Difference between revisions of "Aufgaben:Problem 14"

Revision as of 15:05, 11 June 2015

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) 14:32, 11 June 2015 (CEST)--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 \(L(G)\)?

a)

as \(G\) is abelian, the number of 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 every 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)

group structure of \(\hat{G}\)

The group structure of \(\hat{G}\) is proposed in the exercise and therefore in my view it doesn't need to be proven. But it fallows simply from the fact, that the characters are homomorphous and one-dimensional. It only have to be showed that the product of characters is homomorphous again, then from a) fallows, that it is a character of an irreducable represenation:

$$ \forall x,y \in G :(\chi_a \cdot \chi_b)(xy)=\chi_a(xy)\cdot \chi_b(xy)=\chi_a(x)\cdot\chi_a(y)\cdot \chi_b(x)\cdot \chi_b(y)= (\chi_a\cdot \chi_b)(x)\cdot (\chi_a\cdot \chi_b)(y)$$

The inverse element of \(\chi\) is simply given by: $$ \chi^{-1}(g)=\frac{1}{\chi(g)}=\chi(g^{-1}) $$

and is indeed homomorphous $$ \chi^{-1}(gh)=\frac{1}{\chi(gh)}=\frac{1}{\chi(g)\cdot\chi(h)}=\chi^{-1}(g)\cdot\chi^{-1}(h)$$

And the neutral element is the trivial representation: $$ \tau(g) \cdot \chi(g)= 1 \cdot \chi(g) =\chi(g) = \chi(g) \cdot 1 =\chi(g) \cdot \tau(g) $$

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

Maybe there's a much shorter way to show the linear independence. I used the convolution, but had to prove some properties of the convolution of the characters first. Nevertheless some of the fallowing lemmas could be used for other proofs. $$|\{\chi\}|=|G|=dim(L(G))$$

Lemma 1:

$$\forall \chi \in \hat{G}: \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot |G|$$

where \(\tau(g)=1, \forall g\in G\) 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 \(L(G)\)

dimension equality

\(|G|=|\hat{G}|\), as shown in a). Similarily \(|\hat{G}|=|\hat{\hat{G}}|\)

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

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 \(G\).

$$(\chi \mapsto \chi(x))=(\chi \mapsto \chi(y)) \Leftrightarrow \forall \chi \in \hat{G}:\chi(x)= \chi(y) \Rightarrow$$ Did we have a theorem, that would make this statement trivial?

\(\chi\) is a basis of \(L(G)\), and therefore take i.e a function \(f\) that maps every member of the group to another number. As it is representable as a weighted sum of the irreducible characters, there must at least one character take another value for two different members of \(G\) as \(f(x)\neq f(y) \Leftrightarrow x\neq y\).

$$\Rightarrow x=y$$

bijectivity

A homomorphism between two finite sets of equal size that is injective is automaticaly surjective and therefore bijective, as it holds \(dim(G)=|G|=dim(ker(f))+dim(Im(f))=|Im(f)|=dim(\hat{\hat{G}})=|\hat{\hat{G}}|\)

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) )

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

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 $$f*g \mapsto (\hat{(f*g)}:\chi \mapsto |G|\cdot(f*g,\chi)_G)=(\chi \mapsto |G|\cdot\frac{1}{|G|}\sum\limits_x{(f*g)(x)\cdot\chi^*(x)})$$ $$=(\chi \mapsto\sum\limits_{x,y}{f(xy^{-1})g(y)\chi^*(x)})$$ $$=(\chi \mapsto\sum\limits_{z=xy^{-1}, y}{f(z)g(y)\chi^*(zy)})=(\chi \mapsto\left(\sum\limits_{z}{f(z)\chi^*(z)}\right)\left(\sum\limits_{y}{g(y)\chi^*(y)}\right))$$ $$=(\chi \mapsto |G|\cdot(f,\chi)_G)\cdot(\chi \mapsto |G|\cdot(g,\chi)_G)$$

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

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