Difference between revisions of "Aufgaben:Problem 14"

From Ferienserie MMP2
Jump to: navigation, search
(→‎Solution i))
(Created page with "== 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...")
Line 1: Line 1:
−
=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 Sketch==
 +
--[[User:Brynerm|Brynerm]] ([[User talk: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$$
−
\begin{align}
+
so evrey character of an irreducible representation \(\chi\) can be written as \(\chi:G \rightarrow \mathbb{C}^*\)
−
\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=
+
Further are all homomorphisms \(\rho:G \rightarrow \mathbb{C}^* \) one-dimensional representations and therefore irreducible.
−
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 \( S^1 \times [0,T] \)
 
  
−
''Hint'': consider \(\eta(t):=\int_0^{2\pi} (u(x,t)-v(x,t))^2 dx \)
+
----
  
−
ii) Give an explicit solution of (IVP) for constant initial data \(u_0 \equiv w_0 \in \mathbb{R} \).
+
b)
 +
:'''homomorphism'''
 +
$$x*y \mapsto (\chi \mapsto \chi(x*y))=(\chi \mapsto \chi(x)\cdot\chi(y))=(\chi \mapsto \chi(x))\cdot (\chi \mapsto \chi(y))$$
  
−
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.
+
:'''injectivity'''
 +
$$(\chi \mapsto \chi(x))=(\chi \mapsto \chi(y)) \Leftrightarrow \forall \chi \in \hat{G}:\chi(x)= \chi(y) \Rightarrow ??? \Rightarrow x=y$$
  
−
==Solution i)==
+
----
−
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:
+
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 know also that a combination of \(C^2\)-functions has to be as well in \(C^2\), so 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.
+
:'''homomorphism'''
−
+
$$f*g \mapsto (\hat{(f*g)}:\chi \mapsto |G|\cdot(f*g,\chi)_G)=(\chi \mapsto |G|\cdot\frac{1}{|G|}\sum\limits_x{(f*g)(x)\cdot\chi^*(x)})$$
−
For the following denote: \(u(x,t)=u \),  \(v(x,t)=v\).
+
$$=(\chi \mapsto\sum\limits_{x,y}{f(xy^{-1})g(y)\chi^*(x)})=(\chi \mapsto\sum\limits_{x,y}{f(x)f^{-1}(y)g(y)\chi^*(x)})$$
−
$$  
+
$$=(\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))$$
−
\begin{align}
+
$$=(\chi \mapsto |G|\cdot(f,\chi)_G)\cdot(\chi \mapsto |G|\cdot(g,\chi)_G$$
−
\frac{d}{dt} \eta(t)&=\frac{d}{dt}\int_0^{2\pi} (u-v)^2 dx \\
+
−
&=\int_0^{2\pi}\frac{\partial}{\partial t} (u-v)^2 dx \\
+
−
&=\int_0^{2\pi} 2(u-v) (\frac{\partial}{\partial t}u -\frac{\partial}{\partial t}v) dx \\
+
−
&=\int_0^{2\pi} 2(u-v) ( u^2+\frac{1}{2} \frac{\partial^2}{\partial x^2}u - v^2-\frac{1}{2} \frac{\partial^2}{\partial x^2}v) dx \\
+
−
&=\int_0^{2\pi} 2(u-v) ( u^2- v^2)+2(u-v)(\frac{1}{2} \frac{\partial^2}{\partial x^2}u-\frac{1}{2} \frac{\partial^2}{\partial x^2}v) dx \\
+
−
&=\int_0^{2\pi} 2(u-v)^2 (u+v) dx +\int_0^{2\pi}(u-v)\frac{\partial^2}{\partial x^2}(u-v) dx \\
+
−
\end{align}
+
−
$$
+
  
−
Since \(u,v\) are \(2\pi\)-periodic in \(x\), the second integral using integrating by parts resolves to:
+
:'''injectivity'''
−
$$ \int_0^{2\pi}(u-v)\frac{\partial^2}{\partial x^2}(u-v) dx=-\int_0^{2\pi}\frac{\partial}{\partial x}(u-v)\frac{\partial}{\partial x}(u-v) dx=-\int_0^{2\pi}(\frac{\partial}{\partial x}(u-v))^2 dx \leqslant 0 $$
+
$$\hat{f}=\hat{g} \Rightarrow \forall \chi \in \hat{G}: \hat{f}(\chi)=\hat{g}(\chi)$$
 +
$$ \Rightarrow \forall \chi \in \hat{G}: \sum\limits_{y}{f(y)\chi^*(y)}=\sum\limits_{y}{g(y)\chi^*(y)} \Rightarrow ??? \Rightarrow x=y$$
  
−
So we get:
+
:'''decomposition''' ???
−
$$ \frac{d}{dt}\eta(t) \leqslant 2\int_0^{2\pi}(u-v)^2 (u+v) dx \leqslant 2\int_0^{2\pi}(u-v)^2 (u_{max}+v_{max}) dx = 2 (u_{max}+v_{max})\int_0^{2\pi}(u-v)^2 dx= \phi(t)\eta(t) $$
+
----
−
, where \(\phi(t):=2(u_{max}+v_{max}) \) and \( u_{max}:=max\{|u(x,t)|: (x,t) \in [0,2\pi] \times [0,T]\} \).
+
Maby this is useful
−
(Maximum exists, because \([0,2\pi] \times [0,T] \) is compact and \(u,v\) continuous)
+
  
 +
Lemma: (1) \(\forall \chi_a,\chi_b \in \hat{G}: (\chi_a,\chi_b)_G=\delta_{\chi_a,\chi_b} \) and (2) \((\chi_a*\chi_b)(x)=\chi_a(x)\cdot|G|\cdot\delta_{\chi_a,\chi_b}\)
  
−
Since now all contitions of problem 4a are satisfied, apply proposition 4a:
+
: Sublemma 1:
−
$$
+
$$\forall \chi \in \hat{G}: \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot |G|$$
−
\begin{align}
+
:where \(\tau\) is the trivial representation = neutral element of \(\hat{G}\)
−
\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) ==
+
:Proof: let \(h \in G\) be arbitrary and constant:  
−
A solution to the (IVP) is:
+
$$\sum\limits_g{\chi(g)}=\sum\limits_{i=h^{-1}*g}{\chi(h*i)}=\sum\limits_{i}{\chi(h) \cdot \chi(i)}=\chi(h) \cdot \sum\limits_{i}{\chi(i)}=\chi(h)\cdot\sum\limits_g{\chi(g)}$$
−
$$ u(x,t)=w_0/(1-w_0t)$$
+
−
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) ==
+
$$ \Rightarrow \chi(h)=1\; \forall h \in G \; \text{or} \; \sum\limits_g{\chi(g)}=0$$
−
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 \(w_0 \in (0,\infty) \).
+
$$ \Rightarrow \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot\sum\limits_g{\tau(g)}=\delta_{\chi,\tau}\cdot|G| $$
 +
 
 +
 
 +
: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 \((\chi_a^{-1} \cdot \chi_b) \in \hat{G}\)
 +
 
 +
$$=\chi_a(x)\cdot\delta_{(\chi_a^{-1} \cdot \chi_b),\tau} \cdot|G|=\chi_a(x)\cdot|G|\cdot\delta_{\chi_a,\chi_b}$$

Revision as of 18:07, 9 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) 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 \((\chi_a^{-1} \cdot \chi_b) \in \hat{G}\)

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

Retrieved from "http://ferienserie.ch/index.php?title=Aufgaben:Problem_14&oldid=959"