but $g\notin H$, we have $h+g\notin H$ hence $\chi(h+g)=1\neq \chi h$ in general. –  Davide Giraudo Aug 14 '12 at 15:30
This is not true for nonabelian $G$ (e.g. if we consider ${\bf Z_3}\leq S_3$). I suppose it might make sense to look at $G/H$... –  tomasz Aug 14 '12 at 15:44
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up vote 2 down vote accepted

We can use the following results, from Rudin's book Fourier Analysis on Groups.

Denote $$A(H):=\{\gamma\in\widehat G,\forall h\in H,\gamma(h)=0\}$$ the annihilator of $H$. It's a closed subgroup of $\widehat G$, which ensures us that $\widehat G/A(H)$ is Hausdorff and locally compact.

Theorem 2.1.2. The dual group of $G$ isomorphically homeomorphic to the dual group of $G/H$, and $\widehat G/A(H)$ is isomorphically homeomorphic to the dual group of $H$.

Sketch of proof: Let $\pi$ the projection $\pi\colon G\to G/H$. We define an one-to-one correspondence between $A(H)$ and $\widehat{\left(\frac GH\right)}$ by $$\gamma(x):=\phi(\pi(x)).$$ Furthermore, it's compatible with the group structure.

By Pontryagin duality theorem, and using the fact that for $x_0\notin H$ we can find $\gamma\in \widehat G$ such that $\gamma(x_0)\neq 1$, we deduce the second assertion. To show it's an homeomorphism, we can use the fact that projection is continuous and open.

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You're just too fast :) It's worthwhile to note that $A(H)$ is a closed subgroup of $\widehat{G}$ by definition of the topology on $\widehat{G}$ so that $\widehat{G}/A(H)$ is Hausdorff and locally compact. –  t.b. Aug 14 '12 at 15:51
The problem as stated in Deitmar is to prove $\hat{H} \cong \hat{G}/A(H)$ (and I had just simplified it to the question above), however your sketch here allowed me to prove it! Thanks! –  mck Aug 14 '12 at 16:17
For those also working through Deitmar's book, you can combine the results from earlier in Problem 3.10 to get the last part. We know (from earlier parts) that $\hat{G/H} \cong A(H)$ and $A(A(H)) \cong H.$ Therefore the dual of $\hat{G}/A(H)$ is isomorphic to $A(A(H)) \cong H.$ Now taking the Pontryagin dual of both sides gives the isomorphism in Davide's answer. –  mck Aug 14 '12 at 16:22
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