Aufgaben:Problem 14
Task
Let \(G\) be a finite Abelian group.
a) Prove that the group homomorphisms \(\chi : G → \mathbb{C}^*\) are exactly the characters of irreducible representations of \(G\).
Pointwise multiplication endows the set of irreducible characters of \(G\) with the structure of a finite Abelian group. This group is denoted by \(\hat{G}\). (Remark: \(\hat{G}\) is also called the Pontryagin dual).
b) Show that the map $$G \rightarrow \hat{\hat{G}}$$ $$x \mapsto (\chi \mapsto \chi(x))$$ is an isomorphism of groups.
c) Let \(C(\hat{G})\) denote the \(\mathbb{C}\)-algebra of complex valued functions on \(\hat{G}\) with pointwise multiplication. Prove that the map $$ L(G) \rightarrow C(\hat{G})$$ $$f \mapsto (\hat{f}: \chi \mapsto |G|(f, \chi)_G)$$ is an isomorphism of \(\mathbb{C}\)-algebras (in particuar \(f(x) = \frac{1}{|G|}\sum\limits_{\chi}{\hat{f}(\chi)\chi(x)}\; \forall x \in G\)).
Solution Sketch
--Brynerm (talk) 20:07, 9 June 2015 (CEST) b) and c) are still unsolved (see question marks) Any ideas? a) as \(G\) is abelian, #irreducible representations\(=|C_k| = |G| \Rightarrow \) all irreducible representations have to be one-dimensional, because \(dim(End(\mathbb{C}^G))=|C_k| = |G| = \sum\limits_{\chi}{dim(\chi)}, \chi\) irreducable.
$$\Rightarrow tr(\chi)=\chi$$ so evrey character of an irreducible representation \(\chi\) can be written as \(\chi:G \rightarrow \mathbb{C}^*\)
Further are all homomorphisms \(\rho:G \rightarrow \mathbb{C}^* \) one-dimensional representations and therefore irreducible.
b)
- homomorphism
$$x*y \mapsto (\chi \mapsto \chi(x*y))=(\chi \mapsto \chi(x)\cdot\chi(y))=(\chi \mapsto \chi(x))\cdot (\chi \mapsto \chi(y))$$
- injectivity
$$(\chi \mapsto \chi(x))=(\chi \mapsto \chi(y)) \Leftrightarrow \forall \chi \in \hat{G}:\chi(x)= \chi(y) \Rightarrow ??? \Rightarrow x=y$$
c)
- homomorphism
$$f*g \mapsto (\hat{(f*g)}:\chi \mapsto |G|\cdot(f*g,\chi)_G)=(\chi \mapsto |G|\cdot\frac{1}{|G|}\sum\limits_x{(f*g)(x)\cdot\chi^*(x)})$$ $$=(\chi \mapsto\sum\limits_{x,y}{f(xy^{-1})g(y)\chi^*(x)})=(\chi \mapsto\sum\limits_{x,y}{f(x)f^{-1}(y)g(y)\chi^*(x)})$$ $$=(\chi \mapsto\sum\limits_{z=xy^{-1}, y}{f(z)g(y)\chi^*(zy)})=(\chi \mapsto\left(\sum\limits_{z}{f(z)\chi^*(z)}\right)\left(\sum\limits_{y}{g(y)\chi^*(y)}\right))$$ $$=(\chi \mapsto |G|\cdot(f,\chi)_G)\cdot(\chi \mapsto |G|\cdot(g,\chi)_G$$
- injectivity
$$\hat{f}=\hat{g} \Rightarrow \forall \chi \in \hat{G}: \hat{f}(\chi)=\hat{g}(\chi)$$ $$ \Rightarrow \forall \chi \in \hat{G}: \sum\limits_{y}{f(y)\chi^*(y)}=\sum\limits_{y}{g(y)\chi^*(y)} \Rightarrow ??? \Rightarrow x=y$$
- decomposition ???
Maby this is useful
Lemma: (1) \(\forall \chi_a,\chi_b \in \hat{G}: (\chi_a,\chi_b)_G=\delta_{\chi_a,\chi_b} \) and (2) \((\chi_a*\chi_b)(x)=\chi_a(x)\cdot|G|\cdot\delta_{\chi_a,\chi_b}\)
- Sublemma 1:
$$\forall \chi \in \hat{G}: \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot |G|$$
- where \(\tau\) is the trivial representation = neutral element of \(\hat{G}\)
- Proof: let \(h \in G\) be arbitrary and constant:
$$\sum\limits_g{\chi(g)}=\sum\limits_{i=h^{-1}*g}{\chi(h*i)}=\sum\limits_{i}{\chi(h) \cdot \chi(i)}=\chi(h) \cdot \sum\limits_{i}{\chi(i)}=\chi(h)\cdot\sum\limits_g{\chi(g)}$$
$$ \Rightarrow \chi(h)=1\; \forall h \in G \; \text{or} \; \sum\limits_g{\chi(g)}=0$$
$$ \Rightarrow \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot\sum\limits_g{\tau(g)}=\delta_{\chi,\tau}\cdot|G| $$
- Sublemma 2: $$\forall \chi \in \hat{G}, g \in G: \chi(g)\cdot\chi(g)^*=1$$
- Proof: As \(G\) is finite \(\forall \chi \;\exists n\in \mathbb{N}: \chi^n=id \) (proof: assume that it doesn't hold for all \(n<|G|\ \Rightarrow \forall n<m<|G|: \chi^n \neq \chi^m\) (otherwise \(\chi^{m-n}=id\) ) \( \Rightarrow \chi^{|G|}=id \) because there are no other diffrent elements in \(G\) )
Proof of (1) $$\Rightarrow \exists n\in \mathbb{N}: \chi(g)^n=1 \Rightarrow |\chi(g)|^2=1$$
$$(\chi_a,\chi_b)_G=\frac{1}{|G|}\cdot\sum\limits_g{\chi_a(g) \cdot \chi_b^*(g)}=\frac{1}{|G|} \cdot \sum\limits_g{(\chi_a \cdot \chi_b^*)(g)}=\frac{1}{|G|} \sum\limits_g{\chi_c(g)}$$ with \(\chi_c=\chi_a \cdot \chi_b^* \in \hat{G}\)
$$=\frac{1}{|G|} \cdot \delta_{(\chi_a \cdot \chi_b^*),\tau} \cdot|G|=\delta_{\chi_a,\chi_b}$$
Proof of (2)
$$\chi_a*\chi_b(x)=\sum\limits_g{\chi_a(x*g^{-1}) \chi_b(g)}=\sum\limits_g{\chi_a(x)\chi_a(g^{-1}) \chi_b(g)}$$
$$=\chi_a(x)\sum\limits_g{\chi_a^{-1}(g) \chi_b(g)}=\chi_a(x)\sum\limits_g{(\chi_a^{-1}\cdot\chi_b)(g)}$$ with \((\chi_a^{-1} \cdot \chi_b) \in \hat{G}\)
$$=\chi_a(x)\cdot\delta_{(\chi_a^{-1} \cdot \chi_b),\tau} \cdot|G|=\chi_a(x)\cdot|G|\cdot\delta_{\chi_a,\chi_b}$$
