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) 10:47, 10 June 2015 (CEST)

some arguments are quite tedious. Any ideas for shorter proofs without always relying on \(\{\chi\}\) being a basis of \(L(G)\)?

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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 \(\chi:G \rightarrow \mathbb{C}^*\), \(\mathbb{C}^*=\mathbb{C}\backslash \{0\}\)

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

$$(\chi \mapsto \chi(x))=(\chi \mapsto \chi(y)) \Leftrightarrow \forall \chi \in \hat{G}:\chi(x)= \chi(y) \Rightarrow$$ ??? maybe argue that \(\chi\) is a basis of \(L(G)\) (see c), and therefore take i.e a function \(f\) that maps every member of the group to another number. As it is representable as a sum of the irreducible characters, there must at least one character take another value for two different members of \(G\) as \(f(x)\neq f(y) \Leftrightarrow x\neq y\). Alternatively show that the kernel is the neutral element of the group. That comes to a simular statement, so you have to show that (\(\forall \chi \in \hat{G} \; \chi(x)=1 \Rightarrow x=e\) ),with e=the neutral element of \(G\). Did we have a theorem, that would make this statement trivial? ???

$$\Rightarrow x=y$$

c)

dimension equality

\(dim(L(G))=|G|\) (because \(\{\delta_g| g\in G\}\) is a basis of \(L(G)\) )

\(dim(C(\hat{G}))=|\hat{G}|=|G|\) (see a) )

$$\Rightarrow dim(C(\hat{G}))=dim(L(G))$$

\(\{\chi\}\) is a basis of \(L(G)\) $$|\{\chi\}|=|G|=|L(G)|$$

Lemma 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:

$$\sum\limits_g{\chi(g)}=\sum\limits_{i=h^{-1}*g}{\chi(h*i)}=\sum\limits_{i}{\chi(h) \cdot \chi(i)}=\chi(h) \cdot \sum\limits_{i}{\chi(i)}=\chi(h)\cdot\sum\limits_g{\chi(g)}$$

$$ \Rightarrow \chi(h)=1\; \forall h \in G \; \text{or} \; \sum\limits_g{\chi(g)}=0$$

$$ \Rightarrow \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot\sum\limits_g{\tau(g)}=\delta_{\chi,\tau}\cdot|G| $$

Lemma 2: \((\chi_a*\chi_b)(x)=\chi_a(x)\cdot|G|\cdot\delta_{\chi_a,\chi_b}\)

Proof:

$$\chi_a*\chi_b(x)=\sum\limits_g{\chi_a(x*g^{-1}) \chi_b(g)}=\sum\limits_g{\chi_a(x)\chi_a(g^{-1}) \chi_b(g)}$$

$$=\chi_a(x)\sum\limits_g{\chi_a^{-1}(g) \chi_b(g)}=\chi_a(x)\sum\limits_g{(\chi_a^{-1}\cdot\chi_b)(g)}$$

with \((\chi_a^{-1} \cdot \chi_b) \in \hat{G}\). Use lemma 1

$$=\chi_a(x)\cdot\delta_{(\chi_a^{-1} \cdot \chi_b),\tau} \cdot|G|=\chi_a(x)\cdot|G|\cdot\delta_{\chi_a,\chi_b}$$

Lemma 3: \(\{\chi\}\) are linearly independent

Proof:

Assume

$$\sum_i{\lambda_i \chi_i}=0$$

take any \(\chi_a\) and use Lemma 2

$$ \chi_a*\sum_i{\lambda_i \chi_i}=0$$ $$ \Rightarrow |G|\cdot\lambda_a\cdot\chi_a=0 \Rightarrow \lambda_a=0$$

Therefore \(\{\chi\}\) form a basis of \(L(G)\)

orthonormality of \(\{\chi\}\) (\(\forall \chi_a,\chi_b \in \hat{G}: (\chi_a,\chi_b)_G=\delta_{\chi_a,\chi_b} \)) (maybe Lemma 4 would be enough for the whole proof)

Lemma 4: $$\forall \chi \in \hat{G}, g \in G: \chi(g)\cdot\chi(g)^*=1$$

Proof: As \(G\) is finite \(\forall g \in G \;\exists n\in \mathbb{N}: g^n=e, \) with \(e\) the neutral element of \(G\) (proof: assume that it doesn't hold for all \(n<|G|\ \Rightarrow \forall n<m<|G|: g^n \neq g^m\) (otherwise \(g^{m-n}=e\) ) \( \Rightarrow g^{|G|}=e \) because there are no other diffrent elements in \(G\) )

$$\Rightarrow \forall g \in G \; \exists n\in \mathbb{N}: \chi(g)^n=\chi(g^n)=1 \Rightarrow |\chi(g)|^2=1$$

Proof of orthonormality:

$$(\chi_a,\chi_b)_G=\frac{1}{|G|}\cdot\sum\limits_g{\chi_a(g) \cdot \chi_b^*(g)}=\frac{1}{|G|} \cdot \sum\limits_g{(\chi_a \cdot \chi_b^*)(g)}=\frac{1}{|G|} \sum\limits_g{\chi_c(g)}$$

with \(\chi_c=\chi_a \cdot \chi_b^* \in \hat{G}\). Use Lemma 1

$$=\frac{1}{|G|} \cdot \delta_{(\chi_a \cdot \chi_b^*),\tau} \cdot|G|=\delta_{\chi_a,\chi_b}$$

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 $$kernel(\hat{})=\{f \in L(G): \hat{f}=(\chi \mapsto 0=(f,\chi)_G \}$$ As \(\{\chi\}\) form an orthonormal basis of \(L(G)\), the statement \(0=(f,\chi)_G \; \forall \chi\) is only true for \(f=0\). So the kernel is trivial.

isomorphism

As the dimension match and the map is an injective homomorphism, it is bijective. (use the dimension formula for linear maps to show surjectivity.)

decomposition

As \(\chi\) form a orthonormal basis of \(L(G)\)), it holds

$$f(x)=\sum\limits_{\chi \in \hat{G}}{(f,\chi)_G \cdot \chi(x)}=\frac{1}{|G|}\sum\limits_{\chi \in \hat{G}}{\hat{f}(\chi)\chi(x)}$$