Integration over a compact group with respect to Haar measure

haar-measureintegrationlocally-compact-groupsrepresentation-theory

Let $G$ be a compact abelian group and $\hat G={\rm hom_{\rm continuous}}(G,\mathbb{C}^*)$. Let $\mu$ be the Haar measure on $G$. For $\chi\in\hat G$, show that

$$\int_G\chi\cdot\mu=1\quad\text{if}\quad\chi=\mathbb{1}$$
$$\int_G\chi\cdot\mu=0\quad\text{if}\quad\chi\neq\mathbb{1}$$

The first one is easy and I only need to show the second. My attempt was to use the invariance properties of the Haar measure. First I tried

$$\int_G\chi(g)\cdot\mu(g)=\int_G\chi(g^{-1})\cdot\mu(g)=\int_G(\chi(g))^{-1}\cdot\mu(g)=\int_G\overline{\chi(g)}\cdot\mu(g)$$

But this only gives $\int_G\chi\cdot\mu\in\mathbb{R}$. Then I thought, if there exists a $g_0\in g$ such that $\chi(g_0)=-1$, then

$$\int_G\chi(g)\cdot\mu(g)=\int_G\chi(g_0g)\cdot\mu(g)=\int_G-\chi(g)\cdot\mu(g)\implies\int_G\chi\cdot\mu=0$$

Questions:

(1) Am I on the right track here? If so, how to prove such $g_0$ exists?

(2) This seems to be a problem of the representation theory, but it appeared in an algebraic number theory course while introducing the $L$ function. I am not sure how they are connected. Can anyone provide some backgroud?

(3) I haven't systematically studied the representation theory yet so I am not familiar with some concepts. For example, what is $\hat G$ called and what are the elements of $\hat G$ called? My teacher says elements in $\hat G$ are called characters (or characteristics?), but it doesn't seem to match the definition of a character in any of book on representation theory.

(4) Is there any text on this topic?

Appreciate any help. Thanks in advance.

Best Answer

$\widehat{G}$ is called the dual group of $G$.

Take $g' \in G$ such that $\chi(g')\neq 1$.

$$\chi(g')\cdot \int_G \chi(g) d\mu = \int_G \chi(g'\cdot g)d\mu = \int_G \chi(g)d\mu$$

because multiplication by $\chi(g')$ simply permutes the elements of $G$.

because of the translation invariance of the Haar measure. Therefore

$$(\chi(g') - 1)\int_G \chi(g)d\mu = 0$$

and the result follows.

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