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| − | == Part a) == | + | ==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 \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\). |
| | + | <ol style="list-style-type:lower-latin"> |
| | + | <li>Check that the map \(\circ\) turns \(G\) into a group. Find explicit formulas for the neutral element and inverse.</li> |
| | + | <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> |
| | + | </ol> |
| | + | ==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. |
| | | | |
| − | Prove that, if \(f\) is an elliptic function, then \(f\) is constant if and only if \(f\) has no poles.
| + | 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 |
| | + | $$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$ |
| | | | |
| − | === Solution part a) === | + | ===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) |
| | | | |
| − | '''Definition''' (elliptic function): A function \(f\) is called elliptic if it has the following two properties:
| + | We furthermore show trivially that \(\rho(e) = e\\ |
| | + | \rho(e)(f)(x) = \rho((\mathbb{I},0))(f)(x) = f(\mathbb{I}(x-0)) = f(x)\) |
| | | | |
| − | (a) \(f\) is doubly periodic.
| + | Using the homomorphic property of \(\rho\) and the fact that \(G\) is a group: \(e = \rho(e) = \rho(gg^{-1}) = \rho(g) \rho(g^{-1}) \overset{!}{=} \rho(g) \rho(g)^{-1} \Rightarrow \rho(g) \in GL(V)\) |
| − | | + | |
| − | (b) \(f\) is meromorphic (its only singularities in the finite plane are poles).
| + | |
| − | | + | |
| − | (Taken from Apostol, 1.4)
| + | |
| − | | + | |
| − | '''Liouville's theorem''': If \(f\) is holomorph on \( \mathbb{C} \) and bounded, then \(f\) is constant.
| + | |
| − | | + | |
| − | | + | |
| − | '''To prove''': Let \(f\) be an elliptic function: \(f\) is constant \(\Leftrightarrow\) \(f\) has no poles.
| + | |
| − | | + | |
| − | \( "\Rightarrow" \) Let \(f\) be constant on \( \mathbb{C} \) \( \rightarrow f\) has no poles.
| + | |
| − | | + | |
| − | \( "\Leftarrow" \) Let \(P\) be the fundamental region to \(f\), where \(f\) is elliptic and has no poles.
| + | |
| − | | + | |
| − | \( f(\mathbb{C}) = f(P) \) 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. \(\square\)
| + | |
| − | | + | |
| − | | + | |
| − | ==Part b)==
| + | |
| − | | + | |
| − | \( \vdash: \) 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.
| + | |
| − | | + | |
| − | | + | |
| − | ===Proof 1===
| + | |
| − | | + | |
| − | 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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| − | | + | |
| − | Since \( f \) is a holomorphic function the integral \( \int_{\partial Q } f(z) \, dz = 0 \) by Cauchy's theorem.
| + | |
| − | | + | |
| − | Then:
| + | |
| − | | + | |
| − | $$ \int_{\partial Q } f(z) \, dz = \left( \int_{Q_1 = [0, 1]} + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) f(z) \, dz $$
| + | |
| − | | + | |
| − | By using the parametrizations
| + | |
| − | $$\gamma_1 : [0, 1] \rightarrow Q_2, t \mapsto 1 + it $$
| + | |
| − | $$ \gamma_2 : [0, 1] \rightarrow \tilde{Q_3}, t \mapsto i + t $$
| + | |
| − | $$ \gamma_3 : [0, 1] \rightarrow \tilde{Q_4}, t \mapsto it $$
| + | |
| − | | + | |
| − | (where \( \tilde{Q_3} \) and \( \tilde{Q_4} \) are the edges with negative orientation), we get:
| + | |
| − | | + | |
| − | $$ \begin{align}
| + | |
| − | \int_{\partial Q } f(z) \, dz &= \int_{[0, 1]} f(t) \, dt + \int_{[0, 1]} f(1 + it) \dot{\gamma_1} \, dt - \int_{[0,1]} f(t + i) \dot{\gamma_2} \, dt - \int_{[0, 1]} f(it) \dot{\gamma_3} \, dt \\
| + | |
| − | &= \int_{[0, 1]} f(t) - f(t + i) \, dt + i \cdot \int_{[0, 1]} f(1 + it) - f(it) \, dt
| + | |
| − | \end{align} $$
| + | |
| − | | + | |
| − | And by assumption we have \( f(t) - f(t + i) \in \mathbb{R}_{\leq 0}\) and \( f(1 + it) - f(it) \in \mathbb{R}_{\geq 0} \). But \( 1 \) and \( i \) are linearly independent in the \( \mathbb{R} \) - vector space \( \mathbb{C} \). Since the integrands are real we get \( \forall t \in [0,1] : f(t) - f(t + i) = 0 \), \( f(1 + it) - f(it) = 0 \).
| + | |
| − | | + | |
| − | We now can doubly-periodically and holomorphically continue \( f \) on \( \mathbb{C} \) to an elliptic function with the fundamental pair of periods \( \left( 1, i \right) \) by using the following argument:
| + | |
| − | | + | |
| − | The continuation is obviously continuous, and by the identity principle we have that \( f(z) \equiv f(z + i ) \) on a neighborhood of each \( z \in [0, 1] \) and \( f(z) \equiv f(z + 1) \) for each \( z \in [0, i] \) (''note'': the identity principle holds since every point \( z \in \partial Q \) is a limit point of the coincidence set of \( f ( z + i ) = f (z) \), respectively \( f( z + 1) = f(z) \)).
| + | |
| − | | + | |
| − | We then see easily that the limit \( f'(z) = \lim_{\Delta z \rightarrow 0} \frac{f(z + \Delta z ) - f(z)}{\Delta z} = \lim_{\Delta z \rightarrow 0} \frac{f(z + i + \Delta z ) - f(z + i)}{\Delta z} = f'(z + i) \) for \( z \in [0,1] \) and of course also \( f(z) = f(z + 1) \) for \( z \in [0, i] \) exists. Thus the continuation has an existing derivative of first order at all points in \( \mathbb{C} \) which is, by its periodicity on the boundary, also continuous, ergo the continuation of \( f \) is an entire function. But since \( f \) is entire and elliptic, it follows from part a) that \( f \equiv const \). \( \square \)
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| − | | + | |
| − | ===Proof 2===
| + | |
| − | | + | |
| − | ''You have to be careful here! You can't use Liouville's theorem without proof. If you want to state this proof in the exam you have to learn a proof for Liouville. For all I've seen this is a rather long and complicated proof. If you know a short and easy one, I guess we would all be glad if you stated it here. A.''
| + | |
| − | | + | |
| − | 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:
| + | |
| − | | + | |
| − | $$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$
| + | |
| − | | + | |
| − | $$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$
| + | |
| − | | + | |
| − | 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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| − | | + | |
| − | | + | |
| − | ''Here is a short proof of Liouville's theorem for harmonic function I found. B''
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| − | | + | |
| − | Consider a bounded harmonic function on Euclidean space. Since
| + | |
| − | it is harmonic, its value at any point is its average over any sphere,
| + | |
| − | and hence over any ball, with the point as center. Given two points,
| + | |
| − | choose two balls with the given points as centers and of equal radius.
| + | |
| − | If the radius is large enough, the two balls will coincide except for an
| + | |
| − | arbitrarily small proportion of their volume. Since the function is
| + | |
| − | bounded, the averages of it over the two balls are arbitrarily close,
| + | |
| − | and so the function assumes the same value at any two points. Thus
| + | |
| − | a bounded harmonic function on Euclidean space is a constant.
| + | |
| − | | + | |
| − | ''Mean value property is still to prove but that's not a hard thing to do on \( \mathbb{C} \) : ''
| + | |
| − | | + | |
| − | Since \( \lambda(x,y) \) is harmonic it will be the real part of some holomorphic function, let's call it \( \Lambda \). Let \( r > 0 \). By Cauchy's formula:
| + | |
| − | | + | |
| − | $$ \begin{align}
| + | |
| − | \Lambda (z_0) &= \frac{1}{2\pi i }\oint_{\left| z - z_0 \right| = r}\frac{\Lambda (z)}{z-z_0} \, dz \\
| + | |
| − | &= \frac{1}{2\pi i }\int_{0}^{2 \pi} \frac{\Lambda (z_0 + r e^{it})}{r e^{it}} i r e^{it} \, dt \\
| + | |
| − | \end {align} $$
| + | |
| − | | + | |
| − | And then:
| + | |
| − | | + | |
| − | $$ \begin{align}
| + | |
| − | \lambda (z_0) &= Re \left( \frac{1}{2\pi i} i \int_{0}^{2 \pi} \frac{\Lambda (z_0 + r e^{it})}{r e^{it}} r e^{it} \, dt \right) \\
| + | |
| − | &= Re \left( \frac{1}{2\pi} \int_{0}^{2 \pi} \Lambda (z_0 + r e^{it}) \, dt \right) \\
| + | |
| − | &= \frac{1}{2\pi} \int_{0}^{2 \pi} \lambda (z_0 + r e^{it}) \, dt \\
| + | |
| − | &= \frac{1}{2\pi} \oint_{B_r(z_0)} \lambda (z) \, dz
| + | |
| − | \end{align}$$
| + | |
| | | | |
| − | ''I guess we can see \( \mathbb{C} \) as \( \mathbb{R}^2 \) since \( \lambda (z) \) is a real-valued function. So at least we got this thing proven for \( \mathbb{C} \) which is the case we need. Thanks for the hint. A.''
| + | \(\Rightarrow \rho\) is a representation of \(G\) on \(V\) the space of all real vaued functions on \(\mathbb{R}^3\) |
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 \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\).
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
$$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$
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 \(\mathbb{R}^3\)
