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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(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 1===
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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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Since \( f \) is holomorphic, \( v \) is a harmonic function. We can harmonically and doubly-periodically continue \( v \) to \( v : \mathbb{C} \rightarrow \mathbb{R} \). Furthermore \( v(Q) = v(\mathbb{C}) \) and since \( Q \) is compact we have \( \forall z \in \mathbb{C} \left| v (z) \right| \leq M \) for some \( M \in \mathbb{R} \). We now apply Liouville's theorem for harmonic functions and get \( v \equiv const \). 
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From the Cauchy-Riemann equations it follows immediately that also \( u \) has to be constant. \( \square \)
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'''Liouville's theorem for harmonic functions:''' Let \( \lambda : \mathbb{C} \rightarrow \mathbb{R} \) be a harmonic function and \( \forall z \in \mathbb{C} \left| \lambda (z) \right| \leq M \) for some \( M \in \mathbb{R} \). Then \( \lambda \) is constant.
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''The proof is much too complex to state here. I asked the main assistant by mail if it would be okay to state that without proof. More informations to follow.''
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===Proof 2===
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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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Since \( f \) is a holomorphic function the integral \( \int_{\partial Q } f(z) \, dz = \int_{\partial Q } u + iv \, dz = 0 \). We divide \( \partial Q = Q_1 \cup Q_2 \cup Q_3 \cup Q_4 \) where \( Q_{1, ..., 4} \) are the edges of the square. We imagine \( Q_1 \) to be the interval \( [0, 1] \), and proceed to numerate the edges in positive orientation.
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Then:
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$$ \int_{\partial Q } u + iv \, dz = \int_{\partial Q } u(z) \, dz + i \int_{\partial Q } v(z) \, dz = \int_{\partial Q } u(z) \, dz + i \left( \int_{Q_1 } + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) v(x,y) \, dz $$
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But since the function values of \( v \) on \( Q_1 \) are the same as on \( Q_3 \) and are the same on \( Q_2 \) as on \( Q_4 \) we get
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$$ \int_{\partial Q } v(z) \, dz = 0 $$
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because of the converse orientation of the \( Q_1 \) and \( Q_3 \) part, respectively the \( Q_2 \) and \( Q_4 \) part.
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Thus we have \( \int_{\partial Q } u(z) \, dz = 0 \). We write, since \( \forall y \in \left[ 0, 1 \right] : u(1, y) = u(0, y) + \alpha (y) \) and \( \forall x \in \left[ 0, 1 \right] : u(x, 1) = u(x, 0) + \beta (x) \) where \( \alpha, \beta \in \mathbb{R}_{\geq 0} \):
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$$ \begin{align}
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\int_{\partial Q } u(z) \, dz &=  \int_{\partial Q } u(x, y) \, d(x + iy) \\
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&=  \int_{\partial Q } u(x, y) \, dx + i  \int_{\partial Q } u(x, y) \, dy \\
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&= \left( \int_{Q_1 } + \int_{Q_3} \right) u(x, y) \, dx + i \left(\int_{Q_2 } + \int_{Q_4 } \right) u(x, y) \, dy \\
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&= \int_{[0, 1]} u(x,0) \, dt - \int_{[0, 1]} u(x, 1) \, dt + i \left( \int_{[0, 1]} u(1, t) \, dt - \int_{[0, 1]} u(0,t) \, dt \right) \\
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&= \int_{[0, 1]} \left( - \alpha (t) \right) \, dt + i \left(  \int_{[0, 1]} \beta(t) \, dt \right) = 0
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\end{align}$$
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but since \( \alpha, \beta \) are non-negative it follows that \( \alpha \equiv 0 \) and \( \beta \equiv 0 \) because otherwise the integrals wouldn't disappear. We now can doubly-periodically and holomorphically continue \( f \) on \( \mathbb{Z} + i \mathbb{Z} \) to an elliptic function. But since \( f \) is holomorphic, it follows from part a) that \( f \equiv const \). \( \square \)
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\(\Rightarrow \rho\) is a representation of \(G\) on \(V\) the space of all real vaued functions on \(\mathbb{R}^3\)

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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