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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: |
| − | '''Definition''' (elliptic function): A function \(f\) is called elliptic if it has the following two properties:
| + | $$ |
| − | | + | \begin{align} |
| − | (a) \(f\) is doubly periodic.
| + | \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) \\ |
| − | (b) \(f\) is meromorphic (its only singularities in the finite plane are poles).
| + | &= 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) \\ |
| − | (Taken from Apostol, 1.4)
| + | &= \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) \\ |
| − | '''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.
| + | |
| − | | + | |
| − | 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 \( f \equiv const \). \( \square \)
| + | |
| − | | + | |
| − | ===Proof 2=== | + | |
| − | | + | |
| − | 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 \( v \equiv const \).
| + | |
| − | | + | |
| − | From the Cauchy-Riemann equations it follows immediately that also \( u \) has to be constant. \( \square \)
| + | |
| − | | + | |
| − | '''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.
| + | |
| − | | + | |
| − | | + | |
| − | ''Here is a short proof of Liouville's theorem for harmonic function I found. B''
| + | |
| − | | + | |
| − | 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 this is 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:
| + | |
| − | | + | |
| − | $$ \begin{align}
| + | |
| − | \lambda (z_0) &= Re \left( \frac{1}{2\pi i} i \int_{0}^{2 \pi} \frac{\Lambda (z_0 + r e^{it})}{r e^{it}} r e^{it} \, dt \right) \\
| + | |
| − | &= Re \left( \frac{1}{2\pi} \int_{0}^{2 \pi} \Lambda (z_0 + r e^{it}) \, dt \right) \\
| + | |
| − | &= \frac{1}{2\pi} \int_{0}^{2 \pi} \lambda (z_0 + r e^{it}) \, dt \\
| + | |
| − | &= \frac{1}{2\pi} \oint_{\partial B_r(z_0)} \lambda (z) \, dz
| + | |
| | \end{align}$$ | | \end{align}$$ |
| | + | (Here the dot indicates where the argument of the function goes) |
| | | | |
| − | This is still an integral over the boundary, not quite what we want yet. I thought I could do the next steps elegantly with our good ol' friend Mr. Gauss, but it didn't quite work out. So we have to change to our favorite coordinate system (I'm changing the meaning of \( r \) here, so watch out for that):
| + | We furthermore show trivially that \(\rho(e) = e\\ |
| − | | + | \rho(e)(f)(x) = \rho((\mathbb{I},0))(f)(x) = f(\mathbb{I}(x-0)) = f(x)\) |
| − | $$ \begin{align}
| + | |
| − | \iint_{B_R(x_0, y_0) } \lambda \, dV &= \int_{0}^{R} r \int_{\partial B_r(z_0)} \lambda( z_0 + r e^{i\phi} ) \, d\phi \, dr \\ | + | |
| − | &= 2 \pi \cdot \int_{0}^{R} r \lambda(x_0, y_0) \, dr \\
| + | |
| − | &= \pi \cdot R^2 \lambda(x_0, y_0) \\
| + | |
| − | \end{align} $$
| + | |
| − | | + | |
| − | leading to the formula:
| + | |
| | | | |
| − | $$ \forall (x_0, y_0) \in \mathbb{R}^2 , R > 0 : \lambda(x_0, y_0) = \frac{1}{vol(B_R(x_0, y_0) )} \iint_{B_R(x_0, y_0) } \lambda (x, y) \, dx \, dy $$
| + | Using the homomorphic property of \(\rho\) and the fact that \(G\) is a group: \(e = \rho(e) = \rho(gg^{-1}) = \rho(g) \rho(g^{-1}) \overset{!}{=} \rho(g) \rho(g)^{-1} \Rightarrow \rho(g) \in GL(V)\) |
| | | | |
| − | ''I guess since \( \lambda \) is a real-valued function we can see the complex plane as the two-dimensional euclidean plane. 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 \(\mathbb{R}^3\)
