Pontryagin Duality and Adjoints of Group Automorphisms

Question

Let $G$ be a locally compact abelian group, and let $\hat{G}$ denote the Pontryagin Dual of $G$. I'll write elements of $G$ using latin letters (ex: $g$) and elements of $\hat{G}$ using greek letters (ex: $\gamma$).

Using the duality bracket, I know that every continuous homomorphism $\chi:G\rightarrow\mathbb{T}$ (resp. $\hat{\chi}:\hat{G}\rightarrow\mathbb{T}$) can be written as $\chi\left(g\right)=e^{2\pi i\left\langle g,\gamma\right\rangle },\forall g\in G$ (resp. $\hat{\chi}\left(\gamma\right)=e^{2\pi i\left\langle g,\gamma\right\rangle },\forall\gamma\in\hat{G}$) for some unique $\gamma\in\hat{G}$ (resp. $g\in G$).

Since the duality bracket $\left\langle \cdot,\cdot\right\rangle :G\times\hat{G}\rightarrow\mathbb{R}/\mathbb{Z}$ is $\mathbb{Z}$-bilinear, we can think of it as a kind of “inner product”, and thus, define the “adjoint” of a group homomorphism $A:G\rightarrow G$ as a group homomorphism $A^{T}:\hat{G}\rightarrow\hat{G}$ satisfying the condition: $$e^{2\pi i\left\langle A\left(g\right),\gamma\right\rangle }=e^{2\pi i\left\langle g,A^{T}\left(\gamma\right)\right\rangle },\textrm{ }\forall g\in G,\forall\gamma\in\hat{G}$$

Note: for the purposes of my question, I don't care about whether or not $A^{T}$ is unique.

I strongly suspect that the following statement is true:

“$A$ is an automorphism of $G$ if and only if $A^{T}$ is an automorphism of $\hat{G}$”

but I don't know how to prove it. If this is, as I suspect, a well-known result, an explanation of the proof (or a reference to one) would be most appreciated. Thanks in advance.

Answer 1

Let $A$ be a continuous endomorphism of $G$ and $\hat{G}$ the continuous characters to the circle. $A^T \gamma(g) = \gamma(A g)$ is a continuous endomorphism of $\hat{G}$.

• If $A$ is an automorphism then $A^T$ is an automorphism.

• Automorphism means injective and surjective.

• If $\ker(A)$ is non-trivial then $A^T$ is not surjective because there is some $\gamma$ which is non-trivial on $\ker(A)$ so it is not in the image of $ A^T$.

• If $A$ is not surjective let $H$ its image and take some non-trivial $\gamma\in \widehat{G/H}$ then $\gamma \in \hat{G} \cap \ker(A^T)$.

Whence $A$ is an automorphism iff $A^T$ is an automorphism.

Answer 2

You basically have $\langle A(g),\gamma\rangle=\langle g,A^T\gamma\rangle$ within $\mathbb{R}/\mathbb{Z}$, or $(A^T\gamma)(g)=\gamma(A(g))$ for all $g\in G,\gamma\in\widehat{G}$.

First off, $A^T$ is uniquely defined by $A$, since $A^T\gamma=\gamma\circ A$. Second,

$$ \begin{array}{ll} (A^T(\gamma_1+\gamma_2))(g) & =(\gamma_1+\gamma_2)(A(g)) \\ & =\gamma_1(A(g))+\gamma_2(A(g)) \\ & =(A^T\gamma_1)(g)+(A^T\gamma_2)(g) \\ & = (A^T\gamma_1+A^T\gamma_2)(g) \end{array} $$

for all $g\in G$, hence $A^T(\gamma_1+\gamma_2)=A^T\gamma_1+A^T\gamma_2$ as functions. Third, we may show $A^T$ is not only a homomorphism but an automorphism by exhibiting an inverse: explicitly, we should expect the inverse $(A^T)^{-1}$ to be $(A^{-1})^T$, and checking that is a matter of seeing if $(A^{-1})^T$ is indeed an inverse of $A^T$, which we can do by

$$ \langle (A^{-1})^TA^T\gamma,g\rangle=\gamma(AA^{-1}g)=\gamma(g)=\langle\gamma,g\rangle $$

and similarly for $A^T(A^{-1})^T\gamma$.