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Reading an old paper of Weil's (translation: On certain groups of unitary operators), I'm confused about what should be a rather basic point.

Let $G$ be a locally compact abelian group. Now in addition he assumes that $G$ is isomorphic to its dual group of characters. For $x$ in $G$, let $x^*\in G$ be a character. He says that one denotes $\langle x,x^*\rangle$ the value $x^*(x)$. [I think this bracketed expression is referred to as a pairing more generally, but I'm unsure and would appreciate it if someone could explain this.] He then says one can identify $G$ with $(G^*)^*$ (which I think is just Pontryagin Duality), and says that therefore $$\langle x,x^*\rangle=\langle x^*,x\rangle.$$

Meaning that $x(x^*)=x^*(x)$ for all $x,x^*$? I don't understand how this is obvious. Does this only happen for self-dual groups? Weil seems to suggest that this follows just from $G=(G^*)^*$.

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1 Answer 1

up vote 1 down vote accepted

If $G$ is a locally compact abelian group, let $G^{\vee}$ denote its Pontrjagin dual. The dual pairing $G \times G^{\vee} \to \mathbb{C}$ realizes each element of $G$ as a character of $G^{\vee}$, hence gives a canonical map

$$G \ni x \mapsto (x^{\ast} \mapsto x(x^{\ast}) = x^{\ast}(x)) \in G^{\vee \vee}.$$

Pontrjagin duality asserts that this map is an isomorphism. The equality above is a definition of $x(x^{\ast})$; in other words, the statement that you're confused about doesn't require Pontrjagin duality, and it doesn't require the hypothesis that $G$ is isomorphic to its dual. (I am interpreting $x^{\ast}$ as an arbitrary element of $G^{\vee}$ here.) Pontrjagin duality is only necessary to identify $x(x^{\ast})$ as defined above with the dual pairing $G^{\vee} \times G^{\vee \vee} \to \mathbb{C}$.

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