Difference between revisions of "Aufgaben:Problem 14"

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−
=Problem 4a=
+
== Task ==
−
Let \( T > 0,\eta \in C^{1}([0,T]), \phi \in C^{0}([0,T]) \), and assume that:
+
Let \(G\) be a finite Abelian group.
−
$$ \frac{d}{dt}\eta(t) \leqslant \eta(t)\phi(t), \forall t \in [0,T] $$
+
  
−
Show that:
+
a) Prove that the group homomorphisms \(\chi : G → \mathbb{C}^*\) are exactly the characters of
−
$$\eta(t) \leqslant \eta(0)e^{\int_{0}^{t} \phi(s) ds}, \forall t \in [0,T] $$
+
irreducible representations of \(G\).
  
−
''Hint'': consider  \( \xi(t) := \eta(t) e^{-\int_{0}^{t} \phi(s) ds} \)
+
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).
  
−
==Solution==
+
b) Show that the map
−
Substituting \( \eta(t) = \xi(t) e^{\int_{0}^{t} \phi(s) ds} \) in our assumption, then we get \( \forall t \in [0,T] \):
+
$$G \rightarrow \hat{\hat{G}}$$
 +
$$x \mapsto (\chi \mapsto \chi(x))$$
 +
is an isomorphism of groups.
  
−
$$
+
c) Let \(C(\hat{G})\) denote the \(\mathbb{C}\)-algebra of complex valued functions on \(\hat{G}\) with pointwise
−
\begin{align}
+
multiplication. Prove that the map
−
\frac{d}{dt}\eta(t) &\leqslant \eta(t)\phi(t) \\
+
$$ L(G) \rightarrow C(\hat{G})$$
−
\frac{d}{dt}(\xi(t) e^{\int_{0}^{t} \phi(s) ds}) &\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \\
+
$$f \mapsto (\hat{f}: \chi \mapsto |G|(f, \chi)_G)$$
−
(\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds} + \xi(t) \frac{d}{dt}(e^{\int_{0}^{t} \phi(s) ds})&\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \\
+
is an isomorphism of \(\mathbb{C}\)-algebras (in particuar \(f(x) = \frac{1}{|G|}\sum\limits_{\chi}{\hat{f}(\chi)\chi(x)}\; \forall x \in
−
(\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds} + \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t)&\leqslant \xi(t) e^{\int_{0}^{t} \phi(s) ds}\phi(t) \tag{*}\\
+
G\)).
−
(\frac{d}{dt}\xi(t)) e^{\int_{0}^{t} \phi(s) ds}&\leqslant 0 \\
+
−
\frac{d}{dt}\xi(t) &\leqslant 0 \\
+
−
\end{align}
+
−
$$
+
−
, where we used in the last step that \(\forall x: e^{x}>0 \) and (*) \(\frac{d}{dt}\int_0^t\phi(s)ds) = \phi(t) \) follows from the Fundamental Theorem of Calculus.
+
  
−
Now integrating the result from 0 to t and since \(g(x) \leqslant h(x) \Rightarrow \int g(x) dx \leqslant \int h(x) dx \):
+
==Solution 1==
 +
--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 19:59, 15 June 2015 (CEST)
 +
===a)===
  
−
$$
+
Let \(\rho\) be an irreducable represenation of \(G\) on \(V\Rightarrow\) for any \(g\in G\) (as \(\rho(g) \) is invertible it has only non zero eigenvalues, and every linear transformation over \(\mathbb{C}\) has at least one eigenvalue) there \(\exists \lambda \in \mathbb{C}^*, v \in V\setminus \{0\}\) such that \( \rho(g) v = \lambda v\), so Ker\((\rho(g) - \lambda \mathbb{I}) \neq \{0\}\). \( \forall h \in G\):
−
\begin{align}
+
−
\int_{0}^{t} \frac{d}{dt'}\xi(t') dt' = \xi(t) - \xi(0) &\leqslant 0 = \int_{0}^{t} 0 dx \\
+
−
\eta(t) e^{-\int_{0}^{t} \phi(s) ds}-\eta(0) e^{-\int_{0}^{0} \phi(s) ds} &\leqslant 0 \\
+
−
\eta(t) e^{-\int_{0}^{t} \phi(s) ds}-\eta(0) e^{0} &\leqslant 0 \\
+
−
\eta(t) e^{-\int_{0}^{t} \phi(s) ds}&\leqslant \eta(0) \\
+
−
\eta(t) &\leqslant \eta(0)e^{\int_{0}^{t} \phi(s) ds} \\
+
−
\end{align}
+
−
$$
+
  
−
=Problem 4b=
+
$$ (\rho(g) - \lambda \mathbb{I}) \rho(h)v = \rho(g)\rho(h)v - \lambda \mathbb{I}\rho(h)v$$
−
Consider the initial value problem (IVP)
+
−
$$  
+
−
\begin{align}
+
−
\frac{\partial}{\partial t} u(x,t)-\frac{1}{2} \frac{\partial^2}{\partial x^2} u(x,t) &= (u(x,t))^2 \\
+
−
u(x,0) &= u_0(x) \in C^2(S^1) \\
+
−
\end{align}
+
−
$$
+
  
−
i) Show that, if \(u,v \in C^2(S^1 \times [0,T]) \) are two solutions of (IVP) for the same \(T>0\), then \(u \equiv v \) on \( S^1 \times [0,T] \)
+
\(G\) is abelian \(\Rightarrow \rho(g)\rho(h) = \rho(gh) = \rho(hg) =  \rho(h)\rho(g)\)
  
−
''Hint'': consider \(\eta(t):=\int_0^{2\pi} (u(x,t)-v(x,t))^2 dx \)
+
$$ = \rho(h)\rho(g)v - \lambda\rho(h)v = \rho(h)\lambda v - \lambda\rho(h)v = 0$$
  
−
ii) Give an explicit solution of (IVP) for constant initial data \(u_0 \equiv w_0 \in \mathbb{R} \).
+
\( \Rightarrow Ker(\rho(g) - \lambda \mathbb{I})\) is invariant and therfore equal to  \(V \Rightarrow \rho(g) = \lambda \mathbb{I}\).
  
−
iii) Conclude that, then there is initial data for (IVP) for which no global solution \( u \in C^2(S^1 \times [0,\infty)) \) exists.
+
This implies that every one dimensional subspace of \(V\) is invariant, and thus V has to be one dimensional itself.
  
−
==Solution i)==
+
\(\Rightarrow ch(\rho)(g) = \lambda\) and the forward dircetion is proven.
−
We want to show uniqueness of the solution to the IVP problem.
+
−
First we show that \(\eta(t) \) satisfies the setting of 4a above for every fix \( x \in S^1 \).
+
  
−
Proof:
+
Back: The homomorphisms \(\chi:G \rightarrow \mathbb{C}^* = GL(\mathbb{C}) \) are one-dimensional and therefore irreducible representations of \(G\) on \(\mathbb{C}\).
−
Since u and v are in \( C^2 \) they are continous on our given interval (and limited) and so are combinations of them, we can according to Leibniz Rule change the differentiation and integral, where the Leibniz Rule uses the bounded convergence theorem ([http://en.wikipedia.org/wiki/Dominated_convergence_theorem#Bounded_convergence_theorem]) a corollary of the dominant convergence theorem.
+
 
 +
===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. So I just write this for completeness. The group structure follows simply from the fact, that the characters are homomorphous and one-dimensional. It only has to be shown that the product of characters is homomorphous again, then from a) follows, 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 commutativity results from the commutativity of the pointwise multiplication.
 +
 
 +
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:
 +
$$ \chi(g) \cdot \tau(g)=\tau(g) \cdot \chi(g)= 1 \cdot \chi(g) =\chi(g) $$
 +
 
 +
'''Dimension equality'''
 +
 
 +
\(|G|=|\hat{G}|\), (I hope we can use without proof that \(|\hat{G}|=|\{C_k\}|\). See discussion ). Similarily \(|\hat{G}|=|\hat{\hat{G}}|\)
 +
 
 +
'''Homomorphism'''
 
   
 
   
−
For the following denote: \(u(x,t)=u \),  \(v(x,t)=v\).
+
$$xy \mapsto (\chi \mapsto \chi(xy))=(\chi \mapsto \chi(x)\cdot\chi(y))=(\chi \mapsto \chi(x))\cdot (\chi \mapsto \chi(y))$$
−
$$  
+
 
−
\begin{align}  
+
 
−
\frac{d}{dt} \eta(t)&=\frac{d}{dt}\int_0^{2\pi} (u-v)^2 dx \\
+
'''Injectivity'''
−
&=\int_0^{2\pi}\frac{\partial}{\partial t} (u-v)^2 dx \\
+
 
−
&=\int_0^{2\pi} 2(u-v) (\frac{\partial}{\partial t}u -\frac{\partial}{\partial t}v) dx \\
+
From (a) we know that the characters are homomorphic therefore \( \chi(e) = e = 1\). Because the given map is also homomorphic and \(e\) is maped to \((\chi \mapsto \chi(e)=1) \) the later has to be the trivial element of  \(\hat{\hat{G}}\).  Thus the kernel of our map is \(Z_{\hat{G}}:=\{g\in G:\chi(g)=1\; \forall \chi \in \hat{G}\}\). We want to show that the kernel is trivial.
−
&=\int_0^{2\pi} 2(u-v) ( u^2+\frac{1}{2} \frac{\partial^2}{\partial x^2}u - v^2-\frac{1}{2} \frac{\partial^2}{\partial x^2}v) dx \\
+
 
−
&=\int_0^{2\pi} 2(u-v) ( u^2- v^2)+2(u-v)(\frac{1}{2} \frac{\partial^2}{\partial x^2}u-\frac{1}{2} \frac{\partial^2}{\partial x^2}v) dx \\
+
: '''Lemma 1''':
−
&=\int_0^{2\pi} 2(u-v)^2 (u+v) dx +\int_0^{2\pi}(u-v)\frac{\partial^2}{\partial x^2}(u-v) dx \\
+
$$\forall \chi \in \hat{G}: \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot |G|$$
−
\end{align}
+
::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 fixed:
 +
$$\sum\limits_g{\chi(g)}=\sum\limits_{i=h^{-1}g}{\chi(hi)}=\sum\limits_{i}{\chi(h) \cdot \chi(i)}=\chi(h) \cdot \sum\limits_{i}{\chi(i)}=\chi(h)\cdot\sum\limits_g{\chi(g)}$$
 +
 
 +
$$ \Rightarrow \chi(h)=1\; \forall h \in G \; \text{or} \; \sum\limits_g{\chi(g)}=0$$
 +
 
 +
$$ \Rightarrow \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot\sum\limits_g{\tau(g)}=\delta_{\chi,\tau}\cdot|G| $$
 +
 
 +
 
 +
: '''Lemma 2' ''': \(\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x \in Z_{\hat{G}}} \)
 +
:: let \(\rho \in \hat{G}\) be arbitrary
 +
$$\sum\limits_{\chi}{\chi(x)}=\sum\limits_{\sigma=\rho^{-1}\chi}{\sigma(x)\rho(x)}=\rho(x)\sum\limits_{\sigma \in \hat{G}}{\sigma(x)}$$
 +
$$\Rightarrow \rho(x)=1 \;\forall \rho \in \hat{G}\; or \; \sum\limits_{\chi}{\chi(x)}=0$$
 +
$$\Rightarrow \sum\limits_{\chi}{\chi(x)}=\delta_{x \in Z_{\hat{G}}}\sum\limits_{\chi}{1} =|G|\delta_{x \in Z_{\hat{G}}}$$
 +
 
 +
: '''Lemma 3''': \(Z_{\hat{G}}=\{e\}\)
 +
:: Use Lemma 1
 +
$$ \sum\limits_{\chi}{\sum\limits_{g}{\chi(g)}}=\sum\limits_{\chi}{|G|\delta_{\chi, \tau}}=|G|$$
 +
$$ \sum\limits_{g}{\sum\limits_{\chi}{\chi(g)}}=\sum\limits_{g}{|G|\delta_{g \in Z_{\hat{G}}}}=|G| \cdot |Z_{\hat{G}}| $$
 +
:: \( \Rightarrow |Z_{\hat{G}}|=1 \), and as \(e \in Z_{\hat{G}} \Rightarrow Z_{\hat{G}}=\{e\}\)
 +
 
 +
Therefore It's proven that \(\chi(g)=1, \;\forall \chi \in \hat{G} \Leftrightarrow g=e\) and Lemma 2' becomes '''Lemma 2''':  \(\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x,e} \), which I use in c)
 +
 
 +
'''Bijectivity'''
 +
 
 +
A homomorphism between two finite sets of equal size that is injective is automaticaly surjective and therefore bijective.
 +
<p style="text-align:right;">\(\square\)</p>
 +
 
 +
===c)===
 +
 
 +
Denote:
 +
$$\mathcal{F} : L(G) \rightarrow C(\hat{G}) $$
 +
$$f \mapsto (\hat{f}: \chi \mapsto |G|(f, \chi)_G)$$
 +
 
 +
'''Homomorphism'''
 +
 
 +
An algebra-homomorphism has to be linear and it has to map multiplications to multiplications. The linear part is quite easy, because the inner product is by definition bilinear.
 +
$$\lambda_1 f+ \lambda_2 g \mapsto (\widehat{(\lambda_1 f+\lambda_2 g)}:\chi \mapsto |G|\cdot(\lambda_1 f+\lambda_2 g,\chi)_G)=(\chi \mapsto \lambda_1|G|\cdot(f,\chi)_G)+\lambda_2|G|\cdot(g,\chi)_G))=\lambda_1\hat{f}+\lambda_2\hat{g} $$
 +
 
 +
The multiplication defined on \(L(G)\) is the convolution. It has to be maped to the pointwise multiplication in \(C(\hat{G})\)
 +
$$f*g \mapsto (\widehat{(f*g)}:\chi \mapsto |G|\cdot(f*g,\chi)_G)=(\chi \mapsto |G|\cdot\frac{1}{|G|}\sum\limits_x{(f*g)(x)\cdot\chi(x)^*})$$
 +
$$=(\chi \mapsto\sum\limits_{x,y}{f(xy^{-1})g(y)\chi(x)^*})=(\chi \mapsto\sum\limits_{z=xy^{-1}, y}{f(z)g(y)\chi(zy)^*})$$
 +
$$=(\chi \mapsto\left(\sum\limits_{z}{f(z)\chi(z)^*}\right)\left(\sum\limits_{y}{g(y)\chi(y)^*}\right))=(\chi \mapsto |G|\cdot(f,\chi)_G)\cdot(\chi \mapsto |G|\cdot(g,\chi)_G)=\hat{f}\cdot\hat{g}$$
 +
 
 +
'''Claim''': The inverse of \(\mathcal{F}\) is given by
 +
$$\mathcal{F}^{-1}: C(\hat{G}) \rightarrow  L(G) $$
 +
$$\hat{f} \mapsto (f: x\mapsto \frac{1}{|G|}\sum\limits_{\chi \in
 +
\hat{G}}{\hat{f}(\chi)\chi(x)})$$
 +
 
 +
:Proof of Claim (with use of the character property: \(\chi(g)^* = \chi(g^{-1})\))
 +
$$\mathcal{F}^{-1}(\hat{f})(x) =\frac{1}{|G|}\sum\limits_{\chi}{\hat{f}(\chi)\chi(x)}=\frac{1}{|G|}\sum\limits_{\chi}{\sum\limits_{g}{f(g)\chi(g)^*\chi(x)}}=\frac{1}{|G|}\sum\limits_{g}{f(g)\sum\limits_{\chi}{\chi(g^{-1}x)}}$$
 +
:Use Lemma 2
 +
$$=\frac{1}{|G|}\sum\limits_{g}{f(g)|G|\delta_{g^{-1}x,e}}=\sum\limits_{g}{f(g)\delta_{g,x}}=f(x)$$
 +
 
 +
:Backwards with use of Lemma 1: (alternatively show dimension equality and use the dimension theorem for linear functions (see alternative solution):
 +
$$\mathcal{F}\mathcal{F}^{-1}(\hat{f})(\chi) =\sum\limits_{g}{\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)\rho(g)\chi^*(g)}}=\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)\sum\limits_{g}{(\rho\cdot\chi^{-1})(g)}}$$
 +
 
 +
$$=\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)|G|\delta_{\rho\chi^{-1},\tau}}=\hat{f}(\chi)$$
 +
 
 +
Therefore \(\mathcal{F}\) is an algebra isomorphism
 +
<p style="text-align:right;">\(\square\)</p>
 +
 
 +
==alternative solution==
 +
--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 14:32, 11 June 2015 (CEST)--[[User:Brynerm|Brynerm]] ([[User talk: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 \(|\{\chi\}|=|\{C_k\}| = |G| \Rightarrow \) all irreducible representations have to be one-dimensional, because \(|G| = \sum\limits_{\chi}{dim(\chi)^2} \). (but probably we're not allowed to use the dimension theorem)
 +
 
 +
$$\Rightarrow tr(\chi)=\chi$$
 +
so every character of an irreducible representation \(\chi\) can be written as \(\chi:G \rightarrow \mathbb{C}^*\), \(\mathbb{C}^*=\mathbb{C}\backslash \{0\}=GL(\mathbb{C})\)
 +
 
 +
Further, all homomorphisms \(\rho:G \rightarrow \mathbb{C}^* \) are one-dimensional representations and therefore irreducible.
 +
 
 +
===b)===
 +
 
 +
 
 +
'''\(\{\chi\}\) 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 following 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
 +
$$ \chi_a*\sum_i{\lambda_i \chi_i}=0$$
 +
$$ \Rightarrow |G|\cdot\lambda_a\cdot\chi_a=0 \Rightarrow \lambda_a=0$$
 +
 
 +
: ''')'''
 +
 
 +
I think, Lemma 2,3 are unnecessary: Just put in there Lemma 5 (the orthonormality Lemma from c), ), which you need anyway in this solution. When functions are orthogonal, they can't be linearly dependent.
 +
 
 +
 
 +
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'''
 +
 
 +
$$(\chi \mapsto \chi(x))=(\chi \mapsto \chi(y)) \Leftrightarrow \forall \chi \in \hat{G}:\chi(x)= \chi(y) \Rightarrow$$
 +
 
 +
 
 +
\(\chi\) 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 \(\{\chi\}\) (\(\forall \chi_a,\chi_b \in \hat{G}: (\chi_a,\chi_b)_G=\delta_{\chi_a,\chi_b} \))
 +
 
 +
::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'''
 +
 
 +
An algebra-homomorphism has to be linear and it has to map multiplications to multiplications. The linear part is quite easy, because the inner product is by definition bilinear.
 +
$$\lambda_1 f+ \lambda_2 g \mapsto (\widehat{(\lambda_1 f+\lambda_2 g)}:\chi \mapsto |G|\cdot(\lambda_1 f+\lambda_2 g,\chi)_G)=(\chi \mapsto \lambda_1|G|\cdot(f,\chi)_G)+\lambda_2|G|\cdot(g,\chi)_G))=\lambda_1\hat{f}+\lambda_2\hat{g} $$
 +
 
 +
The multiplication defined on \(L(G)\) is the convolution. It has to be maped to the pointwise multiplication in \(C(\hat{G})\)
 +
$$f*g \mapsto (\widehat{(f*g)}:\chi \mapsto |G|\cdot(f*g,\chi)_G)=(\chi \mapsto |G|\cdot\frac{1}{|G|}\sum\limits_x{(f*g)(x)\cdot\chi^*(x)})$$
 +
$$=(\chi \mapsto\sum\limits_{x,y}{f(xy^{-1})g(y)\chi^*(x)})=(\chi \mapsto\sum\limits_{z=xy^{-1}, y}{f(z)g(y)\chi^*(zy)})$$
 +
$$=(\chi \mapsto\left(\sum\limits_{z}{f(z)\chi^*(z)}\right)\left(\sum\limits_{y}{g(y)\chi^*(y)}\right))=(\chi \mapsto |G|\cdot(f,\chi)_G)\cdot(\chi \mapsto |G|\cdot(g,\chi)_G)=\hat{f}\cdot\hat{g}$$
 +
 
 +
 
 +
'''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.
  
−
Since \(u,v\) are \(2\pi\)-periodic in \(x\), the second integral using integrating by parts resolves to:
 
−
$$ \int_0^{2\pi}(u-v)\frac{\partial^2}{\partial x^2}(u-v) dx=-\int_0^{2\pi}\frac{\partial}{\partial x}(u-v)\frac{\partial}{\partial x}(u-v) dx=-\int_0^{2\pi}(\frac{\partial}{\partial x}(u-v))^2 dx \leqslant 0 $$
 
  
−
So we get:
+
'''isomorphism'''
−
$$ \frac{d}{dt}\eta(t) \leqslant 2\int_0^{2\pi}(u-v)^2 (u+v) dx \leqslant 2\int_0^{2\pi}(u-v)^2 (u_{max}+v_{max}) dx = 2 (u_{max}+v_{max})\int_0^{2\pi}(u-v)^2 dx= \phi(t)\eta(t) $$
+
−
, where \(\phi(t):=2(u_{max}+v_{max}) \) and \( u_{max}:=max\{|u(x,t)|: (x,t) \in [0,2\pi] \times [0,T]\} \).
+
−
(Maximum exists, because \([0,2\pi] \times [0,T] \) is compact and \(u,v\) continuous)
+
  
 +
As the dimension match and the map is an injective homomorphism, it is bijective. (use the dimension formula for linear maps to show surjectivity.)
  
−
Since now all contitions of problem 4a are satisfied, apply proposition 4a:
+
'''decomposition'''
−
$$
+
−
\begin{align}
+
−
\eta(t) &\leqslant \eta(0)e^{\int_{0}^{t} \phi(s) ds} \\
+
−
0 \leqslant \int_0^{2\pi}(u(x,t)-v(x,t))^2 dx &\leqslant \int_0^{2\pi}(u(x,0)-v(x,0))^2 dx e^{\int_{0}^{t} \phi(s) ds} = \int_0^{2\pi}(u_0(x)-v_0(x))^2 dx e^{\int_{0}^{t} \phi(s) ds}=0 \\
+
−
\end{align}
+
−
$$
+
−
,since \(u_0(x)=v_0(x) \). We can now conclude that \(u \equiv v\).
+
  
−
==Solution ii) ==
+
As \(\{\chi\}\) form a orthonormal basis of \(L(G)\)), it holds
−
A solution to the (IVP) is:
+
−
$$ u(x,t)=w_0/(1-w_0t)$$
+
−
since:
+
−
$$
+
−
\begin{align}
+
−
\exists T > 0  : u &\in C^2(S^1 \times [0,T]) \\
+
−
\frac{\partial}{\partial t} u(x,t)&=\frac{\partial}{\partial t} w_0/(1-w_0t)=\frac{w_0^2}{(1-w_0t)^2}=(u(x,t))^2 \\
+
−
\frac{\partial^2}{\partial x^2} u(x,t)&=0 \\
+
−
u(x,0)&=w_0 \\
+
−
\end{align}
+
−
$$
+
−
This must be the unique solution as we have seen in 4b ii).
+
  
−
==Solution iii) ==
+
$$f(x)=\sum\limits_{\chi \in \hat{G}}{(f,\chi)_G \cdot \chi(x)}=\frac{1}{|G|}\sum\limits_{\chi \in \hat{G}}{\hat{f}(\chi)\chi(x)}$$
−
If \(w_0 \in (-\infty,0] \) there is no problem. But for initial data \( w_0 \in (0,\infty) \) we see that \( u(x,t) \) is not defined at \( t=1/w_0 \in (0,\infty)\) and therefore not at every time \( t \in [0,\infty)\).
+
−
So there is no global solution \( u \in C^2(S^1 \times [0,\infty)) \) for \(w_0 \in (0,\infty) \).
+

Latest revision as of 12:19, 4 August 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 1

--Brynerm (talk) 19:59, 15 June 2015 (CEST)

a)

Let \(\rho\) be an irreducable represenation of \(G\) on \(V\Rightarrow\) for any \(g\in G\) (as \(\rho(g) \) is invertible it has only non zero eigenvalues, and every linear transformation over \(\mathbb{C}\) has at least one eigenvalue) there \(\exists \lambda \in \mathbb{C}^*, v \in V\setminus \{0\}\) such that \( \rho(g) v = \lambda v\), so Ker\((\rho(g) - \lambda \mathbb{I}) \neq \{0\}\). \( \forall h \in G\):

$$ (\rho(g) - \lambda \mathbb{I}) \rho(h)v = \rho(g)\rho(h)v - \lambda \mathbb{I}\rho(h)v$$

\(G\) is abelian \(\Rightarrow \rho(g)\rho(h) = \rho(gh) = \rho(hg) = \rho(h)\rho(g)\)

$$ = \rho(h)\rho(g)v - \lambda\rho(h)v = \rho(h)\lambda v - \lambda\rho(h)v = 0$$

\( \Rightarrow Ker(\rho(g) - \lambda \mathbb{I})\) is invariant and therfore equal to \(V \Rightarrow \rho(g) = \lambda \mathbb{I}\).

This implies that every one dimensional subspace of \(V\) is invariant, and thus V has to be one dimensional itself.

\(\Rightarrow ch(\rho)(g) = \lambda\) and the forward dircetion is proven.

Back: The homomorphisms \(\chi:G \rightarrow \mathbb{C}^* = GL(\mathbb{C}) \) are one-dimensional and therefore irreducible representations of \(G\) on \(\mathbb{C}\).

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. So I just write this for completeness. The group structure follows simply from the fact, that the characters are homomorphous and one-dimensional. It only has to be shown that the product of characters is homomorphous again, then from a) follows, 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 commutativity results from the commutativity of the pointwise multiplication.

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: $$ \chi(g) \cdot \tau(g)=\tau(g) \cdot \chi(g)= 1 \cdot \chi(g) =\chi(g) $$

Dimension equality

\(|G|=|\hat{G}|\), (I hope we can use without proof that \(|\hat{G}|=|\{C_k\}|\). See discussion ). Similarily \(|\hat{G}|=|\hat{\hat{G}}|\)

Homomorphism

$$xy \mapsto (\chi \mapsto \chi(xy))=(\chi \mapsto \chi(x)\cdot\chi(y))=(\chi \mapsto \chi(x))\cdot (\chi \mapsto \chi(y))$$


Injectivity

From (a) we know that the characters are homomorphic therefore \( \chi(e) = e = 1\). Because the given map is also homomorphic and \(e\) is maped to \((\chi \mapsto \chi(e)=1) \) the later has to be the trivial element of \(\hat{\hat{G}}\). Thus the kernel of our map is \(Z_{\hat{G}}:=\{g\in G:\chi(g)=1\; \forall \chi \in \hat{G}\}\). We want to show that the kernel is trivial.

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 fixed:

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

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

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


Lemma 2' : \(\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x \in Z_{\hat{G}}} \)
let \(\rho \in \hat{G}\) be arbitrary

$$\sum\limits_{\chi}{\chi(x)}=\sum\limits_{\sigma=\rho^{-1}\chi}{\sigma(x)\rho(x)}=\rho(x)\sum\limits_{\sigma \in \hat{G}}{\sigma(x)}$$ $$\Rightarrow \rho(x)=1 \;\forall \rho \in \hat{G}\; or \; \sum\limits_{\chi}{\chi(x)}=0$$ $$\Rightarrow \sum\limits_{\chi}{\chi(x)}=\delta_{x \in Z_{\hat{G}}}\sum\limits_{\chi}{1} =|G|\delta_{x \in Z_{\hat{G}}}$$

Lemma 3: \(Z_{\hat{G}}=\{e\}\)
Use Lemma 1

$$ \sum\limits_{\chi}{\sum\limits_{g}{\chi(g)}}=\sum\limits_{\chi}{|G|\delta_{\chi, \tau}}=|G|$$ $$ \sum\limits_{g}{\sum\limits_{\chi}{\chi(g)}}=\sum\limits_{g}{|G|\delta_{g \in Z_{\hat{G}}}}=|G| \cdot |Z_{\hat{G}}| $$

\( \Rightarrow |Z_{\hat{G}}|=1 \), and as \(e \in Z_{\hat{G}} \Rightarrow Z_{\hat{G}}=\{e\}\)

Therefore It's proven that \(\chi(g)=1, \;\forall \chi \in \hat{G} \Leftrightarrow g=e\) and Lemma 2' becomes Lemma 2: \(\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x,e} \), which I use in c)

Bijectivity

A homomorphism between two finite sets of equal size that is injective is automaticaly surjective and therefore bijective.

\(\square\)

c)

Denote: $$\mathcal{F} : L(G) \rightarrow C(\hat{G}) $$ $$f \mapsto (\hat{f}: \chi \mapsto |G|(f, \chi)_G)$$

Homomorphism

An algebra-homomorphism has to be linear and it has to map multiplications to multiplications. The linear part is quite easy, because the inner product is by definition bilinear. $$\lambda_1 f+ \lambda_2 g \mapsto (\widehat{(\lambda_1 f+\lambda_2 g)}:\chi \mapsto |G|\cdot(\lambda_1 f+\lambda_2 g,\chi)_G)=(\chi \mapsto \lambda_1|G|\cdot(f,\chi)_G)+\lambda_2|G|\cdot(g,\chi)_G))=\lambda_1\hat{f}+\lambda_2\hat{g} $$

The multiplication defined on \(L(G)\) is the convolution. It has to be maped to the pointwise multiplication in \(C(\hat{G})\) $$f*g \mapsto (\widehat{(f*g)}:\chi \mapsto |G|\cdot(f*g,\chi)_G)=(\chi \mapsto |G|\cdot\frac{1}{|G|}\sum\limits_x{(f*g)(x)\cdot\chi(x)^*})$$ $$=(\chi \mapsto\sum\limits_{x,y}{f(xy^{-1})g(y)\chi(x)^*})=(\chi \mapsto\sum\limits_{z=xy^{-1}, y}{f(z)g(y)\chi(zy)^*})$$ $$=(\chi \mapsto\left(\sum\limits_{z}{f(z)\chi(z)^*}\right)\left(\sum\limits_{y}{g(y)\chi(y)^*}\right))=(\chi \mapsto |G|\cdot(f,\chi)_G)\cdot(\chi \mapsto |G|\cdot(g,\chi)_G)=\hat{f}\cdot\hat{g}$$

Claim: The inverse of \(\mathcal{F}\) is given by $$\mathcal{F}^{-1}: C(\hat{G}) \rightarrow L(G) $$ $$\hat{f} \mapsto (f: x\mapsto \frac{1}{|G|}\sum\limits_{\chi \in \hat{G}}{\hat{f}(\chi)\chi(x)})$$

Proof of Claim (with use of the character property: \(\chi(g)^* = \chi(g^{-1})\))

$$\mathcal{F}^{-1}(\hat{f})(x) =\frac{1}{|G|}\sum\limits_{\chi}{\hat{f}(\chi)\chi(x)}=\frac{1}{|G|}\sum\limits_{\chi}{\sum\limits_{g}{f(g)\chi(g)^*\chi(x)}}=\frac{1}{|G|}\sum\limits_{g}{f(g)\sum\limits_{\chi}{\chi(g^{-1}x)}}$$

Use Lemma 2

$$=\frac{1}{|G|}\sum\limits_{g}{f(g)|G|\delta_{g^{-1}x,e}}=\sum\limits_{g}{f(g)\delta_{g,x}}=f(x)$$

Backwards with use of Lemma 1: (alternatively show dimension equality and use the dimension theorem for linear functions (see alternative solution):

$$\mathcal{F}\mathcal{F}^{-1}(\hat{f})(\chi) =\sum\limits_{g}{\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)\rho(g)\chi^*(g)}}=\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)\sum\limits_{g}{(\rho\cdot\chi^{-1})(g)}}$$

$$=\frac{1}{|G|}\sum\limits_{\rho}{\hat{f}(\rho)|G|\delta_{\rho\chi^{-1},\tau}}=\hat{f}(\chi)$$

Therefore \(\mathcal{F}\) is an algebra isomorphism

\(\square\)

alternative solution

--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 \(|\{\chi\}|=|\{C_k\}| = |G| \Rightarrow \) all irreducible representations have to be one-dimensional, because \(|G| = \sum\limits_{\chi}{dim(\chi)^2} \). (but probably we're not allowed to use the dimension theorem)

$$\Rightarrow tr(\chi)=\chi$$ so every character of an irreducible representation \(\chi\) can be written as \(\chi:G \rightarrow \mathbb{C}^*\), \(\mathbb{C}^*=\mathbb{C}\backslash \{0\}=GL(\mathbb{C})\)

Further, all homomorphisms \(\rho:G \rightarrow \mathbb{C}^* \) are one-dimensional representations and therefore irreducible.

b)

\(\{\chi\}\) 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 following 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

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

)

I think, Lemma 2,3 are unnecessary: Just put in there Lemma 5 (the orthonormality Lemma from c), ), which you need anyway in this solution. When functions are orthogonal, they can't be linearly dependent.


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

$$(\chi \mapsto \chi(x))=(\chi \mapsto \chi(y)) \Leftrightarrow \forall \chi \in \hat{G}:\chi(x)= \chi(y) \Rightarrow$$


\(\chi\) 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 \(\{\chi\}\) (\(\forall \chi_a,\chi_b \in \hat{G}: (\chi_a,\chi_b)_G=\delta_{\chi_a,\chi_b} \))

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

An algebra-homomorphism has to be linear and it has to map multiplications to multiplications. The linear part is quite easy, because the inner product is by definition bilinear. $$\lambda_1 f+ \lambda_2 g \mapsto (\widehat{(\lambda_1 f+\lambda_2 g)}:\chi \mapsto |G|\cdot(\lambda_1 f+\lambda_2 g,\chi)_G)=(\chi \mapsto \lambda_1|G|\cdot(f,\chi)_G)+\lambda_2|G|\cdot(g,\chi)_G))=\lambda_1\hat{f}+\lambda_2\hat{g} $$

The multiplication defined on \(L(G)\) is the convolution. It has to be maped to the pointwise multiplication in \(C(\hat{G})\) $$f*g \mapsto (\widehat{(f*g)}:\chi \mapsto |G|\cdot(f*g,\chi)_G)=(\chi \mapsto |G|\cdot\frac{1}{|G|}\sum\limits_x{(f*g)(x)\cdot\chi^*(x)})$$ $$=(\chi \mapsto\sum\limits_{x,y}{f(xy^{-1})g(y)\chi^*(x)})=(\chi \mapsto\sum\limits_{z=xy^{-1}, y}{f(z)g(y)\chi^*(zy)})$$ $$=(\chi \mapsto\left(\sum\limits_{z}{f(z)\chi^*(z)}\right)\left(\sum\limits_{y}{g(y)\chi^*(y)}\right))=(\chi \mapsto |G|\cdot(f,\chi)_G)\cdot(\chi \mapsto |G|\cdot(g,\chi)_G)=\hat{f}\cdot\hat{g}$$


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

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

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