Difference between revisions of "Aufgaben:Problem 1"

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== Part a) ==
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==Task==
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We endow the Cartesian product \(G=SO\left(3\right)\times\mathbb{R}^{3}\) with the map \(\circ:G\times G\rightarrow G\) defined by \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\).
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<ol style="list-style-type:lower-latin">
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  <li>Check that the map \(\circ\) turns \(G\) into a group. Find explicit formulas for the neutral element and inverse.</li>
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  <li>Verify that the formula \(\left(\rho\left(A,y\right)f\right)\left(x\right):=f\left(A^{-1}\left(x-y\right)\right)\) defines a representation of \(G\) on the space of real valued functions \(f\) on \(\mathbb{R}^{3}\)</li>
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</ol>
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==Solution==
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===Subtask a)===
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Check that \((A,x) \circ (B,y) \in G\):
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\(SO(3)\) is a group under matrix multiplication, so \(AB \in SO(3)\). \(Ay, x \in \mathbb{R}^3\), and since \(\mathbb{R}^3\) is a vector space, \(Ay + x \in \mathbb{R}^3\). So, the statement is proven.
  
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Prove that, if \(f\) is an elliptic function, then \(f\) is constant if and only if \(f\) has no poles.
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We check the associativity of the group multiplication:
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$$
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\begin{align}
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\left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)&=\left(AB,Ay+x\right)\circ\left(C,z\right)\\
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&=\left(\left(AB\right)C,ABz+Ay+x\right)\\
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&=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)\\
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&=\left(A,x\right)\circ\left(BC,Bz+y\right)\\
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&=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)\\
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\end{align}
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$$
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The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\) because
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$$(A,x) \circ (\mathbb{I},0) = (A\mathbb{I},A0 + x) = (A,x) = (\mathbb{I}A, \mathbb{I}x + 0) = (\mathbb{I},0) \circ (A,x)$$ and
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$$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$
  
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=== Solution part a) ===
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===Subtask b)===
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We check that it is indeed a homomorphism:
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$$
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\begin{align}
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\left(\rho [ \left(A,y\right)\circ\left(B,z\right) ] f\right)\left(x\right) &= \left(\rho [ AB,Az+y ] f\right)\left(x\right) \\
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&= f\left(\left(AB\right)^{-1}\left(x-Az-y\right)\right) \\
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&= f\left(B^{-1}A^{-1}\left(x-Az-y\right)\right) \\
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&= f\left(B^{-1}\left(A^{-1}\left(x-y\right)-z\right)\right) \\
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&= \left(\rho\left(A,y\right)\left(f\left(B^{-1}\left(\cdot-z\right)\right)\right)\right)\left(x\right) \\
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&= \left(\rho\left(A,y\right)\rho\left(B,z\right)f\right)\left(x\right) \\
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\end{align}$$
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(Here the dot indicates where the argument of the function goes)
  
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'''Definition''' (elliptic function):  A function \(f\) is called elliptic if it has the following two properties:
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We furthermore show trivially that \(\rho(e) = e\\
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\rho(e)(f)(x) = \rho((\mathbb{I},0))(f)(x) = f(\mathbb{I}(x-0)) = f(x)\)
  
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(a) \(f\) is doubly periodic.
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Using the homomorphic property of \(\rho\) and the fact that \(G\) is a group: \(e = \rho(e) = \rho(gg^{-1}) = \rho(g) \rho(g^{-1}) \overset{!}{=} \rho(g) \rho(g)^{-1} \Rightarrow \rho(g) \in GL(V)\)
  
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(b) \(f\) is meromorphic (its only singularities in the finite plane are poles).
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\(\Rightarrow \rho\) is a representation of \(G\) on \(V\) the space of all real vaued functions on \(\mathbb{R}^3\)
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(Taken from Apostol, 1.4)
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'''Liouville's theorem''': If \(f\) is holomorph on \( \mathbb{C} \) and bounded, then \(f\) is constant.
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'''To prove''': Let \(f\) be an elliptic function: \(f\) is constant \(\Leftrightarrow\) \(f\) has no poles.
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\( "\Rightarrow" \) Let \(f\) be constant on \( \mathbb{C} \) \( \rightarrow f\)  has no poles.
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\( "\Leftarrow" \) Let \(P\) be the fundamental region to \(f\), where \(f\) is elliptic and has no poles.
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\( f(\mathbb{C}) = f(P) \) is compact and thus bounded. Because \(f\) is both holomorph and bounded on \(\mathbb{C}\), Liouville's theorem tells us that \(f\) has to be a constant function. \(\square\)
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==Part b)==
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\( \vdash: \) Let \( Q = \left[ 0, 1\right] \times \left[ 0, 1 \right] \subset \mathbb{C} \) be the unit square, and let \( f \) be a holomorphic function on a neighborhood of \( Q \). Suppose further that \( f( z + i) - f(z) \in \mathbb{R}_{\geq 0} \) for any \( z \in \left[ 0, 1 \right] \) and \( f( z + 1) - f(z) \in \mathbb{R}_{\geq 0} \) for any \( z \in \left[ 0, i \right] \). Then \(f\) is constant.
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===Proof===
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At first, we prove a little Lemma:
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'''Pierre's Lemma:''' If \( f = u + iv \) is a holomorphic function then \( u, v \) are meromorphic.
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\( \color{red}{So \: beautiful! \: See \: discussion} \)
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''Proof of Lemma:''
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If \( a, b \in \mathbb{R} \smallsetminus \{ 0 \} \), then \( a \) and \( b i \) are linearly independent. ( \( \mathbb{C} \) is a two-dimensional \( \mathbb{R} \)-vectorspace with basis \( 1 \) and \( i \). ) If \( v \) is zero then \( f = u \) is constant (see critic in discussion), if \( u = 0 \) then \( v \) is meromorphic anyway and yay. \( \square \)
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Now we can start with the main proof. We write \( z = x + iy \) and \( f(x + iy) = u(x,y) + i v(x,y) \) where \( u, v \) are real valued functions. By assumption we get:
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$$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$
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$$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$
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Therefore
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$$\forall z\in [0,i]: v(z+1)=v(z)$$
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$$\forall z\in[0,1] : v(z+i)=v(z)$$
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And we see that v is doubly-periodic on \(Q\). Pierre's Lemma implies that \(u\) and \(v\) are meromorphic. Because \(f\) has no poles on \(Q\), \(u\) and \(v\) don't have either. Therefore, \(u\) and \(v\) are holomorphic on \(Q\).
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We thus see that \( v \) can be doubly-periodic, analytically continued on \(\mathbb{C}\), so that it's an elliptic function with respect to the lattice \( \Omega = \{ m + in : m, n \in \mathbb{Z} \} \).
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We've shown in part a) that a elliptic, holomorphic function has to be constant. Thus \( v \) is constant.
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From the Cauchy-Riemann equations it follows immediately that also \( u \) has to be constant. \( \square \)
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Latest revision as of 10:35, 30 June 2015

Task

We endow the Cartesian product \(G=SO\left(3\right)\times\mathbb{R}^{3}\) with the map \(\circ:G\times G\rightarrow G\) defined by \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\).

  1. Check that the map \(\circ\) turns \(G\) into a group. Find explicit formulas for the neutral element and inverse.
  2. Verify that the formula \(\left(\rho\left(A,y\right)f\right)\left(x\right):=f\left(A^{-1}\left(x-y\right)\right)\) defines a representation of \(G\) on the space of real valued functions \(f\) on \(\mathbb{R}^{3}\)

Solution

Subtask a)

Check that \((A,x) \circ (B,y) \in G\): \(SO(3)\) is a group under matrix multiplication, so \(AB \in SO(3)\). \(Ay, x \in \mathbb{R}^3\), and since \(\mathbb{R}^3\) is a vector space, \(Ay + x \in \mathbb{R}^3\). So, the statement is proven.

We check the associativity of the group multiplication: $$ \begin{align} \left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)&=\left(AB,Ay+x\right)\circ\left(C,z\right)\\ &=\left(\left(AB\right)C,ABz+Ay+x\right)\\ &=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)\\ &=\left(A,x\right)\circ\left(BC,Bz+y\right)\\ &=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)\\ \end{align} $$ The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\) because $$(A,x) \circ (\mathbb{I},0) = (A\mathbb{I},A0 + x) = (A,x) = (\mathbb{I}A, \mathbb{I}x + 0) = (\mathbb{I},0) \circ (A,x)$$ and $$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$

Subtask b)

We check that it is indeed a homomorphism: $$ \begin{align} \left(\rho [ \left(A,y\right)\circ\left(B,z\right) ] f\right)\left(x\right) &= \left(\rho [ AB,Az+y ] f\right)\left(x\right) \\ &= f\left(\left(AB\right)^{-1}\left(x-Az-y\right)\right) \\ &= f\left(B^{-1}A^{-1}\left(x-Az-y\right)\right) \\ &= f\left(B^{-1}\left(A^{-1}\left(x-y\right)-z\right)\right) \\ &= \left(\rho\left(A,y\right)\left(f\left(B^{-1}\left(\cdot-z\right)\right)\right)\right)\left(x\right) \\ &= \left(\rho\left(A,y\right)\rho\left(B,z\right)f\right)\left(x\right) \\ \end{align}$$ (Here the dot indicates where the argument of the function goes)

We furthermore show trivially that \(\rho(e) = e\\ \rho(e)(f)(x) = \rho((\mathbb{I},0))(f)(x) = f(\mathbb{I}(x-0)) = f(x)\)

Using the homomorphic property of \(\rho\) and the fact that \(G\) is a group: \(e = \rho(e) = \rho(gg^{-1}) = \rho(g) \rho(g^{-1}) \overset{!}{=} \rho(g) \rho(g)^{-1} \Rightarrow \rho(g) \in GL(V)\)

\(\Rightarrow \rho\) is a representation of \(G\) on \(V\) the space of all real vaued functions on \(\mathbb{R}^3\)

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